What is the interval of convergence of the Taylor series of f(x)=cos(3x^2)f(x)=cos(3x2)?

1 Answer
Jun 27, 2018

cos (3x^2) = sum_(n=0)^oo (-1)^n3^(2n)x^(4n)/((2n)!)cos(3x2)=n=0(1)n32nx4n(2n)!

converging for x in (-oo,+oo)x(,+)

Explanation:

Consider the MacLaurin series for cos thetacosθ:

cos theta = sum_(n=0)^oo (-1)^n theta^(2n)/((2n)!)cosθ=n=0(1)nθ2n(2n)!

and let theta = 3x^2θ=3x2:

cos (3x^2) = sum_(n=0)^oo (-1)^n(3x^2)^(2n)/((2n)!)cos(3x2)=n=0(1)n(3x2)2n(2n)!

cos (3x^2) = sum_(n=0)^oo (-1)^n3^(2n)x^(4n)/((2n)!)cos(3x2)=n=0(1)n32nx4n(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->oo) abs (a_(n+1)/a_n) = lim_(n->oo)(9x^4)/((2n+1)(2n+2))

so for any value of x in RR:

lim_(n->oo) abs (a_(n+1)/a_n) = 0

and the series is convergent, which means that the radius of convergence is R=oo.