Getting a little confused on this problem. A function has the following Taylor series:
I'll post the series and my work in few.

Question

Getting a little confused on this problem. A function has the following Taylor series:
I'll post the series and my work in few.

kkutie7 · Answer

$$f(x)=\sum_{k=0}^{\infty}(-1)^{k+1}\frac{ k! }{ (2k)! }(x-4)^{k}$$

kkutie7 · Answer

find (a) f(4)
       (b) f'(4)
       (c) f"(4)

kkutie7 · Answer

I wrote out the first 3 non zero terms:
$$-1+\frac{ x-4 }{ 2 }-\frac{ (x-4)^{2} }{ 24 }$$

kkutie7 · Answer

I they took a shot at the first one like this (Keep in mind I have no clue if I'm even remotely doing this correct. Probably not.) 
$$\frac{ f^{0}(0)x^{0} }{ 0! }=-1\rightarrow f^{0}(0)=\frac{ -1*0! }{ x^{0} }\rightarrow f^{0}(0)=-1$$
For the other two I did the same thing and got zero. Zero is of course wrong.

anonymous · Answer

Just remember that a Taylor series is given by
$$ f(x) = \sum_n a_n(x-x_0)^n = \sum_n \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n$$

So if you know the terms of the series, you should be able to simply read off the derivatives:

$$f(x) = \sum_k(-1)^{k+1} \frac{k!}{(2k)!}(x-4)^k = \sum_k \frac{f^{(k)}(4)}{k!}(x-4)^k$$
which implies that

$$f^{(k)}(4) = (-1)^{k+1} \frac{k!k!}{(2k)!} $$

anonymous · Answer

so 
$$k=0 \rightarrow f(4) = -1$$
$$k=1 \rightarrow f'(4) = \frac{1}{2} $$
$$k=2 \rightarrow f''(4) = -\frac{2!2!}{4!} = -\frac{4}{4!} = -\frac{1}{6} $$

As a side note, the denominator in the third term in the expansion should be 12, not 24.