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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 11 - Power SeriesProblema 11.3.53
Capitolo 11, Problema 11.3.53

Working with binomial series 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


√(1 + x) = 1 + x/2 − x²/8 + x³/16 − ⋯, −1 ≤ x ≤ 1.


√(9 − 9x)

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Start with the given Maclaurin series for \( \sqrt{1 + x} \): \[ \sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{8} + \frac{x^{3}}{16} - \cdots, \quad -1 \leq x \leq 1. \]
Rewrite the function \( \sqrt{9 - 9x} \) by factoring out 9 inside the square root: \[ \sqrt{9 - 9x} = \sqrt{9(1 - x)} = 3 \sqrt{1 - x}. \]
Express \( \sqrt{1 - x} \) in terms of the known series for \( \sqrt{1 + x} \) by substituting \( x \) with \( -x \): \[ \sqrt{1 - x} = 1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \cdots. \] (Note the alternating signs come from substituting \( x \) by \( -x \) in the original series.)
Multiply the series for \( \sqrt{1 - x} \) by 3 to get the series for \( \sqrt{9 - 9x} \): \[ 3 \left(1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \cdots \right) = 3 - \frac{3x}{2} - \frac{3x^{2}}{8} - \frac{3x^{3}}{16} - \cdots. \]
Determine the interval of convergence by considering the substitution \( x \to -x \) in the original interval \( -1 \leq x \leq 1 \). Since the original series converges for \( |x| \leq 1 \), the new series converges for \( | -x | \leq 1 \), which simplifies to \( |x| \leq 1 \).

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Maclaurin Series and Power Series Expansion

A Maclaurin series is a special case of a Taylor series expanded at x = 0, representing a function as an infinite sum of powers of x. It allows approximation of functions near zero by polynomials. Understanding how to write and manipulate these series is essential for finding terms of the expansion for given functions.
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Intro to Power Series

Substitution and Factoring in Power Series

Substitution involves replacing the variable in a known power series with an expression to find a new series. Factoring helps rewrite functions into forms compatible with known expansions. These techniques enable transforming the given function into a form where the Maclaurin series of √(1 + x) can be applied.
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Intro to Power Series

Interval of Convergence and Theorem 11.4

The interval of convergence defines where a power series converges to the function it represents. Theorem 11.4 (usually about radius and interval of convergence) helps determine this interval after substitution. Knowing how to find and interpret this interval ensures the series accurately represents the function within a specific domain.
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Interval of Convergence
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Functions to power series Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series.

f(x) = ln √(1 − x²)

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Combining power series Use the power series representation


f(x ) =ln (1 − x) = −∑ₖ₌₁∞ xᵏ/k, for −1 ≤ x < 1,


to find the power series for the following functions (centered at 0). Give the interval of convergence of the new series.


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Differential equations


a. Find a power series for the solution of the following differential equations, subject to the given initial condition

b. Identify the function represented by the power series.


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The first three Taylor polynomials for f(x)=√(1+x) centered at 0 are p₀ = 1, p₁ = 1+x/2, and p₂ = 1 + x/2 − x²/8. Find three approximations to √1.1.

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{Use of Tech} Maximum error Use the remainder term to find a bound on the error in the following approximations on the given interval. Error bounds are not unique.


sin x ≈ x − x³/6 on [π/4, π/4]

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How would you approximate e⁻⁰ᐧ⁶ using the Taylor series for eˣ?

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