2

I'm trying to show the statement in the title is true for Taylor series centered at any point $\textbf{a}$ in $\mathbb{R}^n$.

It's easy for me to show this for when $\textbf{a}=\vec{0}$ using the fact that the multivariate Taylor expansion for a function $f$ is:

$$f(x) = \sum_{|\alpha|\leq n}\frac{\partial ^{\alpha}f(\textbf{a})}{\alpha!}(\textbf{x} - \textbf{a})^{\alpha} + r_{n,\textbf{a}}(x)$$

where $\alpha$ is the multi-index, and $r$ being the remainder term. Since the multivariate polynomial's variable portion is exactly like $(\textbf{x} - \textbf{a})^{\alpha}$ when $\textbf{a} = \vec{0}$, and the fractional term reduces to the same coefficient as in the original polynomial, we get back the original polynomial. But I am completely lost as to how to proceed in the case where $\textbf{a} = \vec{0}$, a hint would be greatly appreciated it.

lucusk
  • 51
  • 1
    What is $(\mathbf{x} - \mathbf{a})^{\alpha}$ when $\mathbf{x}, \mathbf{a} \in \mathbb{R}^n$ ? – pitchounet Jul 05 '17 at 23:11
  • If $T_a f$ is the Taylor series for $f$ centered at $x=a$ and $S_a f$ is the $f$ shifted by $a$, so that $(S_a f)(x) = f(x+a)$, show that $T_a f = S_a T_0 S_{-a} f$. Then the general case reduces to the case $a = 0$. – Jair Taylor Jul 05 '17 at 23:53
  • 1
    $T_a f = S_{-a} T_0 S_a f$, rather. – Jair Taylor Jul 06 '17 at 01:05
  • @jibounet, $(\textbf{x} - \textbf{a})^{\alpha} = (x_{1} - a_{1})^{\alpha_{1}}(x_{2} - a_{2})^{\alpha_{2}}...(x_{n} - a_{n})^{\alpha_{n}}$ – lucusk Jul 06 '17 at 15:52
  • @JairTaylor, thank you! Here is what I did with Jair's hint. $T_a f = S_{-a} T_0 S_a f = S_{-a} (T_0 S_a f) = S_{-a} S_a f = f$, hence the taylor series of any multivariate polynomial is itself. To show $T_a f = S_{-a} T_0 S_a f$, apply a translation on the $\textbf{x}$ of $f$ which will only affect the partial differential part of the expansion, note $T_0 S_a f$ looks like $T_a f$, except you have $\textbf{x}^{\alpha}$, now translate x in the negative direction which gets you to $(\textbf{x} - \textbf{a})^{\alpha}$, thus getting back the $T_a f$ – lucusk Jul 06 '17 at 18:32
  • Another, more direct approach: You can define the $n$-th Taylor polynomial to be the polynomial of degree $n$ with derivatives that agree with $f$ at the point $x=a$ up to degree $n$. Show that two polynomials with all the same derivatives at a given point are equal. – Jair Taylor Jul 06 '17 at 19:07

0 Answers0