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Interval of convergence?

  1. Aug 18, 2012 #1
    So I know how to find the "Interval of Convergence" for a power series representation of a Function f(x).

    But I Still don't know what that "Interval of Convergence" does for me other than I can choose a number in it and plug it in to the series.

    For Example e[itex]^{x}[/itex]=[itex]\sum^{∞}_{n=0} \frac{x^n}{n!}[/itex] ;when a=0;

    my "Interval of Convergence" is (-∞,∞). SO now lets say i take the # 1 from my "Interval of Convergence" and place it in the series representation of e^x.

    Then i would get back some answer , but what does that answer mean? besides the fact that I got an answer.
  2. jcsd
  3. Aug 18, 2012 #2
    Sum of series exists only for numbers from that interval. If you take other numbers you don't get any "answer".
  4. Aug 19, 2012 #3
    Usually, it is called "radius of convergence" instead of "interval of convergence". A function that is dependent on a series only converges when its argument (in this case, x) has a smaller absolute value than the radius of convergence. The exponential function, the one you gave, is convergent for every x, and hence its radius of convergence is infinity.
  5. Aug 19, 2012 #4


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    "radius of convergence" and "interval of convergence" are two different things. If we have a (real valued) power series of the form [itex]\sum a_n(x- a)^n[/itex] and I know that it has "radius of convergence", r, then I know that the series converges within the interval (a- r, a+ r), it "interval of convergence".

    If we are dealing with complex valued power series, then the "radius of convergence" really is a radius If the series [itex]\sum a_n (z- a)^n[/itex], where a, z, and every [itex]a_n[/itex] are complex numbers, has "radius of convergence" r, then it converges for all z in the interior of the disk with center at a and radius r.
  6. Aug 19, 2012 #5


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    Some nice properties happen within the radius of convergence ,like the fact that you

    can do term-by-term integration and differentiation within it.
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