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Thread: 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...

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

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

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

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

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

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

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



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    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

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


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

    Note that this can be rewritten as

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


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


    Geometric series converge when $\displaystyle |r| < 1$.

    So this function converges when

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


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

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

    Now, you have probably learned, long ago, that the geometric series $\displaystyle \sum ar^n$ will converge as long as |r|< 1. That's where chisigma gets $\displaystyle |\frac{1}{1+x^2}|< 1$. In fact, since $\displaystyle 1+ x^2$ is never negative, $\displaystyle \frac{1}{1+x^2}< 1$ which is equivalent to $\displaystyle 1+ x^2> 1$ or $\displaystyle x^2> 0$. 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...

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

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



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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 $\displaystyle \sum a x^n$ has a "radius of convergence" but other kinds of series may not.

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

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