Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Define Bessel function as: $$J_a(x)=\sum_{n=0}^{\infty}{{(-1)^n}\over{\Gamma(a+n+1)n!}}\left({{x}\over{2}}\right)^{a+2n}.$$ Where $a$ not an integer.

Need to show $$J_{a+1}(x)J_{-a}(x)+J_a(x)J_{-(a+1)}(x)=-{{2\sin(a\pi)}\over{\pi x}}.$$

Note that the Wronskian of $$W(J_a(x),J_{-a}(x))=-{{2\sin(a\pi)}\over{\pi x}}.$$

share|cite|improve this question
So are you allowed to use that result on Wronskian or you need to prove it also? – Norbert Jul 14 '13 at 9:39

We have \begin{align} W(J_a(x),J_{-a}(x))&=J_a(x)J'_{-a}(x)-J'_a(x)J_{-a}(x)=\\ &=J_a(x)\left(J_{-a-1}(x)+\frac{a}{x}J_{-a}(x)\right)-\left(-J_{a+1}(x)+\frac{a}{x}J_a(x)\right)J_{-a}(x)=\\ &=J_a(x)J_{-a-1}(x)+J_{a+1}(x)J_{-a}(x), \end{align} where we have used the differentiation formulas of Bessel functions (see e.g. 10.6.2 here) to pass from the 1st to the second line. Now using the formula for the Wronskian from your question, we easily obtain the seeked identity.

P.S. The differentiation formulas can be proven either directly using series representations or by verifying that e.g. $\displaystyle J'_a(x)-\frac{a}{x}J_a(x)$ satisfies the same differential equation as $-J_{a+1}(x)$ and comparing their behaviour as $x\rightarrow 0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.