1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
  2. jcsd
  3. Oct 28, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Power Series/Taylor Series
  1. Power series and taylor (Replies: 16)

  2. Power series/taylor (Replies: 3)

Loading...