Displaying 10 results from an estimated 10 matches for "confluent".
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" tab...
2009 Jul 22
2
A technical question about the speex preprocessor.
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...
2009 Jul 22
2
A technical question about the speex preprocessor.
..._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) =...
2009 Jul 21
2
A technical question about the speex preprocessor.
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
2009 Jul 22
0
A technical question about the speex preprocessor.
...ithout 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
> 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...
2009 Jul 23
0
A technical question about the speex preprocessor.
...oes 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...
2009 Jul 22
0
A technical question about the speex preprocessor.
...x)
Note that 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
>...
2008 Aug 21
1
Parabolic cylinder function
Dear all,
I need your advice since I am looking for an implementation of the parabolic
cylinder function in R. I found implemantations of the hypergemetric
functions (the Whittaker and the confluent hypogeometric functions) in the
package fAsianOptions but the parabolic cylinder function was unfortunately
not there. Do you know of such implementation?
Thank you very much for your advice.
Cheers,
Zoe
[[alternative HTML version deleted]]
2010 Sep 24
0
Inaccuracy of kummerU (fAsianOptions) (Tricomi function)
Hello,
I need to use the confluent function of second kind, also known as
Tricomi function. It is implemented as kummerU() function in
fAsianOptions package, but I've found very inaccurate values, comparing
with those provided by Mathematica. I think Mathematica values are OK
because kummerU values leads to negative probabil...
2004 Jul 04
1
Rmetrics 191.10057
...tion. The functions included
are: In Part I, the Error Function "erf", the Psi or Digamma
Function "Psi", the Incomplete Gamma Function "igamma", the
Gamma Function fpr complex arguments, and the Pochhammer Symbol
"Pochhammer". In Part II, the Confluent Hypergeometric Functions
of the 1st Kind and 2nd Kind "kummerM" and "kummerU", the
Whittaker Functions "whittakerM" and "whittakerW" and the
Hermite Polynomials "hermiteH"
2004-06-29 fOptions/demo
A new example file named "xmpS...