# What is the radius of convergence of the MacLaurin series expansion for f(x)= sinh x?

Apr 21, 2018

$R = \infty$

#### Explanation:

Let's first find the Maclaurin series expansion for $\sinh x$:

$f \left(x\right) = \sinh x = \frac{{e}^{x} - {e}^{-} x}{2} , f \left(0\right) = \frac{{e}^{0} - {e}^{0}}{2} = 0$

$f ' \left(x\right) = \cosh x = \frac{{e}^{x} + {e}^{-} x}{2} , f ' \left(0\right) = \frac{{e}^{0} + {e}^{0}}{2} = 1$

$f ' ' \left(x\right) = \sinh x , f ' ' \left(0\right) = 0$

$f ' ' ' \left(x\right) = \cosh x , f ' ' ' \left(0\right) = 1$

${f}^{\left(4\right)} \left(x\right) = \sinh x , {f}^{\left(4\right)} \left(0\right) = 0$

${f}^{\left(5\right)} \left(x\right) = \cosh x , {f}^{\left(5\right)} \left(0\right) = 1$

So, we see a pretty consistent pattern of alternating zeroes and ones. Let's write out the first few terms of the series:

The Maclaurin Series expansion is given by

f(x)=sum_(n=0)^oof^((n))(0)x^n/(n!)=f(0)+f'(0)x+f''(0)x^2/(2!)+...

So, for our function, we get

sinhx=0+x+0x^2+x^3/(3!)+0x^4+x^5/(5!)+...

If we ignore the terms involving zero, we see

sinhx=x+x^3/(3!)+x^3/(3!)+...

So, we want odd exponents and odd factorials starting at $1$, so the summation is

sinhx=sum_(n=0)^oox^(2n+1)/((2n+1)!)

To find the radius of convergence, we'll use the Ratio Test, where

a_n=x^(2n+1)/((2n+1)!)

lim_(n->oo)|a_(n+1)/a_n|=lim_(n->oo)|x^(2n+3)/((2n+3)!)*((2n+1)!)/x^(2n+1)|

We're going to want the factorials to cancel out. So, strip some terms out of the larger factorial:

(2n+3)! = (2n+3)(2n+1)(2n+1)!

So we have

lim_(n->oo)|(x^(2n+3)cancel((2n+1)!))/(x^(2n+1)(2n+3)(2n+2)cancel((2n+1)!))|

${x}^{2 n + 3} / {x}^{2 n + 1} = {x}^{2}$

So,

$| {x}^{2} | {\lim}_{n \to \infty} \frac{1}{\left(2 n + 3\right) \left(2 n + 2\right)} < 1$ results in convergence.

The limit goes to $0.$ Thus, this quantity is always $0 < 1$ regardless of what we pick for $x$. We have convergence for all real numbers, IE, $R = \infty$