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I was asked to prove, with the power serie of $J_{\pm \nu}(x)$ that $J_{\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}\sin(x)$ and $J_{-\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}\cos(x)$.

We know that $J_{\frac{1}{2}}=\Sigma_{r=0}^{\infty}\frac{(-1)^r}{r!\Gamma(r+\frac{3}{2})}(\frac{x}{2})^{2r+\frac{1}{2}}$, multypling and diving by $\frac{2}{\sqrt{x}}$ one can obtain $\frac{\sqrt{2}}{\sqrt{x}}\Sigma_{r=0}^{\infty}\frac{(-1)^r}{r!\Gamma(r+\frac{3}{2})}(\frac{x}{2})^{2r+1}$. However, i don't know how to deal with $\Gamma$ (gamma) function, in order to "remove" it, and get a factorial (in order to get then Taylor's polynomial of sine).

With $J_{\frac{1}{2}}=\sqrt{\frac{2}{\pi x}}\sin(x)$ i'll figure out how the other solution will be, there's no need of two explanations.

MatMorPau22
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1 Answers1

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You need the duplication formula for the $\Gamma$-function, $$ \Gamma(z)\Gamma(z+1/2) = 2^{1-2z}\sqrt{\pi} \Gamma(2z), $$ so $$ r!\Gamma(r+3/2) = \Gamma(r+1)\Gamma((r+1)+1/2) = 2^{-1-2r} \sqrt{\pi} \Gamma(2r+2) = 2^{-1-2r} \sqrt{\pi}(2r+1)!, $$ and the result follows. The other one can be done similarly.

Chappers
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  • So, we have $r!\Gamma(r+3/2)=2^{-(2r+1)}\sqrt{\pi}(2r+1)!$. By putting it in our expression we get $\frac{1}{\sqrt{x}}\Sigma_{r=0}^{\infty}\frac{(-1)^r 2^{2r+1}}{\sqrt{\pi}(2r+1)!}(\frac{x}{2})^{2r+1}=\frac{1}{\sqrt{x\pi}}\Sigma_{r=0}^{\infty}\frac{(-1)^r }{(2r+1)!}(x)^{2r+1}=\frac{\sin x}{\sqrt{x\pi}}$. As you can see, there's a $\sqrt{2}$ missing. Can anyone help me? – MatMorPau22 Jun 25 '17 at 15:40
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    You lost it when you divided by $\sqrt{x}$. – Chappers Jun 25 '17 at 15:42