1
$\begingroup$

Let $p(x) = a_kx^k + a_{k-1}x^{k-1} + ... + a_1x+a_0$ be an arbitrary polynomial.

Find the Taylor Series at $0$ for $p$.


Intuitively, I know that the Taylor series is just the polynomial itself $\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}(x)^k$. Which is easy to see by taking the derivatives of the terms in the polynomial. However, I am trying to prove this with a rigid method, if I have to show it rigidly I assume that I have to prove it by induction:

Prove for $k=0$. $$\sum_{k=0}^{0} \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k = \frac{f^{(0)}(0)}{0!}(x)^0=a_0 $$

Now assume that it is true for $k=1$

Have to show that it is true for $k=k+1$ $$\sum_{k=0}^{k+1} \frac{f^{(k)}(0)}{k!}(x)^k = \sum_{k=0}^{k} \frac{f^{(k)}(0)}{k!}(x)^k + a_{k+1}x^{k+1}$$

Where to I go from here in the induction step, and is this the right way to find the Taylor Series?

$\endgroup$
2
  • 1
    $\begingroup$ You can use the uniqueness of the Taylor expansion, noticing that $$\sum_{k=0}^n a_k x^k$$ is a valid Taylor expansion to conclude that this is the only one. $\endgroup$
    – Delta-u
    Apr 11, 2018 at 16:37
  • $\begingroup$ But is this sufficient enough to argue that the Taylor series at 0 for p must be the polynomial itself? $\endgroup$
    – Simbörg
    Apr 11, 2018 at 21:46

1 Answer 1

1
$\begingroup$

You can show by induction that $$f^{(k)}(x)=\sum_{l=k}^n l(l-1)\ldots (l-k+1)a_l x^{l-k}$$ so $$f^{(k)}(0)=k! a_k$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .