Homework: Odd Factoria (1x3x5x) Recurrence Relation Solution

[unscientific]

Homework Statement

The question is stated in the picture attached.

The Attempt at a Solution

I used a slightly different method, by making use of recurrence relations through integration by parts:

J_n = (2/2n+1) J_n+1


J_n = [1x3x5x...]/[2ⁿ] √∏

= [2n-1]! / [2^n-1 * (n-1)! ] = (2n)! / (2ⁿ * n!)


But the answer had 4ⁿ instead at the bottom... Not sure where I went wrong.

 

[uart]

Homework Statement

The question is stated in the picture attached.

The Attempt at a Solution

I used a slightly different method, by making use of recurrence relations through integration by parts:

J_n = (2/2n+1) J_n+1J_n = [1x3x5x...]/[2ⁿ] √∏

= [2n-1]! / [2^n-1 * (n-1)! ] = (2n)! / (2ⁿ * n!)But the answer had 4ⁿ instead at the bottom... Not sure where I went wrong.

Your working is mostly correct unscientific, you just left behind a factor of $2^n$ on the denominator.

Where you had J_n = [1x3x5x...(2n-1)]/[2ⁿ] √∏, this is correct.

Next you can replace the "odd factorial" with,

$$(2n-1)(2n-3) \ldots 1 = \frac{(2n-1)!}{2^{n-1} (n-1)!} = \frac{(2n)!}{2^{n} (n)!}$$

 

[Ray Vickson]

$$1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2m+1) = \frac{(2m+1)!}{2 \cdot 4 \cdot \cdots \cdot 2m} = \frac{(2m+1)!}{2^m m!}.$$
Thus $$\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2m+1) }{2^m} = \frac{(2m+1)!}{4^m m!}.$$

RGV

 

[unscientific]

Your working is mostly correct unscientific, you just left behind a factor of $2^n$ on the denominator.

Where you had J_n = [1x3x5x...(2n-1)]/[2ⁿ] √∏, this is correct.

Next you can replace the "odd factorial" with,

$$(2n-1)(2n-3) \ldots 1 = \frac{(2n-1)!}{2^{n-1} (n-1)!} = \frac{(2n)!}{2^{n} (n)!}$$

Oh damn! I forgot about the 2ⁿ below initially! Can't believe I missed it despite checking numerous times. Sorry, I will check my working more thoroughly next time!

 

[unscientific]

Also, are there any links on this site that teaches one how to use latex? I think it'll make my workings clearer.

 

[cepheid]

Also, are there any links on this site that teaches one how to use latex? I think it'll make my workings clearer.

Dude, read this post, it's awesomely comprehensive:

https://www.physicsforums.com/showthread.php?p=3977517&posted=1#post3977517

 

[Millennial]

Splitting the integral up and setting $x=\sqrt{t}$ yields the integral
$$2\int^{\infty}_{0}\frac{t^n e^{-t}\,dt}{2\sqrt{t}}$$
which gives
$$\int^{\infty}_{0}t^{n-1/2} e^{-t}\,dt = \Gamma(n+1/2)$$
The Lagrange duplication formula can be applied here:
$$\Gamma(n)\Gamma(n+1/2)=2^{1-2n}\sqrt{\pi}\Gamma(2n)$$
For integer n, this simplifies to
$$\Gamma(n+1/2)=\frac{2(2n-1)!\sqrt{\pi}}{4^{n}(n-1)!}$$
which is equivalent to
$$\Gamma(n+1/2)=\frac{(2n)!\sqrt{\pi}}{4^{n}n!}$$
This gives the result in the paper. Usually, I try to work integrals that have a power of a variable and the exponential function in it using the Gamma function. It worked in this case as well.

 

[tooscientific]

I seem to have problems with logging into my main account: unscientific

It always says that I've tried to log in too many times..

 

[tooscientific]

I know I shouldn't be posting here but I have problems with my both accounts (unscientific and this) - It keeps saying that I've tried to log in too many times and can't try for 15mins..

This question is about the green function..

I'm wondering what could be wrong with my working..

I'm getting (d²x/dt²) + x = 0

instead of f(t)...

 

[tooscientific]

I've attached the solutions..I got nearly the same answer just with an additional term of

f(∏)*sin(t)


Unless f(∏) = 0, which they never stated anywhere...

 

[Ray Vickson]

I seem to have problems with logging into my main account: unscientific

It always says that I've tried to log in too many times..

I had that response for a while this morning, too: it said I had exceeded the number of incorrect log-in attempts (not true: 0 is not greater than 4 or whatever), and to wait 15 minutes and try again. It still did not work after waiting 45 minutes, nor after waiting 2 hours. However, it is now working again, obviously.

RGV