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Power series representations

  1. Feb 5, 2005 #1
    Evaluate the indefinite integral as a power series and find radius of convergence. (i don't know how to type the integral and summations signs, sorry)

    (integral sign) (x-tan^-1x)/x^3 dx. ( if you write this out it makes more sense)

    i was able to find the power series of tan^-1x = x^(2n+1) (-1)^n/(2n+1).
    i don't know how to continue on with this. all we have learned is to use the power series of the geometric series 1/(1-x), and some integration/differentiation methods.

    i am rather confused on the whole topic, so if anyone has any ideas, the simplest explanations would be greatly appreciated. thanks
  2. jcsd
  3. Feb 6, 2005 #2
    Assuming that you have the correct power series representation (I'm too lazy to check), the radius of convergence is basically a "circular" area or where the function converges with respect to a given point. In general, the radius of convergence is the absolute value of the given point minus the closest singularity.

    For example, "Find the radius of convergence of 1-sin(x)/ cos(x) evaluated at x=1". The given point is x=1 and the nearest singularity is x=0, because the denominator cannot be 0. Then, the radius of convergence is abs(1-0) which equals 1.

    Another way to find the radius of convergence involves something along the lines of the limit as n approaches infinity evaluated for A(n) / A(n+1). So basically, it's the actual power series divided by the power series again, but replacing each "n" with "n+1". Most terms will cancel out in this way and you'll end up with the same answer as using the technique shown above.

    Tim Nguyen
  4. Feb 6, 2005 #3
    the problem is that i only know how to get the power series rep for the arctan x part of the integral. i can't figure out how to get the representation for the whole thing. once i have that sorted out, the radius of convergence should be simple. any advice on how to solve the rest of the integral?
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