91Ó°ÊÓ

Let the Bessel equation be\({x^2}y'' + xy' + \left( {{x^2} - {n^2}} \right)y = 0\). This equation hastwo linearly independent solutionsfor a fixed value of\(n\).A Bessel equation of the first kind,indicated by\({J_n}(x)\), is one of these solutions that may be derived usingFrobinous approach.

\(\begin{array}{l}{y_1} = {x^a}{J_p}\left( {b{x^c}} \right)\\{y_2} = {x^a}{J_{ - p}}\left( {b{x^c}} \right)\end{array}\)

At\(x = 0\), this solution is regular. The second solution, which is singular at\(x = 0\), is represented by\({Y_n}(x)\)and is calleda Bessel function of the second kind.

\({y_3} = {x^a}\left( {\frac{{cosp\pi {J_p}\left( {b{x^c}} \right) - {J_{ - p}}\left( {b{x^c}} \right)}}{{sinp\pi }}} \right)\)

Let the Bessel’s function of first kind be \({J_v} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!\Gamma (1 + v + n)}}} {\left( {\frac{x}{2}} \right)^{2n + v}}\).

Substitute \(v = - 1/2\) in it yields,

\({J_{ - 1/2}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!\Gamma \left( {1 - \frac{1}{2} + n} \right)}}} {\left( {\frac{x}{2}} \right)^{2n - 1/2}}\)

\({J_{ - 1/2}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{n!\Gamma \left( {\frac{1}{2} + n} \right)}}} {\left( {\frac{x}{2}} \right)^{2n - 1/2}}\) … (1)

Let the gamma function be,

\(\begin{array}{l}n = 0,\Gamma \left( {\frac{1}{2}} \right) = \sqrt \pi \\n = 1,\Gamma \left( {\frac{3}{2}} \right) = \frac{{2!}}{{{2^2}1!}}\sqrt \pi \\n = 2,\Gamma \left( {\frac{5}{2}} \right) = \frac{{4!}}{{{2^4}2!}}\sqrt \pi \\n = 3,\Gamma \left( {\frac{7}{2}} \right) = \frac{{6!}}{{{2^6}3!}}\sqrt \pi \end{array}\)

Then, \(\Gamma \left( {\frac{1}{2} + n} \right) = \frac{{(2n)!}}{{{2^{2n}}n!}}\) … (2)

Substitute the equation (2) into (1).

\(\begin{array}{c}{J_{ - 1/2}} = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{x!\frac{{(2n)!}}{{{2^{2n}}{{\cancel{{}}}^{2n}}}}\sqrt \pi }}} {\left( {\frac{x}{2}} \right)^{2n - 1/2}}\\ = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{2^{2n}}}}{{(2n)!}}} {\left( {\frac{x}{2}} \right)^{2n - 1/2}}\\ = \sqrt \pi \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{2^{2n}}}}{{(2n)!}}} {x^{2n - 1/2}} \cdot {2^{ - 2n + 1/2}}\\ = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{2^{2n}}}}{{(2n)!}}} {x^{2n}} \cdot {x^{ - 1/2}} \cdot {2^{2\pi }} \cdot {2^{1/2}}\\ = \frac{1}{{\sqrt \pi }}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(2n)!}}} {x^{2n}} \cdot \sqrt {\frac{2}{x}} \\ = \sqrt {\frac{2}{{x\pi }}} \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}}}{{(2n)!}}} {x^{2n}}\\ = \sqrt {\frac{2}{{x\pi }}} \cos x\end{array}\)

Hence verified.