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Math Help - Find the region of convergence

  1. #1
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    Find the region of convergence

    Given f_n(x) = (x^2)/(1+x^2)^n, find the region of convergence. I'm not sure what the "region" is. Is it just the same as the radius of convergence?
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    MHF Contributor chisigma's Avatar
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    The 'infinite sum' of functions...

    f(x) = \sum_{n=0}^{\infty} \frac{x^{2}}{(1+x^{2})^{n}} (1)

    ... does converge for all real values of the x variable. For x\ne 0 is...

    |\frac{1}{1+x^{2}}|<1 (2)

    ... and the (1) is a convergent 'geometric series'. For x=0 every term of (1) vanishes so that the 'infinite sum' vanishes also...

    For the complex 'infinite sum' of functions...

     f(z) = \sum_{n=0}^{\infty} \frac{z^{2}}{(1+z^{2})^{n}} (3)

    ... the problem is a little more 'complex'... of course ...



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  3. #3
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    I don't understand..Is that it? Where do I go from that?
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  4. #4
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    Quote Originally Posted by katielaw View Post
    Given f_n(x) = (x^2)/(1+x^2)^n, find the region of convergence. I'm not sure what the "region" is. Is it just the same as the radius of convergence?
    I don't really understand your question.

    Is this an infinite series?

    Is it supposed to be, for example

    \sum_{n = 0}^\infty \frac{x^2}{(1 + x^2)^n}?


    If it is, you're trying to find the values for which the function converges.

    Note that this can be rewritten as

    x^2 \sum_{n = 0}^\infty \frac{1}{(1 + x^2)^n}.


    Can you see that it is a geometric series, with a = 1 and r = \frac{1}{1 + x^2}?


    Geometric series converge when |r| < 1.

    So this function converges when

    \left|\frac{1}{1 + x^2}\right| < 1


    Solve for x.
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  5. #5
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    A "power series", a series of the form \sum a x^n has a "radius of convergence" but other kinds of series may not.

    In particular, you can think of the series you are given, \sum \frac{x^2}{(1+x^2)^n} as a geometric series, \sum ar^n with a= x^2 and r= \frac{1}{1+ x^2}.
    That's chisigma's point.

    Now, you have probably learned, long ago, that the geometric series \sum ar^n will converge as long as |r|< 1. That's where chisigma gets |\frac{1}{1+x^2}|< 1. In fact, since 1+ x^2 is never negative, \frac{1}{1+x^2}< 1 which is equivalent to 1+ x^2> 1 or x^2> 0. That's true for all x except ___ and the region of convergence is "all x except x= ___".
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    MHF Contributor chisigma's Avatar
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    Summarizing...

    \sum_{n=0}^{\infty}\frac{x^{2}}{(1+x^{2})^{n}}

    ... converges \forall x \in \mathbb {R} ...



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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    A "power series", a series of the form \sum a x^n has a "radius of convergence" but other kinds of series may not.

    In particular, you can think of the series you are given, \sum \frac{x^2}{(1+x^2)^n} as a geometric series, \sum ar^n with a= x^2 and r= \frac{1}{1+ x^2}.
    That's chisigma's point.

    Now, you have probably learned, long ago, that the geometric series \sum ar^n will converge as long as |r|< 1. That's where chisigma gets |\frac{1}{1+x^2}|< 1. In fact, since 1+ x^2 is never negative, \frac{1}{1+x^2}< 1 which is equivalent to 1+ x^2> 1 or x^2> 0. That's true for all x except ___ and the region of convergence is "all x except x= ___".
    Actually \frac{1}{1 + x^2} \leq 1.
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  8. #8
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    It is a series of functions. I thought the f_n(x) indicated that. I'm sorry I wasn't clearer.
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  9. #9
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    So is it the same as a sequence of functions as it would be for a power series?
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