{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261013-advance-20261025-r07-partial-derivative-0076", "pilot_number": "R0066-33576", "number": 33576, "family_id": "qf-mod25-partialdeep-slice-difference-quotient-limit", "title": { "zh-Hant": "第 33576 題：偏導密度", "en": "Question 33576: Cancel Only Nonzero h before Taking the Symbolic Slice Limit" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261013-advance-20261025-r07-partial-derivative-0076", "batch_id": "batch-20261013-advance-20261025-r07-partial-derivative-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-partial-derivative" ], "title_zh": "偏導密度", "stem_zh": "固定另一座標的多項式切片。操作=partialdeep-slice-difference-quotient-limit；資料={\"coefficients\":[-12,-2,-3,1,-1,-2],\"x\":-2,\"y\":-3}。請依 [coefficients,x,y,quotient_coefficients,quotient_formula,limit,scope] 欄序回答。", "answer": { "kind": "expression", "value": "([-12, -2, -3, 1, -1, -2], -2, -3, [-3, 1], quotient=constant+h_coefficient*h for h≠0, -3, symbolic_limit_not_finite_difference)", "explanation_zh": "收束波核對=([-12, -2, -3, 1, -1, -2], -2, -3, [-3, 1], quotient=constant+h_coefficient*h for h≠0, -3, symbolic_limit_not_finite_difference)；依完整明示資料重算；精確有限切片不自動升格為一般無限結論；符號消去h後才取極限；不以有限差分證偏導或推出全微分。" }, "difficulty": { "band": "undergraduate_core", "level": 1, "skills": [ "偏導密度" ] }, "variant": { "family_id": "qf-mod25-partialdeep-slice-difference-quotient-limit", "seed": 202610257076, "parameters": { "coefficients": [ -12, -2, -3, 1, -1, -2 ], "x": -2, "y": -3, "domain": "Polynomial_partial_derivative_density", "operation": "partialdeep-slice-difference-quotient-limit", "semantic_template": "qf-mod25-partialdeep-slice-difference-quotient-limit/{\"coefficients\":[-12,-2,-3,1,-1,-2],\"x\":-2,\"y\":-3}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "產生端候選須由主端重算；不構成人工審定、一般無限定理、BVP適定性、全域最佳解、Markov長期性、Lean執行或形式Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-13", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261013-advance-20261025-r07-partial-derivative-0076", "batch_id": "batch-20261013-advance-20261025-r07-partial-derivative-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "200f3e0bd14ea5f3dacc78ef6edd9689a40c6f7e565f5282c8d2aa7cc2f311ec", "family_id": "qf-mod25-partialdeep-slice-difference-quotient-limit", "seed": 202610257076, "semantic_template": "qf-mod25-partialdeep-slice-difference-quotient-limit/{\"coefficients\":[-12,-2,-3,1,-1,-2],\"x\":-2,\"y\":-3}" }, "fields": { "title": "Cancel Only Nonzero h before Taking the Symbolic Slice Limit", "stem": "Use the quadratic candidate coefficients=[-12, -2, -3, 1, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Fix the other coordinate before taking a one-variable derivative. These symbolic polynomial identities do not replace a limit by one finite-difference sample or establish total differentiability of general functions. point x=-2, y=-3; keep y fixed. Give (coefficients,x,y,[constant,h coefficient] of the exact x-difference quotient,quotient=constant+h_coefficient*h for h≠0,limit,symbolic_limit_not_finite_difference).", "answer": { "kind": "expression", "value": "([-12, -2, -3, 1, -1, -2], -2, -3, [-3, 1], quotient=constant+h_coefficient*h for h≠0, -3, symbolic_limit_not_finite_difference)", "explanation": "Expand u(x+h,y)−u(x,y), divide only for nonzero h and then take h→0 in the resulting polynomial. The h coefficient is c_xx and the limit is the constant quotient coefficient, not one sampled finite difference. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-12, -2, -3, 1, -1, -2], -2, -3, [-3, 1], quotient=constant+h_coefficient*h for h≠0, -3, symbolic_limit_not_finite_difference)." }, "skills": [ "Cancel Only Nonzero h before Taking the Symbolic Slice Limit" ], "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." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "固定另一座標的多項式切片。操作=partialdeep-slice-difference-quotient-limit；資料={\"coefficients\":[-12,-2,-3,1,-1,-2],\"x\":-2,\"y\":-3}。請依 [coefficients,x,y,quotient_coefficients,quotient_formula,limit,scope] 欄序回答。", "en": "Use the quadratic candidate coefficients=[-12, -2, -3, 1, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Fix the other coordinate before taking a one-variable derivative. These symbolic polynomial identities do not replace a limit by one finite-difference sample or establish total differentiability of general functions. point x=-2, y=-3; keep y fixed. Give (coefficients,x,y,[constant,h coefficient] of the exact x-difference quotient,quotient=constant+h_coefficient*h for h≠0,limit,symbolic_limit_not_finite_difference)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source roles = [qf-mod25-partialdeep-slice-difference-quotient-limit, [coefficients, domain, operation, x, y]]", "after": "derived = [partialdeep-slice-difference-quotient-limit, Polynomial_partial_derivative_density]", "relation": "implies", "rule_id": "R66_EXACT_source_bank_keys_local_operation_domain", "conditions": [], "explanation": { "zh-Hant": "原bank的operation是LOCAL hyphen，keys與domain各自精確匹配。