Approximating the gamma function near x=-3

[theumbrellaman]

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

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]

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].

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!

 

[mfb]

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

 

[JJacquelin]

[itex]\Gamma(x)≈1/(x(x+1)(x+2)(x+3))[/itex] for x near -3, which doesn't agree with his answer.

It agrees !

 

[theumbrellaman]

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

 

[CellCoree]

I would first like to commend you on your efforts in self-studying and working through the problems in Nearing's book. It is great to see individuals taking initiative and exploring mathematical concepts on their own.

Regarding problem 1.16 about the gamma function, it is important to note that the gamma function is not defined for negative integers. Therefore, the approximations for x near 0, -1, -2, and -3 are not applicable.

However, for x near -3, the approximation given in the problem, \Gamma(x)≈-1/(6(x+3)), is correct. This can be derived using the identity x\Gamma(x)= \Gamma(x+1) and the fact that \Gamma(1)=1. Therefore, for x near -3, we have \Gamma(x)≈\frac{\Gamma(x+1)}{x}≈\frac{1}{x}=-\frac{1}{6(x+3)}.

The graph of the gamma function from -3 through positive values would look something like this:

[insert graph here]

As x approaches -3 from the left, the gamma function approaches negative infinity. As x approaches -3 from the right, the gamma function approaches positive infinity. This is because the gamma function has a pole at x=-3, meaning it is undefined at that point.

I hope this helps clarify the approximation for the gamma function near x=-3 and the graph of the function. Keep up the good work in your self-studying journey!