Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : DEC 2015

$$J_{n+1} =\dfrac {2n}x Jn(x) -J_{n-1} (x) $$

For $n=1, J_2(x) =\dfrac 2xJ_1(x)-J_0(x) \\ n=2, J_3(x) =\dfrac 4x J_2(x) - J_1(x) =\dfrac 4x [\dfrac 2xJ_1(x) -J_0 (x) ] -J_1(x) \\ J_3(x) = [\dfrac 8{x^2} -1] J_1(x)-\dfrac 4x J_0(x)\\ n=3, J_4(x) =\dfrac 6x J_3(x) -J_2(x) =\dfrac 6x [(\dfrac 8{x^2} -1) J_1(x) -\dfrac 4x J_0 (x)] -\dfrac 2xJ_1 (x) +J_0(x) \\ = (\dfrac {48}{x^3}-\dfrac 6x -\dfrac 2x)J_1(x) -\dfrac {24}{x^2}J_0(x) + J_0(x) \\ J_4(x) =(\dfrac {48}{x^3}-\dfrac 8x) J_1(x) -(\dfrac {24}{x^2}-1)J_0(x)$