欄位順序及數字/Boolean/字串角色依來源，所有有理數文字整數省略/1；不改原題與英文、MKO、AST或狀態。", "en": "THIS source uses LOCAL-hyphen operations with exact keys/domain.Keep source field order and numeric/Boolean/string roles;integer rational texts omit/1.Original questions,English,MKOs,ASTs and statuses remain unchanged." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反向核對本步原始角色、完整前提與欄序；只接受這些條件下的結論，不把單向結論還原為唯一前提，也不把有限核對稱一般形式證明。", "en": "Reverse-audit THIS step against its ORIGINAL ordered roles and complete premises.Accept the conclusion only under these conditions;do not invert a one-way implication or call a finite check a general formal proof." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "source roles = [[-12, -2, -3, 1, -1, -2]]", "after": "derived = [1, x, y, x^2, x*y, y^2]", "relation": "implies", "rule_id": "Partial_FULL_polynomial_basis_and_fixed_other_coordinate", "conditions": [], "explanation": { "zh-Hant": "六係數依[1,x,y,x²,xy,y²]完整排列。偏導時另一座標固定；這個完整多項式支持符號恆等式，不從一個差分樣本猜導數或推一般函數全微分。", "en": "The SIX coefficients use the full[1,x,y,x²,xy,y²]basis.Hold the OTHER coordinate fixed.This supplied polynomial supports symbolic identities,not inference of a derivative from one difference sample or total differentiability of arbitrary functions." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反向核對本步原始角色、完整前提與欄序；只接受這些條件下的結論，不把單向結論還原為唯一前提，也不把有限核對稱一般形式證明。", "en": "Reverse-audit THIS step against its ORIGINAL ordered roles and complete premises.Accept the conclusion only under these conditions;do not invert a one-way implication or call a finite check a general formal proof." } }, "arithmetic": null }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "Quotient ORIGINAL twice c_xx := mul(2, 1)", "after": "Quotient ORIGINAL twice c_xx = 2", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步原操作數與次序精確運算；來源有理數的pre-reduction界另行保留，不用浮點或偷偷跨bank放寬。", "en": "Calculate exactly from THIS step’s ORIGINAL ordered operands.Retain the SOURCE rational pre-reduction envelope separately;do not use floating values or silently widen a different bank." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Quotient ORIGINAL twice c_xx，不以小數近似替代。", "en": "Recompute Quotient ORIGINAL twice c_xx with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2", "1" ], "output": "2", "label": "Quotient ORIGINAL twice c_xx" } }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "Quotient ORIGINAL twice c_xx times x := mul(2, -2)", "after": "Quotient ORIGINAL twice c_xx times x = -4", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步原操作數與次序精確運算；來源有理數的pre-reduction界另行保留，不用浮點或偷偷跨bank放寬。", "en": "Calculate exactly from THIS step’s ORIGINAL ordered operands.Retain the SOURCE rational pre-reduction envelope separately;do not use floating values or silently widen a different bank." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Quotient ORIGINAL twice c_xx times x，不以小數近似替代。", "en": "Recompute Quotient ORIGINAL twice c_xx times x with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2", "-2" ], "output": "-4", "label": "Quotient ORIGINAL twice c_xx times x" } }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "Quotient ORIGINAL c_xy times fixed y := mul(-1, -3)", "after": "Quotient ORIGINAL c_xy times fixed y = 3", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步原操作數與次序精確運算；來源有理數的pre-reduction界另行保留，不用浮點或偷偷跨bank放寬。", "en": "Calculate exactly from THIS step’s ORIGINAL ordered operands.Retain the SOURCE rational pre-reduction envelope separately;do not use floating values or silently widen a different bank." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Quotient ORIGINAL c_xy times fixed y，不以小數近似替代。", "en": "Recompute Quotient ORIGINAL c_xy times fixed y with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "-1", "-3" ], "output": "3", "label": "Quotient ORIGINAL c_xy times fixed y" } }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "Quotient ORIGINAL all linear h coefficients := sum(-2, -4, 3)", "after": "Quotient ORIGINAL all linear h coefficients = -3", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步原操作數與次序精確運算；來源有理數的pre-reduction界另行保留，不用浮點或偷偷跨bank放寬。", "en": "Calculate exactly from THIS step’s ORIGINAL ordered operands.Retain the SOURCE rational pre-reduction envelope separately;do not use floating values or silently widen a different bank." