Why Is the Power Series Automatically Centered at x=2?

[nfcfox]

Homework Statement

http://imgur.com/12LbqWL

Part b

Homework Equations

The Attempt at a Solution

Since it says the first four terms, not nonzero, the first four terms would be 0-(1/3-0)+2/9(x-2)-1/9(x-2)^2
I'm confused when it says I need to find these for x=2... Do I just plug in x=2 now and those four terms are them? Those won't be terms they'll be numbers. I'm assuming that the series is already centered at x=2 as that's why is says (x-2)^n so if that's the case, finding f'(2) would be to use the equation for geometric series so A/(1-R) but the first term is 0. Did they mean to say the first four nonzero terms? If someone can confirm this that'd be great.

 

[alivedude]

They want you do write out the derivative $f'(x)$ in the same way as the series you got in the problem, the first four terms and then the general term. After that you should plug in x=2.

 

[nfcfox]

They want you do write out the derivative $f'(x)$ in the same way as the series you got in the problem, the first four terms and then the general term. After that you should plug in x=2.

First four nonzero terms?

 

[alivedude]

First four nonzero terms?

That is what I would do.

 

[nfcfox]

That is what I would do.

So is the power series already centered about x=2 when I take the derivative??

 

[alivedude]

So is the power series already centered about x=2 when I take the derivative??

Well, $f(x)$ is defined by the power series, which is expanded around $x=2$. So taking the derivative of the power series would be to take the derivate of $f(x)$ and its derivative would indeed be a new power series that is indeed expanded around $x=2$ and have the same radius of convergence as $f(x)$ itself.

EDIT: If you look at what you write in your attempted solution I think you can see that the derivative takes to form of a power series.

 

[nfcfox]

Well, $f(x)$ is defined by the power series, which is expanded around $x=2$. So taking the derivative of the power series would be to take the derivate of $f(x)$ and its derivative would indeed be a new power series that is indeed expanded around $x=2$ and have the same radius of convergence as $f(x)$ itself.

EDIT: If you look at what you write in your attempted solution I think you can see that the derivative takes to form of a power series.

Right but I guess my question is why is it automatically centered around x=2.

 

[alivedude]

Right but I guess my question is why is it automatically centered around x=2.

Because it follows from the definition of a power series, i'll show you a general case with a power series expanded around some point $x=c$ and its derivate

$f(x)= \sum_{n=0}^{\infty} a_n(x-c)^n$

let us now take the derivate of this, its a sum so the derivate of the whole thing is just the sum of the derivates of each term. We have that

$f'(x) = \sum_{n=1}^{\infty} na_n(x-c)^{n-1}$

We can now see that $f(x)$ and $f'(x)$ is indeed power series and they are both expanded around $x=c$. I don't think that it could get any clearer than this.