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How can i find $J_{3\over 2}(x)$ and $J_{5\over 2}(x)$ by use this formula :

$$J_{p+{\frac{1}{2}}}(x)=\left(\frac{2}{\pi}\right)^{\frac{1}{2}} \cdot(-1)^p x^{p+{\frac{1}{2}}} \left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^p\left(\frac{\sin x}{x}\right)$$

I know if i want to find $J_{3\over 2}(x)$ substitute by $p=1$ and if I want to find $J_{5\over 2}(x)$ substitute by $p=2$

But i don't know how can i find $\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^2$

alexjo
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Dima
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1 Answers1

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For $p = 2$, the term $\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2$ should be understood as the operator $$\displaystyle \left(\frac{1}{x} \frac{d}{dx}\right)^2=\frac{1}{x} \frac{d}{dx}\frac1x \frac d{dx}.$$ So for $p=2$ $$ \begin{align} J_{{\frac{5}{2}}}(x)&=\left(\frac{2}{\pi}\right)^{\frac{1}{2}} \cdot(-1)^2 x^{{\frac{5}{2}}} \left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\right)^2\left(\frac{\sin x}{x}\right)\\ &=\sqrt\frac{2}{\pi} x^{{\frac{5}{2}}} \left[\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\left(\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\frac{\sin x}{x}\right)\right]\\ &=\sqrt\frac{2}{\pi} x^{{\frac{5}{2}}} \left[\frac{1}{x}\frac{\mathrm d}{\mathrm dx}\left(\frac{1}{x}\frac{x \cos x-\sin x}{x^2}\right)\right]\\ &=\sqrt\frac{2}{\pi} x^{{\frac{5}{2}}} \left[\frac{1}{x}\left(-\frac{3 x \cos x+(x^2-3) \sin x}{x^4}\right)\right]\\ &=-\sqrt\frac{2}{\pi} x^{\frac{5}{2}} \frac{3 x \cos x+(x^2-3) \sin x}{x^5}\\ &=-\sqrt\frac{2}{\pi} \frac{3 x \cos x+(x^2-3) \sin x}{x^{5/2}} \end{align} $$

Check the result with WolframAlpha (alternate forms).

alexjo
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  • ah so the square is really in terms of multiplication denoting composition of functions, i.e. $(\frac{1}{x} \frac{d}{dx})^2 = (\frac{1}{x} \frac{d}{dx}) \circ (\frac{1}{x} \frac{d}{dx})$? That makes a lot of sense actually – Chill2Macht Apr 09 '16 at 23:07
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    Yes it is a composition of operators (the derivative operator). – alexjo Apr 09 '16 at 23:10