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Mathematics
Calculus
An identity with Bessel functions
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[QUOTE="julian, post: 6842479, member: 142346"] It's not exactly "drivng me batty", but let's switch from ##e## to ##x## anyway. Say we have \begin{align*} A J_2 (x) + B Y_2 (x)= \frac{1}{x} \sum_{n=1}^\infty \dfrac{J_n (n x)}{n} \dfrac{J_{n+1} ((n+1) x)}{n+1} \end{align*} We have that ##J_2 (0) = 0## and the RHS is zero at ##x=0## (because of the formula for ##J_n (x)## that I will quote in a moment). We must have ##B=0## because ##Y_2 (x)## is singular at ##x=0##. So we have \begin{align*} A J_2 (x) = \frac{1}{x} \sum_{n=1}^\infty \dfrac{J_n (n x)}{n} \dfrac{J_{n+1} ((n+1) x)}{n+1} \end{align*} We wish to find ##A##. Let's take the 2nd derivative and set ##x=0##. We have the formula: \begin{align*} J_n (x) = \sum_{k=0}^\infty \dfrac{(-1)^k (x/2)^{n+2k}}{k! (n+k)!} \end{align*} In particular \begin{align*} J_2 (x) = \sum_{k=0}^\infty \dfrac{(-1)^k (x/2)^{2+2k}}{k! (2+k)!} \end{align*} Obviously \begin{align*} \left. \frac{d^2}{d x^2} \frac{1}{x} \dfrac{J_n (n x)}{n} \dfrac{J_{n+1} ((n+1) x)}{n+1} \right|_{x=0} = 0 \qquad \text{for } n > 1 \end{align*} So we need only consider \begin{align*} A \left. \frac{d^2}{d x^2} J_2 (x) \right|_{x=0} = \left. \frac{d^2}{d x^2} \frac{1}{x} J_1 ( x) \dfrac{J_2 (2 x)}{2} \right|_{x=0} \end{align*} First \begin{align*} \left. \frac{d^2}{d x^2} J_2 (x) \right|_{x=0} = \left. \frac{d^2}{d x^2} \dfrac{ (x/2)^2}{0! (2+0)!} \right|_{x=0} = \frac{1}{4} . \end{align*} Next \begin{align*} & \left. \frac{d^2}{d x^2} \frac{1}{x} J_1 ( x) \dfrac{J_2 (2 x)}{2} \right|_{x=0} = \nonumber \\ & = \left. \frac{d^2}{d x^2} \left( \frac{1}{x} \sum_{k=0}^\infty \dfrac{(-1)^k (x/2)^{1+2k}}{k! (1+k)!} \right) \left( \frac{1}{2} \sum_{k=0}^\infty \dfrac{(-1)^k (2x/2)^{2+2k}}{k! (2+k)!} \right) \right|_{x=0} \nonumber \\ & = \left. \frac{d^2}{d x^2} \left( \frac{1}{x} \dfrac{(-1)^0 (x/2)}{0! (1+0)!} \right) \left( \frac{1}{2} \dfrac{x^2}{0! (2+0)!} \right) \right|_{x=0} \nonumber \\ & = \left. \frac{d^2}{d x^2} \left( \frac{1}{x} (x/2) \right) \left( \frac{1}{2} \dfrac{x^2}{2!} \right) \right|_{x=0} \nonumber \\ & = \frac{1}{4} \end{align*} implying ##A = 1##. So it is just left to show that: \begin{align*} x^2 \frac{d^2 (RHS)}{d x^2} + x \frac{d (RHS)}{d x} + (x^2 - 4) (RHS) = 0 . \end{align*} [/QUOTE]
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An identity with Bessel functions
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