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This was a task that gave $9$ points in an exam I failed. Since our professor doesn't provide solutions I thought I'd ask here.

Let $f:\mathbb{R^2} \to \mathbb{R}$ be twice continuous partially differentiable and $x_0 \in \mathbb{R^2}$ random. Deduce the following formula for $x \in \mathbb{R^2}$ using Taylor's theorem: $$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}(x-x_0)^TH_f(x_0)(x-x_0)+o(\lVert x-x_0 \rVert^2).$$

Complete the remainder of the second order and show that for the remaining terms $T$ it holds that

$$\lim_{x \to x_0} \frac{T(x)}{\lVert x-x_0 \rVert^2} = 0$$

Can someone tell us how to do this?

The total derivate is defined as:

$$\lim_{x \to \alpha} \frac{\lVert f(x)-f(a)- df_a(x-a)\rVert}{\lVert x-a \rVert} = 0$$

I think from this one can get the difference of the norm which was given an estimate of in the task...

user1234567890
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    (1) random --> arbitrary . (2) be more precise with your language. i have no idea what "remainder of the second order" means. i have no idea what $T$ is. – mathworker21 Apr 10 '20 at 11:51
  • @mathworker21 T are the remaining terms. Remainder of the second order refers to this, I believe: https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder – user1234567890 Apr 10 '20 at 12:56
  • This thread on multivariate Taylor series with the remainder in Peano's form can be helpful. – Alex Ravsky Apr 13 '20 at 10:16

1 Answers1

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Taylor's theorem says that:

If $g: \Bbb R\to \Bbb R$ is $n$ times continuously differentiable on $[x_0,x_0+h]$, then $g(x)$ can be expressed as $$g(x)=g(x_0)+g'(x_0)(x-x_0)+g''(x_0){1\over 2}(x-x_0)^2+\cdots +{g^{(n)}(x_0)\over n!}(x-x_0)^n+h_n(x)(x-x_0)^n$$with $\lim_{x\to x_0} h_n(x)=0$. Hence the term $h_n(x)(x-x_0)^n$ can be denoted by $o[(x-x_0)^n]$.

In this question, since $f(x)$ is twice differentiable anywhere, then so is $g(t)=f(x_0+t(x-x_0))$ (note that $g(0)=f(x_0)$ and $g(1)=f(x)$). Hence $$g(t)=g(0)+g'(0)(t-0)+g''(0){1\over 2}(t-0)^2+o(t^2)$$

from the other side,$$g'(t)=\nabla f(u)^T|_{u=x_0+t(x-x_0)}\cdot (x-x_0)$$hence $$g'(0)=\nabla f(x_0)^T\cdot (x-x_0)$$similarly $$g''(0)={1\over 2}(x-x_0)H_f(x_0)(x-x_0)^T$$hence by substitution, the formula is deduced.

The value of the limit is also a direct consequence of $\lim_{x\to x_0} h_n(x)=0$.

Mostafa Ayaz
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