# Homework Help: Taylor Polynomial

1. Apr 8, 2009

### 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.

2. Apr 8, 2009

### 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?

3. Apr 8, 2009

### 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!

4. Apr 8, 2009

### 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?

5. Apr 8, 2009

### yeahyeah<3

I'm not sure what you mean..?

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

6. Apr 8, 2009

### 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.

7. Apr 8, 2009

### 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...

8. Apr 8, 2009

### 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 $f^{(30)}(3)$?

9. Apr 8, 2009

### yeahyeah<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)

10. Apr 8, 2009

### e(ho0n3

I got that from the definition of the Taylor polynomial.

11. Apr 8, 2009

### yeahyeah<3

okay. thanks so much again!