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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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