Taylor Polynomial Homework: Evaluate f^30(3)

[yeahyeah<3]

Homework Statement

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!

Homework Equations

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:

\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?

 

[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]

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?

 

[yeahyeah<3]

I'm not sure what you mean..?

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

 

[e(ho0n3]

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.

 

[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 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)?

 

[yeahyeah<3]

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

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

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!