# What is the interval of convergence of the Taylor series of f(x)=cos(3x^2)?

Jun 27, 2018

cos (3x^2) = sum_(n=0)^oo (-1)^n3^(2n)x^(4n)/((2n)!)

converging for $x \in \left(- \infty , + \infty\right)$

#### Explanation:

Consider the MacLaurin series for $\cos \theta$:

cos theta = sum_(n=0)^oo (-1)^n theta^(2n)/((2n)!)

and let $\theta = 3 {x}^{2}$:

cos (3x^2) = sum_(n=0)^oo (-1)^n(3x^2)^(2n)/((2n)!)

cos (3x^2) = sum_(n=0)^oo (-1)^n3^(2n)x^(4n)/((2n)!)

Using the ratio test:

lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs ( ( 3^(2(n+1))x^(4(n+1))/((2(n+1))!))/(3^(2n)x^(4n)/((2n)!)))

lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) abs (3^(2(n+1))/3^(2n) * ((2n)!) /((2(n+1))!)* x^(4(n+1))/x^(4n))

lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo)3^(2n+2)/3^(2n) * ((2n)!)/((2n+2)!) * abs(x^(4n+4)/x^(4n))

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \to \infty} \frac{9 {x}^{4}}{\left(2 n + 1\right) \left(2 n + 2\right)}$

so for any value of $x \in \mathbb{R}$:

${\lim}_{n \to \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = 0$

and the series is convergent, which means that the radius of convergence is $R = \infty$.