Understanding Gamma Function and Series Integration

[H2instinct]

Homework Statement

Homework Equations

Gamma(n+1) = n(gamma(n))

The Attempt at a Solution

I have searched around trying to figure out how to do this. So far I have come out with:

Gamma(n) = (n-1)(n-2)...
Gamma(n+1)=n(n-1)(n-2)...2(1) otherwise known as Gamma(n+1)=n!

Problem is... I don't fully understand what I did because I followed a like example. Can someone please help me out with this. Also I have not even attempted the second part, so ignore that for now.

 

[Gregg]

First part.

$$\Gamma(n+1) = n\Gamma(n)$$

$$\Gamma(\frac{5}{2})=\Gamma(1+\frac{3}{2})=\frac{3}{2}\Gamma(1+\frac{1}{2})=\frac{3}{2}\frac{1}{2}\Gamma({\frac{1}{2})=\frac{3\sqrt{\pi}}{4}$$



This part is for the proof.


Assume true for k

$$\Gamma(k+\frac{1}{2})=\frac{1.3.5...(2k-1)\sqrt{\pi}}{2^k}$$

Then prove for k+1

$$\Gamma(k+1+\frac{1}{2})=(k+\frac{1}{2})\left(\frac{1.3.5...(2k-1)\sqrt{\pi}}{2^k}\right)$$


Note the recursion relation, then,


$$\Gamma(k+1+\frac{1}{2})=\frac{1}{2}(2(k+1)-1)\frac{1.3.5...(2k-1)\sqrt{\pi}}{2^k}=\frac{1.3.5...(2k-1)(2(k+1)-1)\sqrt{\pi}}{2^{k+1}}$$


Complete the proof by proving for k=1 and hence all positive integers.

 

[H2instinct]

A little confused because I think you combined both parts of that. Both of those have nothing to do with each other, but otherwise you helped me with the first one.

 

[Gregg]

I just worked out the value of 5/2 for the first one, and for the second I used the relationship we worked out at the beginning and proof by induction to answer the last part of it. I've separated them now too.

 

[H2instinct]

Thank you for the help, but I still don't see the proof from (1.3.5...(2n-1))/2^n == (2n)!/((4^n)(n!))

It's fine. I'm going to move on and get help from a classmate when the time comes. Thanks for the help though!

 

[Gregg]

I'll clarify:

The question asks you to prove that

$$\Gamma(n+\frac{1}{2}) = \frac{1.3.5.7...(2n-1)}{2^n}\sqrt{\pi}$$

We can sort of see this in the 5/2 example, all the numerators are odd numbers and the denominator is always 2, so the odd numbers will multiply successively n-times on the top and 2 will be of the power n on the bottom, and always at the end is the square root of pi. To prove this first "assume" that it is true for any number say, k.

$$\Gamma(k+\frac{1}{2}) = \frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi}$$

We assume this is true, and now we try to prove it for k+1, we want the next odd number to appear on the numerator and for another 2 to be multiplied in the denominator (or half on the numerator) to make the right hand side look like this:


$$\frac{1.3.5.7...(2k-1)(2(k+1)-1)}{2^{k+1}}\sqrt{\pi}$$


Once this happens all we have to do is show that the equation holds for n=1 and then we get it true for n=1,n=1+1 etc.

From the relationship we established,

$$\Gamma(n+1)=n\Gamma(n)$$

We can add 1 quite nicely:

Remember that we have assumed that

$$\Gamma[k+\frac{1}{2}] = \frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi}$$

So for k+1:

$$\Gamma(k+1+\frac{1}{2}) = (k+\frac{1}{2})\Gamma(k+\frac{1}{2})$$

(k+1/2 is n here.)

$$\Gamma(k+1+\frac{1}{2}) = (k+\frac{1}{2})\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi}$$

Now all we need to do is rearrange it

$$(k+\frac{1}{2})\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi} = \frac{1}{2}(2k+1)\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi}$$


Notice that we have a half and a 2k+1, which if you look is actually (2(k+1)-1), our next odd number.

$$\frac{1}{2}(2k+1)\frac{1.3.5.7...(2k-1)}{2^k}\sqrt{\pi} = \frac{1.3.5.7...(2k-1)(2(k+1)-1)}{2^{k+1}}\sqrt{\pi}$$

So if it's true for k it is true for k+1, if you prove it for k=1 you have proved it then for all positive n.

$$\Gamma(1+\frac{1}{2}) = \frac{\sqrt{\pi}}{2} = \frac{1.\sqrt{\pi}}{2^1}$$

$$\Rightarrow \Gamma(n+\frac{1}{2}) = \frac{1.3.5.7...(2n-1)}{2^n}\sqrt{\pi}$$

For positive n.


Edit: The final part of the proof would just be manipulating the expression to match the RHS.

 

[H2instinct]

Thanks, I understood it much better from that.