search for: hypergeom_gain

I got the approximation from a Google book: http://books.google.com/books?id=2CAqsF-RebgC&pg=PA385 Page 392, formula (10.33) Using this formula, you're right, hypergeom_gain() would *not* converge to 1 for large x, but would instead be 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 g...

...b-a))*(x^-a) Which for M(-.25;1;-x) is sqrt(sqrt(x))*1.10326 at x=10 this gives a value of 1.96191 (vs. what I think is the true value of 1.97456). The reason I've gotten into all this is I'm trying to vectorize with SSE intrinsics some of the slower loops in the preprocessor, and the hypergeom_gain function is the only thing stopping me from removing all the branches in the loops. I don't know how critical the accuracy of the function is to the performance of the preprocessor, but if that aforementioned approximation was good enough for all the values of x, it would really speed the l...

...hat 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 use a different solution to the confluent hypergoemetric function when you created the table? John Ridges Jean-Marc Valin wrote: > Hi John, > > M(;;) is the confluent hypergeometric function. > > Jean-Marc > > John Ridges a ?crit : > >> Hi,...

...doc and the code, the code is the one likely to be correct. Jean-Marc John Ridges a ?crit : > I got the approximation from a Google book: > > http://books.google.com/books?id=2CAqsF-RebgC&pg=PA385 > > Page 392, formula (10.33) > > Using this formula, you're right, hypergeom_gain() would *not* converge > to 1 for large x, but would instead be 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...

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 commen...

Something looks odd without your 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 &gt...

...oemetric 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 use a different solution to the > confluent hypergoemetric function when you created the table? > > John Ridges > > > Jean-Marc Valin wrote: > > Hi John, > > > > M(;;) is the confluent hypergeometric function. > > > > Jean-Marc &...

Speex tells me that the decoder is always 5 ms, but it says that the encoder is 5 ms for NB, 8.9375 ms for WB, and 10.90625 ms for UWB. Is there an extra frame of delay in the encoder that isn't otherwise accounted for? John Ridges Jean-Marc Valin wrote: > Quoting John Ridges <jridges at masque.com>: > >> I also need to know the precise delays from Speex but I used