Question

When an elementary function \(f\) is approximated by a second-degree polynomial \(P_{2}\) centered at \(c,\) what is known about \(f\) and \(P_{2}\) at \(c ?\) Explain your reasoning.

Short answer

The function \(f\) and the second-degree Taylor polynomial \(P_2\), centered at \(c\), are equal at \(c\) (i.e., \(f(c) = P_2(c)\)). This is because the other terms in the polynomial \(P_2\) go to zero at \(x = c\), leaving only the \(f(c)\) term.

Step-by-step solution

Understand Concept of Taylor Series

The Taylor series of a real or complex-valued function \(f(x)\) that is infinitely differentiable at a real or complex number \(a\) is the power series: \(f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots \)where \(f'(a)\) denotes the derivative of \(f\) at \(a\), \(f''(a)\) denotes the second derivative of \(f\) at \(a\), and so on.

Analyzing the Specifics of the Problem

In the given problem, we're dealing with a second degree Taylor Polynomial, \(P_2\), centered at \(c\). This can be specifically expressed as:\(P_2(x) = f(c) + f'(c)(x-c) + \frac{f''(c)(x-c)^2}{2!}\)So, we can see that \(P_2(c) = f(c) + f'(c)(c-c) + \frac{f''(c)(c-c)^2}{2!} = f(c)\) because the other terms go to zero because of the \(x-c\) factor. Therefore, the function \(f\) and the Taylor polynomial \(P_2\) are both equal at \(x=c\).

Inference

Thus we see that when the function \(f\) is approximated by a second-degree Taylor polynomial \(P_2\), centered at \(c\), both \(f\) and \(P_2\) have the same value at \(c\) (namely, \(f(c) = P_2(c)\)). This is a fundamental property of Taylor series and provides a notion of the best approximation of the function near point \(c\).