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I have to compute the second order Taylor series of the function

$ f = e^{x^2 + y^2}$ near $(0,0)$.

The Jacobian is:

$$ Df(x,y) = (2\ x\ e^{x^2 + y^2}, 2\ y\ e^{x^2 + y^2}) $$

and the Hessian:

$$ D^2f(x,y) = \left( \begin{array}{cc} 4\ x^2\ e^{x^2 + y^2} & 4\ x\ y\ e^{x^2 + y^2} \\ 4\ x\ y\ e^{x^2 + y^2} & 4\ y^2\ e^{x^2 + y^2} \end{array}\right) $$

both of which are $0$ at $(0,0)$.

So the second order taylor series degenerates to

$$ T_2(x,y) = f(0,0) = 1$$

... which sounds like a rather poor approximation for any $x, y$, e.g. $(0.2,0.2)$. Also it's a bit confusing for me that both Jacobian and Hessian are $0$ at that point.

Do I miss something here, or is the function itself simply not favorable for Taylor approximation?

bonifaz
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    It's very much possible that the Jacobian and the Hessian are zero at some point. Take for example $f(x,y) = x^3$. It just means that to get a good approximation, you need to get more terms. There are pathological cases, though. – Najib Idrissi Jun 20 '13 at 09:33

3 Answers3

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I think you've miscomputed the Hessian. But anyway you can use $$e^{x^2+y^2}=1+(x^2+y^2)+(1/2)(x^2+y^2)^2+\cdots$$ to see what the series out to second degree looks like.

Gerry Myerson
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  • You are right, the Hessian was wrong - did not apply the product rule. That makes a huge difference - thanks a lot! – bonifaz Jun 20 '13 at 09:43
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A related problem. You can go this way $$ e^{x^2+y^2}=1+(x^2+y^2)+\frac{1}{2!}(x^2+y^2)^2+\dots. $$

Another way, you can use the formula

$$ \sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^m y^k}{m! k!}\frac{\partial^{m+k}}{\partial x^m \partial y^k }f(x,y)\Big|_{x=0,y=0} .$$

Community
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Mhenni Benghorbal
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$$2xe^{x^2} \longrightarrow (2x)' e^{x^2} + (2x)(e^{x^2})'$$

This function is analytic, that is, equal to its Taylor series.

not all wrong
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