Hypergeometric functions

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05-21-2007
anaisgcristal

Hypergeometric functions

Hi everyone:
I've been working with hypergeometric series for the last couple of days, and after found a C-function to do this type of mathematical determination, and ensure that the limits of the function and the anomalies weren't include in the calculus..I found the next mess:

Quote:

gsl: hyperg_2F1.c:640: ERROR: domain error
Default GSL error handler invoked.
Abort

and the lines in my code are:

Quote:

aa = -1/(3*w);
bb = (w-1)/(3*w);
cc = 1-(5/(6*w));
dd = -pow(a, -3*w)*(1-omegam)/omegam;
printf("valor de aa= %1.3e, bb = %1.3e, cc = %1.3e y dd = %1.3e\n", aa, bb, cc, dd);
/*lingro = gsl_sf_hyperg_2F1 (aa, bb, cc, -pow(a, -3*w)*(1-omegam)/omegam );*/
lingro = 1.0;

where w = -1...And I evaluated the function (with mathematica) and the result should be 1.

If anyone know wich it's the problem...I really appreciate your help :-)
05-21-2007
zacs7

Quote:

aa = -1/(3*w)
dd = -pow(a, -3*w)*(1-omegam)/omegam;

Is a different from aa?
05-21-2007
anaisgcristal

Quote:

Originally Posted by zacs7

Is a different from aa?

Yes a it's different to aa...and both are define in main
05-21-2007
zacs7

Quote:

Originally Posted by pow usage

If base is negative and exponent is not an integral value, or if base is zero and exponent is negative, a domain error occurs, setting the global variable errno to the value EDOM.

ensure pow(a... isn't negative, and the exponent is an integral (ie not 1/3)
05-21-2007
anaisgcristal

Quote:

Originally Posted by zacs7

ensure pow(a... isn't negative, and the exponent is an integral (ie not 1/3)

Hi again, I checked this before to post :-(, and even print the values of aa, bb, cc, dd to ensure that this variables are correct evaluates... I hink I have some problem in the hypergeometric function..but I'vent used this function before, and I don't have any idea about which it's the problem and how solve it.

but thanks
05-21-2007
Govalant

Make sure that aa,bb ,cc and all others are of type float.

Since in C, if you do 5/2 it evaluates to 2.
but 5.0/2 = 2.5

Also, change every numeric constant and add a .0 (so the type changes to a float)

so your code would be

Code:

aa = -1.0/(3.0*w); bb = (w-1.0)/(3.0*w); cc = 1.0-(5.0/(6.0*w)); dd = -pow(a, -3.0*w)*(1.0-omegam)/omegam; printf("valor de aa= %1.3e, bb = %1.3e, cc = %1.3e y dd = %1.3e\n", aa, bb, cc, dd);
05-21-2007
zacs7

In order to remain completly correct, you should add .0f ;)

ie,

Code:

aa = -1.0f / (3.0f * w);

Also adding some whitespace between your operators so it's readable. Your problem is 99% related to passing incorrect data to the function (as Govalant said)
05-25-2007
anaisgcristal

Hi again!
Well I solve partially the troubles with the hypergeometric function....but I found that I can evaluate the function only with a parameter i.e

Quote:

gsl_sf_hyperg_2F1 (aa, bb, cc, w )

But when I tried to do with:

g

Quote:

sl_sf_hyperg_2F1 (aa, bb, cc, -pow(a, -3*w)*(1-omegam)/omegam )

I found a mistake and the abort the process.

Y tried to pre-declare the variable

Quote:

f = -pow(a, -3*w)*(1-omegam)/omegam
And make gsl_sf_hyperg_2F1 (aa, bb, cc, f )

But this didn't work

Have anyone some idea to how solve this? I've been programming in C since a couple of months and I don't have enought experience to solve this....

Thanks :-)