/*! 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 35 Find the first few terms of the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.
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 \( x \).
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.
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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{x^{3 n}}{n}$$

Connect the midpoints of the sides of an equilateral triangle to form 4 smaller equilateral triangles. Leave the middle small triangle blank, but for each of the other 3 small triangles, draw lines connecting the midpoints of the sides to create 4 tiny triangles. Again leave each middle tiny triangle blank and draw the lines to divide the others into 4 parts. Find the infinite series for the total area left blank if this process is continued indefinitely. (Suggestion: Let the area of the original triangle be 1 ; then the area of the first blank triangle is \(1 / 4 .\) ) Sum the series to find the total area left blank. Is the answer what you expect? Hint: What is the "area" of a straight line? (Comment: You have constructed a fractal called the Sierpi?ski gasket. A fractal has the property that a magnified view of a small part of it looks very much like the original.)

Use the preliminary test to decide whether the following series are divergent or require further testing. Careful: Do not say that a series is convergent; the preliminary test cannot decide this. $$\sum_{n=1}^{\infty} \frac{n !}{(n+1) !}$$

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}}{2^{\ln n}}$$

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\cos \left(e^{x}-1\right)$$

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.