Symbolic Sum of Gamma Function with Negative Integer Values

8 Ansichten (letzte 30 Tage)
Thomas
Thomas am 7 Jan. 2013
I'm trying to define a number of functions with multiple indices, one of which is used to sum factorial terms that may contain negative values. So I replaced the factorial with the gamma function, but now cannot evaluate the resulting sum due to the singularity. These terms should be zero in my functions, here is a simplified example:
syms q k
fun(q) = gamma(-2*q)
fun2(k) = symsum(k/fun(q),q,0,2)
I need this to produce a function so I can then numerically integrate it. Thanks in advance for your help.
  2 Kommentare
Walter Roberson
Walter Roberson am 7 Jan. 2013
To check: when you have a factorial of a negative value, you wish the result to be 0? If so that would not be standard for factorial.
Thomas
Thomas am 7 Jan. 2013
The factorial was in the denominator; the function (or rather that term in the sum) should be zero. When I evaluate the gamma function at negative integers I get infinity, which is correct, but I cannot then sum those terms to get a usable function. Here is the real code:
m = 1;
S_n1q(q) = (2/pi)^0.5/((-2)^q*factorial(q)*gamma(n-2*q-m+1)*zi^(n+m));
S_n_min1_1q(q) = (2/pi)^0.5/((-2)^q*factorial(q)*gamma(n-2*q-m)*zi^(n+m-1));
S_n_plus1_1q(q) = (2/pi)^0.5/((-2)^q*factorial(q)*gamma(n-2*q-m+2)*zi^(n+m+1));
m = 1;
j = 0;
l = 0;
B_n100(k) = symsum(S_n1q(q)*(k*abs(zi))^(n-q+l-0.5)*besselk(n-q-j-0.5,k*abs(zi)),q,0,(n-rem(n,2))/2);
B_n_min1_100(k) = symsum(S_n_min1_1q(q)*(k*abs(zi))^(n-q+l-1.5)*besselk(n-q-j-1.5,k*abs(zi)),q,0,(n-1-rem(n-1,2))/2);
B_n_plus1_100(k) = symsum(S_n_plus1_1q(q)*(k*abs(zi))^(n-q+l+0.5)*besselk(n-q-j+0.5,k*abs(zi)),q,0,(n+1-rem(n+1,2))/2);
I have a bunch of these functions for different values of n, m, j and l, with n=1:36 and I need to integrate them numerically from 0 to Infinity.

Melden Sie sich an, um zu kommentieren.

Antworten (0)

Kategorien

Mehr zu Gamma Functions finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by