# [SOLVED] Incomplete Gamma Function

• Feb 1st 2010, 09:10 AM
[SOLVED] Incomplete Gamma Function
I already posted this in the Applied section but I don't think thats actually the right place so here it is again!

I think I'm pretty close to getting this but can't quite figure out what do next!

From earlier in the question...

$\textrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt \sim 1 - \frac{e^{-x^2}}{\pi} \sum^{\infty}_{n=0} \Gamma \Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$ (*)

Show that erf(x) can be expressed in terms of the incomplete gamma function, and that the above asymptotic expansion agrees with the result obtained in Problem 6. [The formula $\Gamma\Big{(}\frac{1}{2} + n)\Gamma\Big{(}\frac{1}{2} - n\Big{)} = (-1)^n \pi$ (**) for integer n may be useful.]

So I think I have to show the relation... $\gamma\Big{(}\frac{1}{2}, x\Big{)} = \sqrt{\pi} \textrm{erf}(\sqrt{\pi})$ (***) as that's the relation I've seen elsewhere.

The result obtained in question 6 was...

$\gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{\Gamma(\alpha)}{\Gamma(\a lpha - n)}\frac{1}{x^n}$

And also...
$\gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{(\alpha-1)(\alpha-2)\ldots(\alpha-n)}{x^n}$

Right so intro done!

Now what I've done...

From (*)
$\sqrt{\pi} \textrm{erf}(x) = 2 \int^x_0 e^{-t^2} dt \sim \sqrt{\pi} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma\Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$

$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma\Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$
(This step seems to be showing that $\alpha = \frac{1}{2}$ which corresponds to (***))

Using (**)
$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2}$ $\sum^{\infty}_{n=0} \frac{1}{\Gamma\Big{(}\frac{1}{2} - n\Big{)}}\frac{1}{x^{2n+1}}$

$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2} \sum^{\infty}_{n=0} \frac{x^{-(n+1)}}{\Gamma\Big{(}\frac{1}{2} - n\Big{)}}\frac{1}{x^n}$

$= \Gamma\Big{(}\frac{1}{2}\Big{)} - e^{-x^2} \sum^{\infty}_{n=0} \frac{\Gamma\Big{(}\frac{1}{2}\Big{)}}{\Gamma\Big{ (}\frac{1}{2} - n\Big{)}}\frac{x^{-(n+1)}}{x^n}$

But I'm stuck now. I need to do something to the $e^{-x^2} \textrm{ and the } x^{-(n+1)}$ but I'm not sure what... I don't even know if what I've done so far is right to be honest... Any help please!
• Feb 1st 2010, 09:46 AM
Drexel28
Quote:

I already posted this in the Applied section but I don't think thats actually the right place so here it is again!

I think I'm pretty close to getting this but can't quite figure out what do next!

From earlier in the question...

$\textrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt \sim 1 - \frac{e^{-x^2}}{\pi} \sum^{\infty}_{n=0} \Gamma \Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$ (*)

Show that erf(x) can be expressed in terms of the incomplete gamma function, and that the above asymptotic expansion agrees with the result obtained in Problem 6. [The formula $\Gamma\Big{(}\frac{1}{2} + n)\Gamma\Big{(}\frac{1}{2} - n\Big{)} = (-1)^n \pi$ (**) for integer n may be useful.]

So I think I have to show the relation... $\gamma\Big{(}\frac{1}{2}, x\Big{)} = \sqrt{\pi} \textrm{erf}(\sqrt{\pi})$ (***) as that's the relation I've seen elsewhere.

The result obtained in question 6 was...

$\gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{\Gamma(\alpha)}{\Gamma(\a lpha - n)}\frac{1}{x^n}$

And also...
$\gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{(\alpha-1)(\alpha-2)\ldots(\alpha-n)}{x^n}$

Right so intro done!

Now what I've done...

