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## Main Question or Discussion Point

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 [tex]\Gamma(x)= \int_0^\infty t^{x-1}e^{-t}\,dt.[/tex]

The problem asks

I've tried to replicate this technique by approximating [itex]\Gamma(x)[/itex] near -1 as [itex]\Gamma(x)=\frac{\Gamma(x+1)}{x}≈\frac{1/(x+1)}{x}=\frac{1}{x(x+1)}.[/itex] Continuing in this manner leads to [itex]\Gamma(x)≈1/(x(x+1)(x+2)(x+3))[/itex] 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 [itex]x\Gamma(x)= \Gamma(x+1)[/itex]. Earlier in the text he mentions how, since [itex]\Gamma(1)=0!=1[/itex], using the identity above we can make the approximation for x near 0: [itex]\Gamma(x)≈\frac{\Gamma(1)}{x}=\frac{1}{x}[/itex].What is the gamma function for x near 1? 0? -1? -2? -3? Now sketch a graph of the gamma function from -3 through positive values. Ans: Near -3, [itex]\Gamma(x)≈-1/(6(x+3))[/itex]

I've tried to replicate this technique by approximating [itex]\Gamma(x)[/itex] near -1 as [itex]\Gamma(x)=\frac{\Gamma(x+1)}{x}≈\frac{1/(x+1)}{x}=\frac{1}{x(x+1)}.[/itex] Continuing in this manner leads to [itex]\Gamma(x)≈1/(x(x+1)(x+2)(x+3))[/itex] for x near -3, which doesn't agree with his answer. Thanks in advance for the help!