How would I show that
$$J_1(x)+J_3(x)=\frac 4x J_2(x)$$
Using the series definition of the Bessel Function, which is
$$J_p(x)=\sum ^\infty _{n=0} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac x2\right)^{2n+p}$$

I know that it can be achieved using some reindexing, but I am not sure how. I would greatly appreciate any help. Thank you.


2 Answers 2


Yes, there is an index shift. Rewrite

$$J_1(x) = \frac{x}{2} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+1)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$

$$J_3(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n-1)!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1}$$


$$\begin{align}J_1(x)+J_3(x) &= \frac{x}{2} + \sum_{n=1}^{\infty} (-1)^n \left [ \frac{1}{n! (n+1)!} -\frac{1}{(n-1)! (n+2)!} \right ]\left ( \frac{x}{2} \right ) ^{2 n+1}\\ &= \frac{x}{2} + 2 \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+1} \\ &= \frac{4}{x} \left[\frac{1}{2} \left ( \frac{x}{2} \right )^2 + \sum_{n=1}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+2} \right ]\\ &= \frac{4}{x}\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(n+2)!} \left ( \frac{x}{2} \right ) ^{2 n+2} \\\therefore J_1(x)+J_3(x) &= \frac{4}{x} J_2(x) \end{align}$$


Obtain $J'_2(x)$ in two ways using following relations.(These identities can be (easily) established by direct substitution of the provided series for $J_n(x)$ and I have provided these proofs and the end) :

$$ \frac{d}{dx}(x^2J_2(x))=x^2J_1(x)$$ $$ \frac{d}{dx}(x^{-2}J_2(x))=-x^{-2}J_3(x)$$

now expanding the parentheses (differentiation of multiplication) we arrive at :

$$ J'_2(x)=\frac{-2}{x}J_2(x)+J_1(x)$$ $$J'_2(x)=\frac{2}{x}J_2(x)-J_3(x) $$

subtracting the two relations from each other, we arrive at the requested identity.

(Proof of the first identity: $$\frac{d}{dx}(x^2J_2(x))=\sum ^\infty _{n=0} \frac{(-1)^n(n+2)}{\Gamma(n+1)\Gamma(n+3)} (\frac x2)^{2n+3}=$$$$\sum ^\infty _{n=0} \frac{(-1)^n}{\Gamma(n+1)\Gamma(n+2)} (\frac x2)^{2n+3}=x^2J_1(x)$$proof of the second:$$\frac{d}{dx}(x^{-2}J_2(x))=\sum ^\infty _{n=0} \frac{(-1)^nn}{\Gamma(n+1)\Gamma(n+3)}\left(\frac x2\right)^{2n-1}=$$$$\sum ^\infty _{n=1} \frac{(-1)^nn}{\Gamma(n+1)\Gamma(n+3)}\left(\frac x2\right)^{2n-1}=$$$$\sum ^\infty _{n=0} \frac{(-1)^{n+1}}{\Gamma(n+1)\Gamma(n+4)}\left(\frac x2\right)^{2n+1}=-x^{-2}J_3(x)$$)


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