Question

Let f(x) be the function with f(0)= 1/π². What is the nth degree Taylor approximation, denoted as Tₙ(x) for f(x) centered at a=0? a) Tₙ(x) = 1/π²

b) Tₙ(x) = 1 + x/π²

c) Tₙ(x) = 1 - x²/π⁴

d) Tₙ(x) = 1/π² + x/π²

Answer 1

The correct answer is option a. The nth degree Taylor approximation for f(x), centered at a=0, is Tₙ(x) = 1/π², considering we only know the value of f at 0 and not its derivatives.

Step-by-step explanation:

The Taylor series approximation of a function f(x) around a point a is given by the formula:

f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \ldots

Since the only information given about f(x) is the value of f at 0, which is f(0) = 1/\pi^2, and since no information about the derivatives of f at 0 was provided, the best we can say is that the nth degree Taylor approximation, Tn(x), for f(x) centered at a=0, is the constant term which is the value of f at 0. Hence:

Tn(x) = 1/\pi^2

Therefore, the correct option is a) Tn(x) = 1/\pi^2.