Chapter 10.3, Problem 24E

To determine

To find: The Maclaurin series generated by $\sinh x$ and $\cosh x$ .

Answer to Problem 24E

  $\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots +\frac{x^{2n+1}}{(2n+1)!}+\dots$ .

And

  $\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\dots +\frac{x^{2n}}{(2n)!}+\dots$

Given information:

  $\sinh x=\frac{e^x-e^{-x}}{2}$ and cosh $x=\frac{e^x+e^{-x}}{2}$

Calculation:

The given function is, $\sinh x=\frac{e^x-e^{-x}}{2}$ cosh $x=\frac{e^x+e^{-x}}{2}$

McLaurin series generated by function $e^x$ is given by

  $e^x=1+x+\frac{x^2}{2!}+\dots +\frac{x^n}{n!}+\dots$ McLaurin series generated by function $e^{-x}$ is given by

  $e^{-x}=1-x+\frac{x^2}{2!}+\dots +{(-1)}^n\frac{x^n}{n!}+\dots$

Thus, the terms with $n$ even will cancel for $\sinh x=\frac{e^x-e^{-x}}{2},$ and the terms with $n$ odd will cancel for $\cosh x=\frac{e^x+e^{-x}}{2}.$ Therefore, $\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots +\frac{x^{2n+1}}{(2n+1)!}+\dots$ .

  $\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\dots +\frac{x^{2n}}{(2n)!}+\dots$

Explanation of Solution

Given information:

  $\sinh x=\frac{e^x-e^{-x}}{2}$ and cosh $x=\frac{e^x+e^{-x}}{2}$

Calculation:

The given function is, $\sinh x=\frac{e^x-e^{-x}}{2}$ cosh $x=\frac{e^x+e^{-x}}{2}$

McLaurin series generated by function $e^x$ is given by

  $e^x=1+x+\frac{x^2}{2!}+\dots +\frac{x^n}{n!}+\dots$ McLaurin series generated by function $e^{-x}$ is given by

  $e^{-x}=1-x+\frac{x^2}{2!}+\dots +{(-1)}^n\frac{x^n}{n!}+\dots$

Thus, the terms with $n$ even will cancel for $\sinh x=\frac{e^x-e^{-x}}{2},$ and the terms with $n$ odd will cancel for $\cosh x=\frac{e^x+e^{-x}}{2}.$ Therefore, $\sinh x=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\dots +\frac{x^{2n+1}}{(2n+1)!}+\dots$ .

  $\cosh x=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\dots +\frac{x^{2n}}{(2n)!}+\dots$