Multivariable real-valued function: coefficient order, declared domain, slices and point values of a bivariate quadratic

Goal and scope

The coefficient vector in the order [1,x,y,x²,xy,y²] is c=(1,−2,3,1,2,−1), and the declared domain is the closed rectangle R=[−1,2]×[0,3]. (1) Write f and confirm that f:R→ℝ is a function; (2) check the x and y bounds separately to decide whether (2,3), (3,1), (0,−1) lie in R; (3) compute f(2,1) and f(1,2), and prove that g, obtained by swapping the x and y coefficients, is a different function; (4) find the x-slice at y=1 and show that neither the given y=1 slice nor the four specified point values determine f.

Treats only the bivariate integer-coefficient quadratic slice and the closed rectangular domain given by G1 and G2 (MKO assumptions a1, a2). Continuity, differentiability and vector values are not asserted; neither the polynomial identity theorem nor the unisolvence of any point set for quadratics is proved. The canonical MKO and its English edition are not changed.

Givens and symbols

• function_def
• cartesian_product_def
• order_basics
• distributive
• numeric_evaluation

• G1 — The coefficient order is fixed as [1,x,y,x²,xy,y²]; the coefficient vector of f is c=(1,−2,3,1,2,−1) (MKO assumption a1: six integer coefficients).
• G2 — Declared domain R=[−1,2]×[0,3], first coordinate x and second coordinate y; codomain ℝ (MKO assumption a2: only the explicitly declared closed rectangle is checked).
• G3 — Comparison functions (same domain R): g with coefficients (1,3,−2,−1,2,1); h with coefficients (1,−2,2,1,2,0); k with coefficients (1,−3,3,2,2,−1).

• c=(c_0,c_1,c_2,c_3,c_4,c_5) — coefficients in the order [1,x,y,x²,xy,y²]
• f(x,y) — the function given by c=(1,−2,3,1,2,−1)
• R=[-1,2]\times[0,3] — the declared closed rectangular domain
• g — the function with x and y coefficients swapped
• \varphi(x)=f(x,1) — the x-slice at y=1
• h,\ k — comparison functions agreeing with f on a slice or at finitely many points

Rules and sources

• function_def · prerequisite — Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a.
• cartesian_product_def · prerequisite — Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d.
• order_basics · prerequisite — Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
• distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
• numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
• def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
• eq_replace · prerequisite — Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true.
• mul_one · prerequisite — Multiplicative identity: 1·u=u.
• and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
• de_morgan_logic · prerequisite — De Morgan's laws for logic: ¬(P∧Q) ⇔ ¬P∨¬Q and ¬(P∨Q) ⇔ ¬P∧¬Q.
• power_laws · prerequisite — Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1.
• eq_congruence · prerequisite — Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule.
• add_comm · prerequisite — Commutativity of addition: u+v=v+u.
• mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ).
• add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v).
• mko_mvf_def · cited — This entry restricts attention to explicit slices of bivariate real polynomials f:ℝ²→ℝ, with coefficients ordered as [1,x,y,x²,x*y,y²]. Function qualification requires a declared domain and coordinate roles; continuity, differentiability and vector-valued codomains are not automatically asserted. · mko-multivariable-real-function
• p_interval_def · prerequisite — Closed intervals and rectangles: u∈[α,β] ⇔ α≤u≤β; (x,y)∈[α,β]×[γ,δ] ⇔ x∈[α,β] and y∈[γ,δ].

Main derivation

c=(c_0,c_1,c_2,c_3,c_4,c_5)

f(x,y)=c_0+c_1x+c_2y+c_3x^2+c_4xy+c_5y^2

c=(1,-2,3,1,2,-1)

f(x,y)=1-2x+3y+x^2+2xy-y^2

f(x,y)=1-2x+3y+x^2+2xy-y^2,\ (x,y)\in R

f:R\to\mathbb{R}\ \text{is a function}

(2,3)\in R\ ?

(2,3)\in R

(3,1),\ (0,-1)

(3,1)\notin R\ (3>2),\quad (0,-1)\notin R\ (-1<0)

f(2,1)

1-4+3+4+4-1=7

f(1,2)

1-2+6+1+4-4=6

f(2,1)=7,\ f(1,2)=6

f(2,1)\neq f(1,2)

g(x,y)=1+3x-2y-x^2+2xy+y^2

g(x,y)=f(y,x)\quad\text{when }(x,y)\in R\ \land\ (y,x)\in R

g(2,1)=f(1,2)=6,\ f(2,1)=7

g\neq f\ \text{as functions on}\ R

\varphi(x)=f(x,1),\ x\in[-1,2]

\varphi(x)=x^2+3

h(x,y)=1-2x+2y+x^2+2xy

h=f+(y^2-y),\ \ h(x,1)=f(x,1)=x^2+3

h(0,2),\ f(0,2)

h(0,2)=5\neq 3=f(0,2)

k=f+(x^2-x)

k=f\ \text{at}\ (0,0),(1,0),(0,1),(1,1),\ \ k(2,0)=3\neq 1=f(2,0)

\text{S03 to S14}

f=(c,R,\text{coordinate roles});\ \text{the }y=1\text{ slice and the four listed point values each fail to determine }f

Substeps and lemmas

2\in[-1,2]

-1\le 2\le 2

3\in[0,3]

0\le 3\le 3

1,\ -2x,\ 3y,\ x^2,\ 2xy,\ -y^2\ \text{at}\ (2,1)

