What is the 13th Taylor coefficient of f(x) at x=3?

[cathy]

Homework Statement

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

Homework Equations

Well, it looks like I just had to take the derivative, but by the time it gets to the 13th derivative, wouldn't the answer just be zero?

The Attempt at a Solution

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

 

[micromass]

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

 

[cathy]

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

 

[micromass]

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

Well, then the software is wrong.

 

[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

 

[micromass]

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

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} + ...

 

[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?

 

[micromass]

So how do I find the coefficient? Do I have to take the 11th derivative of xe^x, plug 7 in, divide by 11! ?

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.

But then what happens with the (x-7)^11?

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 $(x-7)^{11}$.

 

[cathy]

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.

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

 

[vela]

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).