Problem 18 - Essential Calculus Early Transcendentals

Question

Find the Taylor series for \(f(x)\) centered at the given value of \(a\) . [Assume that \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=\sqrt{x}, \quad a=16$$

Step-by-Step Solution

Answer

The Taylor series for \( \sqrt{x} \) centered at \( a = 16 \) is: \[ 4 + \frac{1}{8}(x-16) - \frac{1}{128}(x-16)^2 + \frac{3}{4096}(x-16)^3 + \cdots \].

Step 1: Find the Derivative of f(x)

To find the Taylor series, we need to determine the derivatives of the function. The function is \( f(x) = \sqrt{x} = x^{1/2} \). The first derivative is \( f'(x) = \frac{1}{2}x^{-1/2} \).

Step 2: Find Higher Derivatives

Continue to find higher derivatives needed for the series. - The second derivative is \( f''(x) = -\frac{1}{4}x^{-3/2} \).- The third derivative is \( f'''(x) = \frac{3}{8}x^{-5/2} \).- The fourth derivative is \( f^{(4)}(x) = -\frac{15}{16}x^{-7/2} \).Continue finding derivatives as needed.

Step 3: Evaluate Derivatives at a = 16

For Taylor series expansion centered at \( a = 16 \), evaluate each derivative at \( x = 16 \):- \( f(16) = \sqrt{16} = 4 \).- \( f'(16) = \frac{1}{2} \, (16)^{-1/2} = \frac{1}{8} \).- \( f''(16) = -\frac{1}{4} \, (16)^{-3/2} = -\frac{1}{128} \).- \( f'''(16) = \frac{3}{8} \, (16)^{-5/2} = \frac{3}{4096} \).- Continue evaluating derivatives as necessary.

Step 4: Write the Taylor Series Expansion

The Taylor series centered at \( a = 16 \) is:\[ f(x) = f(16) + f'(16)(x-16) + \frac{f''(16)}{2!}(x-16)^2 + \frac{f'''(16)}{3!}(x-16)^3 + \cdots \]Substitute the evaluated derivatives:\[ f(x) = 4 + \frac{1}{8}(x-16) - \frac{1}{128}(x-16)^2 + \frac{3}{4096}(x-16)^3 + \cdots \].

Key Concepts

Understanding Derivatives

In mathematics, a derivative represents how a function changes as its input changes. It is a fundamental concept for understanding the behavior of functions. For the function \( f(x) = \sqrt{x} = x^{1/2} \), the process of finding its derivative involves applying basic differentiation rules.

• The first derivative, \( f'(x) = \frac{1}{2}x^{-1/2} \), shows how \( f(x) \) changes with \( x \).
• Each successive derivative gives more insight into the function's behavior, especially around certain points like \( a = 16 \).

Using derivatives, we can generate a Taylor series, a powerful tool in calculus for approximating functions, where each derivative gives us the coefficients for the terms in the series. Derivatives are crucial, as they set the foundation for more complex concepts like Taylor series expansions.

Power Series Expansion Basics

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are constants. In the context of Taylor series, these coefficients are derived from the function's derivatives. When you expand \( f(x) \) around a specific point \( a \), you express it as a sum of powers of \( x-a \). This is useful for approximating functions near that point, especially when precise functional values are difficult to calculate directly.In the exercise, \( f(x) = \sqrt{x} \) is expanded around \( a = 16 \). The powers of \( (x - 16) \) are weighted by evaluated derivatives at \( x = 16 \), which allows us to approximate \( \sqrt{x} \) near \( x = 16 \). The power series expansion offers a way to represent complex functions as a simpler polynomial.

Higher Derivatives and Their Role

Higher derivatives are simply the successive differentiation of a function. For example, the second derivative \( f''(x) \) is the derivative of the first derivative \( f'(x) \), and so on. These higher derivatives are essential in composing a Taylor series.- The second derivative \( f''(x) = -\frac{1}{4}x^{-3/2} \) provides the coefficient for the quadratic term.- The third derivative \( f'''(x) = \frac{3}{8}x^{-5/2} \) affects the cubic term, continuing this pattern for higher terms.Each higher derivative's behavior adds more detail to the approximation around the point \( a \). Evaluating these derivatives at \( x = 16 \), each contributes a term to the series, refining accuracy as more terms are added.

Composing the Taylor Series Expansion

A Taylor series is composed by adding evaluated derivatives at a specific point \( a \), each multiplied by a power of \( (x-a) \) and divided by factorial terms. This pattern continues infinitely but often, a truncated version gives sufficient accuracy.For the function \( f(x) = \sqrt{x} \) centered at \( a = 16 \), the Taylor series looks like this:- Start with \( f(16) = 4 \).- Add the linear term: \( f'(16)(x-16) = \frac{1}{8}(x-16) \).- Add the quadratic term: \( \frac{f''(16)}{2!}(x-16)^2 = -\frac{1}{256}(x-16)^2 \).- Continue with higher terms for greater precision.Each derivative value provides a coefficient for terms, revealing \( \sqrt{x} \)'s behavior near 16. This approach makes the Taylor series a versatile method for function approximation, especially in calculus and analytical settings where finding exact solutions is complex.