Q22P - Mathematical Methods in Physical Sciences

Question

Using (12.2) and (8.1), find, in summation form, the power series for sinhx and coshx. Check the first few terms of your series by computer.

Step-by-Step Solution

Answer

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

Step 1: Given Information.

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

Step 2: Definition of Power Series.

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

Whererepresents the coefficient of the term and represents the constant.

Step 3: Formula’s to be used in solution.

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

Step 4: Find the Power series of and.

$$\begin{gathered} \sin h\left(z\right)=\frac{1}{2}\left[e^z-e^{-z}\right] \\ =\left[\sum _0^{\infty }\frac{z^n}{n!}-\sum _0^{\infty }\frac{{\left(-z\right)}^n}{n!}\right] \\ =\left[\sum _0^{\infty }\frac{z^n}{n!}-\sum _0^{\infty }\frac{{\left(-1\right)}^n{z}^n}{n!}\right] \end{gathered}$$

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

$$\begin{gathered} \sin h\left(z\right)=\frac{1}{2}\left[\sum _0^{\infty }\frac{z^n}{n!}-\sum _0^{\infty }\frac{{\left(-1\right)}^n{z}^n}{n!}\right] \\ =\frac{1}{2}\left[\left(1-1\right)+\left(z+z\right)(\frac{z^2}{2!}-\frac{z^2}{2!})+(\frac{z^3}{3!}-\frac{z^3}{3!})+...\right] \\ =\frac{1}{2}\left[2z+2x\frac{z^3}{3!}+2\times \frac{z^5}{5!}+ ...\right] \\ =\left[z+\frac{z^3}{3!}+\frac{z^5}{5!}+ ...\right] \\ \cos h\left(z\right)=\frac{1}{2}\left[e^z+e^{-z}\right] \\ =\frac{1}{2}\left[\sum _0^{\infty }\frac{z^n}{n!}+\sum _0^{\infty }\frac{{\left(-z\right)}^n}{n!}\right] \\ =\frac{1}{2}\left[\sum _0^{\infty }\frac{z^n}{n!}+\sum _0^{\infty }\frac{{\left(-z\right)}^n}{n!}\right] \end{gathered}$$

 

 Hence,

The Power of 

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