/*! 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 25 In each of Exercises 23-34, deri... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In each of Exercises 23-34, derive the Maclaurin series of the given function \(f(x)\) by using a known Maclaurin series. $$ f(x)=x^{2} \sin (x / 2) $$

Short Answer

Expert verified
The Maclaurin series for \(f(x) = x^2 \sin(x/2)\) is \(\frac{x^3}{2} - \frac{x^5}{48} + \frac{x^7}{3840} - \cdots\).

Step by step solution

01

Recall the Maclaurin series for the known function

The Maclaurin series for \( \sin(x) \) is given by \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \).
02

Substitute \(x/2\) into the series

Since we have \(\sin(x/2)\), substitute \(x/2\) into the Maclaurin series for \(\sin(x)\): \[ \sin(x/2) = \frac{x}{2} - \frac{(x/2)^3}{3!} + \frac{(x/2)^5}{5!} - \frac{(x/2)^7}{7!} + \cdots \].Simplifying each term of the series gives:\[ \sin(x/2) = \frac{x}{2} - \frac{x^3}{48} + \frac{x^5}{3840} - \frac{x^7}{645120} + \cdots \].
03

Multiply by \(x^2\) to get \(f(x)\)

Multiply each term of the series for \(\sin(x/2)\) by \(x^2\) to find the Maclaurin series for \(f(x) = x^2 \sin(x/2)\):\[ f(x) = x^2 \left( \frac{x}{2} - \frac{x^3}{48} + \frac{x^5}{3840} - \cdots \right) = \frac{x^3}{2} - \frac{x^5}{48} + \frac{x^7}{3840} - \cdots \].

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.

Function Series Expansion
Function series expansion is a powerful calculus tool, helping us represent complex functions as infinite sums of simpler terms.
For functions that have derivatives of all orders, such as polynomials, exponentials, or trigonometric functions, we can express them as power series. The Maclaurin series, a specific type of Taylor series, allows us to express functions as expanded series centered at zero.
In general, the Maclaurin series of a function \( f(x) \) can be written as:
  • \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \).
This approach is particularly useful because it transforms difficult functions into series where we can more easily compute values, derivatives, and integrals.
In the given exercise, the series expansion is employed to express the function \( f(x) = x^2 \sin(x/2) \) utilizing the known Maclaurin series expansion of \( \sin(x) \).
By manipulating and expanding \( \sin(x/2) \), we simplify the series to reveal how the complex function can be written as the sum of its simpler components.
Trigonometric Function Series
Trigonometric functions often appear in calculus and can be tricky to handle in their original forms. Functions like sine and cosine can be expanded using series, enabling transformations and simplifications.
The series for \( \sin(x) \) is a well-known trigonometric series represented as:
  • \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \).
This series is derived through the concept of derivatives, considering the odd powers of \( x \) because sine is an odd function.In the exercise, substituting \( x/2 \) into this series, we transformed the original trigonometric function into:
  • \( \sin(x/2) = \frac{x}{2} - \frac{(x/2)^3}{3!} + \frac{(x/2)^5}{5!} - \cdots \).
Our task was to reframe \( \sin(x/2) \) to leverage it in the function \( f(x) \).
This approach highlights how the manipulation of a known trigonometric series can simplify expressing other related trigonometric functions and aid in complex calculus problems.
Calculus Problem Solving
In calculus, solving problems through function series expansion is a remarkable skill. It offers insight into the stability and behavior of a function across different values of \( x \). Using a series expansion involves strategic substitution and manipulation of known series. This approach is not only practical but also elegant.The process begins by identifying a known series that closely resembles the function in question.
In this exercise, we utilized sin(x)'s Maclaurin series as our foundation.
  • Step 1: We substituted to modify for relevant variables (\( x/2 \) in place of \( x \)) and adjusted each term accordingly.
  • Step 2: Rewriting the series by introducing variables, we tailored it to the specific function \( f(x) \).
  • Step 3: We expanded and simplified by multiplying the overall series by another term (\( x^2 \) in this case), adapting to fit the function's entire structure.
By systematically organizing and executing these steps, we unraveled a complex function into a manageable series.
This process not only provides the solution but also deepens understanding of function interactions and the beauty within mathematical problem-solving.

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

In each of Exercises \(45-60,\) determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{1}{\sqrt{n+10}}\)

Determine whether the series converges absolutely, converges conditionally, or diverges. The tests that have been developed in Section 5 are not the most appropriate for some of these series. You may use any test that has been discussed in this chapter. \(\sum_{n=1}^{\infty}(-1)^{n} \frac{n !}{3^{n}}\)

In each of Exercises 49-54, use Taylor series to calculate the given limit. $$ \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} $$

Suppose that every dollar that we spend gives rise (through wages, profits, etc.) to 90 cents for someone else to spend. That 90 cents will generate a further 81 cents for spending, and so on. How much spending will result from the purchase of a \(\$ 16,000\) automobile, the car included? (This phenomenon is known as the multiplier effect.)

Consider the initial value problem $$ \frac{d y}{d x}=2-x-y, \quad y(0)=1 $$ \(\begin{array}{llll}\text { a. Calculate the power series } & \text { expansion }\end{array}\) \(y(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) of the solution up to the \(x^{7}\) term. b. Using the coefficients you have calculated, plot \(S_{3}(x)=\sum_{n=0}^{3} a_{n} x^{n}\) in the viewing rectangle \([-2,2] \times\) [-10,1.7] c. The exact solution to the initial value problem is \(y(x)=3-x-2 e^{-x},\) as can be determined using the methods of Section 7.7 (in Chapter 7 ). Add the plot of the exact solution to the viewing window. From the two plots, we see that the approximation is fairly accurate for \(-1 \leq x \leq 1\), but the accuracy decreases outside this subinterval. d. To see the improvement in accuracy that results from using more terms in a partial sum, replace the graph of \(S_{3}(x)\) with that of \(S_{7}(x)\)

See all solutions

Recommended explanations on Math 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.