Question

How do you find the taylor series for `cos x`, `a=pi/3`?

Answer 1

`cosx = sum_(n=0)^(oo) cos(pi/3+(npi)/2)/(n!)(x-pi/3)^n`

Explanation:

The Taylor series of `f(x)` in `x=x_0` is given by:

`sum_(n=0)^(oo) f^((n))(x_0)/(n!)(x-x_0)^n`

So we must calculate the derivatives of all orders of `cos(x)`.

Now we note that:

`d/(dx)cosx =-sinx =cos(x+pi/2)`

`(d^((2)))/(dx^2)cosx =-cosx =cos(x+pi)`

and in general:

`(d^((n)))/(dx^n)cosx =-cosx =cos(x+(npi)/2)`

which can be proved by mathematical induction.

Thus the Taylor expansion is:

`sum_(n=0)^(oo) cos(pi/3+(npi)/2)/(n!)(x-pi/3)^n`