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The bessel function of the first kind with order $p$ is

$$J_p(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+p+1)}\Big(\frac{x}{2}\Big)^{2n+p}$$

What I'm thinking about is finding a way to get $\sqrt{\frac{2}{\pi x}}$, factor it out, and then reduce the remaining term as the Taylor series representation of the sine function. But I haven't been able to achieve this.

With $p=1/2$ we have that $$\Gamma(n+p+1)=\frac{(2n+1)!!}{(2)^n}\sqrt{\pi}$$ and my overall remaining expression has been so far: $$\frac{(-1)^n (2)^n}{n!(n+1/2)(2n+1)!!\sqrt{\pi}} \left( \frac{x^{2n+1}}{2^{2n+1/2}x^{1/2}}\right)$$

Which looks like a big mess and is not going to reduce to the Taylor series representation of the sine function.

Any ideas? Anyone have a source were they carry out this demonstration?

Thanks a lot.

Dylan
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DLV
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  • are you tied to this particular representation? – tired Apr 25 '15 at 16:32
  • Yeah, I have to show this, and I'm stuck. Apparently it's a popular representation since some sites represent $J_{1/2}$ this way. – DLV Apr 25 '15 at 16:36

2 Answers2

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From the representation you have $$J_{1/2}(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n! \Gamma(n+3/2)} \left(\dfrac{x}2\right)^{2n+1/2} \implies \sqrt{\dfrac{x}2}J_{1/2}(x) = \sum_{n=0}^{\infty} \dfrac{(-1)^n}{n! \Gamma(n+3/2)} \left(\dfrac{x}2\right)^{2n+1} \, (\spadesuit)$$ We have $$\Gamma(n+3/2) = \left(n+\dfrac12\right)\left(n-\dfrac12\right) \cdots \dfrac12 \sqrt{\pi} = \dfrac{(2n+1)(2n-1)\cdots 3 \cdot 1}{2^{n+1}} \sqrt{\pi} = \dfrac{(2n+1)!}{2^{2n+1}n!} \sqrt{\pi}$$ Hence, $$2^{2n+1} n! \Gamma(n+3/2) = (2n+1)! \sqrt{\pi}$$ Plugging this in $(\spadesuit)$, we obtain $$\sqrt{\dfrac{x}2}J_{1/2}(x) = \dfrac1{\sqrt{\pi}}\sum_{n=0}^{\infty} \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} = \dfrac1{\sqrt{\pi}}\sin(x)$$

Also, avoid the double factorial notation. It is a very very poor notation.

Adhvaitha
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  • Thanks. I'll take that double factorial comment into account. – DLV Apr 25 '15 at 16:47
  • I thought I got it but I got stuck on $$\dfrac{(2n+1)(2n-1)\cdots 3 \cdot 1}{2^{n+1}} \sqrt{\pi} = \dfrac{(2n+1)!}{2^{2n+1}n!} \sqrt{\pi}$$ How did you get that? Any suggestions? Thanks. – DLV Apr 25 '15 at 16:55
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Your main tool can be the duplication formula for the Gamma function, $$ \Gamma(z)\Gamma(z+1/2) = 2^{1-2z}\sqrt{\pi}\Gamma(2z), $$ which you can use on the denominator of your expression. It should then fall almost straight out.

Chappers
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