How to prove bessel function J1/2(x) = sqrt(2/πx)sinx;

[leyyee]

I really have no idea.

I started with the frobenius method. Until the recurrence formula.

I got that already. But I just don't know where to plug in the 1/2 into the equation. Can anyone help? I just need to know where to put in the 1/2?

Or can i use the normal bessel function which in general.

$$J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n} = x^m \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}(x^2)^n$$

to prove that function?

need advice thanks..

 

[deiki]

using series from the general case actually works.

to plug the 1/2 into this equation, you will need to study a special case of the gamma function for the factorial. Once you do this, the whole 2^(m+2n)∙n!∙(n+m)! should simplify nicely for m=1/2.

http://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function

to prove that J1/2(x) = sqrt(2/πx)sinx, you would better factorize x^(m-1) out of the sum and remain with x^(2n+1), this should simplify more easily.

 

[leyyee]

the whole thing will become this . but how does it simplify to sinx?


$$J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{1/2+2n}n!(n+1/2)!}x^{1/2+2n} = x^1/2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{1/2+2n}n!(n+1/2)!}(x^2)^n$$

i mean how to simplyfy the (n+1/2)!. thanks

 

[deiki]

just use (n+1/2)! = Γ(n+1/2 + 1 )

 

[leyyee]

with that Γ(n+1/2 + 1 )

am I able to simplify it to sinx ?

 

[leyyee]

thanks for the advice..

i have got the solutions.. thanks