3
$\begingroup$

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?

Regards-

$\endgroup$
8
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ The place for that is meta. Why not post a question about it there? $\endgroup$ – Gerry Myerson Jan 7 '14 at 22:51
  • 1
    $\begingroup$ @GerryMyerson Good idea thanks, I will do it. $\endgroup$ – Seyhmus Güngören Jan 7 '14 at 23:11
  • 6
    $\begingroup$ 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
1
+50
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.