- #1
theumbrellaman
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I've just started self studying James Nearing's "Mathematical Tools for Physicists" (available at http://www.physics.miami.edu/~nearing/mathmethods/mathematical_methods-three.pdf), and I'm having trouble with problem 1.16 about the gamma function, defined for positive x as \Gamma(x)= \int_0^\infty t^{x-1}e^{-t}\,dt.
The problem asks
The problem also suggests to make use of the identity x\Gamma(x)= \Gamma(x+1). Earlier in the text he mentions how, since \Gamma(1)=0!=1, using the identity above we can make the approximation for x near 0: \Gamma(x)≈\frac{\Gamma(1)}{x}=\frac{1}{x}.
I've tried to replicate this technique by approximating \Gamma(x) near -1 as \Gamma(x)=\frac{\Gamma(x+1)}{x}≈\frac{1/(x+1)}{x}=\frac{1}{x(x+1)}. Continuing in this manner leads to \Gamma(x)≈1/(x(x+1)(x+2)(x+3)) for x near -3, which doesn't agree with his answer. Thanks in advance for the help!
The problem asks
The problem also suggests to make use of the identity x\Gamma(x)= \Gamma(x+1). Earlier in the text he mentions how, since \Gamma(1)=0!=1, using the identity above we can make the approximation for x near 0: \Gamma(x)≈\frac{\Gamma(1)}{x}=\frac{1}{x}.
I've tried to replicate this technique by approximating \Gamma(x) near -1 as \Gamma(x)=\frac{\Gamma(x+1)}{x}≈\frac{1/(x+1)}{x}=\frac{1}{x(x+1)}. Continuing in this manner leads to \Gamma(x)≈1/(x(x+1)(x+2)(x+3)) for x near -3, which doesn't agree with his answer. Thanks in advance for the help!
