I want to solve the following problem for x:
\begin{equation} \frac{\mathrm{d}}{\mathrm{d}x}\ e^{-\beta_{1}x}\,{_{1}}F_{1}[-\alpha_{1};-\alpha_{3};\beta_{3}x]=0 \end{equation}
where, $\alpha_{1},\alpha_{3}, \beta_{1}, \beta_{3}, >0$ and $x\geq0$.
Taking the derivative I get
\begin{equation} \frac{\alpha_{1}}{\alpha_{3}}\beta_{3}e^{-\beta_{1}x}\,{_{1}}F_{1}[-\alpha_{1}+1;-\alpha_{3}+1;\beta_{3}x] = \beta_{1}e^{-\beta_{1}x}\,{_{1}}F_{1}[-\alpha_{1};-\alpha_{3};\beta_{3}x] \end{equation}
Rearranging terms,
\begin{equation} \frac{{_{1}}F_{1}[-\alpha_{1};-\alpha_{3};\beta_{3}x]}{{_{1}}F_{1}[-\alpha_{1}+1;-\alpha_{3}+1;\beta_{3}x]} - \frac{\alpha_{1}\beta_{3}}{\alpha_{3}\beta_{1}} =0 \end{equation}
Not 100% sure I know how to move forward. Any help would be appreciated. Thanks.
If $a\neq -1,-2,-3,\dots,$ and $a-b\neq 0,1,2,\dots,$ then the ratio of confluent hypergeometric functions can be expressed as the continued fraction:
\begin{equation} \frac{{_{1}}F_{1}(a,b,z)}{{_{1}}F_{1}(a+1,b+1,z)}=1+\cfrac{u_{1}z}{1+\cfrac{u_{2}z}{1+\cdots}} \end{equation}
where, $u_{2n+1}=\frac{a-b-n}{(b+2n)(b+2n+1)}$, and $u_{2n}=\frac{a+n}{(b+2n-1)(b+2n)}$.
This continued fraction can be truncated at any $u_{n}$ which yields a ratio of polynomials in $z$. Taking the quotient of the polynomials and solving for z will give an approximation to the root in question. Truncating at larger $n$ will yield a closer approximation.