### Gamma Function Subroutine

Hello,
I am new to working in C++ and in order to become better acquainted with it I am working on a program that I have no idea how to start. I am trying to create a subroutine that will calculate the gamma function of a real number, and the way that was suggested was to use the Weierstrass form. I have put the code below but it is not working properly, any suggestions would be greatly appreciated.

 // calculating a gamma function using a subroutine #include #include using std::cout; using std::endl; double PI = 3.141592654; int Gamma( int z, int Nterms ) { int g = 0.577216; int retVal = z*exp(g*z); // apply products for(unsigned n=1; n<=Nterms; ++n) retVal *= (1 + z/n)*exp(-z/n); retVal = 1.0/retVal;// invert return retVal; } int main() { int z,gamma=1; cout<<"Enter the number to calculate gamma: "; cin>> num; for(int i=1;i<=num;i++) { gamma=Gamma(i); } cout<<"The gamma of the given number is: "<
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Well, I'm not sure if this is the programming issue, but the gamma function is continuous and therefore not always an interval. Try making it a double.

I'm new to programming my self, but I am likely starting grad school in physics this fall so I'm glad I'm not completely useless on this site! haha
closed account (D80DSL3A)
You broke it! The function I gave here
http://www.cplusplus.com/forum/beginner/94174/
is a function of 2 variables which returns a double (as noted by willf).

 ...the way that was suggested was to use the Weierstrass form.

I didn't suggest that, you requested it!
 I am trying to create a subroutine that will calculate the gamma function of a real number using the product series expansion of gamma.
That would be the Weierstrass form.

At any rate, I have improved on that function. It should be a function of 1 variable of course. The 2nd variable was for controlling precision.

Building a pre-determined precision into the function is easy:
 ``12345678910111213141516`` ``````double Gamma( double z ) { double g = 0.5772156649;// Euler-Mascheroni constant double retVal = z*exp(g*z); double x = 1.0; unsigned n=0; do { ++n; x = (1 + z/n)*exp(-z/n); retVal *= x; } while( x < (1 - 1e-12) );// the precision is determined here. return 1.0/retVal; }``````

It gives decent results. The check values are from the std::tgamma() function as suggested by Cubbi in the previous thread.
 ``` Gamma(1/2) = 1.77245 check = 1.77245 Gamma(0.999) = 1.00058 check = 1.00058 Gamma(1) = 0.999999 check = 1 Gamma(3) = 2 check = 2 Gamma(5.1) = 27.9317 check = 27.9318 Process returned 0 (0x0) execution time : 1.781 s ```

But it's slow. The higher z is the longer it takes.
A better version takes advantage of the identity Gamma(z+1) = z*Gamma(z)
 ``123456789101112131415161718192021222324`` ``````double Gamma_2( double z ) { double g = 0.5772156649;// Euler-Mascheroni constant double zi = floor(z);// integer part of z double zf = z - zi;// fractional part of z double retVal = zf*exp(g*zf); double x = 1.0; unsigned n=0; do // calculate Gamma for the fractional part of z { ++n; x = (1 + zf/n)*exp(-zf/n); retVal *= x; } while( x < (1 - 1e-12) ); retVal = 1.0/retVal;// invert // build up the value using the identity. for(double y = 0.0; y < zi - 0.5; ++y) retVal *= y + zf; return retVal; }``````

 ``` Gamma_2(1/2) = 1.77245 check = 1.77245 Gamma_2(0.999) = 1.00058 check = 1.00058 Gamma_2(1) = nan check = 1 Gamma_2(3) = nan check = 2 Gamma_2(5.1) = 27.9318 check = 27.9318 Process returned 0 (0x0) execution time : 0.328 s ```

That's almost 6x faster. If the time for only Gamma_2(5.1) was compared it would probably be larger than that, since that's the case that took the big time in the first case.

As you can see there is a small problem at integer z. The reason for the nan results can be seen at line 6: `double retVal = zf*exp(g*zf);` where zf = fractional part of z, which = 0 in these cases. With our initial value for retVal = 0 no amount of multiplication (lines 10-15) can change it, so we arrive at line 17 with retVal still = 0;
Line 17 gives us the nan result (division by 0).
I'll let you modify the function to cope with the case of integer z.
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