From (*)
$\sqrt{\pi} \textrm{erf}(x) = 2 \int^x_0 e^{-t^2} dt \sim \sqrt{\pi} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma\Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$

$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \frac{e^{-x^2}}{\sqrt{\pi}} \sum^{\infty}_{n=0} \Gamma\Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$
(This step seems to be showing that $\alpha = \frac{1}{2}$ which corresponds to (***))

Using (**)
$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2}$ $\sum^{\infty}_{n=0} \frac{1}{\Gamma\Big{(}\frac{1}{2} - n\Big{)}}\frac{1}{x^{2n+1}}$

$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2} \sum^{\infty}_{n=0} \frac{x^{-(n+1)}}{\Gamma\Big{(}\frac{1}{2} - n\Big{)}}\frac{1}{x^n}$

$= \Gamma\Big{(}\frac{1}{2}\Big{)} - \sqrt{\pi} e^{-x^2} \sum^{\infty}_{n=0} \frac{\Gamma\Big{(}\frac{1}{2}\Big{)}}{\Gamma\Big{ (}\frac{1}{2} - n\Big{)}}\frac{x^{-(n+1)}}{x^n}$

But I'm stuck now. I need to do something to the $e^{-x^2} \textrm{ and the } x^{-(n+1)}$ but I'm not sure what... I don't even know if what I've done so far is right to be honest... Any help please!

$\Gamma\left(\frac{1}{2},x\right)=\int_0^x z^{\frac{-1}{2}}e^{-z}\text{ }dz$, let $\sqrt{z}=\xi\implies z=\xi^2\implies dz=2\xi\text{ }d\xi$. So, we see that $\Gamma\left(\frac{1}{2},x\right)=\int_0^{\sqrt{x}} \frac{e^{-\xi^2}}{\xi}\cdot 2\xi\text{ }d\xi=2\int_0^{\sqrt{x}}e^{-\xi^2}\text{ }d\xi=\sqrt{\pi}\text{erf}\left(\sqrt{x}\right)$.
• Feb 1st 2010, 09:53 AM
Thanks for that!

Was my way actually going anywhere or not really..?

Does it not require the knowledge to use $\alpha = \frac{1}{2}$, and doesn't use the $\Gamma(\frac{1}{2} + n) \Gamma(\frac{1}{2} - n) = (-1)^n \pi$ hint which makes me think it is supposed to be solved some other way...

Or is what I've done more to do with the 'and that the above asymptotic expansion agrees with the result obtained in Problem 6.' part?
• Feb 1st 2010, 09:56 AM
Drexel28
Quote:

Thanks for that!

Was my way actually going anywhere or not really..?

Does it not require the knowledge to use $\alpha = \frac{1}{2}$, and doesn't use the $\Gamma(\frac{1}{2} + n) \Gamma(\frac{1}{2} - n) = (-1)^n \pi$ hint which makes me think it is supposed to be solved some other way...

Or is what I've done more to do with the 'and that the above asymptotic expansion agrees with the result obtained in Problem 6.' part?

There is a lack of context here. I am not using your book and so I do not know what it requires or what it needs. Honestly, I just saw that you said to prove that $\Gamma\left(\frac{1}{2},x\right)=\sqrt{\pi}\text{e rf}\left(\sqrt{x}\right)$.

I'll come back later and read it more thoroughly.
• Feb 1st 2010, 10:07 AM
Ah I see!

That's just the relationship I found elsewhere so I had an idea of how to approach this, it's not actually mentioned in the question.

Basically, this is question 7. Question 6 was to find a series expansion of the incomplete gamma function which was shown to be...

Quote:

$\gamma(\alpha,x) \sim \Gamma(\alpha) - e^{-x}x^{\alpha - 1} \sum^{\infty}_{n=0}\frac{\Gamma(\alpha)}{\Gamma(\a lpha - n)}\frac{1}{x^n}$

Then for question 7 you had to show that the error function could be written as...

Quote:

$\textrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int^x_0 e^{-t^2} dt \sim 1 - \frac{e^{-x^2}}{\pi} \sum^{\infty}_{n=0} \Gamma \Big{(}n + \frac{1}{2}\Big{)}\frac{(-1)^n}{x^{2n+1}}$ (*)
Show that erf(x) can be expressed in terms of the incomplete gamma function, and that the above asymptotic expansion agrees with the result obtained in Problem 6. [The formula $\Gamma\Big{(}\frac{1}{2} + n)\Gamma\Big{(}\frac{1}{2} - n\Big{)} = (-1)^n \pi$ (**) for integer n may be useful.]