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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 11.3.27a

Taylor series


a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.


f(x)=sin x, a = π/2

Verified step by step guidance
1
Recall the definition of the Taylor series of a function \(f(x)\) centered at \(a\): \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
Identify the function and center: here, \(f(x) = \sin x\) and \(a = \frac{\pi}{2}\). We will need to compute the derivatives of \(\sin x\) and evaluate them at \(x = \frac{\pi}{2}\).
Calculate the first four derivatives of \(f(x) = \sin x\): - \(f(x) = \sin x\) - \(f'(x) = \cos x\) - \(f''(x) = -\sin x\) - \(f^{(3)}(x) = -\cos x\) - \(f^{(4)}(x) = \sin x\) (which repeats the cycle).
Evaluate each derivative at \(x = \frac{\pi}{2}\): - \(f\left(\frac{\pi}{2}\right) = \sin \frac{\pi}{2} = 1\) - \(f'\left(\frac{\pi}{2}\right) = \cos \frac{\pi}{2} = 0\) - \(f''\left(\frac{\pi}{2}\right) = -\sin \frac{\pi}{2} = -1\) - \(f^{(3)}\left(\frac{\pi}{2}\right) = -\cos \frac{\pi}{2} = 0\)
Write the first four nonzero terms of the Taylor series using the formula: \[f(x) \approx f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f^{(3)}(a)}{3!}(x - a)^3 + \cdots\] Substitute the values found and simplify, including only the nonzero terms.

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Key Concepts

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

Taylor Series Definition

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point. Each term involves the nth derivative evaluated at the center point, multiplied by (x - a)^n and divided by n!. This series approximates the function near the point a.
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Derivatives of sin(x)

To find the Taylor series of sin(x), you need to compute its derivatives at the center point. The derivatives of sin(x) cycle every four steps: sin(x), cos(x), -sin(x), -cos(x), then back to sin(x). Evaluating these at x = π/2 helps determine the coefficients of the series.
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Centering the Series at a = π/2

Centering the Taylor series at a = π/2 means expanding the function around this point, so terms involve powers of (x - π/2). This shifts the approximation focus to values of x near π/2, requiring evaluation of derivatives specifically at π/2 to find accurate coefficients.
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