1,\ -4,\ 3,\ 4,\ 4,\ -1

Reverse audits

• S15 · premise_audit · S03, S10, S13, S14 — Only conjoins conclusions already proved; there is no new inference.
• S14 · premise_audit · function_def, numeric_evaluation — This only refutes 'these four points determine f'; whether some other special point set determines a function within the class of quadratics is not discussed, and no such result is claimed.
• S13 · premise_audit · function_def, numeric_evaluation — One point of the domain suffices to refute h=f.
• S12 · premise_audit · eq_congruence, distributive — Agreement is claimed only on the line y=1; agreement at other y is not claimed, and S13 shows it fails.
• S11 · inverse_equivalence · eq_congruence, add_inverse — The simplification uses only additive inverses and collecting like terms, so it holds both ways.
• S10 · premise_audit · function_def, S09 — Refuting equality of functions needs one point, which must lie in the declared domain; (2,1) is in R.
• S09 · inverse_equivalence · add_comm, mul_comm — Only commutativity of addition and multiplication is used to rearrange terms, so it holds both ways.
• S08 · premise_audit · S06, S07 — Only claims that the values at this pair of points differ; it does not claim that f differs at every swapped pair (for example f(0,0)=f(0,0)).
• S07 · inverse_equivalence · numeric_evaluation — Exact integer evaluation; recomputed by N2.
• S06.1 · inverse_equivalence · numeric_evaluation, power_laws — Each term is a definite integer product, so it holds both ways.
• S06 · inverse_equivalence · numeric_evaluation — Exact integer evaluation, both sides the same number; recomputed by N1.
• S05 · premise_audit · p_interval_def, order_basics — A conjunction is refuted by one failing condition; here an x failure and a y failure are shown separately.
• S04.2 · inverse_equivalence · p_interval_def, order_basics — The same interval definition as S04.1, applied to the second coordinate and the y-interval.
• S04.1 · inverse_equivalence · p_interval_def, order_basics — The interval definition is two-way; using the open interval (−1,2) by mistake would wrongly exclude x=2.
• S04 · inverse_equivalence · p_interval_def, and_intro_elim — Membership in the rectangle is the conjunction of two interval conditions, two-way; both must be checked (MKO paragraph 1).
• S03 · premise_audit · function_def, G2 — The polynomial expression can be evaluated on all of ℝ², but the domain of this function is the declared R (MKO assumption a2); this step asserts neither continuity nor differentiability.
• S02 · inverse_equivalence · eq_replace, S01 — Backward: reading off the coefficients of 1, x, y, x², xy, y² in order recovers (1,−2,3,1,2,−1).
• S01 · premise_audit · mko_mvf_def — This step only fixes how the vector is read; reading the same numbers in another order gives a different function (see S09, S10).

Error controls and negative controls

Swapping the x and y coefficients

The first misconception named by the MKO is swapping the x and y coefficients. The coefficient vector must be read in the order [1,x,y,x²,xy,y²]; read as [1,y,x,y²,xy,x²], the same numbers give g; the identity g(x,y)=f(y,x) holds when both (x,y) and (y,x) lie in R, and g(2,1)=6≠7=f(2,1). Substitution must also respect the coordinate roles: the first coordinate goes to x, the second to y (which is why S06.1 lists the terms one by one).

A few point values cannot establish global equivalence

The second misconception named by the MKO is using a few point values to claim global equivalence, continuity or differentiability. The function k of S14 agrees with f at (0,0), (1,0), (0,1), (1,1) but differs at (2,0); the function h of S13 agrees with f on the whole slice y=1 but differs at (0,2). Checking these four points is insufficient for global equality. With a known polynomial-degree bound, a suitable sufficiently large finite point set can decide an identity; that criterion is not used here.

Check both coordinates of a rectangular domain

For (0,−1), x=0 lies in [−1,2], so checking only x would wrongly put it in R; for (3,1), y=1 lies in [0,3], so checking only y would also err. Also R is a closed rectangle, so boundary points such as (2,3) belong to R.

• f(2,1)=f(1,2), because the same two numbers are used → reject — f(2,1)=7 while f(1,2)=6 (S06, S07).
• k and f agree at (0,0), (1,0), (0,1), (1,1), so k=f → reject — k(2,0)=3≠1=f(2,0), and (2,0)∈R (S14).
• f(3,1)=12 is a value of the declared function f:R→ℝ → reject — (3,1)∉R because x=3>2 (S05); being able to evaluate the expression does not put the point in the domain.
• The y=1 slice x²+3 determines f → reject — h has the same y=1 slice, but h(0,2)=5≠3=f(0,2) (S12, S13).

Result and limitations

f(x,y)=1−2x+3y+x²+2xy−y² is a function on the declared domain R=[−1,2]×[0,3]; (2,3)∈R (f=13), while (3,1) (x>2) and (0,−1) (y<0) are not in R. f(2,1)=7 and f(1,2)=6; the coefficient-swapped g has g(2,1)=f(1,2)=6 at these valid swapped points, so g≠f. The y=1 slice is x²+3, but h has the same slice with h(0,2)=5≠3=f(0,2); k agrees with f at four grid points yet k(2,0)=3≠1=f(2,0).

• This guide does not assert that f is continuous or differentiable; partial derivatives and the gradient belong to a later entry (the MKO's formal next).
• N12 is only a finite check at the two points x∈{0,1}, not an identity; the guide uses it precisely to show that agreement at these four grid points is insufficient for global equality.

Verification and source binding

15 declared checks for this guide; only the listed expressions are covered.

• source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
• adopted source SHA-256: 896db668ee991856b4a5f03263420b664073e8a89c748e47bac90da89b7ed88b
• original source SHA-256: 896db668ee991856b4a5f03263420b664073e8a89c748e47bac90da89b7ed88b
• Lean: NOT_RUN · independent AI: NOT_MEASURED

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