2.5 Q31

[Zhijian Zhu]

hello, could anyone share the process of doing section2.5 #31?
show that $J_{-n}(-u)=(-1)^nJ_n(u)$
thank you guys！

[Yiyi Cheng]

This would be my answer to this question, feel free to ask me any question
$\sum_{-\infty}^{\infty}{J_n(u)z^n}= G(z;u)$
$G(z;u)= e^{(\frac{u}{2})(z-\frac{1}{z})}= G(-\frac{1}{z};u)$
$=\sum_{-\infty}^{\infty} J_n(u)\frac{(-1)^n}{z^n}$
$=\sum_{-\infty}^{\infty} J_{-n}(u)(-1)^n z^n$
Therefore, we get $J_{-n}= (-1)^n J_n(u)$

[Fan Yang]

Hi!
The question you typed is wrong. It should be 𝐽−𝑛(𝑢)=(−1)𝑛𝐽𝑛(𝑢)