Difference in MATLAB function evaluation
2 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
According to the properties of gamma function, for any x > 0
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/160284/image.png)
But when I try to use it in MATLAB, it shows a very large error
gamma(150) - 149*gamma(149)
ans =
-1.5240e+247
Can anybody explain the reason for this? I need it for an approximation algorithm I am trying to develop.
Thanks in advance.
1 commentaire
Star Strider
le 31 Jan 2017
FWIW, I get similar results for these (in R2016b):
n = 149;
q1 = gamma(n+1) - n*gamma(n);
q2 = gamma(n+1) - factorial(n);
q3 = gamma(n+1) - prod(1:n);
q1 =
-15.2402e+246
q2 =
-819.3656e+243
q3 =
-819.3656e+243
I hope this is a problem with my understanding of the gamma function, and not a problem with its implementation in MATLAB.
Réponse acceptée
Honglei Chen
le 31 Jan 2017
For your use case, 150! is a very large number so the numerical precision issue starts to kick in. I'd suggest you to use gammaln instead, for example, you can do
gammaln(150)-(log(149)+gammaln(149))
and the result is indeed 0.
HTH.
0 commentaires
Plus de réponses (1)
Jan
le 31 Jan 2017
Modifié(e) : Jan
le 31 Jan 2017
While the absolute error is huge, the relative error is tiny:
(gamma(150) - 149*gamma(149)) / gamma(150)
>> -3.9868e-014
This is the expected error caused by using the IEEE754 double format. This stores about 16 valid digits.
If an approxmation algorithm is sensitive for this difference, it is instable and of limited use.
0 commentaires
Voir également
Catégories
En savoir plus sur Gamma Functions dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!