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.

The original function was defined as $f(z)=y=\left( \Gamma \left( k,{\frac {z-b}{\theta}} \right) -\Gamma \left( k,{\frac {z-a}{\theta}} \right) \right) \left( \Gamma \left( k \right) \right) ^{-1}$, where the inversed function should in form of $z=f^{-1}(y)$ and $\theta, k$ can be collected in real time. $a, b$ was predefined real values and was constraint by $a < b$.

Is it possible to find the inverse function of it? Or is there good approximation for the inverse function when $k \leq 1$.

share|improve this question
    
You have a function of 5 variables. Inverse function with respect to which one? –  Gerry Myerson May 6 '12 at 11:30
    
it's $z$, the only variable. $a,b$ was constant, $\theta, k$ are from samples. –  Readon Shaw May 6 '12 at 11:56
    
It can be inverted, yes. However, you should know that the inverse is treated as an independent function in its own right (e.g. Mathematica's InverseGammaRegularized[]), and you'll need to resort to numerical methods anyway for evaluation. –  J. M. May 6 '12 at 13:06
    
J.M. I checked the InverseGammaRegularized, it seems only used for inversing gamma function separately, but $f(z)$ is not. it is a linear combination of gamma function. –  Readon Shaw May 6 '12 at 15:00

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.