To find the Taylor series for the function f(x) = sin(x) centered at a = π (p), we first need to recall the Taylor series formula. The Taylor series of a function f(x) about a point a is given by:
T(x) = f(a) + f'(a)(x – a) + (f”(a)/2!)(x – a)^2 + (f”'(a)/3!)(x – a)^3 + …
Where:
- f'(a) is the first derivative of f evaluated at x = a.
- f”(a) is the second derivative of f evaluated at x = a.
- And so on.
Now, let’s calculate the derivatives of f(x) = sin(x):
- f(x) = sin(x) → f(π) = sin(π) = 0
- f'(x) = cos(x) → f'(π) = cos(π) = -1
- f”(x) = -sin(x) → f”(π) = -sin(π) = 0
- f”'(x) = -cos(x) → f”'(π) = -cos(π) = 1
- f””(x) = sin(x) → f””(π) = sin(π) = 0
- f””'(x) = cos(x) → f””'(π) = cos(π) = -1
The pattern repeats every four derivatives. Given these evaluations, we can construct the Taylor series:
T(x) = 0 + (-1)(x – π) + 0(x – π)^2 + (1/3!)(x – π)^3 + 0(x – π)^4 + (-1/5!)(x – π)^5 + …
This can be simplified to:
T(x) = – (x – π) + (x – π)^3 / 6 – (x – π)^5 / 120 + …
The series continues indefinitely and can be expressed in a compact form as:
T(x) = Σ [(-1)^n (x – π)^(2n + 1)] / (2n + 1)! for n = 0 to ∞.
This is the Taylor series expansion for f(x) = sin(x) centered at a = π.