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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 $1+x+y+\tfrac12x^2+xy+y^2$ .

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 $f(a+h)=f(a)+\nabla f(a)\cdot h+\tfrac12h^TH_f(a)h+o(\|h\|^2)$ ).

[L2]

The standard algebra rules compute the displayed derivatives (Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ ).

Proof

technique · calculation

1.1

At $(0,0)$ , $f=1$ , $\nabla f=(1,1)$ , and $H_f=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ .

givenL2algebra

2.1

Substitution into [L1] gives $1+x+y+\tfrac12(x^2+2xy+2y^2)$ , namely the claimed polynomial.

L1step 1.1algebra

3.1

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

step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 59 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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