Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

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, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\left(1-x^{2}\right)^{2 / 3}\)

Step-by-Step Solution

Answer

The terms are \(1 - \frac{2}{3}x^2 + \frac{1}{9}x^4\).

Step 1: Write the Maclaurin Series Formula

The Maclaurin series for a function \(f(x)\) is given by: \[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots\]However, for functions with complex mathematical operations like roots, we often use known series expansions or manipulate existing series.

Step 2: Use Binomial Series Expansion

We can use the generalized binomial series given by:\[(1 + x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \cdots\]For \((1 - x^2)^{2/3}\), consider \(n = 2/3\) and replace \(x\) with \(-x^2\).

Step 3: Substitute and Expand

Using the binomial series formula:\[(1 - x^2)^{2/3} = 1 + \frac{2}{3}(-x^2) + \frac{\frac{2}{3}(\frac{2}{3}-1)(-x^2)^2}{2!} + \frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)(-x^2)^3}{3!} + \cdots\]Calculate each term up to \(x^5\).

Step 4: Calculate Each Term

1. Constant term: - First term is \(1\).2. First order term: - There is no \(x\) term in the expansion.3. Second order term: - \( \frac{2}{3}(-1)x^2 = -\frac{2}{3}x^2 \)4. Fourth order term: - \( \frac{\frac{2}{3}(-\frac{1}{3})(-1)}{2!} x^4 = \frac{1}{9}x^4 \)5. Higher terms: - Fifth term is zero since powers of \(x\) jump from powers of 2 to powers of 4.Summing these, we obtain the expansion up to \(x^5\):\[1 - \frac{2}{3}x^2 + \frac{1}{9}x^4\]

Step 5: Final Answer

The Maclaurin series for \((1-x^2)^{2/3}\), up to the term \(x^5\), is:\[1 - \frac{2}{3}x^2 + \frac{1}{9}x^4\].This is to say that terms of \(x\) and \(x^3, x^5\) are not present.

Key Concepts

Binomial Series Expansion

The binomial series expansion is a powerful tool for handling expressions of the form \((1+x)^n\) when \(n\) is not necessarily a whole number. This series allows us to expand such expressions into an infinite series, which can be very handy for approximations or when handling complex functions in calculus.

• You start with the expression \((1+x)^n\).
• The series is expanded as follows:\[ 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \cdots \]
• Each term involves a coefficient determined by the formula \(n(n-1)(n-2)...\) reapplying for every power of \(x\).

In our example, for \((1-x^2)^{2/3}\), we substituted \(n\) with \(2/3\) and replaced \(x\) with \(-x^2\). This allowed us to systematically calculate each term, helping us find the solution up to the \(x^5\) term.

Taylor Series

The Taylor series is a method of expanding a function into an infinite sum of terms calculated from the values of its derivatives at a single point. When this point is zero, the series is known as the Maclaurin series, which is a specific case of the Taylor series.This expansion can provide an excellent approximation of functions, especially around the point of expansion.

• The general form of a Taylor series is:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \cdots \]
• When \(a = 0\), it is simplified to the Maclaurin series:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]

Taylor series are fundamental in calculus for approximating non-polynomial functions with polynomials, making complex calculations more manageable.

Series Expansion Techniques

There are numerous techniques used for expanding functions into series, each serving different purposes depending on the complexity and type of function you're dealing with. Some commonly used methods include:

• Maclaurin Series: Useful when functions can be expanded around zero.
• Binomial Series: Specifically adapted for powers like \((1+x)^n\) and versatile with fractional and negative exponents.
• Geometric Series: Perfect for expressions in the form \(a/(1-x)\), handling convergence wisely.

Series expansion techniques are at the core of calculus, enabling us to evaluate limits, optimize functions, and solve differential equations. By breaking down complex expressions into simpler polynomial forms, these methods offer a powerful toolkit for mathematical analysis.