{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0001","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"659c336d90e3bf934fecaa9c856dab1fd049224327d90b3f92bacdd67ac1cf32","family_id":"qf-mod50-graddeep-linearity","seed":202611196001,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[1,1],[0,1]],\"gradient_g\":[[0,1],[-1,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[1, 0], grad_g=[0, -1], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([1, 0], [0, -1], 1/2, 1/3, [1/2, -1/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([1, 0], [0, -1], 1/2, 1/3, [1/2, -1/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0002","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44c92a0919cfabf5a406de3f4df12dcedf0643419b84d06c2e0280966f699011","family_id":"qf-mod50-graddeep-linearity","seed":202611196002,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[2,3],\"gradient_f\":[[2,1],[1,1]],\"gradient_g\":[[1,1],[-2,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[2, 1], grad_g=[1, -2], alpha=1/2, beta=2/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([2, 1], [1, -2], 1/2, 2/3, [5/3, -5/6], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([2, 1], [1, -2], 1/2, 2/3, [5/3, -5/6], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0003","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a163af3b482d941e9813e74af68ac22a4dd6c87db34c95a5ec44c38b99dd770","family_id":"qf-mod50-graddeep-linearity","seed":202611196003,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,1],\"gradient_f\":[[3,1],[2,1]],\"gradient_g\":[[2,1],[-3,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[3, 2], grad_g=[2, -3], alpha=1/2, beta=3/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([3, 2], [2, -3], 1/2, 1, [7/2, -2], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([3, 2], [2, -3], 1/2, 1, [7/2, -2], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0004","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e8971567ba255ba93176669d182e3dfa4bc09475852f3ce73b2f13c348c29be","family_id":"qf-mod50-graddeep-linearity","seed":202611196004,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[4,3],\"gradient_f\":[[4,1],[3,1]],\"gradient_g\":[[3,1],[-4,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[4, 3], grad_g=[3, -4], alpha=1/2, beta=4/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([4, 3], [3, -4], 1/2, 4/3, [6, -23/6], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([4, 3], [3, -4], 1/2, 4/3, [6, -23/6], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0005","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91c8bffce7492635d22cc095a78b4d08d142d9c50088dc5d1c9e53a7b4665b76","family_id":"qf-mod50-graddeep-linearity","seed":202611196005,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[5,1],[4,1]],\"gradient_g\":[[4,1],[-5,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[5, 4], grad_g=[4, -5], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([5, 4], [4, -5], 1/2, 1/3, [23/6, 1/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([5, 4], [4, -5], 1/2, 1/3, [23/6, 1/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0006","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff5580fd70653cc0cdb614369ca83a2110388ab089b37b48c4d1dbddae66d9e1","family_id":"qf-mod50-graddeep-linearity","seed":202611196006,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[2,3],\"gradient_f\":[[6,1],[0,1]],\"gradient_g\":[[5,1],[-6,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[6, 0], grad_g=[5, -6], alpha=1/2, beta=2/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([6, 0], [5, -6], 1/2, 2/3, [19/3, -4], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([6, 0], [5, -6], 1/2, 2/3, [19/3, -4], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0007","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4260569f1d8cb67de5fb97a2e573c76071aea009d39287d764ebaad5b2a5da2b","family_id":"qf-mod50-graddeep-linearity","seed":202611196007,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,1],\"gradient_f\":[[7,1],[1,1]],\"gradient_g\":[[6,1],[-7,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[7, 1], grad_g=[6, -7], alpha=1/2, beta=3/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([7, 1], [6, -7], 1/2, 1, [19/2, -13/2], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([7, 1], [6, -7], 1/2, 1, [19/2, -13/2], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0008","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"105861f97fecc47dd13645c38986583e491f02343395f2dc7ef131d6e3b46eb6","family_id":"qf-mod50-graddeep-linearity","seed":202611196008,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[4,3],\"gradient_f\":[[8,1],[2,1]],\"gradient_g\":[[0,1],[-8,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[8, 2], grad_g=[0, -8], alpha=1/2, beta=4/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([8, 2], [0, -8], 1/2, 4/3, [4, -29/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([8, 2], [0, -8], 1/2, 4/3, [4, -29/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0009","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e57c8358247d81ca1a263d6022f3e3e07484e640eb27231c8b28823b63f8190b","family_id":"qf-mod50-graddeep-linearity","seed":202611196009,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[9,1],[3,1]],\"gradient_g\":[[1,1],[-9,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[9, 3], grad_g=[1, -9], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([9, 3], [1, -9], 1/2, 1/3, [29/6, -3/2], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([9, 3], [1, -9], 1/2, 1/3, [29/6, -3/2], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0010","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cf00b06d7b97bd5f12d35acd1b98043f875f52a47c3fe74cfa6907ac2347582","family_id":"qf-mod50-graddeep-linearity","seed":202611196010,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[2,3],\"gradient_f\":[[10,1],[4,1]],\"gradient_g\":[[2,1],[-10,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[10, 4], grad_g=[2, -10], alpha=1/2, beta=2/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([10, 4], [2, -10], 1/2, 2/3, [19/3, -14/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([10, 4], [2, -10], 1/2, 2/3, [19/3, -14/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0011","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1dc67a2a835a7856dfd1cf0ba91020c83e6fee4337ba6e189156da75ca517776","family_id":"qf-mod50-graddeep-linearity","seed":202611196011,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,1],\"gradient_f\":[[11,1],[0,1]],\"gradient_g\":[[3,1],[-11,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[11, 0], grad_g=[3, -11], alpha=1/2, beta=3/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([11, 0], [3, -11], 1/2, 1, [17/2, -11], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([11, 0], [3, -11], 1/2, 1, [17/2, -11], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0012","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"892f4c11b116078b7b92d9c65d94f288ff97d86059916ca0520620f014561d6a","family_id":"qf-mod50-graddeep-linearity","seed":202611196012,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[4,3],\"gradient_f\":[[12,1],[1,1]],\"gradient_g\":[[4,1],[-12,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[12, 1], grad_g=[4, -12], alpha=1/2, beta=4/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([12, 1], [4, -12], 1/2, 4/3, [34/3, -31/2], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([12, 1], [4, -12], 1/2, 4/3, [34/3, -31/2], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0013","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aca5aefd320bb2dd065eef3976cf21d2118464524ae0e7a5c2affe87b26c3e26","family_id":"qf-mod50-graddeep-linearity","seed":202611196013,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[13,1],[2,1]],\"gradient_g\":[[5,1],[-13,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[13, 2], grad_g=[5, -13], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([13, 2], [5, -13], 1/2, 1/3, [49/6, -10/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([13, 2], [5, -13], 1/2, 1/3, [49/6, -10/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0014","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c53cc1137c51a06e8628828e21a09f7f1c2dd3d60310194d763e09f84266a85","family_id":"qf-mod50-graddeep-linearity","seed":202611196014,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[2,3],\"gradient_f\":[[14,1],[3,1]],\"gradient_g\":[[6,1],[-14,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[14, 3], grad_g=[6, -14], alpha=1/2, beta=2/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([14, 3], [6, -14], 1/2, 2/3, [11, -47/6], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([14, 3], [6, -14], 1/2, 2/3, [11, -47/6], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0015","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff06ec83bb5b879e166daa19debdf61ba221c51820bd05f4b59d200917b2973f","family_id":"qf-mod50-graddeep-linearity","seed":202611196015,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,1],\"gradient_f\":[[15,1],[4,1]],\"gradient_g\":[[0,1],[-15,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[15, 4], grad_g=[0, -15], alpha=1/2, beta=3/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([15, 4], [0, -15], 1/2, 1, [15/2, -13], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([15, 4], [0, -15], 1/2, 1, [15/2, -13], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0016","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"959a8d71e08a3929613a69e1223f01c32fac2de0f22b249a8d9a4e51ed64c1df","family_id":"qf-mod50-graddeep-linearity","seed":202611196016,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[4,3],\"gradient_f\":[[16,1],[0,1]],\"gradient_g\":[[1,1],[-16,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[16, 0], grad_g=[1, -16], alpha=1/2, beta=4/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([16, 0], [1, -16], 1/2, 4/3, [28/3, -64/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([16, 0], [1, -16], 1/2, 4/3, [28/3, -64/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0017","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd0a28bdcc854b683e60e4d22189ad571e0c88303c4966c879cb22098e3b637f","family_id":"qf-mod50-graddeep-linearity","seed":202611196017,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[17,1],[1,1]],\"gradient_g\":[[2,1],[-17,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[17, 1], grad_g=[2, -17], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([17, 1], [2, -17], 1/2, 1/3, [55/6, -31/6], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([17, 1], [2, -17], 1/2, 1/3, [55/6, -31/6], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0018","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b997c75017e23f4f28598b1bd62cf6e506f3ac8349512b9e5dbb7bd7f8d7e94d","family_id":"qf-mod50-graddeep-linearity","seed":202611196018,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[2,3],\"gradient_f\":[[18,1],[2,1]],\"gradient_g\":[[3,1],[-18,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[18, 2], grad_g=[3, -18], alpha=1/2, beta=2/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([18, 2], [3, -18], 1/2, 2/3, [11, -11], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([18, 2], [3, -18], 1/2, 2/3, [11, -11], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0019","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a709680275b07d725c27b91f4ef34237a39c1386b74ecb3b661033de42e9f63","family_id":"qf-mod50-graddeep-linearity","seed":202611196019,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,1],\"gradient_f\":[[19,1],[3,1]],\"gradient_g\":[[4,1],[-19,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[19, 3], grad_g=[4, -19], alpha=1/2, beta=3/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([19, 3], [4, -19], 1/2, 1, [27/2, -35/2], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([19, 3], [4, -19], 1/2, 1, [27/2, -35/2], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0020","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28b10a44325945d4139a2d783e5d5b856d163a0e60b57a6b854e7d1b2918cbb8","family_id":"qf-mod50-graddeep-linearity","seed":202611196020,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[4,3],\"gradient_f\":[[20,1],[4,1]],\"gradient_g\":[[5,1],[-20,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[20, 4], grad_g=[5, -20], alpha=1/2, beta=4/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([20, 4], [5, -20], 1/2, 4/3, [50/3, -74/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([20, 4], [5, -20], 1/2, 4/3, [50/3, -74/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0021","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8fae58fe901ebff348a4da773f0b586d97267fd618a5dc10457348e5050b840f","family_id":"qf-mod50-graddeep-linearity","seed":202611196021,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[21,1],[0,1]],\"gradient_g\":[[6,1],[-21,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[21, 0], grad_g=[6, -21], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([21, 0], [6, -21], 1/2, 1/3, [25/2, -7], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([21, 0], [6, -21], 1/2, 1/3, [25/2, -7], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0022","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a0aa5d89a7152aeb1786cc08c471fc55cabf105f9780222b16363036fd9aed6e","family_id":"qf-mod50-graddeep-linearity","seed":202611196022,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[2,3],\"gradient_f\":[[22,1],[1,1]],\"gradient_g\":[[0,1],[-22,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[22, 1], grad_g=[0, -22], alpha=1/2, beta=2/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([22, 1], [0, -22], 1/2, 2/3, [11, -85/6], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([22, 1], [0, -22], 1/2, 2/3, [11, -85/6], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0023","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56476f77ad1cf1f51a8674b32e216d03de00c4b26cb4eb362d86c2910f16dbe1","family_id":"qf-mod50-graddeep-linearity","seed":202611196023,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,1],\"gradient_f\":[[23,1],[2,1]],\"gradient_g\":[[1,1],[-23,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[23, 2], grad_g=[1, -23], alpha=1/2, beta=3/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([23, 2], [1, -23], 1/2, 1, [25/2, -22], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([23, 2], [1, -23], 1/2, 1, [25/2, -22], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0024","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8cf79ccabeab5d52d15fe53d9a801696f1064c91bd9dd0f49adb60bbb0a9e7a","family_id":"qf-mod50-graddeep-linearity","seed":202611196024,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[4,3],\"gradient_f\":[[24,1],[3,1]],\"gradient_g\":[[2,1],[-24,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[24, 3], grad_g=[2, -24], alpha=1/2, beta=4/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([24, 3], [2, -24], 1/2, 4/3, [44/3, -61/2], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([24, 3], [2, -24], 1/2, 4/3, [44/3, -61/2], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0025","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"faa6d78869bb3c95f1684bffd3610e454382d58e726aa764b4be62c73085c04f","family_id":"qf-mod50-graddeep-linearity","seed":202611196025,"semantic_template":"graddeep-linearity|{\"alpha\":[1,2],\"beta\":[1,3],\"gradient_f\":[[25,1],[4,1]],\"gradient_g\":[[3,1],[-25,1]]}"},"fields":{"title":"Gradient Linearity Preserves the Fixed Coordinate Order","stem":"Use gradients of differentiable f and g at the same point: grad_f=[25, 4], grad_g=[3, -25], alpha=1/2, beta=1/3. Give (gradient_f, gradient_g, alpha, beta, combined_gradient, coordinate_scope).","answer":{"kind":"expression","value":"([25, 4], [3, -25], 1/2, 1/3, [27/2, -19/3], pointwise_gradient_linearity_in_fixed_coordinates)","explanation":"Compute alpha*grad_f+beta*grad_g component by component in the fixed Euclidean basis. The supplied derivative vectors and differentiability premise are not inferred from a finite difference. This identity is not an extremum or convergence certificate. The exact result, retaining the requested order, is ([25, 4], [3, -25], 1/2, 1/3, [27/2, -19/3], pointwise_gradient_linearity_in_fixed_coordinates)."