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.

  • 1
    $\begingroup$ Are you aware of the Kummer transformation, by any chance? Nevertheless, it is not possible in general to represent such a quotient by another hypergeometric function, but you might be interested in its continued fraction expansion. $\endgroup$ Jul 14, 2016 at 14:44
  • $\begingroup$ @J.M I do not necessarily need to represent the quotient as another hypergeometric function. I am just looking to simplify it. Since the hypergeometric function is a polynomial, isn't their a way to represent the quotient of two of them as a power series? Also, I have the NIST handbook and am taking a look at the continued fraction bit you posted. $\endgroup$ Jul 14, 2016 at 15:32
  • $\begingroup$ "the hypergeometric function is a polynomial" - then it would seem to me that you had neglected to mention that the numerator parameter is a nonpositive integer. That is a special case that requires a different way of handling. $\endgroup$ Jul 14, 2016 at 15:42
  • $\begingroup$ Please excuse me for the confusion, the parameters $\alpha_{1},\alpha_{3},\beta_{1},\beta_{3}$ are all positive. All four parameters can be integers or of the for $(2n+1)/2$. $\endgroup$ Jul 14, 2016 at 15:46
  • 1
    $\begingroup$ I am currently far away from my books, but look up The Special Functions and Their Approximations. $\endgroup$ Aug 2, 2016 at 15:46

1 Answer 1


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.


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