Question

Prove the function \(f(x) = \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \) is a solution of the differential equation \({f^{\prime \prime }}(x) + f(x) = 0\).

Short answer

The resultant solution of the differential equation is \({f^{\prime \prime }}(x) + f(x) = 0\).

Step-by-step solution

Concept of Differentiation.

Differentiation is a technique for calculating the instantaneous rate of change of a function depending on one of its variables.

Differentiate the function.

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

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

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

\(\begin{aligned}{f^{\prime \prime }}(x) &= \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(2n)(2n - 1){x^{2n - 2}}}}{{(2n)!}}} \\{f^{\prime \prime }}(x) &= \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(2n)(2n - 1){x^{2n - 2}}}}{{(2n)(2n - 1)!}}} \\{f^{\prime \prime }}(x) &= \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}(2n - 1){x^{2n - 2}}}}{{(2n - 1)(2n - 2)!}}} \\{f^{\prime \prime }}(x) &= \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{2n - 2}}}}{{(2n - 2)!}}} \end{aligned}\)

Assess the value of \(n\).

To start from \(n = 0\) instead of \(n = 1\) replace the equation with \(n + 1\)

\(\begin{aligned}{f^{\prime \prime }}(x) &= \sum\limits_{n = 1}^\infty {\frac{{{{( - 1)}^n}{x^{2n - 2}}}}{{(2n - 2)!}}} \\{f^{\prime \prime }}(x) &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}{x^{2(n + 1) - 2}}}}{{(2(n + 1) - 2)!}}} \\{f^{\prime \prime }}(x) &= \sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^{n + 1}}{x^{2n}}}}{{((2n + 2) - 2)!}}} \\{f^{\prime \prime }}(x) &= ( - 1)\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \end{aligned}\)

In comparison, consider the function.

\(\begin{aligned}{f^{\prime \prime }}(x) &= ( - 1)\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^{2n}}}}{{(2n)!}}} \\{f^{\prime \prime }}(x) &= - f(x)\end{aligned}\)

By the above simplification

f"(x) = -f(x)

f"(x) - f(x) = 0

As a result, the resultant solution of the differential equation is \({f^{\prime \prime }}(x) + f(x) = 0\).