},"skills":["Gradient Linearity Preserves the Fixed Coordinate Order"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0026","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f753246d72d54c422194328ce884622e6215401e1f2aaa18e771647ef0f5523","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196026,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[1,1],[-1,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[1, -1] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([1, -1], [-1, 1], -2, 2, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([1, -1], [-1, 1], -2, 2, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0027","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ff44f4f8a103eb7d1c4f5e48714c25a7c773ce6a1eb4be3c70f22f90caf2699","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196027,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[2,1],[-2,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[2, -2] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([2, -2], [-2, 2], -8, 8, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([2, -2], [-2, 2], -8, 8, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0028","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7da84686e17f670df3bc3802426fb329615c4c1c718ddf79d20b67a10e29dc35","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196028,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[3,1],[-3,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[3, -3] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([3, -3], [-3, 3], -18, 18, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([3, -3], [-3, 3], -18, 18, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0029","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc89fbb676ea4178f9feabac619ee904ee6bc45246c6fffacbfcb12ec760288b","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196029,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[4,1],[-4,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[4, -4] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([4, -4], [-4, 4], -32, 32, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([4, -4], [-4, 4], -32, 32, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0030","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c41c067eb2e7f6f1423e8ea17895ffd1b43a384bb770100646cdd6efde17b0b","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196030,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[5,1],[-5,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[5, -5] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([5, -5], [-5, 5], -50, 50, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([5, -5], [-5, 5], -50, 50, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0031","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7047a21710a5280aa17414af0660b885057da68e2bbc3209531abc7286950a99","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196031,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[6,1],[-6,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[6, -6] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([6, -6], [-6, 6], -72, 72, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([6, -6], [-6, 6], -72, 72, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0032","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4384d82d6bc895118c3ddef3e75d0dc710efdb0504b87d054124126a451a7dc9","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196032,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[7,1],[-7,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[7, -7] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([7, -7], [-7, 7], -98, 98, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([7, -7], [-7, 7], -98, 98, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0033","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d1d88ed9794793bf1861d86de26713ee5ee25b9d8290c2ee422b2ab1a966593","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196033,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[8,1],[-8,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[8, -8] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([8, -8], [-8, 8], -128, 128, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([8, -8], [-8, 8], -128, 128, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0034","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"84904e7130b810897cfd251dcb5525d5d39565e07eda73de1cb4b1f7a2aa66c9","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196034,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[9,1],[-9,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[9, -9] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([9, -9], [-9, 9], -162, 162, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([9, -9], [-9, 9], -162, 162, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0035","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"60721cbc0d36a453ed3c9df1adb47576aedb79235579c9c9294eca8ae236b8c9","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196035,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[10,1],[-10,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[10, -10] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([10, -10], [-10, 10], -200, 200, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([10, -10], [-10, 10], -200, 200, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0036","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f059e708f8c11b889ea4c905dd5f3cc844acbaabadbb8b89fa6ca45a0ff712c","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196036,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[11,1],[-11,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[11, -11] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([11, -11], [-11, 11], -242, 242, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([11, -11], [-11, 11], -242, 242, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0037","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf8d532fa917af8124e50e83f44bc1d13f6a64a0d26dbb02cd9ee06ef23c83a5","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196037,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[12,1],[-12,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[12, -12] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([12, -12], [-12, 12], -288, 288, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([12, -12], [-12, 12], -288, 288, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0038","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"057a9a1c7c8f93230b73e679d9ee501cdde8b13891d915e2f7223a161103e28c","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196038,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[13,1],[-13,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[13, -13] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([13, -13], [-13, 13], -338, 338, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([13, -13], [-13, 13], -338, 338, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0039","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3659054f3c586e3d217843c66dc1ce572a2953cc048095eb30bf1ca8b743e8b5","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196039,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[14,1],[-14,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[14, -14] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([14, -14], [-14, 14], -392, 392, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([14, -14], [-14, 14], -392, 392, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0040","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f3d61a43795361e0c6e2812a68e9d6f83eacb47fc175c2218f1141cfd852cfd","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196040,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[15,1],[-15,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[15, -15] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([15, -15], [-15, 15], -450, 450, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([15, -15], [-15, 15], -450, 450, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0041","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"76a9e223e641ff71877fc9cc4c574f52c4f89a36dec5acdf44ede61b69e556e0","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196041,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[16,1],[-16,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[16, -16] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([16, -16], [-16, 16], -512, 512, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([16, -16], [-16, 16], -512, 512, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0042","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2cd9fd89413df403757ccb6a2896b29bf47920cce82bde596aa729d469b15f6","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196042,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[17,1],[-17,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[17, -17] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([17, -17], [-17, 17], -578, 578, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([17, -17], [-17, 17], -578, 578, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0043","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37206ed2a1d7c15358bbaad484ba2c6002382bfb089a54e1e1bbca31942c60de","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196043,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[18,1],[-18,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[18, -18] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([18, -18], [-18, 18], -648, 648, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([18, -18], [-18, 18], -648, 648, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0044","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5eb3524a7d7d4c99ca8f43336dd35e81a0d39ef93f84911d0ad69ee9eb5f107","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196044,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[19,1],[-19,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[19, -19] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([19, -19], [-19, 19], -722, 722, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([19, -19], [-19, 19], -722, 722, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0045","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ece30d36fbf26f02d7fc3bf00fb5e4ce235db19ac2f5d5c671385da34d17e329","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196045,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[20,1],[-20,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[20, -20] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([20, -20], [-20, 20], -800, 800, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([20, -20], [-20, 20], -800, 800, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0046","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6160a91e1c78c2bf29b768e8816e3c61660325534090d195be9dca1519117a79","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196046,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[21,1],[-21,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[21, -21] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([21, -21], [-21, 21], -882, 882, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([21, -21], [-21, 21], -882, 882, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0047","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24703b1f4c699de63e84418905817c0ba9d6fe71efeebce61d9c979db03a7c3c","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196047,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[22,1],[-22,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[22, -22] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([22, -22], [-22, 22], -968, 968, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([22, -22], [-22, 22], -968, 968, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0048","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99978e0267a24dddc9bdb2a8369efdfebeffaa332a691de0bd10a82ce45abeb2","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196048,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[23,1],[-23,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[23, -23] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([23, -23], [-23, 23], -1058, 1058, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([23, -23], [-23, 23], -1058, 1058, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0049","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19adbfc7cd1734e9b8cd990dbb42a2b361f2f52dce4db0a61b02df5508803a9f","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196049,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[24,1],[-24,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[24, -24] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([24, -24], [-24, 24], -1152, 1152, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([24, -24], [-24, 24], -1152, 1152, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0050","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf929609b1211589c418d8da800306e5e08eada67efe24139e04e61ab0e7b709","family_id":"qf-mod50-graddeep-negative-direction","seed":202611196050,"semantic_template":"graddeep-negative-direction|{\"gradient\":[[25,1],[-25,1]]}"},"fields":{"title":"A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product","stem":"Use nonzero Euclidean gradient=[25, -25] at a differentiable point. Give (gradient, negative_gradient, directional_inner_product, gradient_norm_squared, strictly_negative, first_order_scope).","answer":{"kind":"expression","value":"([25, -25], [-25, 25], -1250, 1250, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)","explanation":"Set the unnormalized direction to minus the gradient and compute its inner product with the gradient as minus the squared norm. This is a first-order directional fact, not a guarantee that every positive finite step decreases the objective. The exact result, retaining the requested order, is ([25, -25], [-25, 25], -1250, 1250, Yes, negative_gradient_has_strictly_negative_directional_inner_product_at_a_nonstationary_point)."},"skills":["A Nonzero Negative Gradient Has Strictly Negative Directional Inner Product"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0051","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f42699dad76bb524c5ff9654d4dcdc59cba1aab786d2c3b8b7847acd134a63d4","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196051,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[1,1],[1,1]],[[1,1],[2,1]]],\"x\":[[0,1],[1,1]],\"y\":[[1,1],[0,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[1, 1], [1, 2]], x=[0, 1], y=[1, 0], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[1, 1], [1, 2]], [0, 1], [1, 0], [-1, 1], [0, 1], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[1, 1], [1, 2]], [0, 1], [1, 0], [-1, 1], [0, 1], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0052","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1494e8ff2c8a23ab0242c5c25ae34c2ece11c6790759b4142d0866b15a6f8a00","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196052,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[2,1],[1,1]],[[1,1],[3,1]]],\"x\":[[1,1],[1,1]],\"y\":[[1,1],[-1,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[2, 1], [1, 3]], x=[1, 1], y=[1, -1], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[2, 1], [1, 3]], [1, 1], [1, -1], [0, 2], [2, 6], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[2, 1], [1, 3]], [1, 1], [1, -1], [0, 2], [2, 6], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0053","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b40e57ab256725c1f0746090b6dd0357a2ad29b4175c854a8c404fa54d5d6e92","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196053,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[3,1],[1,1]],[[1,1],[4,1]]],\"x\":[[2,1],[1,1]],\"y\":[[1,1],[-2,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[3, 1], [1, 4]], x=[2, 1], y=[1, -2], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[3, 1], [1, 4]], [2, 1], [1, -2], [1, 3], [6, 13], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[3, 1], [1, 4]], [2, 1], [1, -2], [1, 3], [6, 13], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0054","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"daaac49473dba5fff15f968f3da2e522aecc53e7b341d40778e351253f0e60ef","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196054,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[4,1],[1,1]],[[1,1],[5,1]]],\"x\":[[3,1],[1,1]],\"y\":[[1,1],[-3,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[4, 1], [1, 5]], x=[3, 1], y=[1, -3], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[4, 1], [1, 5]], [3, 1], [1, -3], [2, 4], [12, 22], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[4, 1], [1, 5]], [3, 1], [1, -3], [2, 4], [12, 22], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0055","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55826ee8764cc41fea8e44603257cbfdcfe6e6882affea7a4a72f46d7f07f61c","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196055,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[5,1],[1,1]],[[1,1],[6,1]]],\"x\":[[4,1],[1,1]],\"y\":[[1,1],[-4,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[5, 