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 Tommy's Gamma trick ? tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 10/14/2015, 09:37 PM (This post was last modified: 10/16/2015, 12:19 PM by tommy1729.) Here I Will explain the Gamma trick. Tet(x) is computed from interpolating 1/tet(x). So we need An interpolation for 1/tet(x). Let Cs be the continuum Sum. 1/tet(x) = Cs ( 1/tet(x) ) - Cs ( 1/ln(tet(x)) ) Obvious. So we need to interpolate Cs( 1/tet(x) ). We use Cs( 1/tet(x) + ln(x) ) - Cs( ln(x) ). Cs ln = lngamma. Use the generalized formula < replacing product by comm operator > described below to interpolate the Cs f(x) = Cs( 1/tet(x) + ln(x) ). Lim n -> oo Gamma(z) = n ! n^Z / [ z (z+1) (z+2) ... (z+n)]. A Well known limit btw. In this case the comm operator is A • B = f^[-1](f(A) + f(B)). So we have a limit form for the Cs(f) and therefore also Cs(1/tet(x)). From there we get the continuüm product Cp (1/tet(x)). And finally Cp(tet(x)). However to compute f(A) we need the values for tet(x). But we know tet ' (x) = Cp(tet(x)). We use that to set Up a recursion equation. The details are complicated ... Notice this is different from mike3 method for Cs and the euler formula for Cs. Also uniqueness is not yet achieved and a few more details need to be added. Forgive me for the incompleteness Im still considering. Regards Tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread Tommy's Gamma trick ? - by tommy1729 - 10/14/2015, 09:37 PM RE: Tommy's Gamma trick ? - by tommy1729 - 10/16/2015, 12:28 PM RE: Tommy's Gamma trick ? - by tommy1729 - 10/17/2015, 12:22 AM RE: Tommy's Gamma trick ? - by tommy1729 - 10/22/2015, 12:34 PM RE: Tommy's Gamma trick ? - by tommy1729 - 11/02/2015, 01:43 AM RE: Tommy's Gamma trick ? - by tommy1729 - 11/04/2015, 12:06 AM RE: Tommy's Gamma trick ? - by tommy1729 - 11/04/2015, 01:26 PM RE: Tommy's Gamma trick ? - by tommy1729 - 11/07/2015, 01:02 PM

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