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The Beta function has recursive relation: $$ Beta(x,y) = Beta(x, y+1) + Beta(x+1, y) \! $$

Is there some least redundant condition, if added, will make the recursive relation an equivalent definition of the Beta function?

The purpose of my question is to understand special functions from very basic view. Thanks!

Tim
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  • If you are interested in positive integer values for $x,y;$ you can use $$B(1,y)=\frac{1}{y},\quad B(x,1)=\frac{1}{x}$$ for the base cases. – gammatester Feb 05 '14 at 14:24
  • @gammatester: Thanks! (1) Do $B(x,y)=B(x,y+1) + B(x+1,y), B(1,y)=1/y$, and $B(x,1)=1/x$ fully define $B$ on $\mathbb{N} \times \mathbb{N}$? (2) Are $B(1,y)=1/y$, and $B(x,1)=1/x$ also true when $x$ and $y$ are any complex numbers with positive real parts? – Tim Feb 06 '14 at 12:56
  • If guess the answer is yes, use the recursive definition: $$B(1,y) = \frac{1}{y}\ B(x,1) = \frac{1}{y}\ B(x,y) = B(x-1,y) - B(x-1,y+1)\ B(x,y) = B(x,y-1) - B(x+1,y-1) $$ A Pascal implementation works (rounding errors and execution time are obviously larger than for standard functions), but I cannot prove mathematically that recursive function is well defined. – gammatester Feb 06 '14 at 13:19
  • $B(1,y)=1/y$ is valid for complex $y$ as long as $y$ is no negative integer or 0. This follows from the relation to the $\Gamma$ function: $$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$ $$B(1,y)=\frac{\Gamma(1)\Gamma(y)}{\Gamma(1+y)}=\frac{\Gamma(y)}{y\Gamma(y)}= \frac{1}{y} $$Same for $x$ via symmetry. – gammatester Feb 06 '14 at 13:27

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The easier definition to leverage is probably: B(x+1,y) = B(x,y) * x/(x+y)

For example, here is some python code that will compute the beta function for specified positive integer values of x and y. It also compares the result to Scipy

from scipy import special

x,y = 5,50
Beta = 1/y
for i in range(1,x):
    Beta = Beta * i / (i + y)

print Beta
print special.beta(x,y)
Bscan
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