Word Count : 470

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Sure! Let’s solve this step by step, manually, using the series definition of Bessel functions: The series definition of Bessel function of order $m$ is: $$ J_m(x) = \sum_{k=0}^{\infty} (-1)^k \frac{1}{k! \, \Gamma(m+k+1)} \left( \frac{x}{2} \right)^{2k+m}. $$ We want to show that: $$ J_2(x) = \frac{2}{x} J_1(x) - J_0(x). $$ --- ### Step 1: Write series for $J_0(x)$, $J_1(x)$, $J_2(x)$ 1. Order 0: $$ J_0(x) = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(k!) \, \Gamma(0+k+1)} \left( \frac{x}{2} \right)^{2k}. $$ Since $\Gamma(k+1) = k!$, we get: $$ J_0(x) = \sum_{k=0}^{\infty} (-1)^k \frac{1}{(k!)^2} \left( \frac{x}{2} \right)^{2k}. $$ --- 2. Order 1: $$ J_1(x) = \sum_{k=0}^{\infty} (-1)^k \frac{1}{k! \, \Gamma(1+k+1)} \left( \frac{x}{2} \right)^{2k+1}. $$ $\Gamma(k+2) = (k+1)!$, so $$ J_1(x) = \sum_{k=0}^{\infty} (-1)^k \frac{1}{k! _______ _____ ______ __________ _____ ______.
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