1], [1, 6]], x=[4, 1], y=[1, -4], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[5, 1], [1, 6]], [4, 1], [1, -4], [3, 5], [20, 33], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[5, 1], [1, 6]], [4, 1], [1, -4], [3, 5], [20, 33], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0056","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"181dc69b70532959d6bd4f68cd0263452fc630f71cfabb8ac043b0d78ecc16d5","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196056,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[6,1],[1,1]],[[1,1],[2,1]]],\"x\":[[5,1],[1,1]],\"y\":[[1,1],[-5,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[6, 1], [1, 2]], x=[5, 1], y=[1, -5], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[6, 1], [1, 2]], [5, 1], [1, -5], [4, 6], [30, 16], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[6, 1], [1, 2]], [5, 1], [1, -5], [4, 6], [30, 16], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0057","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31a1a6eb9e8f17e68064846a60d1b903c6df073028439a1d512a8755b8b2734e","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196057,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[7,1],[1,1]],[[1,1],[3,1]]],\"x\":[[6,1],[1,1]],\"y\":[[1,1],[-6,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[7, 1], [1, 3]], x=[6, 1], y=[1, -6], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[7, 1], [1, 3]], [6, 1], [1, -6], [5, 7], [42, 26], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[7, 1], [1, 3]], [6, 1], [1, -6], [5, 7], [42, 26], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0058","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"553676df0cf20c903c7431cb4fa9f899a34998bf5b8d1a81301c4e4dc0b817ad","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196058,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[8,1],[1,1]],[[1,1],[4,1]]],\"x\":[[7,1],[1,1]],\"y\":[[1,1],[-7,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[8, 1], [1, 4]], x=[7, 1], y=[1, -7], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[8, 1], [1, 4]], [7, 1], [1, -7], [6, 8], [56, 38], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[8, 1], [1, 4]], [7, 1], [1, -7], [6, 8], [56, 38], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0059","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec3d27a98293010e8a10902b5222adef4c6caddf5284fcabc4efaffcdae5aaba","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196059,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[9,1],[1,1]],[[1,1],[5,1]]],\"x\":[[8,1],[1,1]],\"y\":[[1,1],[-8,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[9, 1], [1, 5]], x=[8, 1], y=[1, -8], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[9, 1], [1, 5]], [8, 1], [1, -8], [7, 9], [72, 52], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[9, 1], [1, 5]], [8, 1], [1, -8], [7, 9], [72, 52], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0060","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f829d72aa94fa0a32721ddc848d702ff6bb7c896d4bfdb6498974a3f27d8bf9a","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196060,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[10,1],[1,1]],[[1,1],[6,1]]],\"x\":[[9,1],[1,1]],\"y\":[[1,1],[-9,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[10, 1], [1, 6]], x=[9, 1], y=[1, -9], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[10, 1], [1, 6]], [9, 1], [1, -9], [8, 10], [90, 68], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[10, 1], [1, 6]], [9, 1], [1, -9], [8, 10], [90, 68], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0061","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dbf54cf9fbd8e482b035fb260c28e793f2f5a854d2ff8d111b5c3d634482a29d","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196061,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[11,1],[1,1]],[[1,1],[2,1]]],\"x\":[[10,1],[1,1]],\"y\":[[1,1],[-10,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[11, 1], [1, 2]], x=[10, 1], y=[1, -10], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[11, 1], [1, 2]], [10, 1], [1, -10], [9, 11], [110, 31], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[11, 1], [1, 2]], [10, 1], [1, -10], [9, 11], [110, 31], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0062","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b0294757a215eec12d6b570776e584fc645cdac519f4adc8afacb9463df1e10","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196062,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[12,1],[1,1]],[[1,1],[3,1]]],\"x\":[[11,1],[1,1]],\"y\":[[1,1],[-11,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[12, 1], [1, 3]], x=[11, 1], y=[1, -11], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[12, 1], [1, 3]], [11, 1], [1, -11], [10, 12], [132, 46], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[12, 1], [1, 3]], [11, 1], [1, -11], [10, 12], [132, 46], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0063","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d40ada5a5d7d620d61cf0590b8d5d629e4fb1c2e2f1551e36bf24b8561e7f830","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196063,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[13,1],[1,1]],[[1,1],[4,1]]],\"x\":[[12,1],[1,1]],\"y\":[[1,1],[-12,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[13, 1], [1, 4]], x=[12, 1], y=[1, -12], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[13, 1], [1, 4]], [12, 1], [1, -12], [11, 13], [156, 63], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[13, 1], [1, 4]], [12, 1], [1, -12], [11, 13], [156, 63], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0064","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a899bd4a83ee733e730eeb218f6af9c8dc488c4adf7e61aab9bbeeed3808b0df","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196064,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[14,1],[1,1]],[[1,1],[5,1]]],\"x\":[[13,1],[1,1]],\"y\":[[1,1],[-13,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[14, 1], [1, 5]], x=[13, 1], y=[1, -13], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[14, 1], [1, 5]], [13, 1], [1, -13], [12, 14], [182, 82], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[14, 1], [1, 5]], [13, 1], [1, -13], [12, 14], [182, 82], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0065","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72ac4167742c1b26b8ad00b3c5d6f5a55812a043743525c628264ebf07a7f15d","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196065,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[15,1],[1,1]],[[1,1],[6,1]]],\"x\":[[14,1],[1,1]],\"y\":[[1,1],[-14,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[15, 1], [1, 6]], x=[14, 1], y=[1, -14], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[15, 1], [1, 6]], [14, 1], [1, -14], [13, 15], [210, 103], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[15, 1], [1, 6]], [14, 1], [1, -14], [13, 15], [210, 103], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0066","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca9c0fb985a011a9914882496bfa396150ad4e68a60cf29fe56a19258d66f55f","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196066,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[16,1],[1,1]],[[1,1],[2,1]]],\"x\":[[15,1],[1,1]],\"y\":[[1,1],[-15,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[16, 1], [1, 2]], x=[15, 1], y=[1, -15], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[16, 1], [1, 2]], [15, 1], [1, -15], [14, 16], [240, 46], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[16, 1], [1, 2]], [15, 1], [1, -15], [14, 16], [240, 46], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0067","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f66e6be9ef7f5d5829be268dc03d4b006f9685a5d3b8a6bb9aa6fa95abcc444c","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196067,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[17,1],[1,1]],[[1,1],[3,1]]],\"x\":[[16,1],[1,1]],\"y\":[[1,1],[-16,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[17, 1], [1, 3]], x=[16, 1], y=[1, -16], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[17, 1], [1, 3]], [16, 1], [1, -16], [15, 17], [272, 66], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[17, 1], [1, 3]], [16, 1], [1, -16], [15, 17], [272, 66], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0068","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9bf2cbc465bd8c4d6e81a4bcda084896b6d6c386d39e7ca12f927b7907b30b7","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196068,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[18,1],[1,1]],[[1,1],[4,1]]],\"x\":[[17,1],[1,1]],\"y\":[[1,1],[-17,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[18, 1], [1, 4]], x=[17, 1], y=[1, -17], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[18, 1], [1, 4]], [17, 1], [1, -17], [16, 18], [306, 88], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[18, 1], [1, 4]], [17, 1], [1, -17], [16, 18], [306, 88], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0069","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01c03cd801c460c314de00c1ec666cbeb7d215ad151fc85f94a108476e27a385","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196069,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[19,1],[1,1]],[[1,1],[5,1]]],\"x\":[[18,1],[1,1]],\"y\":[[1,1],[-18,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[19, 1], [1, 5]], x=[18, 1], y=[1, -18], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[19, 1], [1, 5]], [18, 1], [1, -18], [17, 19], [342, 112], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[19, 1], [1, 5]], [18, 1], [1, -18], [17, 19], [342, 112], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0070","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49c93799ccc5dafc76cd93da3ba8971668274a8867b28dbb521543580fc12425","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196070,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[20,1],[1,1]],[[1,1],[6,1]]],\"x\":[[19,1],[1,1]],\"y\":[[1,1],[-19,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[20, 1], [1, 6]], x=[19, 1], y=[1, -19], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[20, 1], [1, 6]], [19, 1], [1, -19], [18, 20], [380, 138], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[20, 1], [1, 6]], [19, 1], [1, -19], [18, 20], [380, 138], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0071","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5eb201fdce0174f249d007bf6db17810f2e541ed264c10ecfda5e9dc59f65c88","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196071,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[21,1],[1,1]],[[1,1],[2,1]]],\"x\":[[20,1],[1,1]],\"y\":[[1,1],[-20,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[21, 1], [1, 2]], x=[20, 1], y=[1, -20], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[21, 1], [1, 2]], [20, 1], [1, -20], [19, 21], [420, 61], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[21, 1], [1, 2]], [20, 1], [1, -20], [19, 21], [420, 61], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0072","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fb46e4ff67a089e2acc98eabddccab0594aaf0064c57d18d7b9568b06da0b57","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196072,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[22,1],[1,1]],[[1,1],[3,1]]],\"x\":[[21,1],[1,1]],\"y\":[[1,1],[-21,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[22, 1], [1, 3]], x=[21, 1], y=[1, -21], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[22, 1], [1, 3]], [21, 1], [1, -21], [20, 22], [462, 86], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[22, 1], [1, 3]], [21, 1], [1, -21], [20, 22], [462, 86], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0073","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9435b2b6d25f53b2d32646d1ccde100f7c833624812eae95d80400a690138b0d","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196073,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[23,1],[1,1]],[[1,1],[4,1]]],\"x\":[[22,1],[1,1]],\"y\":[[1,1],[-22,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[23, 1], [1, 4]], x=[22, 1], y=[1, -22], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[23, 1], [1, 4]], [22, 1], [1, -22], [21, 23], [506, 113], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[23, 1], [1, 4]], [22, 1], [1, -22], [21, 23], [506, 113], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0074","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"354d997726ceeb8031a984002c946a6324ca2f9c4d28fb5ba0410cd54ce5f0b9","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196074,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[24,1],[1,1]],[[1,1],[5,1]]],\"x\":[[23,1],[1,1]],\"y\":[[1,1],[-23,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[24, 1], [1, 5]], x=[23, 1], y=[1, -23], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[24, 1], [1, 5]], [23, 1], [1, -23], [22, 24], [552, 142], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[24, 1], [1, 5]], [23, 1], [1, -23], [22, 24], [552, 142], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0075","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"56997860f8d12f916fb3af1f3aa50efce0d93698e60b9d508357e046ca3bc53a","family_id":"qf-mod50-graddeep-quadratic-difference","seed":202611196075,"semantic_template":"graddeep-quadratic-difference|{\"symmetric_matrix\":[[[25,1],[1,1]],[[1,1],[6,1]]],\"x\":[[24,1],[1,1]],\"y\":[[1,1],[-24,1]]}"},"fields":{"title":"A Symmetric Quadratic Has Gradient Difference Q Times Point Difference","stem":"Use symmetric square Q=[[25, 1], [1, 6]], x=[24, 1], y=[1, -24], for f(u)=1/2*u^T*Q*u. Give (symmetric_matrix, x, y, point_difference, gradient_difference, quadratic_identity).","answer":{"kind":"expression","value":"([[25, 1], [1, 6]], [24, 1], [1, -24], [23, 25], [600, 173], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)","explanation":"Retain x−y and compute Q*(x−y) in exact coordinate order. Symmetry supplies gradient Q*u. The identity does not assert positive definiteness, convexity, a minimum or convergence. The exact result, retaining the requested order, is ([[25, 1], [1, 6]], [24, 1], [1, -24], [23, 25], [600, 173], grad_f_x_minus_grad_f_y_equals_Q_times_x_minus_y_for_symmetric_quadratic)."},"skills":["A Symmetric Quadratic Has Gradient Difference Q Times Point Difference"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0076","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0e12d9102a389dda4e36c9d53d978eeaff698770061a3d6042c7a1b5ea20249","family_id":"qf-mod50-graddeep-boundary","seed":202611196076,"semantic_template":"graddeep-boundary|{\"mode\":\"zero_gradient\",\"model\":{\"dimension\":1}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=zero_gradient, complete model={\"dimension\":1}. Give (mode, dimension, zero_gradient, stationary, minimum_claim, boundary).","answer":{"kind":"expression","value":"(zero_gradient, 1, [0], Yes, No, zero_gradient_only_certifies_stationarity)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (zero_gradient, 1, [0], Yes, No, zero_gradient_only_certifies_stationarity)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0077","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af711e15ebe43c38c55686f9cc080c6248e5af6a77363714dea8bb5ed3378539","family_id":"qf-mod50-graddeep-boundary","seed":202611196077,"semantic_template":"graddeep-boundary|{\"mode\":\"positive_gradient_direction\",\"model\":{\"gradient\":[[2,1],[-1,1]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=positive_gradient_direction, complete model={\"gradient\":[2,-1]}. Give (mode, gradient, positive_direction, directional_inner_product, positive, descent_claim, boundary).","answer":{"kind":"expression","value":"(positive_gradient_direction, [2, -1], [2, -1], 5, Yes, No, positive_gradient_is_first_order_ascent_not_descent)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (positive_gradient_direction, [2, -1], [2, -1], 5, Yes, No, positive_gradient_is_first_order_ascent_not_descent)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0078","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf7c0b489017bf5fb71474ec7436ebd66316f11bf0b79c3f089882d9ebbc3844","family_id":"qf-mod50-graddeep-boundary","seed":202611196078,"semantic_template":"graddeep-boundary|{\"mode\":\"non_euclidean_metric\",\"model\":{\"coordinate_gradient\":[[3,1],[1,1]],\"metric\":[[[2,1],[0,1]],[[0,1],[3,1]]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=non_euclidean_metric, complete model={\"coordinate_gradient\":[3,1],\"metric\":[[2,0],[0,3]]}. Give (mode, coordinate_partials_covector, positive_definite_metric, metric_gradient, metric_independence_claim, boundary).","answer":{"kind":"expression","value":"(non_euclidean_metric, [3, 1], [[2, 0], [0, 3]], [3/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (non_euclidean_metric, [3, 1], [[2, 0], [0, 3]], [3/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0079","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"562d01c975a1832c20fc720dc1ddda30fea508b58b54006c2f97391656766147","family_id":"qf-mod50-graddeep-boundary","seed":202611196079,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[4,1],\"step\":[1,5]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":4,\"step\":[1,5]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 4, 1/5, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 4, 1/5, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0080","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c10733d4169360e6308613157221da8496dd1d515401a1b2392e2c8b99887127","family_id":"qf-mod50-graddeep-boundary","seed":202611196080,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[5,1],\"step\":[1,6]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":5,\"step\":[1,6]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 5, 1/6, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 5, 1/6, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0081","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d2d72927c779cb10fdfaa5f8f9553bce6aaf23b2578112fffc25068c7008dc2","family_id":"qf-mod50-graddeep-boundary","seed":202611196081,"semantic_template":"graddeep-boundary|{\"mode\":\"positive_gradient_direction\",\"model\":{\"gradient\":[[6,1],[-1,1]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=positive_gradient_direction, complete model={\"gradient\":[6,-1]}. Give (mode, gradient, positive_direction, directional_inner_product, positive, descent_claim, boundary).","answer":{"kind":"expression","value":"(positive_gradient_direction, [6, -1], [6, -1], 37, Yes, No, positive_gradient_is_first_order_ascent_not_descent)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (positive_gradient_direction, [6, -1], [6, -1], 37, Yes, No, positive_gradient_is_first_order_ascent_not_descent)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0082","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"833162a34c54b1444014d17f1156b1ebad983bffb91de454108c8de087636c3c","family_id":"qf-mod50-graddeep-boundary","seed":202611196082,"semantic_template":"graddeep-boundary|{\"mode\":\"non_euclidean_metric\",\"model\":{\"coordinate_gradient\":[[7,1],[1,1]],\"metric\":[[[2,1],[0,1]],[[0,1],[3,1]]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=non_euclidean_metric, complete model={\"coordinate_gradient\":[7,1],\"metric\":[[2,0],[0,3]]}. Give (mode, coordinate_partials_covector, positive_definite_metric, metric_gradient, metric_independence_claim, boundary).","answer":{"kind":"expression","value":"(non_euclidean_metric, [7, 1], [[2, 0], [0, 3]], [7/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (non_euclidean_metric, [7, 1], [[2, 0], [0, 3]], [7/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0083","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86031f518e6a1def8f5a6be1466adb349793e90f8d84efaf6261661248879482","family_id":"qf-mod50-graddeep-boundary","seed":202611196083,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[8,1],\"step\":[1,9]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":8,\"step\":[1,9]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 8, 1/9, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 8, 1/9, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0084","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cddcce04c44df368fd43c786fa58ddc20e8002637c92cf7070107cc2e94d4e51","family_id":"qf-mod50-graddeep-boundary","seed":202611196084,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[9,1],\"step\":[1,10]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":9,\"step\":[1,10]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 9, 1/10, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 9, 1/10, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0085","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"009d3b9777ecc57ec2c6eacfaacc81060f8b7258340d5e99299a7b99fedc2b4e","family_id":"qf-mod50-graddeep-boundary","seed":202611196085,"semantic_template":"graddeep-boundary|{\"mode\":\"positive_gradient_direction\",\"model\":{\"gradient\":[[10,1],[-1,1]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=positive_gradient_direction, complete model={\"gradient\":[10,-1]}. Give (mode, gradient, positive_direction, directional_inner_product, positive, descent_claim, boundary).","answer":{"kind":"expression","value":"(positive_gradient_direction, [10, -1], [10, -1], 101, Yes, No, positive_gradient_is_first_order_ascent_not_descent)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (positive_gradient_direction, [10, -1], [10, -1], 101, Yes, No, positive_gradient_is_first_order_ascent_not_descent)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0086","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b896b6db0a0c5d3e1324fe85354e3a4db0a26bc18f78db2361fe47168405103a","family_id":"qf-mod50-graddeep-boundary","seed":202611196086,"semantic_template":"graddeep-boundary|{\"mode\":\"non_euclidean_metric\",\"model\":{\"coordinate_gradient\":[[11,1],[1,1]],\"metric\":[[[2,1],[0,1]],[[0,1],[3,1]]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=non_euclidean_metric, complete model={\"coordinate_gradient\":[11,1],\"metric\":[[2,0],[0,3]]}. Give (mode, coordinate_partials_covector, positive_definite_metric, metric_gradient, metric_independence_claim, boundary).","answer":{"kind":"expression","value":"(non_euclidean_metric, [11, 1], [[2, 0], [0, 3]], [11/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (non_euclidean_metric, [11, 1], [[2, 0], [0, 3]], [11/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0087","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2440e8edd8eaabb2ff7cff5a5d4dfccb7443e02517b5f894c4a69854b5a3625","family_id":"qf-mod50-graddeep-boundary","seed":202611196087,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[12,1],\"step\":[1,13]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":12,\"step\":[1,13]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 12, 1/13, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 12, 1/13, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0088","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"168066a7d2b9d9fc2e650086237d614d195d57e267baa0f056bc2172a4829642","family_id":"qf-mod50-graddeep-boundary","seed":202611196088,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[13,1],\"step\":[1,14]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":13,\"step\":[1,14]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 13, 1/14, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 13, 1/14, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0089","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"18aabcaa45aee4581727345ff40b3f5b299387fc221716b82928746d8343fbb1","family_id":"qf-mod50-graddeep-boundary","seed":202611196089,"semantic_template":"graddeep-boundary|{\"mode\":\"positive_gradient_direction\",\"model\":{\"gradient\":[[14,1],[-1,1]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=positive_gradient_direction, complete model={\"gradient\":[14,-1]}. Give (mode, gradient, positive_direction, directional_inner_product, positive, descent_claim, boundary).","answer":{"kind":"expression","value":"(positive_gradient_direction, [14, -1], [14, -1], 197, Yes, No, positive_gradient_is_first_order_ascent_not_descent)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (positive_gradient_direction, [14, -1], [14, -1], 197, Yes, No, positive_gradient_is_first_order_ascent_not_descent)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0090","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea75c0e8c41692866298505a17ffc8e5273ca83cdcc742d939896adf0e4acf64","family_id":"qf-mod50-graddeep-boundary","seed":202611196090,"semantic_template":"graddeep-boundary|{\"mode\":\"non_euclidean_metric\",\"model\":{\"coordinate_gradient\":[[15,1],[1,1]],\"metric\":[[[2,1],[0,1]],[[0,1],[3,1]]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=non_euclidean_metric, complete