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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 11, Problem 11.2.78

Inverse sine Given the power series
1/√(1 − x²) = 1 + (1/2)x² + (1 ⋅ 3)/(2 ⋅ 4) x⁴ + (1 ⋅ 3 ⋅ 5)/(2 ⋅ 4 ⋅ 6) x⁶ +⋯
for −1<x<1, find the power series for f(x) = sin ⁻¹ x centered at 0.

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1
Recall that the function \( f(x) = \sin^{-1} x \) is the integral of its derivative. Since \( \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} \), we can use the given power series for \( \frac{1}{\sqrt{1 - x^2}} \) to find the series for \( \sin^{-1} x \).
Write the given power series explicitly as: \[ \frac{1}{\sqrt{1 - x^2}} = 1 + \frac{1}{2} x^2 + \frac{1 \cdot 3}{2 \cdot 4} x^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^6 + \cdots \]
Integrate the series term-by-term with respect to \( x \) from 0 to \( x \) to find \( \sin^{-1} x \): \[ \sin^{-1} x = \int_0^x \frac{1}{\sqrt{1 - t^2}} dt = \int_0^x \left(1 + \frac{1}{2} t^2 + \frac{1 \cdot 3}{2 \cdot 4} t^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} t^6 + \cdots \right) dt \]
Integrate each term individually: - The integral of 1 with respect to \( t \) is \( t \). - The integral of \( t^2 \) is \( \frac{t^3}{3} \). - The integral of \( t^4 \) is \( \frac{t^5}{5} \), and so on. So the series becomes: \[ \sin^{-1} x = x + \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \cdot \frac{x^5}{5} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \frac{x^7}{7} + \cdots \]
Simplify the coefficients to write the power series for \( \sin^{-1} x \) centered at 0. This series converges for \( -1 < x < 1 \) and represents the inverse sine function as a power series.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power Series Representation

A power series expresses a function as an infinite sum of terms involving powers of the variable, typically centered at a point (here, 0). It allows complex functions to be approximated by polynomials within a certain interval of convergence, facilitating analysis and computation.
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Intro to Power Series

Relationship Between a Function and Its Derivative

The derivative of the inverse sine function, sin⁻¹(x), is 1/√(1 − x²). Knowing the power series for the derivative enables us to find the power series for the original function by integrating term-by-term within the interval of convergence.
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Derivatives of Other Trig Functions

Term-by-Term Integration of Power Series

Integrating a power series term-by-term involves integrating each term individually, which is valid within the radius of convergence. This process is used to find the power series of a function when the series for its derivative is known, as in finding sin⁻¹(x) from 1/√(1 − x²).
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Intro to Power Series
Related Practice
Textbook Question

{Use of Tech} Newton's derivation of the sine and arcsine series Newton discovered the binomial series and then used it ingeniously to obtain many more results. Here is a case in point.

a. Referring to the figure, show that x = sin s or s = sin ⁻¹ x.

b. The area of a circular sector of radius r subtended by an angle θ is 1/2r²θ. Show that the area of the circular sector APE is s/2, which implies that

s = 2 ∫₀ˣ √(1 − t²) dt − x √(1 −x²)

c. Use the binomial series for f(x) = √(1 − x²) to obtain the first few terms of the Taylor series for s=sin ⁻¹ x.

d. Newton next inverted the series in part (c) to obtain the Taylor series for x=sin s. He did this by assuming sin s = ∑ aₖ sᵏ and solving x = sin(sin ⁻¹ x) for the coefficients aₖ. Find the first few terms of the Taylor series for sin s using this idea (a computer algebra system might be helpful as well).

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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.


tan x ≈ x on [−π/6, π/6]

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Textbook Question

Exponential function In Section 11.3, we show that the power series for the exponential function centered at 0 is


eˣ = ∑ₖ₌₀∞ (xᵏ)/k!, for −∞ < x < ∞


Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series.


f(x) = e⁻³ˣ

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Radius and interval of convergence Determine the radius and interval of convergence of the following power series.


∑ₖ₌₁∞ (3x + 2)ᵏ/k

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Limits Evaluate the following limits using Taylor series.

lim ₓ→₁ (x 1)/(ln x)

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{Use of Tech} Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.


f(x) = ∜x with a=16; approximate ∜13.

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