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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
All textbooksBriggs 3rd EditionCh. 11 - Power SeriesProblem 11.R.2
Chapter 11, Problem 11.R.2

Taylor polynomials Find the nth-order Taylor polynomial for the following functions centered at the given point a.
ƒ(x) = sin 2x, n = 3, a = 0

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1
Identify the function and the point of expansion: here, the function is \(f(x) = \sin(2x)\) and the center is \(a = 0\).
Recall the formula for the nth-order Taylor polynomial centered at \(a\): \[T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x - a)^k,\] where \(f^{(k)}(a)\) is the \(k\)th derivative of \(f\) evaluated at \(x = a\).
Compute the derivatives of \(f(x) = \sin(2x)\) up to the 3rd order: - \(f(x) = \sin(2x)\) - \(f'(x) = 2 \cos(2x)\) - \(f''(x) = -4 \sin(2x)\) - \(f^{(3)}(x) = -8 \cos(2x)\)
Evaluate each derivative at \(x = 0\): - \(f(0) = \sin(0) = 0\) - \(f'(0) = 2 \cos(0) = 2\) - \(f''(0) = -4 \sin(0) = 0\) - \(f^{(3)}(0) = -8 \cos(0) = -8\)
Construct the 3rd-order Taylor polynomial using the formula: \[T_3(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3,\] and substitute the values found in the previous step.

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Key Concepts

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

Taylor Polynomial

A Taylor polynomial approximates a function near a point a by using a finite sum of derivatives of the function at a. The nth-order Taylor polynomial includes terms up to the nth derivative, providing a polynomial that closely matches the function's behavior near a.
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Taylor Polynomials

Derivatives of Trigonometric Functions

To construct the Taylor polynomial for ƒ(x) = sin(2x), you need to compute derivatives of sin(2x) at the point a. Knowing the derivatives of sine and cosine functions, and applying the chain rule for the inner function 2x, is essential for finding these values.
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Derivatives of Other Inverse Trigonometric Functions

Centering at a Point (a = 0)

Centering the Taylor polynomial at a = 0 means the polynomial is expanded around x = 0, also called a Maclaurin polynomial. This simplifies calculations since derivatives are evaluated at zero, and powers of (x - 0) become powers of x.
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Related Practice
Textbook Question

Approximating real numbers Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. There is more than one way to choose the center of the series.


sin 20°

Textbook Question

Write out the first three terms of the Maclaurin series for the following functions.

ƒ(x) = (1 + x)^(1/3)"

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Textbook Question

Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.

ƒ(x) = 1/(4 + x²), a = 0

Textbook Question

Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.

ƒ(x) = cos x, a = π/2

1
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Textbook Question

Convergence Write the remainder term Rₙ(x) for the Taylor series for the following functions centered at the given point a. Then show that lim ₙ → ∞ |Rₙ(x)| = 0, for all x in the given interval.

ƒ(x) = sinh x + cosh x, a = 0, - ∞ < x < ∞

Textbook Question

Approximating ln 2 Consider the following three ways to approximate

ln 2.

a. Use the Taylor series for ln (1 + x) centered at 0 and evaluate it at x = 1 (convergence was asserted in Table 11.5). Write the resulting infinite series.