# Taylor series coefficient

1. Apr 14, 2014

### cathy

1. The problem statement, all variables and given/known data

F(x)=7x
Determine the 13th taylor coefficient of the taylor series generated by f at x=3

2. Relevant equations

Well, it looks like I just had to take the derivative, but by the time it gets to the 13th derivative, wouldnt the answer just be zero?

3. The attempt at a solution

I thought so, but zero isn't the answer. I thought it would be 0/13!, but that isnt correct :/
Can you tell me what I'm doing wrong?

2. Apr 14, 2014

### micromass

Staff Emeritus
Well, zero is the correct answer. Why do you think it is false?

3. Apr 14, 2014

### cathy

i'm entering it into my homework and it's marking it wrong.

4. Apr 14, 2014

### micromass

Staff Emeritus
Well, then the software is wrong.

5. Apr 14, 2014

### cathy

Maybe I'm going about the problem wrong? I seem to be getting a lot of them wrong. f(x)=3xe^x. Determine its 11^{th} Taylor coefficient of the Taylor Series generated by f at x = 7.
For this, I am getting 3*7^14/13! and that is wrong too

6. Apr 14, 2014

### micromass

Staff Emeritus
Yeah, that one is indeed wrong. You seem to be finding a Taylor series of the form

$$f(x) = f(0) + f^\prime(0)x + \frac{f^{\prime\prime}(0)}{2!}x^2 + .... + \frac{f^{(11)}(0)}{11!} x^{11} + ...$$

However, this is the Taylor expansion around $0$. You want the Taylor expansion around $7$ which should yield something of the form:

$$f(x) = f(7) + f^\prime(7)(x-7) + \frac{f^{\prime\prime}(7)}{2!}(x-7)^2 + .... + \frac{f^{(11)}(7)}{11!} (x-7)^{11} + ...$$

7. Apr 14, 2014

### cathy

So how do I find the coefficient? Do I have to take the 11th derivative of xe^x, plug 7 in, divide by 11! ?
But then what happens with the (x-7)^11?

8. Apr 14, 2014

### micromass

Staff Emeritus
That is one possibility, but I would not do that since it requires many computations.
What I would do is first find the general Taylor series of $e^x$ in $7$. This should be very easy. So you have
$$P(x) = a_0 + a_1 (x-7) + a_2 (x-7)^2 + ...$$
as Taylor series of $e^x$. Then to find the Taylor series of $3xe^x$, I would do:
$$f(x) = 3xe^x = 3(x-7)e^x + 3\cdot 7 e^x = 3(x-7)P(x) + 21P(x)$$
now you can substitute the series of $P(x)$ into the above and easily find the Taylor series you want without much computations.

What do you mean? Nothing happens to it. It's part of the Taylor series. Note also that you are asked to give the coefficients of the $11$th degree. So you should find the number that comes before $(x-7)^{11}$.

9. Apr 14, 2014

### cathy

I'm still a bit confused. What do you mean by substituing the series into the above?

10. Apr 14, 2014

### vela

Staff Emeritus
You might find it simpler to let u=x-7 and find the series about u=0. First, express f(x) in terms of u to get
$$f(x) = 3xe^x = 3(u+7)e^{u+7}.$$ Find the Taylor series for f in powers of u, and finally, use the substitution to get a series in terms of powers of (x-7).