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Quotient ORIGINAL all linear h coefficients，不以小數近似替代。", "en": "Recompute Quotient ORIGINAL all linear h coefficients with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "-2", "-4", "3" ], "output": "-3", "label": "Quotient ORIGINAL all linear h coefficients" } }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "source roles = [[-12, -2, -3, 1, -1, -2], -2, -3]", "after": "derived = [u(x+h,y)-u(x,y)=(-3)*h+(1)*h^2]", "relation": "implies", "rule_id": "Partial_SYMBOLIC_difference_expand_ALL_six_terms", "conditions": [], "explanation": { "zh-Hant": "展開全部原六項並相減：常數/y/y²項消去，x項給c_x*h，x²項給2*c_xx*x*h+c_xx*h²，xy項給c_xy*y*h。這是每個h的符號恆等式。", "en": "Expand and subtract ALL six original terms:constant/y/y² cancel,the x term gives c_x*h,the x² term gives2*c_xx*x*h+c_xx*h²,and xy gives c_xy*y*h.This is a symbolic identity for every h." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反向核對本步原始角色、完整前提與欄序；只接受這些條件下的結論，不把單向結論還原為唯一前提，也不把有限核對稱一般形式證明。", "en": "Reverse-audit THIS step against its ORIGINAL ordered roles and complete premises.Accept the conclusion only under these conditions;do not invert a one-way implication or call a finite check a general formal proof." } }, "arithmetic": null }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "source roles = [-3, 1, -2, -3]", "after": "derived = [[-3, 1], -3, h!=0 before division;h->0 afterwards]", "relation": "implies", "rule_id": "Partial_CANCEL_h_ONLY_nonzero_THEN_polynomial_limit", "conditions": [], "explanation": { "zh-Hant": "只在h≠0除以h，差商成constant+c_xx*h，再由多項式連續性取h→0得constant。h=0時原差商未定義，不能先代0作0/0；有限差分不是此極限證明。", "en": "Divide by h ONLY for h≠0,obtaining constant+c_xx*h.THEN polynomial continuity gives limit constant as h→0.The original quotient is undefined at h=0;never substitute0 into0/0.A finite difference is not this limit proof." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "反向核對本步原始角色、完整前提與欄序；只接受這些條件下的結論，不把單向結論還原為唯一前提，也不把有限核對稱一般形式證明。", "en": "Reverse-audit THIS step against its ORIGINAL ordered roles and complete premises.Accept the conclusion only under these conditions;do not invert a one-way implication or call a finite check a general formal proof." } }, "arithmetic": null } ], "math_model": { "result": [ [ -12, -2, -3, 1, -1, -2 ], -2, -3, [ -3, 1 ], "quotient=constant+h_coefficient*h for h≠0", -3, "symbolic_limit_not_finite_difference" ] }, "conclusion": { "canonical_value": "([-12, -2, -3, 1, -1, -2], -2, -3, [-3, 1], quotient=constant+h_coefficient*h for h≠0, -3, symbolic_limit_not_finite_difference)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261013-advance-20261025-r07-partial-derivative-foundations", "batch_path": "public/data/questions/batches/batch-20261013-advance-20261025-r07-partial-derivative-foundations/questions.jsonl", "batch_sha256": "6b2bb9d942c4d9785602fed035b922edb653a7317cb9355963393ee09ee97224", "original_line": 76, "original_line_sha256": "76b80ee695165fa185852bc86d150e2806ec859843d80cadaefbea89ef2a0b68", "question_compact_sha256": "200f3e0bd14ea5f3dacc78ef6edd9689a40c6f7e565f5282c8d2aa7cc2f311ec", "english_row_compact_sha256": "f5ade53fc0e46a3df1db3face2b60b911781c9037ca22fc0d5e6dda8a0732fb6", "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261025-r07-partial-derivative-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261025-r07-partial-derivative-foundations.en.jsonl" }, "mko_bindings": { "mko-partial-derivative": { "sha256": "7159526c8e30cc8cb9086f9690d279fd3852328abe371206b3441ecf6ef69ce8", "english_sha256": "11ec37b647f687ce00aba1864579424c7da8dccdbac8051d553f0a9763c182fe", "original_path": "public/data/mko/partial-derivative.json", "english_path": "public/data/i18n/editions/mko-partial-derivative.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "df58fcce25614251047f7c41d31ff93b5bfec19cd9bbcaa5fbb42a4daceb0310", "original_candidate_sha256": "938952faae26f73d7a247df90b0203838d50ca1c56b07aed57410f6a94304738", "original_candidate_line_sha256": "938952faae26f73d7a247df90b0203838d50ca1c56b07aed57410f6a94304738" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-012-p67", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "99b4f114b2ec93164b6aa5681e0107033a240e7a3f0a8988633ba65320ede267", "question_first_witness_sha256": "d9bd3a1940da8a73f790ee3fe460c9a8523791bdbe72c8e0304e3740b66710cc" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 8, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "a074276bc795be2e8fd9a1d261e17d8181c2a2abff153dd2b08335373bca9921", "arithmetic_steps": 4 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }