Problem 46 - 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. $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16}-\cdots, \text { for }-1 < x \leq 1$$ $$\sqrt{4+x}$$

Step-by-Step Solution

Answer

The first four nonzero terms of the Taylor series centered at 0 for the function $$\sqrt{4+x}$$ are: $$\sqrt{4+x} = 1+\frac{3}{2} x-\frac{1}{8} x^2+\frac{1}{16}x^3+\cdots$$ The interval of convergence for the series is $$(-5, -3]$$.

Step 1: Substitution

Substitute $$4+x$$ in place of $$x$$ in the Taylor series for $$\sqrt{1+x}$$ to get a series for $$\sqrt{4+x}$$. $$\sqrt{4+(4+x)} = 1+\frac{4+x}{2}-\frac{(4+x)^{2}}{8}+\frac{(4+x)^{3}}{16}-\cdots$$

Step 2: Simplification

Simplify the series and adjust the terms to obtain the first four nonzero terms: $$\sqrt{4+x} = 1+2+\frac{x}{2}-\frac{(4+x)^{2}}{8}+\frac{(4+x)^{3}}{16}-\cdots$$ $$\sqrt{4+x} = 3+\frac{x}{2}-\frac{16+8x+x^2}{8}+\frac{64+48x+12x^2+x^3}{16}-\cdots$$

Step 3: Combine Similar Terms

Combine similar terms in the expression to obtain a simplified power series: $$\sqrt{4+x} = 3+\frac{x}{2}-2-\frac{x}{2}-\frac{x^{2}}{8}+\frac{4}{2}+\frac{3}{2}x+\frac{3}{4}x^{2}+\frac{1}{16}x^{3}-\cdots$$

Step 4: Simplify the Series

Collect similar terms in the power series and get the first four nonzero terms of the Taylor series: $$\sqrt{4+x} = 1+\frac{3}{2} x-\frac{1}{8} x^2+\frac{1}{16}x^3+\cdots$$

Step 5: Interval of Convergence

Since the interval of convergence for the given Taylor series of $$\sqrt{1+x}$$ is $$-1 < x \leq 1$$, we will find the corresponding interval for $$4 + x$$: $$-1 < 4 + x \leq 1$$ $$-5 < x \leq -3$$ So, the interval of convergence for the Taylor series of $$\sqrt{4+x}$$ is $$(-5, -3]$$. In conclusion, the first four nonzero terms of the Taylor series centered at 0 for the function $$\sqrt{4+x}$$ are given by: $$\sqrt{4+x} = 1+\frac{3}{2} x-\frac{1}{8} x^2+\frac{1}{16}x^3+\cdots$$ And the interval of convergence for the series is $$(-5, -3]$$.

Key Concepts

Power Series

A power series is a special type of mathematical series where the terms are powers of the variable, usually centered at a particular point. In our case, the series is centered at 0. This allows us to represent functions as a series of polynomials, which can make complex functions easier to work with or approximate.

For example, consider the function \( \sqrt{1 + x} \), its Taylor series expansion at 0 is expressed as a power series:

• \( \sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \cdots \)

Power series are particularly valuable in calculus as they help approximate functions near the center point, leading to simpler calculations in complex analysis and engineering.

When using power series, each term typically involves a coefficient, a power of \( x \), and a factorial in the denominator, making it easy to add or remove terms to adjust for the degree of accuracy needed.

Interval of Convergence

An interval of convergence specifies the range of values for which a power series converges to a finite value. In simpler terms, it's the set of all \( x \) values where the series accurately represents the function.

For the original Taylor series of \( \sqrt{1+x} \), the interval of convergence is \((-1, 1]\). However, when transformation involves substitution, like replacing \(x\) with \(4 + x\), the interval changes. By analyzing the series, we map the original convergence interval through substitution, calculating this new range step-by-step:

• Initially, \(-1 < x \leq 1\).
• Apply substitution: \(-1 < 4 + x \leq 1\).
• Solve for \(x\): \(-5 < x \leq -3\).

Thus, the interval of convergence for the transformed series is \((-5, -3]\). Understanding this concept is crucial for knowing which \(x\) values the series representation holds true for.

Series Expansion

Series expansion refers to rewriting a function as an infinite sum of terms in a series. This can simplify functions by breaking them into more manageable parts, expressed typically as a power series for ease of calculation near a specific point.

The process of series expansion can generate approximations for functions like \(\sqrt{1+x}\), leading to insights or easier analysis for complex scenarios:

• Starts with a known series: \(\sqrt{1+x}\) with its power series format.
• Substitute and expand the series based on transformations, like \(\sqrt{4+x}\).
• Simplify and combine similar terms, as shown in the solution steps, to achieve the first non-zero terms.

Series expansions provide a methodical approach for smoothing out complex function behaviors into comprehensible polynomial-like expressions, aiding in analytical calculations and revealing patterns or trends otherwise hidden.

Substitution in Series

Substitution in series involves replacing variables within a power series to adapt or modify the series for new functions. This process helps in deriving series for complex or related functions based on known expansions.

In our solution, the task involved substituting \(4 + x\) into the Taylor series of \(\sqrt{1+x}\):

• Initially, the base series is \( \sqrt{1+x} \), with known terms.
• The substitution \(x = 4+x\) transforms the series into one that approximates \(\sqrt{4+x}\).
• Simplification follows, adjusting terms to conserve the necessary powers of \(x\) and coefficients.

Substitution is crucial when expanding functions or adapting series for similar functions not directly covered by initial expansions, illustrating the fluidity and adaptability of power series in broader mathematical contexts.