Taylor Polynomial

yeahyeah<3

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.

e(ho0n3

By (-1)^n+1 [(x-3)^n2]/n!, do you mean:

[tex]\frac{(-1)^{n+1}}{n!}(x-3)^{2n}[/tex]

In any case, ask yourself this: What does the term with [itex]f^{(30)}(3)[/itex] in the Taylor expansion look like?

yeahyeah<3

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!

e(ho0n3

See here.

That's what it looks like in p(x), but if you didn't know of p(x), what would the term look like?

yeahyeah<3

I'm not sure what you mean..?

like f'(30) (x-3)^30 kind of thing?
30!

e(ho0n3

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.

yeahyeah<3

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...

e(ho0n3

You wrote, sort of, that the term that contains [itex]f^{(30)}(3)[/itex] is

[tex]\frac{f^{(30)}(3)}{30!}(x-3)^{30}[/tex]

and this should equal

[tex]\frac{(x-3)^{30}}{15!}[/tex]

right? So what is [itex]f^{(30)}(3)[/itex]?

yeahyeah<3

so it is E 30!
15!
Thanks so much!

just one last question..
I don't understand how you got
[tex]
\frac{f^{(30)}(3)}{30!}(x-3)^{30}
[/tex]

only because I thought that it would be the 30th derivative of f(3) not f^30 (3)

e(ho0n3

I got that from the definition of the Taylor polynomial.

yeahyeah<3

okay. thanks so much again!