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Chapter 11: Problem 18

Find the Taylor series for \(f(x)\) centered at the given value of a. [Assume that \(f\) hat \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0 . ]\) \(f(x)=\sin x, \quad a=\pi / 2\)

Short Answer

Expert verified
The Taylor series for \( \sin x \) centered at \( \frac{\pi}{2} \) is \( 1 - \frac{1}{2}(x - \frac{\pi}{2})^2 + \frac{1}{24}(x - \frac{\pi}{2})^4 - \cdots \).

Step by step solution

01

Identify the Function and Center

The given function is \( f(x) = \sin x \) and the center is \( a = \frac{\pi}{2} \). We are to find the Taylor series expansion of \( \sin x \) around \( \frac{\pi}{2} \).
02

Calculate the Function and its Derivatives

Calculate \( f(a) = f(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) = 1 \). Now find the derivatives at \( a = \frac{\pi}{2} \):- \( f'(x) = \cos x \), so \( f'(\frac{\pi}{2}) = \cos(\frac{\pi}{2}) = 0 \).- \( f''(x) = -\sin x \), so \( f''(\frac{\pi}{2}) = -1 \).- \( f'''(x) = -\cos x \), so \( f'''(\frac{\pi}{2}) = 0 \).- \( f''''(x) = \sin x \), so \( f''''(\frac{\pi}{2}) = 1 \).Note the repeating cycle: \( f^{(n)}(x) = \sin x, \cos x, -\sin x, -\cos x \) depending on whether \( n \equiv 0, 1, 2, 3 \mod 4 \).
03

Construct the Taylor Series

The Taylor series for \( f(x) \) about \( x = a \) (where \( a = \frac{\pi}{2} \)) is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n.\]Substitute the derivatives calculated:\[\sin x = 1 - \frac{1}{2!}(x - \frac{\pi}{2})^2 + \frac{1}{4!}(x - \frac{\pi}{2})^4 - \frac{1}{6!}(x - \frac{\pi}{2})^6 + \cdots\]This series is derived from the repeated cycle obtained in the previous step.
04

Simplify the Series

Every second derivative contributes to the series, creating alternating series terms:\[\sin x \approx 1 - \frac{1}{2}(x - \frac{\pi}{2})^2 + \frac{1}{24}(x - \frac{\pi}{2})^4 - \cdots\]The series terms involve factorials in the denominator of odd-numbered terms.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power Series Expansion
The power series expansion is a fundamental concept for understanding Taylor series. A power series is an infinite sum of terms, each in the form of a power of a variable. It looks like this:
  • It's expressed as \( \sum_{n=0}^{\infty} a_n (x - c)^n \), where \( a_n \) are coefficients, and \( c \) is the center of the series.
  • With Taylor series, the coefficients \( a_n \) are specific derivatives of a function evaluated at a point \( c \).
In our example, we are expanding \( \sin x \) around \( \pi/2 \). Here, \( c = \pi/2 \) serves as the center. For each term, we calculate the derivative of \( \sin x \) at \( \pi/2 \) and divide it by \( n! \), then multiply by \((x-\pi/2)^n\). This expansion allows us to approximate \( \sin x \) using a polynomial, which is often easier to use for calculations.
Derivatives of Trigonometric Functions
Understanding derivatives, especially of trigonometric functions, is critical in building a Taylor series for functions like \( \sin x \). Here are important points about the derivatives:
  • The derivatives of \( \sin x \) follow a cyclical pattern:
    • \( f'(x) = \cos x \)
    • \( f''(x) = -\sin x \)
    • \( f'''(x) = -\cos x \)
    • \( f''''(x) = \sin x \), and so on.
  • This repetitive cycle can be summarized using modular arithmetic: for \( n \equiv 0, 1, 2, 3 \mod 4 \), the derivative cycle maintains location and sign.
  • Evaluating these derivatives at \( x = \pi/2 \) determines the coefficients of our power series.
By understanding these derivatives, you can see the pattern and easily construct terms for the Taylor series.
Alternating Series
An alternating series is a series where the terms alternate in sign. This happens in our Taylor series for \( \sin x \) around \( \pi/2 \):
  • The series starts with a positive term, then negative, then positive, and so on. It forms: \[ \sin x \approx 1 - \frac{1}{2}(x - \frac{\pi}{2})^2 + \frac{1}{24}(x - \frac{\pi}{2})^4 - \cdots \]
  • This alternating pattern often results from trigonometric functions' derivatives, which shift signs cyclically as calculated.
  • Alternating series are importantly convergent, which means they tend toward a limit as more terms are added.
Recognizing and understanding alternating series can help in appraising their accuracy and predict convergence behavior, especially in terms of trigonometric expansions.
sin x Expansion
The expansion of \( \sin x \) using its Taylor series is a classical example in calculus. It reveals how trigonometric functions can be expressed as infinite polynomials:
  • For \( \sin x \) centered at \( \pi/2 \), the expansion starts with the value of \( \sin(\pi/2) = 1 \).
  • Subsequent terms derive from incremented even powers of \((x - \pi/2)\), using cycles of \( \sin \) and \( \cos \) derivatives to determine sign and behavior.
  • We see how important even-power terms in this expansion cancel or bridge polarities with factorial denominators: \[ 1 - \frac{1}{2}(x - \frac{\pi}{2})^2 + \frac{1}{24}(x - \frac{\pi}{2})^4 - \cdots \]
  • With each additional term, the approximation of \( \sin x \) becomes more accurate, illustrating its periodic nature and smooth curves even with polynomial representation.
Understanding \( \sin x \) expansion not only aids in grasping deeper calculus concepts but also enhances practical application in physics and engineering, where wave and oscillatory functions feature prominently.

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Most popular questions from this chapter

Use the Alternating Series Estimation Theorem or Taylor's Inequality to estimate the range of values of \(x\) for which the given approximation is accurate to within the stated error. Check your answer graphically. $$\arctan x \approx x-\frac{x^{3}}{3}+\frac{x^{5}}{5} \quad(|\operatorname{error}|<0.05)$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=\sec x, \quad a=0, \quad n=2, \quad-0.2 \leqslant x \leqslant 0.2$$

Show that the sequence defined by $$ a_{1}=1 \quad a_{n+1}=3-\frac{1}{a_{n}} $$ is increasing and \(a_{n}<3\) for all \(n\) . Deduce that \(\left\\{a_{n}\right\\}\) is convergent and find its limit.

(a) A sequence \(\left\\{a_{n}\right\\}\) is defined recursively by the equation \(a_{n}=\frac{1}{2}\left(a_{n-1}+a_{n-2}\right)\) for \(n \geqslant 3,\) where \(a_{1}\) and \(a_{2}\) can be any real numbers. Experiment with various values of \(a_{1}\) and \(a_{2}\) and use your calculator to guess the limit of the sequence. (b) Find \(\lim _{n \rightarrow \infty} a_{n}\) in terms of \(a_{1}\) and \(a_{2}\) by expressing \(a_{n+1}-a_{n}\) in terms of \(a_{2}-a_{1}\) and summing a series.

Test the series for convergence or divergence. $$\sum_{k=1}^{\infty} \frac{k \ln k}{(k+1)^{3}}$$
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