Can we solve the following equation for $x$?

Question

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.)

Answer 1

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.

Answer 2

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 $\int 1/(x^{2n} +1)dx$? .