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Thread: Analytic continuation

  1. #1
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    Analytic continuation

    This seems to be a hard problem for me. Can someone help?
    Suppose $\displaystyle F(z)=\sum_{n=1}^{\infty} d(n)z^n, for\mid{z}\mid <1$ $\displaystyle d(n)$ is the number of divisors of n. The radius of convergence of this series is 1. Show that $\displaystyle \sum_{n=1}^{\infty} d(n)z^n=\sum_{n=1}^{\infty}\frac{z^n}{1-z^n}$. Then use this to show that if $\displaystyle z=r$ with $\displaystyle 0<r<1$, then $\displaystyle \mid{F(r)}\mid\geq c\frac{1}{1-r}log(1/(1-r))$ as $\displaystyle r\rightarrow 1$.
    Smilarly, if $\displaystyle \theta=2\pi p/q$ $\displaystyle p,q$ are positive integers and $\displaystyle z=re^{i\theta}$, then $\displaystyle \mid{F(re^{i\theta})}\mid\geq c_{p/q}\frac{1}{1-r}log(1/(1-r))$ as $\displaystyle r\rightarrow 1$. Conclude that there can not be any regular point on the unit disc.
    Last edited by namelessguy; Oct 8th 2008 at 06:14 PM.
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  2. #2
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    I can perhaps start it:

    $\displaystyle \frac{q^n}{1-q^n}=\sum_{k=1}^{\infty} \left(q^{n}\right)^k$

    So that

    $\displaystyle \sum_{n=1}^{\infty}\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\left(q ^n\right)^k$

    $\displaystyle =q+q^2+q^3+q^4+\cdots$
    $\displaystyle +q^2+q^4+q^6+q^8+\cdots$
    $\displaystyle +q^3+q^6+q^9+q^{12}+\cdots$
    $\displaystyle +q^4+q^8+q^{12}+q^{16}+\cdots$
    $\displaystyle +q^5+q^{10}+q^{15}+q^{20}+\cdots$

    $\displaystyle +\vdots$

    Grouping like-powers, it becomes obvious that the number of terms for each power is the divisor function $\displaystyle d(nk)$ so that that sum can be written as:

    $\displaystyle \sum_{n=1}^{\infty}\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}d(n)q^n$

    The rest I assume may be approached from the general theory of Lacunary series and the Hadamard Gap Theorem although I wouldn't know how.
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  3. #3
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    Quote Originally Posted by shawsend View Post
    I can perhaps start it:

    $\displaystyle \frac{q^n}{1-q^n}=\sum_{k=1}^{\infty} \left(q^{n}\right)^k$

    So that

    $\displaystyle \sum_{n=1}^{\infty}\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\left(q ^n\right)^k$

    $\displaystyle =q+q^2+q^3+q^4+\cdots$
    $\displaystyle +q^2+q^4+q^6+q^8+\cdots$
    $\displaystyle +q^3+q^6+q^9+q^{12}+\cdots$
    $\displaystyle +q^4+q^8+q^{12}+q^{16}+\cdots$
    $\displaystyle +q^5+q^{10}+q^{15}+q^{20}+\cdots$

    $\displaystyle +\vdots$

    Grouping like-powers, it becomes obvious that the number of terms for each power is the divisor function $\displaystyle d(nk)$ so that that sum can be written as:

    $\displaystyle \sum_{n=1}^{\infty}\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}d(n)q^n$

    The rest I assume may be approached from the general theory of Lacunary series and the Hadamard Gap Theorem although I wouldn't know how.
    Thanks shawsend. I don't think I have learned either of the theorems you mentioned yet.
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  4. #4
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    Ok. Then I probably misunderstand the problem. Don't want to lead you down the wrong path.

    I think I was confusing convergence of the series on the disc of convergence and the ability to analytically continue the series past the boundary. These two I think are not related. My understanding of the problem then is that there are no point on the boundary that the series can be analytically continued to a disc containing a point on the boundary. Or no?
    Last edited by shawsend; Oct 9th 2008 at 10:40 AM.
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  5. #5
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    You can get the result for z=r by an elementary calculation:

    $\displaystyle \begin{aligned}F(r) = \frac r{1-r} + \frac{r^2}{1-r^2} + \frac{r^3}{1-r^3} + \ldots &= \frac1{1-r}\left(r + \frac {r^2}{1+r} + \frac{r^3}{1+r+r^2} + \ldots\right)\\ &\geqslant \frac1{1-r}\left(r + \frac {r^2}2 + \frac{r^3}3 + \ldots\right)\\ &= \frac1{1-r}\ln(1/(1-r)).\end{aligned}$

    But the result for $\displaystyle z = re^{2\pi pi/q}$ doesn't look as though it will come out easily from that method.
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