- #1

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## Main Question or Discussion Point

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.

[tex]

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

[/tex]

to prove that function?

need advice thanks..

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.

[tex]

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

[/tex]

to prove that function?

need advice thanks..