Problem 41 - Calculus: Early Transcendentals

Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. \( f(x) = \frac {x}{\sqrt {4 + x^2}} \)

Step-by-Step Solution

Answer

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

Step 1: Recognize the Function

The function given is \( f(x) = \frac{x}{\sqrt{4 + x^2}} \). To find the Maclaurin series, we need to manipulate this into a form that's easier to expand as a series.

Step 2: Rewrite the Function

Recognize that \( \frac{1}{\sqrt{4 + x^2}} \) can be rewritten using the binomial series: \( (4 + x^2)^{-1/2} \). This is a key step in recognizing how the Maclaurin series for a binomial expression can be applied here.

Step 3: Binomial Series Expansion

The binomial series for \( (1 + u)^n \) is given by \( 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \ldots \). For \( (4 + x^2)^{-1/2} \), rewrite it as \( 4^{-1/2}(1 + \frac{x^2}{4})^{-1/2} \) to match the binomial form.

Step 4: Simplify for Maclaurin Series

Use the binomial series on \( (1 + \frac{x^2}{4})^{-1/2} = 1 - \frac{1}{8}x^2 + \frac{3}{128}x^4 - \ldots \), simplify using \( \frac{1}{2} \) is \( -1/2 \). Recognize that the function can be simplified to an involved series.

Step 5: Multiply Simplified Series by x

Multiply the entire series by \( x \) to find \( f(x) \), so that each \( x^n \) term becomes \( x^{n+1} \). Therefore, the series becomes \( x - \frac{1}{8}x^3 + \frac{3}{128}x^5 - \ldots \).

Step 6: Combine and Write the Series

The final Maclaurin series for \( f(x) = \frac{x}{\sqrt{4 + x^2}} \) is \[ x - \frac{1}{8}x^3 + \frac{3}{128}x^5 - \ldots \]. Recognizing terms contribute to a rapid convergence with further higher degree terms diminishing quickly.

Key Concepts

Binomial Series

The binomial series is a powerful tool in calculus, especially when dealing with functions that involve exponents and roots. The general form of a binomial series expansion is for expressions of the type \((1 + u)^n\), where \(n\) can be any real number. This is expressed as:

• \(1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots\)

In the context of a Maclaurin series, which is a type of Taylor series centered at zero, the binomial series allows us to handle complex expressions which include terms that can "become zero" in the polynomial series form. For example, transforming a function like \((4 + x^2)^{-1/2}\) requires us to rewrite it using the binomial theorem for ease of expansion. Here, it is simplified to \(4^{-1/2}(1 + \frac{x^2}{4})^{-1/2}\) by extracting 4 as a base term. This manipulation sets the stage for obtaining a series expansion, which is essential for calculations where exact results may be hard to compute.

Function Expansion

Function expansion is a mathematical technique used for breaking down complex functions into simpler terms or series.By expressing a function as a series, it becomes easier to approximate, differentiate, or integrate.In particular, the Maclaurin series expansion is an expression of a function as an infinite sum of terms calculated from the derivatives of that function at a single point, often zero.In our example, the function \(f(x) = \frac{x}{\sqrt{4 + x^2}}\) is challenging to handle directly.To proceed, we use the concept of series expansion, starting by rewriting the complicated square root form into a fraction with a binomial expression.

• The function is outlined as \(\frac{x}{4^{-1/2}(1 + \frac{x^2}{4})^{-1/2}}\).
• Next, this is expanded using the known binomial series results to signal the Maclaurin series expansion.
• Each term in the expansion is multiplied by \(x\) to adjust the powers accordingly: \(x - \frac{1}{8}x^3 + \frac{3}{128}x^5 - \ldots\).

This approach helps in situations where direct computation may be nontrivial or when specific details about the function's behavior around a point are needed.

Series Convergence

Series convergence is an important concept when dealing with infinite series, such as the Maclaurin series. Convergence refers to whether the sum of an infinite series approaches a finite limit. When working with a series expansion, it is crucial to establish that as more terms are added, the series becomes a close approximation of the function.In the given example, the Maclaurin series for \(f(x) = \frac{x}{\sqrt{4 + x^2}}\) results in a series: \[x - \frac{1}{8}x^3 + \frac{3}{128}x^5 - \ldots\].Several points about series convergence include:

• Each additional term in the series provides a better approximation of \(f(x)\).
• For the series to converge, coefficients of higher terms should decrease rapidly, which ensures the summed series approaches a finite value.
• Convergence depends on the chosen value of \(x\); often, the series converges best within a particular interval.

Recognizing convergence ensures that our series can reliably represent the function within a certain domain, making the expansion practical for real-world applications.