Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-18, find the terms through \(x^{5}\) in the Maclaurin series for \(f(x)\). Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=(1+x)^{3 / 2} $$

Step-by-Step Solution

Answer

The Maclaurin series is \( 1 + \frac{3}{2}x + \frac{3}{8}x^2 - \frac{1}{16}x^3 + \frac{3}{128}x^4 - \frac{1}{128}x^5 \).

Step 1: Recall the Maclaurin Series Formula

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). For this problem, we will need to find the derivatives of \((1+x)^{3/2}\) and evaluate them at \(x=0\).

Step 2: Compute the Function and its Derivatives

First, compute \(f(0)\). Since \( f(x) = (1+x)^{3/2} \), we have \( f(0) = 1^{{3/2}} = 1 \). For the first derivative: \( f'(x) = \frac{3}{2}(1+x)^{1/2} \). Evaluating at \( x = 0 \), \( f'(0) = \frac{3}{2}(1)^{1/2} = \frac{3}{2} \).

Step 3: Calculate Higher-Order Derivatives

Find the second derivative: \( f''(x) = \frac{3}{4}(1+x)^{-1/2} \). Evaluate at \( x = 0\): \( f''(0) = \frac{3}{4}(1)^{-1/2} = \frac{3}{4} \). Third derivative: \( f'''(x) = -\frac{3}{8}(1+x)^{-3/2} \), and \( f'''(0) = -\frac{3}{8} \). Fourth derivative: \( f^{(4)}(x) = \frac{9}{16}(1+x)^{-5/2} \), so \( f^{(4)}(0) = \frac{9}{16} \). Fifth derivative: \( f^{(5)}(x) = -\frac{45}{32}(1+x)^{-7/2} \), and \( f^{(5)}(0) = -\frac{45}{32} \).

Step 4: Substitute into the Maclaurin Series Expansion

Insert the derivatives into the series formula:- Constant term (\(x^0\)): \( f(0) = 1 \)- Linear term (\(x^1\)): \( f'(0) = \frac{3}{2} \)- Quadratic term (\(x^2\)): \( \frac{f''(0)}{2!} = \frac{3}{8} \)- Cubic term (\(x^3\)): \( \frac{f'''(0)}{3!} = -\frac{1}{16} \)- Quartic term (\(x^4\)): \( \frac{f^{(4)}(0)}{4!} = \frac{3}{128} \)- Quintic term (\(x^5\)): \( \frac{f^{(5)}(0)}{5!} = -\frac{1}{128} \).Thus, the Maclaurin series up to \( x^5 \) is: \[ f(x) = 1 + \frac{3}{2}x + \frac{3}{8}x^2 - \frac{1}{16}x^3 + \frac{3}{128}x^4 - \frac{1}{128}x^5 \].

Key Concepts

Taylor Series

The Taylor series is a powerful tool in calculus used to represent functions as infinite sums of polynomials based on their derivatives at a specific point. When the series is centered at zero, it is specifically called a Maclaurin series.
This method approximates complex functions using simple polynomial expressions, making it easier to perform calculations or approximate values.

• The general form of a Taylor series for a function \( f(x) \) about a point \( a \) is given by:
  \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \, \cdots \]
• A Maclaurin series is a special case of the Taylor series where \( a = 0 \), which simplifies the formula.

In practice, finding a Taylor series involves calculating the necessary derivatives at point \( a \), then substituting those into the series formula. This expansion helps in approximating the function's value near that point.

Derivatives

Derivatives are a core concept in calculus that describe the rate at which a function is changing at any given point. They are crucial in forming Taylor and Maclaurin series because each term in these series relies on the function's derivatives.
The process of finding derivatives includes differentiating the function repeatedly to get higher-order derivatives.

• The first derivative \( f'(x) \) gives the slope of the tangent line to the function at a point.
• Higher order derivatives, such as the second derivative \( f''(x) \) and third derivative \( f'''(x) \), give insights into the function's concavity and rate of change of the rate of change.

Understanding derivatives allows you to construct polynomial approximations of functions, as each derivative contributes a term to the series expansion.

Polynomial Approximation

Polynomial approximation is the process of estimating complex functions with simpler polynomial expressions. This method is grounded in the use of Taylor or Maclaurin series to express functions in terms of their derivatives.
Approximating a function as a polynomial entails:

• Capturing the function's behavior near a certain point by using derivatives.
• Constructing a polynomial with terms through a specified degree, such as \( x^5 \) in this context.

Each additional term in the polynomial typically improves the approximation's accuracy over a wider range near the center point. For example, the Maclaurin series of \( f(x) = (1+x)^{3/2} \) through \( x^5 \) provides a detailed polynomial representation with terms indicating various degrees of the function's change.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of several variables. While Taylor series can be single-variable, they can also be adapted to multivariable functions to analyze changes over different variables simultaneously.
In a multivariable context:

• The concept of partial derivatives arises, involving differentiation with respect to one variable while keeping others constant.
• Polynomials can approximate functions of several variables, aiding in understanding and computing multi-dimensional changes.

Though our original problem is focused on a single variable Maclaurin series, transitioning to multivariable calculus involves extending these principles to functions like \( f(x, y, z) \) or similar. This often requires calculating and utilizing partial derivatives and constructing more complex polynomial expressions.