Is it possible to approximate (or even find) a solution for the following equation: $$\frac{e^{-a}(v+u)(a^x+e^{a}x\Gamma(x,a))}{\Gamma(x+1)}-u\approx0,$$

where $x\ge 0$ and integer, and the parameters $0<a<1$ and $u>0$, $v>0$.

It should be solved for $x$, can it be expressed in terms of the parameters?

Numerical methods show that such solution exists. For example, for $x=1$, $a=0.7$, $v=96.5543$ and $u=523$ the term value is $0.0246$.

Note: $x$ can be taken as a real number and then use the floor or ceiling functions to define the final expression.

  • 1
    $\begingroup$ Within you equation there is a $p$ used but you did not defined it. $\endgroup$ – mrtaurho Nov 30 '18 at 22:20
  • 2
    $\begingroup$ Can you give a numerical example? I'm not even sure which variable you're solving for. Also, what would be considered a good approximation? $\endgroup$ – Maxim Dec 1 '18 at 5:11
  • 1
    $\begingroup$ Using $v=96.52509765$ while keeping all of the other values the same gets you much closer to zero. Also, all solutions for $x$ will only depend on $a$ and the ratio $r=v/u$. To see that divide both sides of the equation by $u$. $\endgroup$ – JimB Dec 1 '18 at 23:54
  • 1
    $\begingroup$ Not sure there's much more than noting that the ratio $u/v$ can be written in terms of $x$ and $a$: $\frac{-a^x+e^a \Gamma (x+1)-e^a x \Gamma (x,a)}{a^x+e^a x \Gamma (x,a)}$. Plotting $a$ vs $r$ for various integer values of $x$ might be informative. $\endgroup$ – JimB Dec 2 '18 at 2:31
  • 1
    $\begingroup$ If $a$ and $q = u/v$ are fixed and you're solving for $x$, it seems the lhs as a function of $x$ is monotonically increasing on $(-1, \infty)$ and always has a root there, finding which should be straightforward (the root may be negative). If you're allowing $x < -1$, the function has multiple zeroes on $(-\infty, -1)$, finding the zero closest to an integer will be more tricky. $\endgroup$ – Maxim Dec 2 '18 at 3:41

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.