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Homework Help: Power Series/Taylor Series

  1. Oct 28, 2008 #1
    What exactly the connection between a power series and taylor series? I know all taylor series are power series with the coefficients = f^n(a)/n!, but what can we say about power series? My graduate student instructor said that all convergent power series are taylor series and I don't quite understand that. To me, taylor series just seems like a special case of power series where the coefficients has a special formula. Anyone care to explain? Thanks in advance.
     
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  3. Oct 28, 2008 #2

    Dick

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    I think you've already said it. All taylor series are power series. If the power series converges on a finite interval to some function f(x), then you can regard the power series as a taylor series for f(x). It's nothing really deep.
     
  4. Nov 1, 2008 #3
    so can we say taylor series represents function in a more precise way than power series does?
     
  5. Nov 1, 2008 #4

    Dick

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    If a power series 'represents' a function (by being equal to it on a open interval surrounding a point), then it IS the taylor series.
     
  6. Nov 1, 2008 #5

    Hurkyl

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    Exercise: Let
    [tex]\sum_{n = 0}^{+\infty} a_n x^n[/tex]
    be a power series that converges for [itex]x \in (-R, R)[/itex]. Find a function [itex]f(x)[/itex] whose Taylor series is that power series.
     
  7. Nov 1, 2008 #6

    HallsofIvy

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    Not just a "special case". If a power series converges to a function, the it is a Taylor's series for that function.

    Problem: Find the Taylor's series for 1/(1- x) around x= 0 and find its radius of convergence. Yes, you could calculate the derivatives and evaluate at x= 0. That's not very difficult.

    But it is much easier to recall that, if |r|< 1, then the geometric series
    [tex]\sum_{n=0}^\infty a r^n= \frac{a}{1- r}[/tex]
    and just take a= 1, x= r: The Taylor's series for 1/(1- x) about x= 0 is
    [tex]\sum x^n[/tex].
    Because that is a power series that converges to 1/(1-x), it is the Taylor's series. Also, the fact that the geometric series converges for |r|< 1 and diverges for |r|> 1 tells us that the radius of convergence is 1.
     
  8. Nov 2, 2008 #7
    i donno...seems i still not getting the point at all ==;

    it's like if im given a question asking me to find a taylor series of a function, i will definitely lost. and the "center of convergence" how am i going to find it? using convergence test to find the interval of x and pick the middle point?
     
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