Problem 49 - 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. $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \text { for }-1 < x \leq 1$$ $$\sqrt{a^{2}+x^{2}}, a > 0$$

Step-by-Step Solution

Answer

The first four non-zero terms of the Taylor series for the function \(\sqrt{a^2 + x^2}\) are: $$\sqrt{a^2 + x^2} = 1 + \frac{a^2 - 1}{2} + \frac{x^2}{2} - \frac{a^4 - 2a^2}{8} - \frac{x^4}{8} + \frac{a^2x^2}{4}$$ The interval of convergence for this series is \(0 < a^2 + x^2 \leq 2\).

Step 1: Substitute and rearrange the given Taylor series

We are given the Taylor series for \(\sqrt{1+x}\) as: $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \text { for }-1 < x \leq 1$$ We need to find the Taylor series for \(\sqrt{a^2 + x^2}\), where \(a > 0\). Let's substitute \(a^2 + x^2 - 1\) for \(x\) in the given Taylor series: $$\sqrt{1+(a^2 + x^2 -1)}=1+\frac{a^2 + x^2 - 1}{2}-\frac{(a^2 + x^2 - 1)^{2}}{8}+\frac{(a^2 + x^2 - 1)^{3}}{16}-\cdots$$ Now, notice that \(\sqrt{1+(a^2 + x^2 - 1)} = \sqrt{a^2 + x^2}\), which is our desired function.

Step 2: Expand and find the Taylor series

Now, we need to expand the terms and find the Taylor series for our function, considering only the non-zero terms up to the first four: $$\sqrt{a^2 + x^2} = 1+\frac{a^2 + x^2 - 1}{2}-\frac{(a^2 + x^2 - 1)^{2}}{8}+\frac{(a^2 + x^2 - 1)^{3}}{16} - \cdots$$ Expanding and keeping only the first four non-zero terms: $$\sqrt{a^2 + x^2} = 1 + \frac{a^2 - 1}{2} + \frac{x^2}{2} - \frac{(a^4 - 2a^2 + 1 + 2a^2x^2 + x^4 - 2x^2)}{8}+ \cdots$$ Combine like terms: $$\sqrt{a^2 + x^2} = 1 + \frac{a^2 - 1}{2} + \frac{x^2}{2} - \frac{a^4 - 2a^2}{8} - \frac{x^4}{8} + \frac{a^2x^2}{4} + \cdots$$ So the Taylor series for our function, up to the first four non-zero terms, is: $$\sqrt{a^2 + x^2} = 1 + \frac{a^2 - 1}{2} + \frac{x^2}{2} - \frac{a^4 - 2a^2}{8} - \frac{x^4}{8} + \frac{a^2x^2}{4}$$

Step 3: Find the interval of convergence

By substituting \((a^2 + x^2 - 1)\) for \(x\) in the given Taylor series for \(\sqrt{1+x}\), we have not changed the interval of convergence for the series. The interval of convergence for \(\sqrt{1+x}\) is \(-1 < x \leq 1\), so the interval of convergence for the Taylor series for our function \(\sqrt{a^2 + x^2}\) will be the same: $$-1 < a^2 + x^2 - 1 \leq 1$$ $$0 < a^2 + x^2 \leq 2$$ In summary, the first four non-zero terms of the Taylor series for the function \(\sqrt{a^2 + x^2}\) are as follows: $$\sqrt{a^2 + x^2} = 1 + \frac{a^2 - 1}{2} + \frac{x^2}{2} - \frac{a^4 - 2a^2}{8} - \frac{x^4}{8} + \frac{a^2x^2}{4}$$ And the interval of convergence for this series is \(0 < a^2 + x^2 \leq 2\).

Key Concepts

Power series

A power series is a way to represent a function as an infinite sum of terms. Each term in a power series is a multiple of some variable raised to a power. Power series are incredibly useful, as they allow us to express complex functions in a simpler, polynomial-like form.
For the Taylor series discussed in the exercise, the function is expressed through a power series with terms involving powers of \(x\). The series given for \(\sqrt{1+x}\) is a perfect example, where the function is expanded and each term involves increasing powers of \(x\).
Power series are especially important because they can be manipulated like polynomials, which makes them very handy in calculus and analysis. They can be added, subtracted, differentiated, or even integrated term by term.

• This manipulation provides a flexible tool for solving problems in mathematical analysis.
• In the given exercise, the power series allows us to approximate the square root function in a region around 0.

Interval of convergence

The interval of convergence is crucial when working with power series. It tells us the range of values for which the series accurately represents a function. Outside of this interval, the series might not converge or give correct results.
In the exercise, the initial series for \(\sqrt{1+x}\) converges for \(-1 < x \leq 1\). When we substitute to find the series for \(\sqrt{a^2 + x^2}\), we adjust the interval accordingly: \(0 < a^2 + x^2 \leq 2\).
Understanding convergence is essential, as it ensures the approximations made by the series are valid.

• The concept helps prevent errors when using the series outside the valid range.
• Checking convergence helps evaluate whether the mathematical manipulation of the series remains accurate.

Substitution technique

The substitution technique is a method used to derive new power series from existing ones. By substituting a new expression into a known series, we can easily find a series for complex functions without having to start the expansion from scratch.
In this exercise, we take the known Taylor series of \(\sqrt{1+x}\) and substitute \(a^2 + x^2 - 1\) in place of \(x\). This transformation allows us to find the Taylor series for \(\sqrt{a^2 + x^2}\).
Substitution helps simplify the process of finding series expansions, especially when dealing with complex expressions.

• It saves time and effort, reducing the amount of direct computation needed for complex series.
• This method lets us leverage existing solutions to solve similar problems quickly.

Factoring

Factoring is the process of breaking down an expression into simpler components, usually to reveal underlying structure or to simplify calculations. It plays an important role in simplifying terms within power series expansions.
In the provided solution, factoring is involved when rearranging and combining terms. For instance, expressions are reorganized to group like terms together, making the series easier to analyze and understand.
Factoring helps in reducing the complexity of terms, which can significantly simplify series computations.

• It allows series components to be written elegantly, making subsequent calculations and manipulations more straightforward.
• This technique helps in identifying patterns or symmetries within series expansions.