model={\"coordinate_gradient\":[15,1],\"metric\":[[2,0],[0,3]]}. Give (mode, coordinate_partials_covector, positive_definite_metric, metric_gradient, metric_independence_claim, boundary).","answer":{"kind":"expression","value":"(non_euclidean_metric, [15, 1], [[2, 0], [0, 3]], [15/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (non_euclidean_metric, [15, 1], [[2, 0], [0, 3]], [15/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0091","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd9ca6cf78eabb61b106e68e4fac7f2860e244b1bf5f87556f9cba74b4e04d44","family_id":"qf-mod50-graddeep-boundary","seed":202611196091,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[16,1],\"step\":[1,17]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":16,\"step\":[1,17]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 16, 1/17, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 16, 1/17, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0092","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3897015c77c13beb195d3e330885c0e6087b04e871f7d22207cff0036733efa","family_id":"qf-mod50-graddeep-boundary","seed":202611196092,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[17,1],\"step\":[1,18]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":17,\"step\":[1,18]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 17, 1/18, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 17, 1/18, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0093","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0833dc74a6291e9f2d372c0319080e3ec563d559ab937ec89231308f99529387","family_id":"qf-mod50-graddeep-boundary","seed":202611196093,"semantic_template":"graddeep-boundary|{\"mode\":\"positive_gradient_direction\",\"model\":{\"gradient\":[[18,1],[-1,1]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=positive_gradient_direction, complete model={\"gradient\":[18,-1]}. Give (mode, gradient, positive_direction, directional_inner_product, positive, descent_claim, boundary).","answer":{"kind":"expression","value":"(positive_gradient_direction, [18, -1], [18, -1], 325, Yes, No, positive_gradient_is_first_order_ascent_not_descent)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (positive_gradient_direction, [18, -1], [18, -1], 325, Yes, No, positive_gradient_is_first_order_ascent_not_descent)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0094","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e231fc1a87bf8ca7c6309416d15e933031956ac5c494858b6e3a1c08208c1b2","family_id":"qf-mod50-graddeep-boundary","seed":202611196094,"semantic_template":"graddeep-boundary|{\"mode\":\"non_euclidean_metric\",\"model\":{\"coordinate_gradient\":[[19,1],[1,1]],\"metric\":[[[2,1],[0,1]],[[0,1],[3,1]]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=non_euclidean_metric, complete model={\"coordinate_gradient\":[19,1],\"metric\":[[2,0],[0,3]]}. Give (mode, coordinate_partials_covector, positive_definite_metric, metric_gradient, metric_independence_claim, boundary).","answer":{"kind":"expression","value":"(non_euclidean_metric, [19, 1], [[2, 0], [0, 3]], [19/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (non_euclidean_metric, [19, 1], [[2, 0], [0, 3]], [19/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0095","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"138459db628e51d646532de0d23f62f9a9200d4d14ff152b7be5902615b9b604","family_id":"qf-mod50-graddeep-boundary","seed":202611196095,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[20,1],\"step\":[1,21]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":20,\"step\":[1,21]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 20, 1/21, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 20, 1/21, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0096","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1f3eece1455fa36126f58fe9af53f523c0b5d65c4f98c510454f85743c06099","family_id":"qf-mod50-graddeep-boundary","seed":202611196096,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[21,1],\"step\":[1,22]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":21,\"step\":[1,22]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 21, 1/22, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 21, 1/22, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0097","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1361fb70a7289e81bc422c8df1b06b2e5c27821ccac5da014a1f85939a940916","family_id":"qf-mod50-graddeep-boundary","seed":202611196097,"semantic_template":"graddeep-boundary|{\"mode\":\"positive_gradient_direction\",\"model\":{\"gradient\":[[22,1],[-1,1]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=positive_gradient_direction, complete model={\"gradient\":[22,-1]}. Give (mode, gradient, positive_direction, directional_inner_product, positive, descent_claim, boundary).","answer":{"kind":"expression","value":"(positive_gradient_direction, [22, -1], [22, -1], 485, Yes, No, positive_gradient_is_first_order_ascent_not_descent)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (positive_gradient_direction, [22, -1], [22, -1], 485, Yes, No, positive_gradient_is_first_order_ascent_not_descent)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0098","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c6c6f9aad61fd2ee91cc5cddb8adea5d054c5d014569d92fedc7d71b08eb7b49","family_id":"qf-mod50-graddeep-boundary","seed":202611196098,"semantic_template":"graddeep-boundary|{\"mode\":\"non_euclidean_metric\",\"model\":{\"coordinate_gradient\":[[23,1],[1,1]],\"metric\":[[[2,1],[0,1]],[[0,1],[3,1]]]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=non_euclidean_metric, complete model={\"coordinate_gradient\":[23,1],\"metric\":[[2,0],[0,3]]}. Give (mode, coordinate_partials_covector, positive_definite_metric, metric_gradient, metric_independence_claim, boundary).","answer":{"kind":"expression","value":"(non_euclidean_metric, [23, 1], [[2, 0], [0, 3]], [23/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (non_euclidean_metric, [23, 1], [[2, 0], [0, 3]], [23/2, 1/3], No, coordinate_partials_become_a_gradient_vector_only_after_the_metric_identification)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0099","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"016099b47bed00878de293b0ca7df074eed6cf5b467ff198c57613c3af5e7c58","family_id":"qf-mod50-graddeep-boundary","seed":202611196099,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[24,1],\"step\":[1,25]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":24,\"step\":[1,25]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 24, 1/25, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 24, 1/25, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-pulse-20261119-r06-0100","batch_id":"batch-20261015-advance-20261119-r06-gradient-vector-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d59ef5361d163f1dc5a0e7ffa834f4496f3bdf2f8432ab2f8334094a4df6c619","family_id":"qf-mod50-graddeep-boundary","seed":202611196100,"semantic_template":"graddeep-boundary|{\"mode\":\"finite_difference_confusion\",\"model\":{\"difference_quotient\":[25,1],\"step\":[1,26]}}"},"fields":{"title":"Stationarity, Metric Identification and Finite Differences Have Different Responsibilities","stem":"Use mode=finite_difference_confusion, complete model={\"difference_quotient\":25,\"step\":[1,26]}. Give (mode, finite_difference_quotient, nonzero_step, derivative_or_gradient_certificate_claim, boundary).","answer":{"kind":"expression","value":"(finite_difference_confusion, 25, 1/26, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)","explanation":"A zero gradient certifies stationarity only. The positive gradient has positive directional inner product and is first-order ascent. For a declared positive-definite metric M, identify the coordinate-partial covector by solving M*grad=partials rather than assuming the Euclidean vector. A single nonzero-step finite difference is not a derivative or gradient certificate. The exact result, retaining the requested order, is (finite_difference_confusion, 25, 1/26, No, one_finite_difference_quotient_is_not_a_derivative_or_gradient_certificate)."},"skills":["Stationarity, Metric Identification and Finite Differences Have Different Responsibilities"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}