Chapter 8: Problem 18
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 \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=\sqrt{x}, \quad a=16$$
Short Answer
Expert verified
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 by step solution
01
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 \( f'(x) = \frac{1}{2}x^{-1/2} \).
02
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.
03
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.
04
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:\[ f(x) = 4 + \frac{1}{8}(x-16) - \frac{1}{128}(x-16)^2 + \frac{3}{4096}(x-16)^3 + \cdots \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( a = 16 \).
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.