# Power series Representation

• nfcfox
In summary, the conversation is about finding the first four nonzero terms of a power series for the derivative of a given function, centered around x=2. The derivative can be found by taking the derivative of the given power series and plugging in x=2. The reason why it is automatically centered around x=2 is because it follows from the definition of a power series.

## Homework Statement

http://imgur.com/12LbqWL

Part b

## 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.

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.

• SammyS
alivedude said:
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?

nfcfox said:
First four nonzero terms?

That is what I would do.

alivedude said:
That is what I would do.
So is the power series already centered about x=2 when I take the derivative??

nfcfox said:
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.

alivedude said:
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.

nfcfox said:
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.