# Toronto Math Forum

## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Zhijian Zhu on November 13, 2018, 10:49:00 PM

Title: 2.5 Q31
Post by: Zhijian Zhu on November 13, 2018, 10:49:00 PM
hello, could anyone share the process of doing section2.5 #31?
show that $J_{-n}(-u)=(-1)^nJ_n(u)$
thank you guys！

Title: Re: 2.5 Q31
Post by: Yiyi Cheng on November 13, 2018, 10:59:00 PM
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 $J_{-n}= (-1)^n J_n(u)$
Title: Re: 2.5 Q31
Post by: Fan Yang on November 13, 2018, 11:00:15 PM
Hi!
The question you typed is wrong. It should be 𝐽−𝑛(𝑢)=(−1)𝑛𝐽𝑛(𝑢)