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I am trying to find the minimum of this function analytically: $$ G(x)=\frac{c(a.x^3+b)}{x}+2(a.x^3+b)\sum_{m=1}^{\infty}\frac{e^{-\beta^2m^2(T-\frac{c}{x})}-e^{-\beta^2m^2T}}{\beta^2m^2} $$ where $0<x<1$ and all of the other variables have positive value. But the derivative is so complex and it is not solvable analytically. I tried to use the solution of Basal problem to approximate the main function as follows: $$ \sum_{m=1}^{\infty}\frac{e^{-\beta^2m^2(T-\frac{c}{x})}-e^{-\beta^2m^2T}}{\beta^2m^2}<\frac{e^{-\beta^2(T-\frac{c}{x})}}{\beta^2}.\frac{\pi^2}{6} $$ so I can write: $$ G(x)<\frac{c(a.x^3+b)}{x}+2(a.x^3+b)\frac{e^{-\beta^2(T-\frac{c}{x})}}{\beta^2}.\frac{\pi^2}{6} $$ Then, I tried to solve the new formula analytically so that I can earn the minimum of upper bound of the $G(x)$. But I couldn't do it again.

Is there any way to get rid of exponential part or do something else to earn the minimum of G(x)?

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I don't know if it helps, but I think this can be written as: $$ G(x) = (ax^3 + b)\left(\frac{c}{x} + 2\int_0^{\frac{c}{x}}\psi\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt\right) \quad \left(x > \frac{c}{T}\right), $$ or, perhaps a little more simply: $$ F(y) = \left(\frac{ac^3}{y^3} + b\right)\left(y + 2\int_0^y\psi\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt\right) \quad (0 < y < T), $$ where $\psi$ is the function defined on p.15 of H. M. Edwards, Riemann's Zeta Function (1974, repr. Dover 2001): $$ \psi(s) = \sum_{m=1}^\infty e^{-m^2{\pi}s} \quad (s > 0). $$ Perhaps someone who knows something about theta functions (I don't, I'm afraid) can do something with this?

In terms of this notation, which seems to be standard: $$ \vartheta(s) = \sum_{m=-\infty}^\infty e^{-m^2{\pi}s} \quad (s > 0), $$ the formulae are so much neater: \begin{gather*} G(x) = (ax^3 + b)\int_0^{\frac{c}{x}}\vartheta\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt \quad \left(x > \frac{c}{T}\right), \\ F(y) = \left(\frac{ac^3}{y^3} + b\right)\int_0^y\vartheta\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt \quad (0 < y < T) \end{gather*} that I increasingly suspect this has only got us back to where the problem came from in the first place! If so, I'm sorry!

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  • $\begingroup$ Thank you for your answer, but as you told, this only got us back to the original problem. ;-) $\endgroup$ – Mostafa Nov 13 '18 at 14:16
  • $\begingroup$ I do not understand the last simplication. how did you write it? Probably you dropped a $y$ from the original formulation. $\endgroup$ – Mostafa Nov 13 '18 at 16:35
  • $\begingroup$ Very likely! Can you give your own version of the problem in its original form? That would be more useful than my (imperfectly) reconstructed version. I could probably then delete my "answer" - no point in cluttering up the place! :) $\endgroup$ – Calum Gilhooley Nov 13 '18 at 17:33
  • $\begingroup$ By the way, if you are going to show the original form of the problem, it is probably best to literally draw a line under the present version of the question (blank line, followed by at least three dashes, followed by another blank line), and then add a new section describing how you arrived at the reformulation. (See Markdown Editing Help.) $\endgroup$ – Calum Gilhooley Nov 13 '18 at 18:39
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    $\begingroup$ (Double take.) I see what you mean now, about dropping the $y$. Doesn't it correspond (as I imagined it did) to the $m = 0$ term in the expression for $\vartheta(s)$? $\endgroup$ – Calum Gilhooley Nov 13 '18 at 18:51

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