Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to know if there is a (preferably closed-form) solution for

$-B \ln y -A y \ln y + A y- A =0$ for $y$

Where $A, B \in \mathbb{R}^{+}$. I have reasons to think there isn't a closed form solution, but I am also sure that there are solutions other than $y=1$

Is there any theorem saying there are no closed-form solutions to this equation? If there is, do you think an approximate solution might be possible? If there isn't, any ideas how to solve it?

(I already asked this question before, but with a sign change that made the solution almost trivial. It was a typo, so I'm asking again)


Note: My question is motivated by the need for a solution to this problem:

$1-\frac{A}{x} \gamma (2,\frac{x}{B}) = 0$ for $x$

The solution needn't be exact, but it'd be great if it was in closed form (I'm trying to avoid numerical approximations). Especially, I'm interested in the case $r$ around $B$, but also would like convergence for $B \rightarrow 0$.

My approach was to perform an asymptotic expansion on the incomplete lower gamma function following http://dlmf.nist.gov/8.11, truncating the series after 2 terms. After simplifying, Mathematica gives me:

$x + x \frac{A}{B} e^{-\frac{x}{B}}+A e^{-\frac{x}{B}} -A =0$

So, after the substitution $x=-B \ln y$, I get: $-B \ln y -A y \ln y + A y- A =0$ for $y$

Maybe there is a better approach to this problem?

Thanks!

share|improve this question
1  
If you express $y$ as $e^x$, you should be able to solve it with this: en.wikipedia.org/wiki/Lambert_W_function (I only know this function because it appears quite often in stackexchange posts so I might be completely wrong) –  xavierm02 Apr 6 '13 at 18:06

2 Answers 2

Hint:

$$0=-B\log y-Ay\log y+Ay-A=\log\frac{1}{y^B}+\log\frac{1}{y^{Ay}}+Ay+A=\log\frac{1}{y^{Ay+B}}+Ay+A$$

This doesn't look, as you say, like having a closed form but perhaps the above, together or not with xavier's comment (Lambert function), can give you some good approximation.

share|improve this answer
    
Thank you, I will definitely try this. Do you happen to know what theorem proves there isn't a closed form for this sort of equations? –  misi Apr 7 '13 at 5:22
    
@misi, read this: en.wikipedia.org/wiki/… –  DonAntonio Apr 10 '13 at 23:01

Changing variable $log(y)=x$ the equation can be taken in the form:

$e^x= -\frac{B}{A} \frac{ x +\frac{A}{B}}{ x -1}$

Such equation can be formally solved by Lagrange series (related to Lambert W function):

$x(a,b)=t- (t-s) \sum_{n=1} \frac{L_n' (n(t-s))}{n} e^{-nt} a^n$

where:

$L_n(x)$ is n-th Laguerre polynomial and $L_n'(x)$ its first derivative

$s=-\frac{A}{B}$

$t=1$

$a=-\frac{B}{A}$

For details see:

"On the generalization of the Lambert W function with applications in theoretical physics" István Mező, Árpád Baricz http://arxiv.org/abs/1408.3999

share|improve this answer
    
This problem seems linked with: math.stackexchange.com/questions/1033398/… –  giorgiomugnaini Feb 21 at 14:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.