Toronto Math Forum
MAT3342018F => MAT334Lectures & Home Assignments => Topic started 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！

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 $J_{n}= (1)^n J_n(u)$

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