hypergeom without symbolic math toolbox

Hi, is there any way to use the hypergeom function without having the symbolic math toolbox? I have tried some of the posted in the file exchange, but they all tend to give different results for large arguments compared to hypergeom. Thanks

[Dyuman Joshi]

"I have tried some of the posted in the file exchange, but they all tend to give different results for large arguments compared to hypergeom."

The functions posted in FEX are numerical where as the hypergeom function is symbolical. There will be a difference in the precision of the values obtained.

"is there any way to use the hypergeom function without having the symbolic math toolbox?"

No, atleast not symbolically.

Why do you not want to use the symbolic math toolbox? Or do you not have access to it?

[Star Strider]

Please describe the problem you want to solve.

[Eze Far]

I don't have the symbolic math toolbox.

I'm fitting a function to some data. The function involves the confluent hypergeometric function of the first kind 1F1(a,b,z), where in my case z is real and can go to large values. I'm thinking to use the integral expression, but I guess it will take much longer than hypergeom. Thank you

[Torsten]

I'm thinking to use the integral expression, but I guess it will take much longer than hypergeom.

Just test it:

https://uk.mathworks.com/matlabcentral/answers/261559-confluent-hypergeometric-function-of-the-first-kind?s_tid=prof_contriblnk

F = @(a,b,z) gamma(b)/(gamma(b-a)*gamma(a))*integral(@(t) exp(z.*t).*t.^(a-1).*(1-t).^(b-a-1),0,1);
tic
for i = 1:1000
    F(1,2,-pi);
end
toc

Elapsed time is 0.119686 seconds.

tic
for i = 1:1000
    hypergeom(1,2,-pi);
end
toc

Elapsed time is 8.940058 seconds.

[Dyuman Joshi]

@Torsten, that function won't work when a==b, and it does not provide a solution for large(ish) values.

(One could argue if 100 is large or not)

F = @(a,b,z) gamma(b)/(gamma(b-a)*gamma(a))*integral(@(t) exp(z.*t).*t.^(a-1).*(1-t).^(b-a-1),0,1);


%1
F(1,2,3)

ans = 6.3618

hypergeom(1,2,3)

ans = 6.3618

%2
F(1e2,1e1,1e0)

Warning: Inf or NaN value encountered.

ans = NaN

hypergeom(1e2,1e1,1e0)

ans = 2.4862e+03

%3
F(10,5,1)

Warning: Minimum step size reached near x = 1. There may be a singularity, or the tolerances may be too tight for this problem.

ans = 0

hypergeom(10,5,1)

ans = 6.4804

[Torsten]

Yes, it was a lousy attempt ... But at least it shows that numerical evaluation can be much faster - if it works :-)

[Eze Far]

It works just fine for a=1/2 and b=3/2. Thanks

[Walter Roberson]

hypergeomic function has a tendancy towards large intermediate values.

[Dyuman Joshi]

"It works just fine for a=1/2 and b=3/2."

The FEX submissions didn't work for these values?

[the cyclist]

For some values of parameters, hypergeometric variates can also be well approximated by either binomial or poisson variates. (Web search turns these up, if that sounds useful.)