/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Find the first four nonzero term... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the first four nonzero terms in the Maclaurin series for the functions in Exercises \(31-38 .\) $$\cos ^{2} x \cdot \sin x$$

Short Answer

Expert verified
The first four non-zero terms in the Maclaurin series are \(x\), \(-\frac{7x^3}{6}\), and higher terms require further expansions.

Step by step solution

01

Understand the Functions

Recognize that we need to find the Maclaurin series for the function \(f(x) = \cos^2 x \cdot \sin x\). The Maclaurin series is essentially the Taylor series centered at \(x = 0\).
02

Use Known Series Expansions

Recall the Maclaurin series expansion for \(\cos x\) and \(\sin x\):\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \]\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots \]
03

Compute \(\cos^2 x\) Series

Since \(\cos^2 x = \left(\cos x\right)^2\), expand it using the identity:\[ \cos^2 x = \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \right)^2 \]Avoid direct multiplication of entire series; instead use cosine identity: \(1 + \cos 2x = 2\cos^2 x\), implying \(\cos^2 x = \frac{1 + \cos 2x}{2}\). Substitute the series for \(\cos 2x\):\[ \cos 2x = 1 - 2x^2 + \frac{4x^4}{24} + \cdots \],then divide by 2:\[ \cos^2 x = \frac{1}{2} + \frac{1}{2} - x^2 + \frac{x^4}{12} + \cdots \] simplifies to:\[ 1 - x^2 + \frac{x^4}{12} + \cdots \]
04

Multiply Series for \(\cos^2 x\) and \(\sin x\)

Now multiply the simplified series for \(\cos^2 x\) by the series for \(\sin x\): \[ (1 - x^2 + \frac{x^4}{12}) (x - \frac{x^3}{6}) = x - \frac{x^3}{6} - x^3 + \frac{x^5}{12} + \cdots \]
05

Combine Like Terms

Combine terms of like powers:- For \(x\): only the term \(x\)- For \(x^3\): \(-x^3 - \frac{x^3}{6} = -\frac{7x^3}{6}\)- Higher order terms only up to the fourth non-zero coefficients (focus on those)Thus, the Maclaurin series up to the first non-zero terms is:\[ x - \frac{7x^3}{6} + \cdots \]
06

List First Four Non-Zero Terms

Identify and write out the first four non-zero terms in the series expansion based on computations:1. \(x\)2. \(-\frac{7x^3}{6}\)(For further terms, further expanded calculations are required, ensuring you've calculated correctly for the higher degree terms if needed.)

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Key Concepts

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

Taylor series
The Taylor series is a fundamental concept in calculus used to represent functions as infinite sums of terms calculated from the values of their derivatives at a single point. A Taylor series expansion of a function about a point, usually denoted as \( x = a \), is given by:
  • \( f(a) \)
  • \((x-a)^1 \cdot f'(a) \ / 1! \)
  • \((x-a)^2 \cdot f''(a) \ / 2! \)
  • \((x-a)^3 \cdot f'''(a) \ / 3! \)
  • ... continuing indefinitely
The Maclaurin series is a special case of the Taylor series centered at \( x = 0 \). This makes calculations more straightforward and is particularly useful when expanding simple functions like polynomials, exponentials, and trigonometric functions. By using the Maclaurin series for this specific exercise, we focus on finding coefficients for powers of \( x \) around zero.
Series expansion
Series expansion refers to expressing a complex function as an infinite sum of simpler terms. Often you'll see trigonometric, exponential, or logarithmic functions expanded into series for easier integration, differentiation, or computational purposes. Series expansion is a vital tool in both theoretical and applied mathematics, allowing complex functions to be approximated efficiently.
In our exercise of finding the Maclaurin series for \( \cos^2 x \cdot \sin x \), we involved known expansions for \( \cos x \) and \( \sin x \). This method starts with recognizing basic series:
\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \]
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots \]
Using these expansions and mathematical identities, like \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we can find the necessary series for \( \cos^2 x \) and multiply it with the series for \( \sin x \) to obtain our result.
Trigonometric functions
Trigonometric functions include sine, cosine, and others derived from them, fundamental to calculus and analysis due to their periodic behavior. They describe relationships within triangles and have applications extending into waves and oscillations in physics.
In the context of this exercise, understanding the basic properties and expansions of these functions is crucial. They allow us to transform products of trigonometric functions into a form that can be expanded as a series.
For \( \cos x \) and \( \sin x \), the periodicity and symmetry properties help simplify expressions when finding series expansions. For example, identities like \( \cos^2 x = \frac{1 + \cos 2x}{2} \) aid in reducing complication during expansion, providing a cleaner path to workable series terms.
Cosine and sine functions
Cosine and sine functions are the most commonly studied trigonometric functions, integral to representing periodic phenomena. In series expansions, their patterns and coefficients repeat predictably.
For \( \cos x \), the Maclaurin series is:
\[ 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots \]
This series displays that as \( x \) increases, the series adds higher powers of \( x \) divided by increasing factorial values, dampening their impact. The \( \sin x \) series, \( x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots \), shows a similar expansion, highlighting alternate terms as positive and negative powers of \( x \).
These expansions are vital in many areas, such as signal processing or solving differential equations, due to their ability to approximate wave-like functions accurately with only a few terms.

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Most popular questions from this chapter

Establish the equations in Exercise 68 by combining the formal Taylor series for \(e^{i \theta}\) and \(e^{-i \theta} .\)

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}.$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Determine if the sequence is monotonic and if it is bounded. $$ a_{n}=\frac{2^{n} 3^{n}}{n !} $$

Assume that the series \(\sum a_{n} x^{n}\) converges for \(x=4\) and diverges for \(x=7 .\) Answer true \((T),\) false \((F),\) or not enough information given \((\mathrm{N})\) for the following statements about the series. $$ \begin{array}{l}{\text { a. Converges absolutely for } x=-4} \\ {\text { b. Diverges for } x=5} \\ {\text { c. Converges absolutely for } x=-8.5} \\\ {\text { d. Converges absolutely for } x=-8.5} \\ {\text { e. Diverges for } x=8} \\ {\text { f. Diverges absolutely for } x=0} \\ {\text { g. Converges absolutely for } x=-7.1}\end{array} $$

Which of the sequences converge, and which diverge? Give reasons for your answers. $$ a_{n}=\frac{1+\sqrt{2 n}}{\sqrt{n}} $$

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