## Taylor Polynomial

1. The problem statement, all variables and given/known data
The Taylor polynomial of degree 100 for the function f about x=3 is given by
p(x)= (x-3)^2 - (x-3)^4/2! +... + (-1)^n+1 [(x-3)^n2]/n! +... - (x-3)^100/50!
What is the value of f^30 (3)?

D) 1/15! or E)30!/15!

2. Relevant equations

3. The attempt at a solution
I know the bottom of the answer is 15! because n=15 (for the exponent to be 30) but i'm not sure what the top does.

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 By (-1)^n+1 [(x-3)^n2]/n!, do you mean: $$\frac{(-1)^{n+1}}{n!}(x-3)^{2n}$$ In any case, ask yourself this: What does the term with $f^{(30)}(3)$ in the Taylor expansion look like?
 how do you get the math problem to look like that -.- and that is the question I'm asking for help on... i think the term looks like (x-3)^30 15! ? but how does that answer the question? Thanks!

## Taylor Polynomial

 Quote by yeahyeah<3 how do you get the math problem to look like that -.-
See here.

 i think the term looks like (x-3)^30 15! ?
That's what it looks like in p(x), but if you didn't know of p(x), what would the term look like?

 I'm not sure what you mean..? like f'(30) (x-3)^30 kind of thing? 30!

 Quote by yeahyeah<3 like f'(30) (x-3)^30 kind of thing? 30!
Yes, that kind of thing. So now you know what the general form of the term looks like and what it actually is. I leave the rest to you.

 i still don't understand =/ I know the bottom of the term is 15! but I don't know how to get what the top is...
 You wrote, sort of, that the term that contains $f^{(30)}(3)$ is $$\frac{f^{(30)}(3)}{30!}(x-3)^{30}$$ and this should equal $$\frac{(x-3)^{30}}{15!}$$ right? So what is $f^{(30)}(3)$?
 so it is E 30! 15! Thanks so much! just one last question.. I don't understand how you got $$\frac{f^{(30)}(3)}{30!}(x-3)^{30}$$ only because I thought that it would be the 30th derivative of f(3) not f^30 (3)
 I got that from the definition of the Taylor polynomial.
 okay. thanks so much again!