I wonder if the following infinite sum is always negative for all (finite) $A,d>0$ and $B<0$. Any counterexample also suffice. Here is the sum: $$\frac{\partial}{\partial d}\sum_{n=1}^\infty n \int_{(-\infty,B),(A,\infty)} f^{n}(x;d)\mbox{d}x$$ with $$f^n(x;d)=\int_{B}^A f^{n-1}(x-w;d)f^1(w;d)\mbox{d}w$$ where $$f^1(x;d)=\frac{1}{\sqrt{2\pi d^2}}e^{\frac{-(x+d^2/2)^2}{2d^2}}$$

One can calculate $f^2$ as $$f^2(x;d)=\frac{e^{\frac{-(x+d^2)^2}{4d^2}}(\mbox{Erf}(-\frac{-2A+x}{2d})-\mbox{Erf}(-\frac{-2B+x}{2d}))}{4d\sqrt{\pi}}$$

The first term of the sum can be calculates as

$$\int_{(-\infty,B),(A,\infty)} f^{1}(x;d)\mbox{d}x=\int_{-\infty}^{B}\frac{1}{\sqrt{2\pi d^2}}e^{\frac{-(x+d^2/2)^2}{2d^2}}\mbox{d}x+\int_{A}^{\infty}\frac{1}{\sqrt{2\pi d^2}}e^{\frac{-(x+d^2/2)^2}{2d^2}}\mbox{d}x$$ $$\quad\quad\quad\quad\quad\quad\quad\quad=1+\frac{1}{2}\left(\mbox{Erf}\left(\frac{B+d^2/2}{d\sqrt{2}}\right)-\mbox{Erf}\left(\frac{A+d^2/2}{d\sqrt{2}}\right)\right)$$

From here taking the derivative with respect to $d$ we get the first term (including the derivation) as $$\frac{\partial}{\partial d}\left(1+\frac{1}{2}\left(\mbox{Erf}\left(\frac{B+d^2/2}{d\sqrt{2}}\right)-\mbox{Erf}\left(\frac{A+d^2/2}{d\sqrt{2}}\right)\right)\right)$$ $$=\frac{(A-d^2/2)e^{\frac{-(A+d^2/2)^2}{2d^2}}-(B-d^2/2)e^{\frac{-(B+d^2/2)^2}{2d^2}}}{\sqrt{2\pi }d^2}$$

Accordingly the sum can be rewritten as

$$\frac{\partial}{\partial d}\sum_{n=1}^\infty n \int_{(-\infty,B),(A,\infty)} f^{n}(x;d)\mbox{d}x$$ $$=1*\left(\frac{(A-d^2/2)e^{\frac{-(A+d^2/2)^2}{2d^2}}-(B-d^2/2)e^{\frac{-(B+d^2/2)^2}{2d^2}}}{\sqrt{2\pi }d^2}\right)$$ $$+2*\left(\frac{\partial}{\partial d}\int_{(-\infty,B),(A,\infty)}\frac{e^{\frac{-(x+d^2)^2}{4d^2}}(\mbox{Erf}(-\frac{-2A+x}{2d})-\mbox{Erf}(-\frac{-2B+x}{2d}))}{4d\sqrt{\pi}}\mbox{d}x\right)$$ $$+3*\left(\frac{\partial}{\partial d}\int_{(-\infty,B),(A,\infty)}f^{3}(x;d)\mbox{d}x\right)+4*\left(\frac{\partial}{\partial d}\int_{(-\infty,B),(A,\infty)}f^{4}(x;d)\mbox{d}x\right)\ldots$$

I am not able to evaluate the following integrals because the integral of the error function doesnt have a closed form.

I also dont think that I can get any solution to this problem. Therefore, I will be happy to see any comment: if I have to think from another point of view? use some tricks? or simply leave the question because it is not solvable?


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    $\begingroup$ Putting a bounty notice in the title does not actually set a bounty on a question. $\endgroup$ – Gerry Myerson Jan 7 '14 at 21:26
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    $\begingroup$ If you ask me, I think the whole bounty system should be changed. $50$ pts for listing and extra $xyz$ (user defined) pts for bounty. If the bounty should be granted or should be given back to the owner of the question should be decided by at least $5$ persons whose overall reputation is over $abcd$. $\endgroup$ – Seyhmus Güngören Jan 7 '14 at 22:02
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    $\begingroup$ The place for that is meta. Why not post a question about it there? $\endgroup$ – Gerry Myerson Jan 7 '14 at 22:51
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    $\begingroup$ @GerryMyerson Good idea thanks, I will do it. $\endgroup$ – Seyhmus Güngören Jan 7 '14 at 23:11
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    $\begingroup$ IMHO, I don't think there is anything wrong with the bounty arrangement proposed in the title (as long as one keep the promise). I have seen so many bounties wasted on good questions w/o getting any answers / attracts really bad answers / and bounty automatically awarded to bad answers. I don't mind someone try something new. $\endgroup$ – achille hui Jan 8 '14 at 11:08

This is not necessarily an answer but in the spirit of having another viewpoint or some additional tricks, note that in the limit as $A \to + \infty$ and $B \to - \infty$ that your expression for ${f^n}(x;d)$ appears in the form of a convolution.

Therefore, applying the convolution theorem, we can write, ${f^n}(x;d)$, as $${f^n}(x;d) = {\Im ^{ - 1}}(\Im ({f^{\,n - 1}})\Im ({f^{\,1}}))$$ where ${\Im()}$ denotes the Fourier transform and ${\Im ^{ - 1}}$ is the inverse transform.

Note especially that the transform of ${f^1}(x;d)$ is given by: $$F(\omega ) = \Im ({f^{\,1}}) = {\textstyle{1 \over {\sqrt {2\pi } }}}{e^{ - {\textstyle{1 \over 2}}{d^2}\omega (\omega + {\rm{i}})}},\,\,{{\rm{i}}^2} = - 1$$

which might lead to some simplification.

I have not worked out the details but if you substitute this into your recurrence relation, and then assuming that the final ${\Im ^{ - 1}()}$ is tractable (but maybe it doesn’t need to be to see how the sum works out), then this might give you give another “calculation-handle” on your problem.

  • $\begingroup$ when $A\rightarrow \infty$ and $B\rightarrow -\infty$, and if $d>0$ is finite the summation of all the terms (without the derivative operation) goes to infinity. Therefore I am not sure if I can use it for my problem. $\endgroup$ – Seyhmus Güngören Jan 7 '14 at 21:09
  • $\begingroup$ The limit as $A \to + \infty$ and $B \to - \infty$ is required to strictly satisfy the definition of the convolution operation. If your ${f^n}(x;d)$ vanishes outside some interval, $[ - {\textstyle{T \over 2}}, + {\textstyle{T \over 2}}]$, then you don’t necessarily have to let $A \to + \infty$ and $B \to - \infty$, don’t know if that helps. $\endgroup$ – Bruce Dean Jan 7 '14 at 21:22

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