ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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A second-order Taylor polynomial computed from gradient and Hessian data

Statement

For $f (x, y) = e^{x} (1 + y + y^{2})$ , the second-order Taylor polynomial at the origin is $1 + x + y + \frac{1}{2} x^{2} + x y + y^{2}$ .

Facts & Assumptions

Given: $f (x, y) = e^{x} (1 + y + y^{2})$ .

[L1]

The second-order Taylor polynomial is determined by the value, gradient, and Hessian (Second-order Taylor expansion $f (a + h) = f (a) + \nabla f (a) \cdot h + \frac{1}{2} h^{T} H_{f} (a) h + o (∥ h ∥^{2})$ ).

[L2]

The exponential is smooth and $(e^{x})^{'} = e^{x}$ (The exponential function is smooth and $(\exp)^{'} = \exp$ ); the product and sum rules compute the displayed derivatives (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · calculation

1.1givenL2algebra

At $(0, 0)$ , $f = 1$ , $\nabla f = (1, 1)$ , and $H_{f} = (\begin{array}{cc} 1 & 1 \\ 1 & 2 \end{array})$ .

2.1L1step 1.1algebra

Substitution into [L1] gives $1 + x + y + \frac{1}{2} (x^{2} + 2 x y + 2 y^{2})$ , namely the claimed polynomial.

3.1step 2.1algebra∎

Thus the computed value, gradient, and Hessian yield the stated second-order approximation.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Taylor's theorem in several variables (standard reference, not scraped)