Question

1. (i). Write the value of J3/2(x).​

Answer 1

SOLUTION

TO DETERMINE

The value of $\displaystyle \sf{J_{ \frac{3}{2} }(x)}$

CONCEPT TO BE IMPLEMENTED

$\displaystyle \sf{J_{ n }(x)}$ is Bessel Function of order n

Now $\displaystyle \sf{J_{n }(x)}$ is defined as

$\displaystyle \sf{J_{ n + 1 }(x) = \frac{2n}{x} J_{ n }(x) - J_{ n - 1 }(x)}$

EVALUATION

Here $\displaystyle \sf{J_{ n }(x)}$ is Bessel Function of order n

Now $\displaystyle \sf{J_{n }(x)}$ is defined as

$\displaystyle \sf{J_{ n + 1 }(x) = \frac{2n}{x} J_{ n }(x) - J_{ n - 1 }(x)}$

Now Putting $\displaystyle \sf{n = \frac{1}{2} }$ we get

$\displaystyle \sf{J_{ \frac{3}{2} }(x) = \frac{1}{x} J_{ \frac{1}{2} }(x) - J_{ - \frac{1}{2} }(x)} \: \: \: - - - - (1)$

We know that

$\displaystyle \sf{ J_{ \frac{1}{2} }(x) = \sqrt{ \frac{2}{\pi x} } \sin x}$

$\displaystyle \sf{ J_{ - \frac{1}{2} }(x) = \sqrt{ \frac{2}{\pi x} } \cos x}$

Putting these values in Equation 1 we get

$\displaystyle \sf{J_{ \frac{3}{2} }(x) = \frac{1}{x} J_{ \frac{1}{2} }(x) - J_{ - \frac{1}{2} }(x)}$

$\displaystyle \sf{ \implies \: J_{ \frac{3}{2} }(x) = \frac{1}{x} \sqrt{ \frac{2}{\pi x} } \sin x - \sqrt{ \frac{2}{\pi x} } \cos x}$

$\displaystyle \sf{ \implies \: J_{ \frac{3}{2} }(x) =\sqrt{ \frac{2}{\pi x} } \: \bigg( \: \frac{ \sin x}{x} - \cos x \: \bigg)}$

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