91Ó°ÊÓ

Derivatives are a fundamental concept in calculus, and they play a crucial role in constructing Taylor series, which are essentially power series representations of functions. To determine the Taylor series for a particular function like \(f(x) = \sqrt{x}\), you need to find its derivatives. Derivatives measure how the function changes at a particular point, indicating slope or rate of change.

When computing derivatives for \(f(x) = x^{1/2}\):

• The first derivative \(f'(x) = \frac{1}{2}x^{-1/2}\) tells us the slope of \(f(x)\) at different points.
• The second derivative \(f''(x) = -\frac{1}{4}x^{-3/2}\) gives us the curvature or concavity of the function.
• The third derivative \(f'''(x) = \frac{3}{8}x^{-5/2}\) tells us how the curvature changes.

These derivatives are calculated up to the desired degree to have enough information for the Taylor series expansion.
In the given exercise, each of these derivatives is evaluated at \(x = 16\), forming the coefficients in the Taylor polynomial. This transforms our understanding of the function around the point \(a = 16\).

Power series are infinite series that can represent functions as sums of their derivatives at a given point. A Taylor series is a specific kind of power series centered at a point \(a\). In the context of the exercise, the function \(f(x) = \sqrt{x}\) can be expressed as a Taylor series at \(x = 16\) as follows:

• The general expression for a Taylor series is \(\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n\), where \(f^{(n)}(a)\) are the function's derivatives evaluated at \(a\).
• For \(f(x) = \sqrt{x}\) at \(a = 16\), the Taylor series begins with the constant term \(f(16) = 4\), followed by the linear and higher-order terms based on the derivatives.

This series approximation allows us to understand the behavior of \(f(x)\) near \(x = 16\) as it breaks the function down into simpler polynomial form. Such approximations are especially useful for complex functions that are hard to analyze directly.

The radius of convergence is a critical aspect of power series like the Taylor series, determining the interval around the center \(a\) where the series converges to the function. For the function \(f(x) = \sqrt{x}\) centered at \(a = 16\), the concept of the radius of convergence determines the interval where the Taylor series accurately represents the function.

In this example:

• The radius of convergence for \(\sqrt{x}\) when expanded as a series is linked to the form \((1+u)^n\) with \(u = x/16 - 1\).
• This relationship reveals that the binomial series converges when \(|u| < 1\), implying \(|x - 16| < 16\).
• Thus, the radius of convergence \(R\) is determined to be 16.

Knowing the radius is vital because it helps in understanding how well the power series approximates the function across different x-values. Outside of this interval, the series may diverge and fail to represent the function accurately.