91Ó°ÊÓ

The given expression is sinh (x) , cosh (x) ..

A Power Series which is in one variable is an infinite series written in the form of$\underset{\mathbf{n}\mathbf{=}\mathbf{0}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathbf{a}}_{\mathbf{n}}{\left(x-c\right)}^{\mathbf{n}}\mathbf{=}{\mathbf{a}}_{\mathbf{0}}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\left(x-c\right)\mathbf{+}{\mathbf{a}}_{\mathbf{2}}{\left(x-c\right)}^{\mathbf{2}\mathbf{}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{Where}\mathbf{}{\mathbf{a}}_{\mathbf{n}}$

Whererepresents the coefficient of the term and represents the constant.

$$\begin{gathered} \mathrm{Let}’\mathrm{s}\mathrm{state}\mathrm{the}\mathrm{notation}\mathrm{to}\mathrm{be}\mathrm{used}\mathrm{in}\mathrm{the}\mathrm{solution}. \\ \cosh \left(\mathrm{ix}\right)=\cosh \left(\mathrm{iz}\right) \\ \sinh \left(\mathrm{ix}\right)=\sinh \left(\mathrm{iz}\right) \\ {\mathrm{e}}^{\mathrm{z}}=\underset{0}{\overset{∞}{∑\frac{{\mathrm{z}}^{\mathrm{n}}}{\mathrm{n}!}}} \\ \sinh \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{z}}-{\mathrm{e}}^{-\mathrm{z}}}{2} \\ \cosh \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{\mathrm{z}}-{\mathrm{e}}^{-\mathrm{z}}}{2} \end{gathered}$$

$$\begin{gathered} \sin h\left(z\right)=\frac{1}{2}\left[e^z-e^{-z}\right] \\ =\left[\underset{0}{\overset{∞}{∑}}\frac{z^n}{n!}-\underset{0}{\overset{∞}{∑}}\frac{{\left(-z\right)}^n}{n!}\right] \\ =\left[\underset{0}{\overset{∞}{∑}}\frac{z^n}{n!}-\underset{0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^n{z}^n}{n!}\right] \end{gathered}$$

If n is odd then the notation has a value, but if n is even then the notation is zero.

$$\begin{gathered} \sin h\left(z\right)=\frac{1}{2}\left[\underset{0}{\overset{∞}{∑}}\frac{z^n}{n!}-\underset{0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^n{z}^n}{n!}\right] \\ =\frac{1}{2}\left[\left(1-1\right)+\left(z+z\right)(\frac{z^2}{2!}-\frac{z^2}{2!})+(\frac{z^3}{3!}-\frac{z^3}{3!})+...\right] \\ =\frac{1}{2}\left[2z+2x\frac{z^3}{3!}+2×\frac{z^5}{5!}+...\right] \\ =\left[z+\frac{z^3}{3!}+\frac{z^5}{5!}+...\right] \\ \cos h\left(z\right)=\frac{1}{2}\left[e^z+e^{-z}\right] \\ =\frac{1}{2}\left[\underset{0}{\overset{∞}{∑}}\frac{z^n}{n!}+\underset{0}{\overset{∞}{∑}}\frac{{\left(-z\right)}^n}{n!}\right] \\ =\frac{1}{2}\left[\underset{0}{\overset{∞}{∑}}\frac{z^n}{n!}+\underset{0}{\overset{∞}{∑}}\frac{{\left(-z\right)}^n}{n!}\right] \end{gathered}$$

Hence,

The Power of

$$\begin{gathered} \sinh \left(\mathrm{x}\right)=\left[\mathrm{x}+\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}+...\right]\mathrm{And} \\ \sinh \left(\mathrm{x}\right)=\left[1+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^4}{4!}+...\right]. \end{gathered}$$