what is the series expansion of cos2x ?

Answer 1

cos x = sum from n=0 to inf of (-1)^n * x^(2n)/(2n)!

cos 2x =
= sum from n=0 to inf of (-1)^n * (2x)^(2n)/(2n)! =
= sum from n=0 to inf of (-1)^n *2^2n * x^(2n)/2n! =
= sum from n=0 to inf of (-1)^n * 4^n * x^(2n)/2n! =
= 1 - 4x^2/2 + 16x^4/24 - 64x^6/720 +- ...

Answer 2

Let f(x)=cos 2x. =>f(0)=cos 0=1

Differentiating successively with respect to x,

f'(x)=-2sin 2x =>f'(0)=-2(0)=0

f"(x)=-4cos 2x =>f"(0)=-4(1)=-4

f"'(x)=8sin 2x =>f"'(0)=8(0)=0

f""(x)=16cos 2x =>f""(0)=16(1)=16

According to McLaurin's series,

f(x)=f(0)+(x/1!)(f'(0)+[(x^2)/2!](f"(0)+[(x^3)/3!](f"'(0)+[(x^4)/4!](f""(0))+......................

cos 2x=1+(x/1!)(0)+[(x^2)/2!](-4)+[(x^3)/3!](0)+[(x^4)/4!](16)+..............

cos 2x=1-4[(x^2)/2!]+16[(x^4)/4!]+...............

cos 2x=1-4[(x^2)/2]+16[(x^4)/24]+...............

cos 2x=1-2(x^2)+(2/3)(x^4)+...............