# Approximating the gamma function near x=-3

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 $$\Gamma(x)= \int_0^\infty t^{x-1}e^{-t}\,dt.$$

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, $\Gamma(x)≈-1/(6(x+3))$

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!

mfb
Mentor
You are on the right track, and nearly done:
Close to -3, what is x(x+1)(x+2) approximately?

$\Gamma(x)≈1/(x(x+1)(x+2)(x+3))$ for x near -3, which doesn't agree with his answer.
It agrees !

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haha wow can't believe I didn't notice that. thanks!