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I have the series

$\sum_{n=0}^{\infty}(-1)^{n}a_{n}(\nu)\frac{\sin[\nu\,(m-n)]}{\nu\,(m^2-n^2)}=\frac{1}{m}$,

where $m$ is an integer. Is it possible to compute the coefficients $a_{n}(\nu)$? An exact solution is found for $\nu=\pi$:

$a_0=1, a_{n}=2\times(-1)^n$.

It is related to a well known problem of Pierce's electrodes in physics. My problem also deals with the Pierce electrodes but in a different geometry. I found a numerical solution for the most wanted value $\nu=2\pi/3$ using SVD pseudoinverse but is not satisfactory in a certain sense. Since the above equation for $a_{n}$ looks simple it is hoped that a simple solution to the problem might exist.

Attached sequence of numerical solutions of the truncated system

$\sum_{n=0}^{N}(-1)^{n}a_{n}(\nu)\frac{\sin[\nu\,(m-n)]}{\nu\,(m^2-n^2)}=\frac{1}{m}$,

seems to indicate that the numerical solution slowly converges to an oscillating function.

Numerical solution for $N=60$ equations: Numerical solution for <span class=$N=60$ equations">

Numerical solution for $N=120$ equations: Numerical solution for <span class=$N=120$ equations">

Numerical solution for $n=240$ equations: Numerical solution for <span class=$N=240$ equations">

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  • $\begingroup$ Are there some additional conditions on $a_n(\nu)$ or am I misreading something? Why couldn't you just take all $a_n(\nu)$ to be zero except $a_0$. $\endgroup$ – j.c. Oct 19 '13 at 7:57
  • $\begingroup$ @j.c.:Indeed, that is the obvious solution: $a_n=\delta_{n,0}$ with $\nu\rightarrow 0$. $\endgroup$ – Jon Oct 19 '13 at 12:04
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    $\begingroup$ @j.c: $a_n$ may not depend on $m$. It is assumend that $m$ runs from 1 to $\infty$, so we have infinite number of equations. $\endgroup$ – Igor Kotelnikov Oct 20 '13 at 11:44
  • $\begingroup$ Looks like weakly-* converging to something strange... For further numerical investigation I would suggest not to use the pure pseudo-inverse but some regularization in the solution of the truncated linear system. Probably ordinary Tikhonov-regularization, i.e. solving $(A^*A + \alpha I)a = A^*b$ (where $A$ is the coefficient matrix and $b$ is the right hand side $b_m = 1/m$) with some small $\alpha$, would be helpful to see what the limit may be. Moreover, one may want to couple $\alpha$ to $N$ in some clever way. Also: Is it clear that a solution exists? $\endgroup$ – Dirk Oct 21 '13 at 7:25
  • $\begingroup$ @Dirk, I thought that SVD pseudoinverse is a sort of regularization method so that Tikhonov-regularization is not needed. $\endgroup$ – Igor Kotelnikov Oct 23 '13 at 8:48
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Here is an approach that leads to an integral equation for $$A(\nu,t) := \sum_{n=0}^{\infty}(-1)^{n}\frac{a_{n}(\nu)}{\nu} e^{Int}.$$

First we notice that $$(-1)^{n}\frac{a_{n}(\nu)}{\nu} = \frac{1}{2\pi} \int_0^{2\pi} A(\nu,t) e^{-Int}\,dt.$$

The given identity multiplied by $e^{2mz}$ can be restated as \begin{split} \frac{e^{2mz}}{m} &= \sum_{n=0}^{\infty}(-1)^{n}a_{n} (\nu)\frac{\sin[\nu\,(m-n)]}{\nu\,(m^2-n^2)}e^{2mz} \\ &= (-1)^{m}\frac{a_m(\nu)e^{2mz}}{2m} + \sum_{n=0}^{m-1} (-1)^{n}\frac{a_{n} (\nu)}{\nu}\sin(\nu\,(m-n)) \frac{e^{(m-n)z}}{m-n}\frac{e^{(m+n)z}}{m+n} + \sum_{n=m+1}^{\infty} (-1)^{n}\frac{a_{n} (\nu)}{\nu}\sin(\nu\,(n-m)) \frac{e^{-(n-m)z}}{n-m}\frac{e^{(m+n)z}}{m+n}. \end{split}

For the first sum, we have

\begin{split} &\sum_{n=0}^{m-1} \frac{1}{2\pi}\int_0^{2\pi}dt\, A(\nu,t)e^{-Int} \Im e^{I\nu(m-n)} \int_{-\infty}^z dx\,e^{(m-n)x} \int_{-\infty}^z dy\,e^{(m+n)y} \\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, e^{(I\nu+x+y)m} \sum_{n=0}^{m-1} e^{(-It-I\nu-x+y)n}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, e^{(I\nu+x+y)m} \frac{1-e^{(-It-I\nu-x+y)m}}{1-e^{-It-I\nu-x+y}}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, \frac{e^{(I\nu+x+y)m}-e^{(-It+2y)m}}{1-e^{-It-I\nu-x+y}}. \end{split}

Similarly, for the second sum \begin{split} &\sum_{n=m+1}^{\infty} \frac{1}{2\pi}\int_0^{2\pi}dt\, A(\nu,t)e^{-Int} \Im e^{I\nu(n-m)} \int_{-\infty}^{-z} dx\,e^{(n-m)x} \int_{-\infty}^z dy\,e^{(m+n)y} \\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, e^{(-I\nu-x+y)m} \sum_{n=m+1}^{\infty} e^{(-It+I\nu+x+y)n}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, e^{(-I\nu-x+y)m} \frac{e^{(-It+I\nu+x+y)(m+1)}}{1-e^{-It+I\nu+x+y}}\\ =&\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, e^{-It+I\nu+x+y} \frac{e^{(-It+2y)m}}{1-e^{-It+I\nu+x+y}}. \end{split}

Finally, $$(-1)^{m}\frac{a_m(\nu)e^{2mz}}{2m} = \frac{\nu}{2\pi}\int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dy\, e^{(-It+2y)m}.$$


Summing over $m=1,2,\dots$, we get \begin{split} -\log(1-e^{2z}) &= \frac{\nu}{2\pi}\int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dy\, \frac{e^{-It+2y}}{1-e^{-It+2y}} \\ &+\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^z dx\, \int_{-\infty}^z dy\, \frac{ \frac{e^{I\nu+x+y}}{1-e^{I\nu+x+y}}-\frac{e^{-It+2y}}{1-e^{-It+2y}}}{1-e^{-It-I\nu-x+y}}\\ &+\frac{1}{2\pi}\Im \int_0^{2\pi}dt\, A(\nu,t) \int_{-\infty}^{-z} dx\, \int_{-\infty}^z dy\, \frac{e^{-2It+I\nu+x+3y}}{(1-e^{-It+I\nu+x+y})(1-e^{-It+2y})}. \end{split}

Evaluating integrals over $x$ and $y$ (which is a techincal task) will give an integral equation for $A(\nu,t)$.

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