Problem 50 - Calculus Early Transcendentals

Question

Use properties of power series, substitution, and factoring to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Give the interval of convergence for the new series. Use the Taylor series. $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \text { for }-1 < x \leq 1$$ $$\sqrt{4-16 x^{2}}$$

Step-by-Step Solution

Answer

Answer: The first four nonzero terms of the Taylor series for the function \(\sqrt{4-16x^2}\) are \(2-4x^2+16x^4-\frac{256x^6}{8}+\cdots\). The interval of convergence for this series is \(-\frac{1}{2} < x \leq \frac{1}{2}\).

Step 1: Determine a relationship between the given functions

Since we want to find the Taylor series for the function \(\sqrt{4-16x^2}\) and we are given the series for the function \(\sqrt{1+x}\), we need to first determine a relationship between the two functions. To do this, we can rewrite the function we want to find the Taylor series for in terms of the given function: $$\sqrt{4-16x^2} = \sqrt{4(1-4x^2)} = 2\sqrt{1-4x^2}$$

Step 2: Make a substitution into the given Taylor series

Now that we have expressed the new function in terms of the given function, we can make a substitution into the Taylor series for \(\sqrt{1+x}\): $$2\sqrt{1-4x^2} = 2\left(1+\frac{-4x^2}{2}-\frac{(-4x^2)^2}{8}+\frac{(-4x^2)^3}{16}+ \cdots \right)$$

Step 3: Simplify the expression

Now we can simplify the expression by multiplying the terms inside the parenthesis by 2: $$2\sqrt{1-4x^2} = 2-4x^2+\frac{(-4x^2)^2}{4}-\frac{(-4x^2)^3}{8}+ \cdots$$

Step 4: Find the first four nonzero terms

The first four nonzero terms of the Taylor series for the function \(\sqrt{4-16x^2}\) are: $$2\sqrt{4-16x^2} = 2-4x^2+16x^4-\frac{256x^6}{8}+ \cdots$$

Step 5: Determine the interval of convergence for the new series

To find the interval of convergence for the new series, we need to determine the range of x values where the new series converges. Since the original Taylor series converges for \(-1 < x \leq 1\), we need to find the interval for the 4x² term: $$-1 < -4x^2 \leq 1$$ Dividing by -4: $$\frac{1}{4} \geq x^2 > -\frac{1}{4}$$ Taking the square root of both sides: $$\frac{1}{2} \geq x > -\frac{1}{2}$$ Thus, the interval of convergence for the new Taylor series is \(-\frac{1}{2} < x \leq \frac{1}{2}\).

Key Concepts

Power Series

A power series is a way of representing a function as an infinite sum of terms calculated from the values of its derivatives at a single point, usually 0. This representation makes it easier to analyze and manipulate functions, especially for calculus applications. A power series can be expressed in the form:

• \[ \sum_{n=0}^{\infty} c_n (x-a)^n \]

Here, \(c_n\) are coefficients and \(a\) is the center of the series, which is often 0 for simplicity.
These series are versatile since they provide a framework to approximate complex functions with polynomial-like precision.
By focusing on the series’ first few terms, students can effectively approximate functions and gain insight into their behavior without complex calculations.
Power series are foundational in studying and applying Taylor series, where functions are expressed as a sum of derivatively-calculated terms centered around a specific point.

Interval of Convergence

The interval of convergence is the set of \(x\)-values for which a power series converges, or sums to a well-defined, finite value. This is a crucial concept as it defines the domain over which the series accurately represents the function.
To find the interval of convergence, we apply tests like the ratio test, checking series' behavior as \(n\) approaches infinity.

• If the series converges at a point inside the interval, it typically converges for all points until the boundary, where we need to test separately.
• Understanding this interval ensures we use series expansions within valid domains only, preserving accuracy.

In the exercise, the interval of convergence for \( \sqrt{4-16x^2} \) results in \(-\frac{1}{2} < x \leq \frac{1}{2}\), indicating that the series accurately represents the function within this range.

Substitution Method

The substitution method in power series involves replacing terms in a given series with equivalent expressions to suit a new context. This process allows us to adapt standard series for different functions without starting from scratch.
In the original exercise, the known Taylor series for \(\sqrt{1+x}\) was adapted into the series for \(\sqrt{4-16x^2}\) by substituting \(-4x^2\) for \(x\).

• Substitution simplifies complex manipulations, offering a direct path to tailor existing series for new functions.
• This method is not only efficient but also reinforces the connections between related mathematical expressions.

By applying substitution, students learn how to leverage calculus tools effectively and understand the underlying relationships in functions.

Factoring in Calculus

Factoring in calculus is a technique used to simplify expressions or functions to make them easier to work with, especially when identifying solutions or further transformations. It often reveals simpler components that might involve powers, roots, or trigonometric identities.
In the exercise, the expression \(\sqrt{4-16x^2}\) is factored as \(2\sqrt{1-4x^2}\), allowing the direct application of a known Taylor series.

• This technique reduces complexity, facilitating substitution and expansion of the series for easier calculations.
• Factoring often uncovers symmetries or other properties within functions that simplify subsequent analysis.

Using factoring alongside techniques like substitution and series expansions is a hallmark of effective calculus problem-solving, aiding in clearer, more manageable transformations.