I have following equation to solve for $x$ $$k-ax^2\sum_{n=0}^{\infty}\frac{(-x^mb)^n}{(1+tn)}(1-u^{-n})=0$$ where $k$ is a real number, $a,b,t,u$ are positive constants. Any ideas will be very helpful. Thanks in advance. (Also please add the appropriate tags for this question.)
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Where does this come from? – Jacky Chong Nov 02 '16 at 03:49
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its is from one of the research paper. But I want to solve this equation – Frank Moses Nov 02 '16 at 03:51
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Do you mean indeed $-(x^mb)^n$ or perhaps $(-x^mb)^n$ ? – user90369 Nov 02 '16 at 14:07
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@user90369 thank you. I have corrected it. How do you know that right place for minus sign? Is it just coincidence or this is some famous function? – Frank Moses Nov 02 '16 at 14:11
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I only thought that this single minus doesn't make sense. With the formula of Claude Leibovici you have to change now $\tanh$ to $\tan$ . The general function to create a closed term for your series is $\int \frac{dx}{1+x^t}$ used as a independently function. For $t=2$ you get $\arctan$ (or written often as $\tan^{-1}$). – user90369 Nov 02 '16 at 14:20
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The sum is an incomplete beta function which may have an inverse with inverse beta regularized, a Mathematica function. – Тyma Gaidash Jun 21 '23 at 02:23
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Since you used the tag hypergeometric-function, you could notice that $$t\sum_{n=0}^{\infty}\frac{(x^mb)^n}{(1+tn)}\left(1-\frac 1 {u^{n}}\right)=\Phi \left(b x^m,1,\frac{1}{t}\right)-t \, _2F_1\left(1,\frac{1}{t};1+\frac{1}{t};\frac{b x^m}{u}\right)$$ where appears the Hurwitz-Lerch transcendent function.
This leaves you with the problem of solving $$\frac{k t}{a x^2}=t \, _2F_1\left(1,\frac{1}{t};1+\frac{1}{t};\frac{b x^m}{u}\right)-\Phi \left(b x^m,1,\frac{1}{t}\right)$$ which looks to be a nice monster.
Probably, only numerical methods could be considered for solving for $x$.
For $t=1$ , this would reduce to $$\frac{k }{a x^2}=\frac{ \log \left(1-b x^m\right)-u \log \left(1-\frac{b x^m}{u}\right)}{bx^m}$$ For $t=2$ , this would reduce to $$\frac{k }{a x^2}=\frac{ \sqrt{u} \tanh ^{-1}\left(\frac{\sqrt{b} }{\sqrt{u}}x^{m/2}\right)-\tanh ^{-1}\left(\sqrt{b} x^{m/2}\right)}{x^{m/2}\sqrt{b}}$$ I cannot see any further "simple" expression.
Claude Leibovici
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thank you for your answer. I actually want to avoid numerical answer. Though your provided expression looks much nice than of original question – Frank Moses Nov 02 '16 at 05:49
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@FrankMoses. I really do not see how you could avoid numerical methods (except, may be, for some very specific values of the parameters). – Claude Leibovici Nov 02 '16 at 06:18
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do you have any idea about those "specific" values? I will be very happy if I could write an analytically solvable expression for even a single set of parameters – Frank Moses Nov 02 '16 at 06:28
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Just as a joke : $u=1$ or $m=0$ ! Could you give me an idea about the ranges for $a,b,t,u,m$ ? – Claude Leibovici Nov 02 '16 at 06:33
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Hahaha. Nice one. $t>1, m>2$ and $0\leq u\leq 1$. These are some ranges. Other constants has same ranges as given in my OP. – Frank Moses Nov 02 '16 at 06:39
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In any manner I should start defining $b x^{m}=y$; the equation would look a bit nicer. – Claude Leibovici Nov 02 '16 at 10:04
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First some definitions for $f_{inv}, f:=f(x), g:=g(x), \nu, \mu$ which are depending on $t$ :
$arc…$ is used for inverse trigonometric function and $f^{-1}$ is the general inverse function of $f$ but $-1$ can be misunderstood. To get a neutral symbol I want to use here $f_{inv}$ for the inverse ("$inv$") function of $f$ which can be developed as a Taylor series around $x=0$ .
Second: In this case I choose $t\ge 1$ but the conditions for the derivations can be changed and therefore the value range for $t$.
Be $f’:=-f^{2-t}g^{t-1}$ and $g’:=f$ with $f(0):=1$, $g(0):=0$, $f(\nu):=0$, $g(\nu):=1$, $\displaystyle (\frac{g}{f})(\mu):=1$ .
It follows that $\enspace f^{t-1}f’+ g^{t-1}g’=-fg^{t-1}+ g^{t-1}f=0 \enspace$ and therefore $\enspace f^t+ g^t=1$ .
Third: $\enspace$ With $\enspace\displaystyle (\frac{g}{f})’=1+(\frac{g}{f})^t \enspace $ one gets $\enspace\displaystyle (\frac{g}{f})_{inv}(x)=\int\limits_0^x \frac{ d\tau }{1+\tau^t}$ .
Notes:
$(1)\enspace\displaystyle \mu=(\frac{g}{f})_{inv}(1)=\int\limits_0^1 \frac{ d\tau }{1+\tau^t}$
$(2)\enspace$ Because of $\enspace\displaystyle (\frac{1}{f})^t=1+(\frac{g}{f})^t\enspace$ it’s $\enspace\displaystyle \nu=\int\limits_0^1 \frac{ d\tau }{\sqrt[t]{1+\tau^t}}=\lim\limits_{x\to\infty}(\frac{g}{f})_{inv}(x)$ .
$(3)\enspace$ Your equation chances to $$\frac{1}{\sqrt[t]{b}}(\frac{g}{f})_{inv}( \sqrt[t]{bx^m} )- \sqrt[t]{ \frac{u}{b} } ( \frac{g}{f})_{inv}( \sqrt[t]{ \frac{b}{u} x^m} ) =\frac{k}{a}x^{\frac{m}{t}-2} \, .$$
$(4)\enspace$ I don't know how to call $\enspace\displaystyle (\frac{g}{f})_{inv}\enspace$ but maybe $f$ and $g$ could be called Fermat functions, if this expression is not used for other functions. Under elliptic functions you will perhaps find some useful informations.
For natural $t:=n$ see the formula for $\int \frac{ dx }{x^n+1}$ : What is the integration of $\int 1/(x^{2n} +1)dx$? .
