Textbook Question
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)
97
views
Ch. 11 - Power Series
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 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)
Verified step by step guidance1
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 \( |x| \leq 1 \).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
Recommended video:
05:58
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.
Recommended video:
05:58Intro 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.
Recommended video:
08:44Interval of Convergence
Related Practice
Textbook Question
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.
f(3x) = ln (1 − 3x)
83
views
Textbook Question
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.
y′(t) − 3y = 10, y(0) = 2
43
views
Textbook Question
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.
69
views
Textbook Question
{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]
87
views
Textbook Question
How would you approximate e⁻⁰ᐧ⁶ using the Taylor series for eˣ?
86
views