search for: kummer

Displaying 10 results from an estimated 10 matches for "kummer".

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2009 Jul 22
2
A technical question about the speex preprocessor.
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 use a different solution to the confluent hypergoemetric...
2009 Jul 21
2
A technical question about the speex preprocessor.
...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 Is it possible that the comment is out of date and the values of Kummer's function used to make the table were different? Any help would be greatly appreciated. John Ridges
2009 Jul 22
2
A technical question about the speex preprocessor.
...; 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 >> "hypergeom_gain" table. Did you use a d...
2009 Jul 22
2
A technical question about the speex preprocessor.
...idges <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 >>>> &...
2009 Jul 22
0
A technical question about the speex preprocessor.
...e 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 > "hypergeom_gain" table. Did you use a different solution to the...
2009 Jul 22
0
A technical question about the speex preprocessor.
...> > > > 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 > >> "hypergeom_gai...
2009 Jul 23
0
A technical question about the speex preprocessor.
...;>> >>>> >>>> >>>>> 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 th...
2002 Jun 03
1
Powerpoint and modification time
...ce security mode = 00 directory mask = 0700 force directory mode = 0700 directory security mask = 0700 force directory security mode = 00 ... ... ... If I forgot giving you any other important information please let me know. Thanks for your help in advance. Markus Kummer -- GMX - Die Kommunikationsplattform im Internet. http://www.gmx.net
2009 Jun 30
3
Delays estimation in Speex algorithms
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
2008 Sep 09
3
sort by value fails
hi all, i want use sort_by_value to sort by "username". so in the indexing process i add the value i want sort by as value (lang is php) $document->add_value(0, 'value'); when i sort i got all specialchars at the beginning ? ? ? alf the values be indexed as utf8. what do i wrong? any suggestions? regards, sven