How do you find the power series for $f(x)=cosh(x^2)$ and determine its radius of convergence?

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Answer 1 · 2018-04-16T13:06:25

$cosh (x^2) = sum_(n=0)^oo x^(4n)/((2n)!)$

with radius of convergence $R=oo$ .

Start from the MacLaurin series for the exponential function:

$e^t = sum_(k=0)^oo t^k/(k!)$

and using the exponential formula for the hyperbolic cosine:

$cosh t = 1/2(e^t+e^(-t))$

$cosh t = 1/2(sum_(k=0)^oo t^k/(k!) + sum_(k=0)^oo (-t)^k/(k!) )$

$cosh t = 1/2sum_(k=0)^oo t^k/(k!) (1+ (-1)^k )$

We can see that for $k$ even $(1+ (-1)^k ) = 2$ while for $k$ odd $(1+ (-1)^k ) = 0$ .

So all the odd index terms are null, and if we let $k = 2n$ :

$cosh t = sum_(n=0)^oo t^(2n)/((2n)!)$

Finally let $t=x^2$

$cosh (x^2) = sum_(n=0)^oo(x^2)^(2n)/((2n)!) = sum_(n=0)^oo x^(4n)/((2n)!)$

Using the ratio test we can see that:

$lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( (x^(4(n+1))/((2(n+1))!) )/ (x^(4n)/((2n)!) ))$

$lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( x^(4n+4)/x^(4n) ((2n)!)/((2n+2)!)$

$lim_(n->oo) abs(a_(n+1)/a_n) = x^4 lim_(n->oo) 1/((2n+2)(2n+1)) = 0$

whatever is the value of $x$ , and we can conclude that the series has radius of convergence $R=oo$ .

How do you find the power series for f(x)=cosh(x^2)f(x)=cosh(x^2) and determine its radius of convergence?