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Chapter 10: Problem 19

Find the Taylor series for \(\sin (x)\), expanding around \(\pi / 2\).

Short Answer

Expert verified
The Taylor series for \( \sin(x) \) expanded about \( \pi / 2 \) is \( 1 - (2! / 2!) (x - \pi / 2)^2 + (4! / 4!) (x - \pi / 2)^4 - (6! / 6!) (x - \pi / 2)^6 + (8! / 8!) (x - \pi / 2)^8 - \cdots \).

Step by step solution

01

Identify the function and its derivatives

The function is \( \sin(x) \). The first few derivatives at \( \pi / 2 \) are: \( f'(x) = \cos(x) \), \( f''(x) = -\sin(x) \), \( f'''(x) = -\cos(x) \), and \( f''''(x) = \sin(x) \). Notice the pattern: the derivatives cycle every four terms.
02

Evaluate the derivatives at \( \pi / 2 \)

Evaluate each of the derivatives at \( \pi / 2 \): \( f(\pi / 2) = 1 \), \( f'(\pi / 2) = 0 \), \( f''(\pi / 2) = -1 \), \( f'''(\pi / 2) = 0 \), and \( f''''(\pi / 2) = 1 \).
03

Write out the Taylor series

Following the formula for the Taylor series, the result is: \( \sin(x) = 1 - (2! / 2!) (x - \pi / 2)^2 + (4! / 4!) (x - \pi / 2)^4 - (6! / 6!) (x - \pi / 2)^6 + (8! / 8!) (x - \pi / 2)^8 - \cdots \).

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Key Concepts

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

Trigonometric Functions
Trigonometric functions are fundamental in calculus. They include sine (\( \sin(x) \)) and cosine (\( \cos(x) \)), among others. These functions are essential in describing phenomena in physics and engineering.
  • Sine and cosine have periodic behavior, making them cyclical functions, repeating their values in regular intervals.
  • Sine starts its cycle at zero, while cosine begins at one. This phase difference is important.
The periodic nature of trigonometric functions means their derivatives also show a cyclical pattern. This is crucial in tasks like finding Taylor series expansions. Understanding those cycles can help predict behavior without calculating each value anew.
Derivatives and Differentiation
Differentiation is a process in calculus where we find the rate at which a function changes. The derivative of a function gives us a new function that tells us the slope of the given function at any point.
  • The first derivative, \( f'(x) \), tells us the slope of the function \( f(x) \).
  • The second derivative, \( f''(x) \), tells us about the concavity of \( f(x) \).
  • The pattern of differentiation is important, especially for trigonometric functions like \( \sin(x) \).
For \( \sin(x) \), the derivatives cycle every four terms: \( \sin(x) \), \( \cos(x) \), \( -\sin(x) \), \( -\cos(x) \), then repeat. This cyclic nature simplifies problems like Taylor series by revealing intricate patterns.
Pattern Recognition in Calculus
Calculus often involves recognizing patterns to solve problems more efficiently. In the process of finding Taylor series, pattern recognition serves as a powerful tool.
  • For the function \( \sin(x) \), recognizing the cyclic pattern of its derivatives at specific points is vital.
  • Evaluating these derivatives at \( \pi/2 \) shows another pattern: alternate derivatives are zero, and the signs change periodically.
Recognizing these patterns allows for constructing Taylor series expansions efficiently and correctly for trigonometric functions. This method reduces potential calculation errors and provides insights into function behavior, aiding in understanding within calculus.

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Most popular questions from this chapter

Find the first few terms of the two-variable Maclaurin series representing the function \(f(x, y)=\sin (x+y)\).

Using the Maclaurin series, show that $$ \int_{0}^{x_{1}} \cos (a x) \mathrm{d} x=\left.\frac{1}{a} \sin (x)\right|_{0} ^{x_{1}}=\frac{1}{a} \sin \left(a x_{1}\right) . $$

Find two different Taylor series to represent the function $$ f(x)=\frac{1}{x} $$ such that one series is $$ f(x)=a_{0}+a_{1}(x-1)+a_{2}(x-1)^{2}+\cdots $$ and the other is $$ f(x)=b_{0}+b_{1}(x-2)+b_{3}(x-2)^{2}+\cdots $$ Show that \(b_{n}=a_{n} / 2^{n}\) for any value of \(n\). Find the interval of convergence for each series (the ratio test may be used). Which series must you use in the vicinity of \(x=3\) ? Why?

Test the following series for convergence $$ \sum_{n=0}^{\infty}\left((-1)^{n} n / n !\right) $$ Note: \(n !=n(n-1)(n-2) \cdots(2)(1)\) for positive integral values of \(n\) and \(0 !=1\).

Find the first few coefficients for the Maclaurin series for the function $$ f(x)=\sqrt{1+x} . $$
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