Question

How do you find the first three terms of the Taylor series for `f(x)=cos(5x)` for `x=0`?

Answer 1

`f(x)~~1+0(x-0)+(-25)/(2*1)(x-0)^2=1-25/2x^2`.

Method:
The Taylor series centered at `a` for `f(x)` is:

`f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ * * * +(f^(n+1)(a))/((n+1)!)(x-a)^(n+1)+ * * *`

The first three terms will involve `f(x)=cos5x`, `f'(x)=-5sin5x`, and `f''(x)-25cos5x`, each evaluated at `a=0`

We find: `f(0)=1`, `f'(0)=0`, and `f''(0)=-25`.

Substitute into the series and simplify is necessary.

`f(x)~~1+0(x-0)+(-25)/(2*1)(x-0)^2=1-25/2x^2`.