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Power Series

  1. Apr 29, 2007 #1
    1. The problem statement, all variables and given/known data
    f(x) = x^4 / (2 - x^4). Specify radius of convergence.

    2. Relevant equations
    Power Series
    f`(x) = c2 + 2c2(x-a) + 3c3(x-a)^2 + .... = (infinity)sigma(n=1) [n * cn * (x-a)^(n-1)]

    3. The attempt at a solution
    I'm not sure what to do. Usually, most problems are like x^3 / x^4, so I'm not sure what to do. Using L'Hoptials, differentiating top and bottom doesn't do much.

    Like if it were x / (9+x^2). I could easily pull out an x and a 9

    (x/9) * [1/ [1-(-x/3)^2] ]

    Then look at just the -(x/3)^2 as a geometric series

    |-(x/3)^2| < 1 = |x^2| / 9 < 1

    |x^2| < 9
    -3 < x < 3

    R = 3 and I = -3, 3

    But not sure what to do with the one above in how to manipulate it.
  2. jcsd
  3. Apr 29, 2007 #2
    hmm if you mean radius of convergence about x=0 it´s pretty straightforward i think.
    Just look at the denominator it´s 2-x^4 now just look what´s the nearest pole of that in the complex plane. Sure it´s 0 = 2- x^4 -> r = (2)^(1/4), that´s your radius of convergence.
    Another way would be this:
    that f(x) and compute the series:
    you get : f(x) = 1/2*x^4 + 1/4*x^8+ 1/8x^12 ....
    now take the formula for the radius of convergence
    (a_n)^(1/n) and compute the limit n-> infinity
    a_n is surely 2^(-1/4*n)
    now compute the limit and you get 2^(-1/4) take the reciprocal and you got r=2^(1/4)

  4. Apr 29, 2007 #3


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    It doesn't make sense to talk about the "radius of convergence" of function! The radius of convergence is defined for a power series. Of course, a function like this can be written as a power series by taking its Taylor's series. But then the radius of convergence depends upon the center: what is a in (x-a)n?

    Assuming, as Mr. Brown did, that you mean "what is the radius of convergence of the MacLaurin series (i.e. the Taylor's series about a= 0) for x4/(2-x4)", then you don't really have to find the Taylor's series. The series will converge as long as there is nothing stopping it! In other words as long as the function is "analytic" (Actually, the definition of "analytic" is that the Taylor's series converges to the given function in some neighborhood of the point!). Here, that means as long as the function is "smooth"- has as many derivatives as you want. It is obvious that the function x4/(2- x4) is smooth as long as the denominator is not 0: the "problem" occurs when 2- x4= 0. That is, as long as x is not [itex]^4\sqrt{2}[/itex]. What we really should do is look at the "complex plane"- the "radius" of convergence really is a "radius"- the radius of the largest circle around the given point (here, (0,0)) that does not include a point where the function is not continuous. This function is not continuous at any complex number where 2- x4= 0. That is at [itex]^4\sqrt{2}[/itex], [itex]-^4\sqrt{2}[/itex], [itex]i^4\sqrt{2}[/itex], and [itex]-i^4\sqrt{2}[/itex]. Fortunately, all of those lie on the circle about 0 with radius [itex]^4\sqrt{2}[/itex]. The "radius of convergence" of the MacLaurin series for x2/(2- x2) is,as Mr. Brown said, [itex]^4\sqrt{2}[/itex].
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