Question

(a) Show the function \({J_1}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \)which satisfies the differential equation\({x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x) = 0.\)

(b) Show that the \(J_0^\prime (x) = - {J_1}(x)\).

Short answer

(a) The function \({J_1}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \) which satisfies the differential equation \({x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x) = 0\).

(b) The condition satisfied \({J^\prime }_0(x) = - {J_1}(x)\).

Step-by-step solution

Concept of differential equation.

Differential equations take a form similar to \(f(x) + f'(x) = 0\), where \(f'\)is “f-prime,” the derivative of \(f'\), such an equation relates a function \(f(x)\) to its derivative. In order to solve the differential equation, you must first find the function \(f(x)\).

Differentiate the function and substitute the values.

(a)

The function is \({J_1}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

Let \({J_1}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

\(J_1^\prime (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}(2n + 1){x^{2n}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

\({J^{\prime \prime }}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}(2n)(2n + 1){x^{2n - 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

Substitute, the values of \(J_1^{\prime \prime }(x),J_1^\prime (x)\) and \({J_1}(x)\) in \({x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x) = 0\)

\(\left( {\begin{aligned}{{x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x)}\\{ + \left( {{x^2} - 1} \right){J_1}(x)}\end{aligned}} \right)\)\( = \left( {\begin{aligned}{{x^2}\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(2n)(2n + 1){x^{2n - 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} + x\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}(2n + 1){x^{2n}}}}{{n!(n + 1)!{2^{2n + 1}}}}} }\\{ + \left( {{x^2} - 1} \right)\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} }\end{aligned}} \right)\)

\(\left( {\begin{aligned}{{x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x)}\\{ + \left( {{x^2} - 1} \right){J_1}(x)}\end{aligned}} \right)\)\( = \left( {\begin{aligned}{{x^2}\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(2n)(2n + 1){x^{2n - 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} + x\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}(2n + 1){x^{2n}}}}{{n!(n + 1)!{2^{2n + 1}}}}} }\\{ + \left( {{x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} - \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} } \right)}\end{aligned}} \right)\)

\(\left( {\begin{aligned}{{x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x)}\\{ + \left( {{x^2} - 1} \right){J_1}(x)}\end{aligned}} \right)\)\( = \left( {\begin{aligned}{\sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(2n)(2n + 1){x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} + \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}(2n + 1){x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} }\\{ + \left( {\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 3}}}}{{n!(n + 1)!{2^{2n + 1}}}}} - \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} } \right)}\end{aligned}} \right)\)

\(\left( {\begin{aligned}{{x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x)}\\{ + \left( {{x^2} - 1} \right){J_1}(x)}\end{aligned}} \right)\)\( = \sum\limits_{n = 0}^\infty {\left\{ {\frac{{\begin{aligned}{*{20}{l}}{{{( - 1)}^n}2n(2n + 1){x^{2n + 1}} + {{( - 1)}^n}(2n + 1){x^{2n + 1}}}\\{ + {{( - 1)}^n}{x^{2n + 1}} - {{( - 1)}^n}{x^{2n + 1}}}\end{aligned}}}{{n!(n + 1)!{2^{2n + 1}}}}} \right\}} \)

Simplify the value.

\({x^2}{J^{\prime \prime }}_1(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x)\)\( = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {(2n)(2n + 1) + (2n + 1) + {x^2} - 1} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

\({x^2}{J^{\prime \prime }}_1(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x)\)\( = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {4{n^2} + 2n + 2n + 1 + {x^2} - 1} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

\({x^2}{J^{\prime \prime }}_1(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x)\)\( = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {4{n^2} + 4n + {x^2}} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

\({x^2}{J^{\prime \prime }}_1(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x)\)\( = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {4{n^2} + 4n} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} + {x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \)

Explain the term of the series.

x²J"₁(x)+xJ'₁(x)+(x²-1)J₁(x) = \(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}(4{n^2} + 4n)}}{{n!(n + 1)!{2^{2n + 1}}}} + {x^2}{J_1}(x)} \)

Consider the summation part \(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {4{n^2} + 4n} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} \),

\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {4{n^2} + 4n} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} = 0 - \frac{{{x^3}}}{2} + \frac{{{x^5}}}{{16}} - \frac{{{x^7}}}{{384}} + \frac{{{x^9}}}{{18432}} \cdots \)

Compare into the \({J_1}(x)\)

\(\begin{aligned}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} &= \frac{x}{2} - \frac{{{x^3}}}{{16}} + \frac{{{x^5}}}{{384}} - \frac{{{x^7}}}{{18432}} + \cdots \\\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} &= - {x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \\\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} &= - {x^2}{J_1}(x)\end{aligned}\)\(\begin{aligned}{l}{x^2}J_1^{\prime \prime }(x) + x{J_1}(x) + \left( {{x^2} - 1} \right){J_1}(x) &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}\left( {4{n^2} + 4n} \right)}}{{n!(n + 1)!{2^{2n + 1}}}}} + {x^2}\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \\{x^2}J_1^{\prime \prime }(x) + x{J_1}(x) + \left( {{x^2} - 1} \right){J_1}(x) &= - {x^2}{J_1}(x) + {x^2}{J_1}(x)\\{x^2}J_1^{\prime \prime }(x) + x{J_1}(x) + \left( {{x^2} - 1} \right){J_1}(x) &= 0\end{aligned}\)

Therefore. the function \({J_1}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n + 1}}}}{{n!(n + 1)!{2^{2n + 1}}}}} \) satisfies the differential equation. \({x^2}J_1^{\prime \prime }(x) + xJ_1^\prime (x) + \left( {{x^2} - 1} \right){J_1}(x) = 0.\)

Differentiate the function \({J_0}(x)\).

(b)

Let, \({J_0}(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{{2^{2n}}{{(n!)}^2}}}} \),

\(J_0^\prime (x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}2n{x^{2n - 1}}}}{{{2^{2n}}{{(n!)}^2}}}} \)

Expand the function.

Function is \(J_0^\prime (x)\),

\(\begin{aligned}J_0^\prime (x) &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}2n{x^{2n - 1}}}}{{{2^{2n}}{{(n!)}^2}}}} \\J_0^\prime (x) &= 0 - \frac{x}{2} + \frac{{{x^3}}}{{16}} - \frac{{{x^5}}}{{385}} + \cdots \\ - {J_1}(x) &= - \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}2n{x^{2n - 1}}}}{{{2^{2n}}{{(n!)}^2}}}} \\ - {J_1}(x) &= \left( {\frac{x}{2} + \frac{{{x^3}}}{{16}} - \frac{{{x^5}}}{{385}} + \cdots } \right)\end{aligned}\)

Therefore, the condition \(J_0^\prime (x) = - {J_1}(x)\) is proved.