Question

What is the Taylor series for `f(x)= cosx` centered on `x= pi/3`?

Answer 1

`f(x) = sum_(n=0)^oo ((-1)^nsin(pi/6 + (npi)/2))/(n!)(x-pi/3)^n`

Explanation:

The Taylor series of a function `f(x)` centered about `a` takes the form
`f(x) = sum_(n=0)^oo f^((n))(a)/(n!)(x-a)^n`

Looking at the derivatives of cosine, we have, for `k in ZZ^+`
`f^((4k))(pi/3) = cos(pi/3) = 1/2`
`f^((4k+1))(pi/3) = -sin(pi/3) = -sqrt(3)/2`
`f^((4k+2))(pi/3) = -cos(pi/3) = -1/2`
`f^((4k+3))(pi/3) = sin(pi/3) = sqrt(3)/2`

So we get

`f(x) = (1/2)/(0!)(x-pi/3)^0 + (-sqrt(3)/2)/(1!)(x-pi/3)^1 + (-1/2)/(2!)(x-pi/3)^2 + ...`

or, using the sine function to create a closed form,

`f(x) = sum_(n=0)^oo ((-1)^nsin(pi/6 + (npi)/2))/(n!)(x-pi/3)^n`