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 The inverse gamma function. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/11/2014, 08:28 PM Because of recent events here, I felt the need to talk about the inverse gamma function. Im talking about the entire approximation of exp^[1/2] and the Taylor coefficients being O ( 1/ Gamma( n (arc2sinh(n) - 1) ) ). Cleary from the viewpoint of numerical analysis it seems usefull to have a good approximation of the inverse gamma function. The functional inverse that is. Now my honesty forces me to quote the source of an intresting approximation of the inverse gamma function : David W Cantrell http://mathforum.org/kb/message.jspa?messageID=342552 In case that page gets removed I post the main formula here : A is the positive zero of the digamma function ( 1.4616... ) B = sqrt(2 pi) / e - Gamma(A). ( 0.0365... ) L(x) = ln( (x+c) / sqrt(2 pi) ) W(x) is the Lambert W function. ( the functional inverse of x exp(x)) Gamma^[-1](x) = approx = 1/2 + L(x)/W(L(x)/e) The error term goes to 0 as x goes to +oo. Together with the approximation for the Lambert W : LambertW(x) = ln(x) - ln(ln(x)) + ln(ln(x)) / ln(x) this gives a practical way to compute the inverse gamma function. I think there is also a page by wolfram about the inverse gamma but I cant find it ? ( And no, I dont mean the inverse regularized gamma or the statistical inverse gamma distribution ) I would like to see an integral representation of the inverse gamma function too. regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/11/2014, 10:12 PM (05/11/2014, 08:28 PM)tommy1729 Wrote: Because of recent events here, I felt the need to talk about the inverse gamma function. Im talking about the entire approximation of exp^[1/2] and the Taylor coefficients being O ( 1/ Gamma( n (arc2sinh(n) - 1) ) ). Cleary from the viewpoint of numerical analysis it seems usefull to have a good approximation of the inverse gamma function. The functional inverse that is. Now my honesty forces me to quote the source of an intresting approximation of the inverse gamma function : David W Cantrell http://mathforum.org/kb/message.jspa?messageID=342552 In case that page gets removed I post the main formula here : A is the positive zero of the digamma function ( 1.4616... ) B = sqrt(2 pi) / e - Gamma(A). ( 0.0365... ) L(x) = ln( (x+c) / sqrt(2 pi) ) W(x) is the Lambert W function. ( the functional inverse of x exp(x)) Gamma^[-1](x) = approx = 1/2 + L(x)/W(L(x)/e) The error term goes to 0 as x goes to +oo. Together with the approximation for the Lambert W : LambertW(x) = ln(x) - ln(ln(x)) + ln(ln(x)) / ln(x) this gives a practical way to compute the inverse gamma function. I think there is also a page by wolfram about the inverse gamma but I cant find it ? ( And no, I dont mean the inverse regularized gamma or the statistical inverse gamma distribution ) I would like to see an integral representation of the inverse gamma function too. regards tommy1729 I always like posting my own results ^_^, for $\Re(z) > 0$ $\phi(z) = \frac{1}{\G(z)}\int_0^\infty \frac{w^{z-1}}{w!}\,dw$ then by some of my fractional calculus theorems. For $\Re(w) > 0$ and $\sigma > 0$ $\frac{1}{w!} = \frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma+i \infty} \G(\xi)\phi(\xi)w^{-\xi}\,d\xi$ Is that a good integral expression? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/12/2014, 11:06 PM (05/11/2014, 10:12 PM)JmsNxn Wrote: (05/11/2014, 08:28 PM)tommy1729 Wrote: Because of recent events here, I felt the need to talk about the inverse gamma function. Im talking about the entire approximation of exp^[1/2] and the Taylor coefficients being O ( 1/ Gamma( n (arc2sinh(n) - 1) ) ). Cleary from the viewpoint of numerical analysis it seems usefull to have a good approximation of the inverse gamma function. The functional inverse that is. Now my honesty forces me to quote the source of an intresting approximation of the inverse gamma function : David W Cantrell http://mathforum.org/kb/message.jspa?messageID=342552 In case that page gets removed I post the main formula here : A is the positive zero of the digamma function ( 1.4616... ) B = sqrt(2 pi) / e - Gamma(A). ( 0.0365... ) L(x) = ln( (x+c) / sqrt(2 pi) ) W(x) is the Lambert W function. ( the functional inverse of x exp(x)) Gamma^[-1](x) = approx = 1/2 + L(x)/W(L(x)/e) The error term goes to 0 as x goes to +oo. Together with the approximation for the Lambert W : LambertW(x) = ln(x) - ln(ln(x)) + ln(ln(x)) / ln(x) this gives a practical way to compute the inverse gamma function. I think there is also a page by wolfram about the inverse gamma but I cant find it ? ( And no, I dont mean the inverse regularized gamma or the statistical inverse gamma distribution ) I would like to see an integral representation of the inverse gamma function too. regards tommy1729 I always like posting my own results ^_^, for $\Re(z) > 0$ $\phi(z) = \frac{1}{\G(z)}\int_0^\infty \frac{w^{z-1}}{w!}\,dw$ then by some of my fractional calculus theorems. For $\Re(w) > 0$ and $\sigma > 0$ $\frac{1}{w!} = \frac{1}{2\pi i}\int_{\sigma - i\infty}^{\sigma+i \infty} \G(\xi)\phi(\xi)w^{-\xi}\,d\xi$ Is that a good integral expression? Dear James. For starters if you are trying to find the integral I asked for : 1) I asked for the functional inverse of the Gamma function. Not the reciprocal. The whole OP was about the functional inverse of the Gamma function. Although I could have stated that more clearly when I asked about the integral representation ... 2) ... Also defining f(x) By M^[-1] M^[1] f(x) seems a bit lame. That looks similar to saying x = sqrt(x)^2 or x = exp(ln(x)). 3) despite 1) and 2) why do you wonder if that is OK ? You know the mellin inversion theorem. Thanks anyway. Maybe a second attempt. Im not sure such an integral representation exists btw. regards tommy1729 JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/13/2014, 02:18 PM (05/12/2014, 11:06 PM)tommy1729 Wrote: Dear James. For starters if you are trying to find the integral I asked for : 1) I asked for the functional inverse of the Gamma function. Not the reciprocal. The whole OP was about the functional inverse of the Gamma function. Although I could have stated that more clearly when I asked about the integral representation ... 2) ... Also defining f(x) By M^[-1] M^[1] f(x) seems a bit lame. That looks similar to saying x = sqrt(x)^2 or x = exp(ln(x)). 3) despite 1) and 2) why do you wonder if that is OK ? You know the mellin inversion theorem. Thanks anyway. Maybe a second attempt. Im not sure such an integral representation exists btw. regards tommy1729 oooooo functional inverse. That's tricky... « Next Oldest | Next Newest »

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