Chapter 1: Problem 35
Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\frac{x}{\sin x}$$
Short Answer
Expert verified
The first few terms of the Maclaurin series for \( \frac{x}{\sin x} \) are: \( 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \text{...} \).
Step by step solution
01
Recall the Maclaurin series formula
The Maclaurin series for a function is given by the expression:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 + \text{...} \]We need to find these terms for the function \( \frac{x}{\sin x} \).
02
Use standard series expansions
First, write the Maclaurin series expansion for \( \sin x \):\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \text{...} \]Next, take the reciprocal of this series to find the series for \( \frac{1}{\sin x} \).
03
Find series for \( \frac{1}{\sin x} \)
Compute the reciprocal series to first few terms:\[ \frac{1}{\sin x} = \frac{1}{x - \frac{x^3}{6} + \frac{x^5}{120} - \text{...}} \approx \frac{1}{x} + \frac{x}{6} + \frac{7x^3}{360} + \text{...} \]
04
Multiply by \( x \) to get \( \frac{x}{\sin x} \)
Multiply the series found in the previous step by \( x \):\[ \frac{x}{\sin x} = x \left( \frac{1}{x} + \frac{x}{6} + \frac{7x^3}{360} + \text{...} \right) = 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \text{...} \]
05
Verify results using a computer
Use a computational tool to verify the series expansion. The function \( \frac{x}{\sin x} \) should have a series expansion matching the derived terms up to the desired degree of accuracy.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Series Expansion
A series expansion is a way to represent a function as a sum of terms calculated from the values of its derivatives at a single point. The Maclaurin series is a specific type of series expansion centered at zero. It allows us to approximate complex functions using polynomial terms. In the Maclaurin series of a function \( f(x) \), each term involves a derivative of \( f \) evaluated at zero and a power of \( x \). The formula looks like this:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \text{...} \]
Here, we needed to find the Maclaurin series for the function \( \frac{x}{\sin x} \). We started by using the standard series expansion for \( \sin x \):
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \text{...} \]
Applying this and further steps helped simplify the problem into a more manageable form.
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \text{...} \]
Here, we needed to find the Maclaurin series for the function \( \frac{x}{\sin x} \). We started by using the standard series expansion for \( \sin x \):
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \text{...} \]
Applying this and further steps helped simplify the problem into a more manageable form.
Reciprocal of a Series
Finding the reciprocal of a series is a crucial step when dealing with functions like \( \frac{1}{\sin x} \). It involves inverting each term's coefficient methodically to obtain the desired series. For instance, given the series
\[ \frac{1}{\sin x} = \frac{1}{x - \frac{x^3}{6} + \frac{x^5}{120} - \text{...}} \]
To find its reciprocal, we approximate it step-by-step:
\[ \frac{1}{\sin x} = \frac{1}{x} + \frac{x}{6} + \frac{7x^3}{360} + \text{...} \]
Each new term is determined based on the prior terms and follows a pattern that simplifies the division of terms. In this context, this reciprocal series helped us find the series for \( \frac{x}{\sin x} \) by multiplying it back with \( x \).
\[ \frac{1}{\sin x} = \frac{1}{x - \frac{x^3}{6} + \frac{x^5}{120} - \text{...}} \]
To find its reciprocal, we approximate it step-by-step:
\[ \frac{1}{\sin x} = \frac{1}{x} + \frac{x}{6} + \frac{7x^3}{360} + \text{...} \]
Each new term is determined based on the prior terms and follows a pattern that simplifies the division of terms. In this context, this reciprocal series helped us find the series for \( \frac{x}{\sin x} \) by multiplying it back with \( x \).
Trigonometric Functions
Trigonometric functions, such as \( \sin x \), are fundamental in mathematics. Their series expansions help us express these functions as polynomials, which are easier to manipulate, differentiate, and integrate. The Maclaurin series of \( \sin x \) is given by:
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \text{...} \]
This series is derived from the derivatives of \( \sin x \) at zero. Each consecutive term involves higher powers and factorials of the variable, but they alternate in sign. Utilizing this series enabled us to transform \( \frac{x}{\sin x} \) into a more straightforward polynomial form. This understanding is essential not only in pure mathematics but also in physics and engineering for modeling waves and oscillations.
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \text{...} \]
This series is derived from the derivatives of \( \sin x \) at zero. Each consecutive term involves higher powers and factorials of the variable, but they alternate in sign. Utilizing this series enabled us to transform \( \frac{x}{\sin x} \) into a more straightforward polynomial form. This understanding is essential not only in pure mathematics but also in physics and engineering for modeling waves and oscillations.
Computational Verification
Computational verification plays a vital role in confirming the accuracy of derived series expansions. After calculating the theoretical Maclaurin series of \( \frac{x}{\sin x} \), we use a computer or a graphing calculator to verify our results. With computational tools like MATLAB, Mathematica, or even online calculators, we can plot the function \( \frac{x}{\sin x} \) and its series expansion. This ensures that the series computed matches the actual function up to the desired degree of accuracy. For instance, by checking the function's values at various points, we validate that the series
\[ \frac{x}{\sin x} = 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \text{...} \]
indeed approximates \( \frac{x}{\sin x} \) correctly. This step not only confirms our mathematical calculations but also provides insight into the convergence and usefulness of series approximations.
\[ \frac{x}{\sin x} = 1 + \frac{x^2}{6} + \frac{7x^4}{360} + \text{...} \]
indeed approximates \( \frac{x}{\sin x} \) correctly. This step not only confirms our mathematical calculations but also provides insight into the convergence and usefulness of series approximations.