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Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function. $$y=\frac{x}{\sin x}$$

Short Answer

Expert verified
The series is \( 1 + \frac{x^2}{6} \).

Step by step solution

01

Recall the Maclaurin Series for Sine

The Maclaurin series for \( \sin x \) is \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \). This series expansion is essential for finding \( y = \frac{x}{\sin x} \).
02

Invert the Series of \( \sin x \)

To find \( \frac{1}{\sin x} \), we first truncate the series to get the dominant terms: \( \sin x \approx x - \frac{x^3}{6} \). We seek a series that when multiplied by \( x - \frac{x^3}{6} \) gives 1. For simplicity, we propose a series \( a_0 + a_1 x + a_2 x^2 \) and solve for coefficients.
03

Determine the Coefficients of the Inverse Series

Assume \( \frac{1}{\sin x} \approx a_0 + a_1 x + a_2 x^2 \). Multiplying by the truncated series gives: \[ (a_0 + a_1 x + a_2 x^2)(x - \frac{x^3}{6}) = x - \frac{x^3}{6} \]. We set terms to equate the coefficient of each power of \( x \) to find \( a_0 = 1/x, a_1 = 0, a_2 = 1/6 \).
04

Multiply by \( x \) to Find \( \frac{x}{\sin x} \)

By multiplying backward \[ \frac{x}{\sin x} = x(a_0 + a_1 x + a_2 x^2) \], we substitute \( a_0 = 1/x \), \( a_1 = 0 \), \( a_2 = 1/6 \): \[ = x \left(\frac{1}{x} + 0 + \frac{x^2}{6}\right) = 1 + \frac{x^2}{6} \].
05

Conclusion

The first three non-zero terms of the Maclaurin series for \( \frac{x}{\sin x} \) are \( 1 + \frac{x^2}{6} \).

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Key Concepts

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

Power Series
A power series is an infinite series of terms in the form \( a_nx^n \), where \( a_n \) represents coefficients and \( n \) is a non-negative integer. These series are powerful tools in mathematics as they simplify complex functions into a sum that can be worked with easily. A common example of a power series is the Maclaurin series, which is a type of Taylor series centered at zero.

In practice, we often truncate power series to a finite number of terms for approximate calculations. This is particularly useful when finding the series representation of more complicated functions, such as \( \frac{x}{\sin x} \). Here, the sine function is replaced by its Maclaurin series, \( \sin x = x - \frac{x^3}{3!} + \ldots \). By using just the first few terms, we obtain an approximation that makes further calculations more manageable. Such simplifications are crucial in engineering and physics, where complex problems often need tractable solutions.
Multiplication and Division of Series
When dealing with power series, multiplication and division become handy techniques for function manipulation. Multiplying and dividing series allow us to discover new functions or simplify complex expressions.

To use these techniques effectively, consider the example of finding the series for \( \frac{x}{\sin x} \). The power series for \( \sin x \) is inverted to approximate \( \frac{1}{\sin x} \). By proposing a series \( a_0 + a_1x + a_2x^2 \), and multiplying it by the truncated \( \sin x \) series, you equate coefficients for a match to 1. This process determines the coefficients of the inverted series.

Next, you multiply the resulting series by \( x \) to provide \( \frac{x}{\sin x} \). By combining these algebraic manipulations, we find that the first few non-zero terms of the resulting Maclaurin series are \( 1 + \frac{x^2}{6} \). This process illustrates how multiplication and division of series allows us to decompose and reconstruct functions in a more understandable form.
Trigonometric Functions
Trigonometric functions like sine and cosine often appear in power series, enabling their use in a deeper analysis of mathematical concepts. These functions, when expanded into a series, provide more accessible representations that are crucial in calculus and applied mathematics.

For example, the sine function's Maclaurin series expansion, \( \sin x = x - \frac{x^3}{3!} + \cdots \), illustrates how we can transform a transcendental function into a polynomial form. This is particularly useful because polynomial forms are easier to manipulate through standard arithmetic operations like addition, subtraction, multiplication, and division.

Such expansions of trigonometric functions allow engineers and scientists to model waveforms, oscillatory motion, and other phenomena that inherently involve trigonometric expressions. Moreover, analyzing these functions in a simplified series form offers insights into detailed behavior, such as approximating function values near zero or understanding oscillation patterns.

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Most popular questions from this chapter

Find a power series representation for the function and determine the interval of convergence. $$ f(x)=\frac{2}{3-x} $$

Evaluate the indefinite integral as an infinite series. $$\int x \cos \left(x^{3}\right) d x$$

The meaning of the decimal representation of a number $$0 . d_{1} d_{2} d_{3} \ldots\( (where the digit \)d_{i}\( is one of the numbers \)0,1\( \)2, \ldots, 9 )\( is that \)\quad 0 . d_{1} d_{2} d_{3} d_{4} \ldots=\frac{d_{1}}{10}+\frac{d_{2}}{10^{2}}+\frac{d_{3}}{10^{3}}+\frac{d_{4}}{10^{4}}+\cdots$$ Show that this series always converges.

The Cantor set, named after the German mathematician Georg Cantor \((1845-1918),\) is constructed as follows. We start with the closed interval \([0,1]\) and remove the open interval \(\left(\frac{1}{3}, \frac{2}{3}\right) .\) That leaves the two intervals \(\left[0, \frac{1}{3}\right]\) and \(\left[\frac{2}{3}, 1\right]\) and we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in \([0,1]\) after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is \(1 .\) Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side \(1,\) then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is 1. This implies that the Sierpinski carpet has area \(0 .\)

Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int \frac{t}{1+t^{3}} d t $$

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