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$\displaystyle e^x=I_0(x)+2\sum_{n=1}^\infty I_n(x)$

Using the following expansion

$\displaystyle e^{i\rho\cos \varphi}=\sum_{m=-\infty}^{\infty}J_m(\rho)[ie^{i\varphi}]^m$ with $\varphi=0 , \rho = -ix$

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In homework-type problems, you should first give your own attempts and thoughts. – GEdgar Jun 12 '13 at 0:35

Split $\displaystyle e^{i\rho\cos \varphi}$ into real and imaginary parts, and using the relations here, the fact that $J_{2n}(-ix) = J_{2n}(ix) = i^{2n} I_{2n}(x)$, $J_{2n-1}(-ix) = -J_{2n-1}(ix) = -i^{2n-1} I_{2n-1}(x)$ and substituting $\rho = -ix$ and $\varphi = 0$ we get:

$\cos(\rho\cos(\varphi)) = \cosh(x) = I_0(x) + 2\sum_1^{\infty}(-1)^ni^{2n}I_{2n}(x)$.


$i\sin(\rho\cos(\varphi)) = \sinh(x) = 2\sum_1^{\infty}(-1)^{n}i^{2n}I_{2n-1}(x)$.

Summing the two and the result $\displaystyle e^x=I_0(x)+2\sum_{n=1}^\infty I_n(x)$ follows.

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