search for: hypergoemetr

...gamma(1.25)/sqrt(sqrt(x)) which would approach zero. Now if the formula for the hypergeometric gain were instead gamma(1.5) * M(-.5;1;-x) / sqrt(x) that *would* approach 1, but that's just me noodling around with the formula to get something that approaches 1. Since I don't know how the hypergoemetric gain was derived (or even really what it means) I don't know if that's useful or not. Can you tell me what the source was for the original table values? John Ridges Jean-Marc Valin wrote: > Something looks odd without your values (or the doc) because hypergeom_gain() > should r...

By my reckoning the confluent hypergoemetric functions should have the following values: M(-.25;1;-.5) = 1.11433 M(-.25;1;-1) = 1.21088 M(-.25;1;-1.5) = 1.29385 M(-.25;1;-2) = 1.36627 M(-.25;1;-2.5) = 1.43038 M(-.25;1;-3) = 1.48784 M(-.25;1;-3.5) = 1.53988 M(-.25;1;-4) = 1.58747 M(-.25;1;-4.5) = 1.63134 M(-.25;1;-5) = 1.67206 M(-.25;1;-5....

Thanks for the confirmation Jean-Marc. I kind of suspected from the comments that it was the confluent hypergoemetric function, which I was trying to evaluate using Kummer's equation, namely: M(a;b;x) is the sum from n=0 to infinity of (a)n*x^n / (b)n*n! where (a)n = a(a+1)(a+2) ... (a+n-1) But when I use Kummer's equation, I don't get the values in the "hypergeom_gain" table. Did you u...

...(sqrt(x)) which > would approach zero. Now if the formula for the hypergeometric gain were > instead gamma(1.5) * M(-.5;1;-x) / sqrt(x) that *would* approach 1, but > that's just me noodling around with the formula to get something that > approaches 1. Since I don't know how the hypergoemetric gain was derived > (or even really what it means) I don't know if that's useful or not. Can > you tell me what the source was for the original table values? > > John Ridges > > > Jean-Marc Valin wrote: >> Something looks odd without your values (or the doc) b...

...r values (or the doc) because hypergeom_gain() should really approach 1 as x goes to infinity. But in the end, an approximation is probably OK because denoising is anything but an exact science :-) Jean-Marc Quoting John Ridges <jridges at masque.com>: > By my reckoning the confluent hypergoemetric functions should have the > following values: > > M(-.25;1;-.5) = 1.11433 > M(-.25;1;-1) = 1.21088 > M(-.25;1;-1.5) = 1.29385 > M(-.25;1;-2) = 1.36627 > M(-.25;1;-2.5) = 1.43038 > M(-.25;1;-3) = 1.48784 > M(-.25;1;-3.5) = 1.53988 > M(-.25;1;-4) = 1.58747 > M(-.25;...

Hi, I've been trying to re-create the table in the function "hypergeom_gain" in preprocess.c, and I just simply can't get the same values. I get the same value for the first element, so I know I'm computing gamma(1.25)^2 correctly, but I can't get the same numbers for M(-.25;1;-x), which I assume is Kummer's function. Is it possible that the comment is out of

...at the table data has an interval of .5 for the x axis. How far are your results from the data in the table? Cheers, Jean-Marc Quoting John Ridges <jridges at masque.com>: > Thanks for the confirmation Jean-Marc. I kind of suspected from the > comments that it was the confluent hypergoemetric function, which I was > trying to evaluate using Kummer's equation, namely: > > M(a;b;x) is the sum from n=0 to infinity of (a)n*x^n / (b)n*n! > where (a)n = a(a+1)(a+2) ... (a+n-1) > > But when I use Kummer's equation, I don't get the values in the > "hyperg...