Q. 91 - Calculus

Question

Use the Maclaurin series for $e^x$ and $e^{-x}$ to prove that

$\sinh x=\sum _{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}.$

Step-by-Step Solution

Answer

That, which is $\sinh x=\sum _{i=0}^x\frac{1}{(2k+1)!}{x}^{2\lambda ·1}$ is proved.

Step 1 Given Information

Consider the hyperbolic sine function: $\sinh x=\frac{e^x-e^{-x}}{2}$

Step 2 Proof

Since,

$\begin{array}{rcl}{e}^x& =& \sum _{k=0}^{\infty }\frac{1}{k!}{x}^k\\ & =& 1+x+\frac{{\displaystyle x^2}}{{\displaystyle 2!}}+\frac{{\displaystyle x^3}}{{\displaystyle 3!}}+\frac{{\displaystyle x^4}}{{\displaystyle 4!}}+\frac{{\displaystyle x^3}}{{\displaystyle 5!}}+\dots \end{array}$

And also.

$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\frac{x^5}{5!}+\cdots$

Now, subtract the series for $e^x$ and $e^{-x}$

$e^x-e^{-x}=2x+\frac{2x^3}{3!}+\frac{2x^5}{5!}+\dots$

So,

$\begin{array}{rcl}\frac{e^x-e^{-x}}{2}& =& x+\frac{x^3}{3!}+\frac{x^3}{5!}+\dots \\ & =& \sum _{k=0}^{\infty }\frac{{\displaystyle 1}}{{\displaystyle (2k+1)!}}{x}^{2k+1}\\ & =& \sin hx\end{array}$