Answer
We prove that $f\left( x \right) = \sinh x = \dfrac{{{{\rm{e}}^x} - {{\rm{e}}^{ - x}}}}{2}$ is represented by the Maclaurin series $\dfrac{1}{2}\mathop \sum \limits_{n = 0}^\infty \dfrac{{1 - {{\left( { - 1} \right)}^n}}}{{n!}}{x^n}$, for all $x$.
Work Step by Step
We have $f\left( x \right) = \sinh x = \dfrac{{{{\rm{e}}^x} - {{\rm{e}}^{ - x}}}}{2}$.
In Example 3 it is shown that ${{\rm{e}}^x}$ is represented by the Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}}$. Also, it is known in Exercise 62 that ${{\rm{e}}^{ - x}}$ is represented by the Maclaurin series $\mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}$ for all $x$. Therefore, $f\left( x \right) = \sinh x = \dfrac{{{{\rm{e}}^x} - {{\rm{e}}^{ - x}}}}{2}$ is represented by the Maclaurin series $\dfrac{1}{2}\left( {\mathop \sum \limits_{n = 0}^\infty \dfrac{{{x^n}}}{{n!}} - \mathop \sum \limits_{n = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}} \right) = \dfrac{1}{2}\mathop \sum \limits_{n = 0}^\infty \dfrac{{1 - {{\left( { - 1} \right)}^n}}}{{n!}}{x^n}$, for all $x$.
