How to find radius of convergence for the taylor series of #(sinx)^2#?

Question

12705 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-27T08:36:11

#sin^2x = sum_(k=1)^oo (-1)^(k+1) 2^(2k-1)/((2k)!) x^(2k) #

with #R=oo#.

Using the trigonometric identity:

#sin^2x = (1-cos2x)/2#

start from the MacLaurin series:

#cost = sum_(k=0)^oo (-1)^k t^(2k)/((2k)!)#

that has radius of convergence #R=oo# and let #t= 2x#:

#cos(2x) = sum_(k=0)^oo (-1)^k (2x)^(2k)/((2k)!) = sum_(k=0)^oo (-1)^k 2^(2k)/((2k)!) x^(2k)#

Extract from the series the term for #k=0#:

#cos(2x) = 1+ sum_(k=1)^oo (-1)^k 2^(2k)/((2k)!) x^(2k)#

and then:

#sin^2x = 1/2(1-1-sum_(k=1)^oo (-1)^k 2^(2k)/((2k)!) x^(2k)) #

#sin^2x = -1/2sum_(k=1)^oo (-1)^k 2^(2k)/((2k)!) x^(2k) #

#sin^2x = sum_(k=1)^oo (-1)^(k+1) 2^(2k-1)/((2k)!) x^(2k) #

still with #R=oo#.