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Power series Representation

  1. Apr 29, 2016 #1
    1. The problem statement, all variables and given/known data
    http://imgur.com/12LbqWL

    Part b

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Apr 29, 2016 #2
    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.
     
  4. Apr 29, 2016 #3
    First four nonzero terms?
     
  5. Apr 29, 2016 #4
    That is what I would do.
     
  6. Apr 29, 2016 #5
    So is the power series already centered about x=2 when I take the derivative??
     
  7. Apr 29, 2016 #6
    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.
     
  8. Apr 29, 2016 #7
    Right but I guess my question is why is it automatically centered around x=2.
     
  9. Apr 29, 2016 #8
    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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