Power series Representation

  • #1
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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.
 

Answers and Replies

  • #2
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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.
 
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  • #3
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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?
 
  • #5
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That is what I would do.
So is the power series already centered about x=2 when I take the derivative??
 
  • #6
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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.
 
  • #7
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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.
 
  • #8
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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.
 

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