Question

If \({\rm{f(x) = }}{{\rm{e}}^{{{\rm{x}}^{\rm{2}}}}}\)show that \({{\rm{f}}^{{\rm{(2n)}}}}{\rm{(0) = }}\frac{{{\rm{(2n)!}}}}{{{\rm{n!}}}}\).

Short answer

Hence, it is shown that\({f^{(2n)}}(0) = \frac{{(2n)!}}{{n!}}\).

Step-by-step solution

Definition of Function

A function is a relationship between a group of inputs that each have one output. A function, in simple terms, is a relationship between inputs in which each input is associated to only one output.

Show that \({e^{{x^2}}} = \sum\limits_{n = 0}^\infty  {\frac{{{x^{2n}}}}{{n!}}} \)

But we also know that for every function f(x), the power series is given by

\(f(x) = \sum\limits_{n = 0}^\infty {\left( {\frac{{{f^{(n)}}(0)}}{{n!}}} \right)} {x^n}\)

Hence \(\frac{{{f^{(2n)}}(0)}}{{(2n)!}}\)is the coefficient of \({x^{2n}}\)in the power series of \(f(x)\)centered at 0 .

We also know that

\({e^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \)

Therefore

\(\begin{array}{c}f(x) = {e^{{x^2}}}\\ = \sum\limits_{n = 0}^\infty {\frac{{{{\left( {{x^2}} \right)}^n}}}{{n!}}} \\ = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n}}}}{{n!}}} \end{array}\)

We can say that the coefficient of \({x^{2n}}\)in power series of \(f(x)\)is \(\frac{1}{{n!}}\)

Hence,

\(\frac{{{f^{(2n)}}(0)}}{{(2n)!}} = \frac{1}{{n!}}\)

\({f^{(2n)}}(0) = \frac{{(2n)!}}{{n!}}\)

Hence, we can show that \({e^{{x^2}}} = \sum\limits_{n = 0}^\infty {\frac{{{x^{2n}}}}{{n!}}} \).