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

1 Answer
Apr 21, 2018

R=ooR=

Explanation:

Let's first find the Maclaurin series expansion for sinhxsinhx:

f(x)=sinhx=(e^x-e^-x)/2, f(0)=(e^0-e^0)/2=0f(x)=sinhx=exex2,f(0)=e0e02=0

f'(x)=coshx=(e^x+e^-x)/2, f'(0)=(e^0+e^0)/2=1

f''(x)=sinhx, f''(0)=0

f'''(x)=coshx, f'''(0)=1

f^((4))(x)=sinhx, f^((4))(0)=0

f^((5))(x)=coshx, f^((5))(0)=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^(2n+3)/x^(2n+1)=x^2

So,

|x^2|lim_(n->oo)1/((2n+3)(2n+2))<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=oo