Taylor polynomial help

Question

About Taylor polynomials:

I have $$f(x)=(x-4)^9$$

And I need to find the 10th order taylor polynomial about $x=4$

Now, I tried solving it: All but the last term is equal to $0$ since for example $$f''(a)\frac{(x-a)^2}{2!}= (4-4)\frac{x-4}2=0$$

So I'm left with the last term of $$36288\frac{(x-a)^9}{10!}=\frac{(x-4)^9}{10}$$

But this is the wrong answer, can anyone help me out here?

$$f(x)=(x-4)^9=\sum_{n=0}^\infty \delta_{n,9}(x-4)^n$$ – obataku Oct 13 '13 at 17:38 — obataku, Oct 13 '13 at 17:38

score 1 · Answer 1 · edited Oct 13 '13 at 17:59

1

It already is the 10th order Taylor polynomial about $x=4$. Every term in the taylor series will be zero except for that one.

The problem you have is that you wrote $ 9! (x-4)^9/10!$, the factorial on the bottom is wrong it is supposed to be $9!$ not $10!$. The way to remember this is that the factorial downstairs must always agree with the power of $x$.

edited Oct 13 '13 at 17:59

dfeuer

9,069

answered Oct 13 '13 at 17:44

Spencer

12,271

Taylor polynomial help

1 Answers1