Question

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

Answer 1

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)!)`