How can you find the taylor expansion of f(x) =sinx about x=0?

1 Answer
Sep 28, 2015

See the explanation.

Explanation:

f(x-x_0)=sum_(k=0)^n f^((k))(x_0) (x-x_0)^k/(k!) + f^((k+1))(epsilon)(x-x_0)^(k+1)/((k+1)!)

In our case x_0=0 so:

f(x)=sum_(k=0)^n f^((k))(0) x^k/(k!) + f^((k+1))(epsilon)x^(k+1)/((k+1)!)

f(x)=sinx
f(0)=0
f'(x)=cosx => f'(0)=1
f''(x)=-sinx => f''(0)=0
f^((3))(x)=-cosx => f^((3))(0)=-1
f^((4))(x)=sinx => f^((4))(0)=0
and so on....

sinx= 0 + 1*x/(1!) + 0 - 1*x^3/(3!) + 0 + 5*x/(5!) +0 - 1*x^7/(7!)...

sinx = sum_(k=0)^oo (-1)^k x^(2k+1)/((2k+1)!)