Ch. 11 - Power Series

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 11 - Power Series

Problema 11.3.67b

Capitolo 11, Problema 11.3.67b

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
b. The function f(x) = csc x has a Taylor series centered at π/2.

Guida verificata passo dopo passo

Recall that a Taylor series of a function \(f(x)\) centered at \(a\) is given by the infinite sum \(\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n\), where \(f^{(n)}(a)\) denotes the \(n\)-th derivative of \(f\) evaluated at \(x = a\).

To determine if \(f(x) = \csc x\) has a Taylor series centered at \(x = \frac{\pi}{2}\), first check if \(f\) is differentiable at \(x = \frac{\pi}{2}\) and in some open interval around it.

Evaluate \(f\left(\frac{\pi}{2}\right) = \csc\left(\frac{\pi}{2}\right) = 1\), which is finite, so the function is defined at \(x = \frac{\pi}{2}\).

Next, consider the behavior of \(\csc x\) near \(x = \frac{\pi}{2}\). Since \(\sin x\) is nonzero and smooth around \(\frac{\pi}{2}\), \(\csc x = \frac{1}{\sin x}\) is also smooth and infinitely differentiable there.

Therefore, \(f(x) = \csc x\) is infinitely differentiable at \(x = \frac{\pi}{2}\), so it does have a Taylor series centered at \(\frac{\pi}{2}\). The Taylor series converges to \(f(x)\) in some interval around \(\frac{\pi}{2}\) where \(\sin x \neq 0\).

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Taylor Series and Center of Expansion

A Taylor series represents a function as an infinite sum of derivatives at a specific point called the center. The series converges to the function near this point if the function is sufficiently smooth and differentiable there. Understanding the center is crucial to determine if the series exists and converges.

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Taylor Series

Domain and Differentiability of csc(x)

The function csc(x) = 1/sin(x) is undefined where sin(x) = 0, causing vertical asymptotes. For a Taylor series to exist at a point, the function must be defined and infinitely differentiable there. Checking if csc(x) is defined and smooth at π/2 is essential.

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Percorso guidato

5:10

Finding the Domain and Range of a Graph

Radius of Convergence and Singularities

The radius of convergence of a Taylor series is limited by the nearest singularity to the center point. Since csc(x) has singularities where sin(x) = 0, the distance from π/2 to these points affects the convergence. Identifying these singularities helps determine if the Taylor series is valid.

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Radius of Convergence

Pratica correlata

Domanda del libro di testo

{Use of Tech} Bessel functions Bessel functions arise in the study of wave propagation in circular geometries (for example, waves on a circular drum head). They are conveniently defined as power series. One of an infinite family of Bessel functions is

J₀(x) = ∑ₖ₌₀∞ (−1)ᵏ/(2²ᵏ(k!)²) x²ᵏ

b. Find the radius and interval of convergence of the power series for J₀.

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Domanda del libro di testo

Taylor series and interval of convergence

b. Write the power series using summation notation.

f(x) = ln (x − 2), a = 3

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Domanda del libro di testo

Taylor series and interval of convergence

b. Write the power series using summation notation.

f(x)=2/(1−x)³, a=0