- #1
matematikuvol
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Homework Statement
Show
[tex]e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n[/tex]
Homework Equations
[tex]J_k(x)=\sum^{\infty}_{n=0}\frac{(-1)^n}{(n+k)!n!}(\frac{x}{2})^{2n+k}[/tex]
The Attempt at a Solution
Power series product
[tex](\sum^{\infty}_{n=0}a_n)\cdot (\sum^{\infty}_{n=0} b_n)=\sum^{\infty}_{n=0}c_n[/tex]
where
[tex]\sum^n_{i=0}a_ib_{n-i}[/tex]
[tex](\sum^{\infty}_{n=0}\frac{1}{n!}(\frac{x}{2})^nt^n)\cdot (\sum^{\infty}_{n=0} \frac{(-1)^n}{n!}(\frac{x}{2})^nt^{2i-n})=\sum^{\infty}_{n=0}c_n[/tex]
where
[tex]c_n=\sum^{n}_{i=0}\frac{(-1)^{n-i}}{i!(n-i)!}(\frac{x}{2})^nt^{2i-n}[/tex]
So
[tex]e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=0}\sum^{n}_{i=0} \frac {(-1)^{n-i}}{i!(n-i)!}(\frac{x}{2})^nt^{2i-n}[/tex]
I don't see how to get from here
[tex]e^{\frac{x}{2}(t-\frac{1}{t})}=\sum^{\infty}_{n=-\infty}J_n(x)t^n[/tex]