Question

Use properties of power series. substitution, and factoring to find the first four nonzero terms of the Maclaurin series for the following functions. Give the interval of convergence for the new series (Theorem 11.4 is useful). Use the Maclaurin series . $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \quad \text { for }-1 \leq x \leq 1$$ $$\sqrt{1+x^{2}}$$

Short answer

Question: Find the first four nonzero terms of the Maclaurin series for the function \(\sqrt{1+x^2}\) based on the given Maclaurin series for \(\sqrt{1+x}\), and determine its interval of convergence. Answer: The first four nonzero terms of the Maclaurin series for the function \(\sqrt{1+x^2}\) are: \[1, \frac{x^2}{2}, -\frac{x^4}{8}, \frac{x^6}{16}\] The interval of convergence for this series is \(-1 \leq x \leq 1\).

Step-by-step solution

Substitute \(x^2\) into the given Maclaurin series

Replace \(x\) with \(x^2\) in the given series for \(\sqrt{1+x}\). $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{(x^2)^2}{8}+\frac{(x^2)^3}{16}-\cdots$$

Simplify the series

Simplify each term by distributing the exponents and combining any constants. $$\sqrt{1+x^2}=1+\frac{x^2}{2}-\frac{x^4}{8}+\frac{x^6}{16}-\cdots$$

Find the first four nonzero terms

The first four nonzero terms are clearly visible in the simplified series: $$1, \frac{x^2}{2}, -\frac{x^4}{8}, \frac{x^6}{16}$$

Determine the interval of convergence

The original series for \(\sqrt{1+x}\) converges for \(-1 \leq x \leq 1\). Since we replaced \(x\) with \(x^2\), our new interval of convergence must satisfy \(-1 \leq x^2 \leq 1\). Let's solve for \(x\) using these inequalities: - For \(-1 \leq x^2\): All real numbers satisfy this inequality since the square of any real number is positive or zero. - For \(x^2 \leq 1\): - We find the roots: \(x^2 = 1 \implies x = \pm 1\). - Since \(x^2\) is a parabolic function with a vertex at the origin and that opens upwards, the inequality \(x^2 \leq 1\) will be true between the roots of the equation \(x^2 = 1\). - Therefore, the solution to \(x^2 \leq 1\) is \(-1 \leq x \leq 1\). Thus, the interval of convergence for the Maclaurin series of the function \(\sqrt{1+x^2}\) is also \(-1 \leq x \leq 1\).

Key concepts

Power Series

A power series is a mathematical series that represents a function as an infinite sum of terms. Each term is a power of a variable, typically denoted by \(x\), which is raised to increasing integer exponents. The general form of a power series is given by:

• \(\sum_{n=0}^{\infty} a_n (x-c)^n\)

This series expands around a point \(c\), known as the center of the series. In a Maclaurin series, specifically, the center is \(c = 0\). Consequently, each term in the series has the simple form \(a_n x^n\). The series makes it easier to work with complex functions, especially when calculating approximate values or derivatives.
Power series is quite useful in calculus, especially for functions that are not easily differentiated or integrated using other methods. It allows a function to be expressed in an easily manipulated infinite sum, which makes understanding and computing a function's behavior simpler.

Interval of Convergence

The interval of convergence for a power series is the set of \(x\)-values for which the series converges to a finite sum. Determining this interval is crucial because a power series only provides a valid representation of a function within this interval.
For the Maclaurin series of \(\sqrt{1+x}\), the interval of convergence is \(-1 \leq x \leq 1\). This means within this range, substituting values of \(x\) into the series expansion yields correct calculations of \(\sqrt{1+x}\).
When substituting \(x^2\) for \(x\) in the series, we derive a new interval of convergence by solving the inequality \(-1 \leq x^2 \leq 1\). This results in the interval \(-1 \leq x \leq 1\) because \(x^2\) itself can take values from 0 to 1 when \(x\) is within this interval. This shows us that the new series accurately represents \(\sqrt{1+x^2}\) for \(-1 \leq x \leq 1\). Understanding the concept of the interval of convergence helps identify where a power series approximation remains valid.

Substitution

Substitution in power series involves replacing the variable \(x\) with another expression to derive series expansion for different functions. It is a powerful tool when dealing with transformations of existing series.

• In the problem, we replace \(x\) by \(x^2\) in the given series \(\sqrt{1+x}\), simplifying our process of finding the series for \(\sqrt{1+x^2}\).
• This substitution allowed us to directly modify the terms of the given series with minimal calculation.

Substitution within series manipulation provides efficiency when transforming one known series into another. This is particularly effective when dealing with functions that are composition of familiar forms. Applying substitution requires careful attention to ensure that the new series remains valid within its interval of convergence, reflecting how the change affects the variable across all terms in the sequence.

Factoring

Factoring in the context of power series often refers to simplifying or combining expressions after substitution or transformation, allowing an easier representation of complex terms.
In this exercise, after substituting \(x^2\) into the series for \(\sqrt{1+x}\), each term like \(\frac{(x^2)^2}{8}\) was simplified by properly distributing the exponent, resulting in \(\frac{x^4}{8}\).

• Factoring ensures that the new terms are expressed in their simplest form.
• This clarity is useful for both presenting the series clearly and performing calculations accurately later on.

Factoring is often paired with other manipulations such as substitution or differentiation/integration of series terms. It helps in managing expressions by reducing unnecessary complexity in calculations, contributing to a broader understanding of the behavior of the function represented in its power series form.