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In Problems 1-18, find the terms through \(x^{5}\) in the Maclaurin series for \(f(x)\). Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x)\). $$ f(x)=\frac{\cos x-1+x^{2} / 2}{x^{4}} $$

Short Answer

Expert verified
The Maclaurin series for \(f(x)\) through \(x^5\) is \(\frac{1}{24}\).

Step by step solution

01

Recap of Known Maclaurin Series for Basic Functions

Recall the Maclaurin series for the basic functions: - \(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots\)We need these formulas to construct the series for \(f(x)\).
02

Simplify the Numerator

The numerator is \( \cos x - 1 + \frac{x^2}{2} \). Plug in the Maclaurin series for \( \cos x \):\[\cos x - 1 + \frac{x^2}{2} = \left(1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots\right) - 1 + \frac{x^2}{2} \]This simplifies to:\[\frac{x^4}{24}\]
03

Divide by \(x^4\)

We now have the numerator \(\frac{x^4}{24}\). Divide this by \(x^4\) to find:\[f(x) = \frac{x^4}{24x^4} = \frac{1}{24}\]
04

Find Terms Through \(x^5\)

Since the series reduces to a constant \(\frac{1}{24}\), its expansion contains only that single term, and no further terms are needed up to \(x^5\). The complete Maclaurin series through \(x^5\) is just \(\frac{1}{24}\).

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Key Concepts

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

Calculus
Calculus plays a pivotal role in understanding and applying the concept of the Maclaurin series, which is a type of series expansion. It helps us explore how functions behave at certain points and how they can be expressed as a series of increasingly complex terms. The Maclaurin series is particularly useful in calculus because it allows us to approximate functions using simple polynomial terms.

In calculus, we derive the Maclaurin series from the Taylor series by setting the expansion point to zero. This process involves determining derivatives of a function at a single point and expressing the function as an infinite sum of its derivatives evaluated at that point:
  • The first derivative represents the slope or rate of change of the function.
  • Higher-order derivatives provide more detailed information about the function's curvature and behavior.
Understanding derivatives within calculus is essential to constructing accurate series expansions for various functions, which is critical in many scientific and engineering applications.
Series Expansion
A series expansion is a powerful mathematical tool used to express a function as a sum of terms. These terms might be simple polynomials or more complex functions. In the case of the Maclaurin series, we're dealing with polynomials centered around zero.

The process of creating a series expansion consists of several steps:
  • Identify the function you wish to expand.
  • Calculate the derivatives of the function.
  • Evaluate these derivatives at the point of expansion, which is zero for the Maclaurin series.
  • Formulate the series by combining these evaluated derivatives with a factorial term.
In the problem given, simplifying terms and organizing them into a series is crucial. By using series expansions, we can approximate complex functions like trigonometric ones, making it easier to work with them in various contexts.
Trigonometric Functions
Trigonometric functions like sine and cosine are essential in mathematics for modeling periodic phenomena such as sound waves or circular motion. In series expansion, these functions can be approximated by polynomials. For example, the Maclaurin series for \( ext{cos }x\):
  • Starts at 1 since \( ext{cos}(0) = 1\),Continuing with terms \(- \frac{x^2}{2} + \frac{x^4}{24}\), implying a gradual approximation of \( ext{cos }x\) around zero.
  • Using known series for sine and cosine functions is advantageous for forming new expressions, like in the original exercise where \(\frac{\sin x}{\cos x}\) was used to find \(\tan x\).
This transformation relies on recognizing patterns in trigonometric identities and their series expansions, allowing for more straightforward calculations when deriving or simplifying expressions.
Polynomial Approximation
Polynomial approximation simplifies complex functions, transforming them into easily understandable and manipulable sums of polynomial terms. The Maclaurin series provides a foundational approach to this approximation by expressing functions as sums where each term's degree increases.

This method is beneficial for numerical analysis and computational applications because:
  • Polynomials are generally easier to compute than intricate functions.
  • They provide approximate solutions that can be as accurate as needed by including more terms in the series.
  • They help grasp fundamental behaviors of functions by examining lower-degree terms before higher ones.
In the given exercise, the Maclaurin series expansion was reduced to a constant \(\frac{1}{24}\), highlighting that sometimes simplification yields a result needing no further polynomial terms beyond the constant.

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Most popular questions from this chapter

In Problems \(1-8\), find the convergence set for the given power series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}} $$

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(\frac{1}{2-\frac{1}{2}}, \frac{2}{3-\frac{1}{3}}, \frac{3}{4-\frac{1}{4}}, \frac{4}{5-\frac{1}{5}}, \ldots\)

Let \(f(x)=(1+x)^{1 / 2}+(1-x)^{1 / 2}\). Find the Maclaurin series for \(f\) and use it to find \(f^{(4)}(0)\) and \(f^{(51)}(0)\).

A famous sequence \(\left\\{f_{n}\right\\}\), called the Fibonacci Sequence after Leonardo Fibonacci, who introduced it around A.D. 1200 , is defined by the recursion formula $$ f_{1}=f_{2}=1, \quad f_{n+2}=f_{n+1}+f_{n} $$ (a) Find \(f_{3}\) through \(f_{10}\) - (b) Let \(\phi=\frac{1}{2}(1+\sqrt{5}) \approx 1.618034\). The Greeks called this number the golden ratio, claiming that a rectangle whose dimensions were in this ratio was "perfect." It can be shown that $$ \begin{aligned} f_{n} &=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] \\\ &=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n} \phi^{-n}\right] \end{aligned} $$ Check that this gives the right result for \(n=1\) and \(n=2\). The general result can be proved by induction (it is a nice challenge). More in line with this section, use this explicit formula to prove that \(\lim _{n \rightarrow \infty} f_{n+1} / f_{n}=\phi\). (c) Using the limit just proved, show that \(\phi\) satisfies the equation \(x^{2}-x-1=0\). Then, in another interesting twist, use the Quadratic Formula to show that the two roots of this equation are \(\phi\) and \(-1 / \phi\), two numbers that occur in the explicit formula for \(f_{n}\).

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ \frac{x-2}{1^{2}}+\frac{(x-2)^{2}}{2^{2}}+\frac{(x-2)^{3}}{3^{2}}+\frac{(x-2)^{4}}{4^{2}}+\cdots $$

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