/*! 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 The expression \(\frac{x}{1+a x^... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The expression \(\frac{x}{1+a x^{2}}\) is to be used as an approximation to \(\frac{1}{2} \ln [(1+x) /(1-x)]\) on \(-1

Short Answer

Expert verified
Set \( a = -\frac{1}{3} \) to nullify the first error term. The coefficient of \( x^5 \) becomes \(-\frac{4}{45}\).

Step by step solution

01

Expand the Logarithmic Expression

Use the Taylor series expansion for \( \ln(1+x) \) and \( \ln(1-x) \) to express the function \( \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \). The series are: \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \) and \( \ln(1-x) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots \). When subtracted, \( \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) = x + \frac{1}{3}x^3 + \frac{1}{5}x^5 + \cdots \).
02

Expand the Rational Expression

The expression \( \frac{x}{1+ax^2} \) can be expanded using the geometric series expansion \( \frac{1}{1-(-ax^2)} = 1 + ax^2 + a^2x^4 + \ldots \). Thus, \( \frac{x}{1+ax^2} = x - ax^3 + a^2x^5 - \cdots \).
03

Find the Difference Between the Two Expressions

Subtract the second expanded expression from the first to find the series: \( \frac{x}{1+a x^{2}} - \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right) = (-\left(\frac{1}{3} + a\right)x^3 - \left(\frac{1}{5} - a^2\right)x^5 - \cdots) \).
04

Determine Value of Constant a to Minimize Error

Since the error is dominated by the first term \(-\left(\frac{1}{3} + a\right)x^3\), set \( \frac{1}{3} + a = 0 \) to make this term zero. Solving gives \( a = -\frac{1}{3} \).
05

Determine Coefficient of x^5 with Chosen a

Substitute \( a = -\frac{1}{3} \) into the second term of the error series: \( -\left(\frac{1}{5} - a^2\right)x^5 = -\left(\frac{1}{5} - \left(-\frac{1}{3}\right)^2\right)x^5 \). Calculate this value to evaluate the coefficient of \( x^5 \): \( \left(\frac{1}{9} - \frac{1}{5}\right)x^5 = -\frac{4}{45}x^5 \).
06

Determine Radius of Convergence R

The domain requirement \(|x|<1\) is checked, so the series converges for any \(|x|<1\). Thus, \( R = 1 \).
07

Plot the Graphs

For the graphs, you verify that \( y=\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\), \(y=x+\frac{1}{3}x^3\), and \( y=\frac{3x}{3-x^2}\) all result in similar values between \(0 \leq x \leq 1\). Plot each function for this range to compare their shapes.

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.

Taylor Series
The Taylor Series is a powerful mathematical tool used to approximate functions by expressing them in the form of an infinite sum of terms calculated from the values of its derivatives at a single point. In essence, it allows us to represent complex functions as an expansion of simpler polynomials. This is particularly useful for functions that are otherwise difficult to handle analytically. The Taylor Series can be written as:
  • For a function \( f(x) \), the Taylor series centered at \( a \) is: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]
In the exercise, we use the Taylor series to expand both \( \ln(1+x) \) and \( \ln(1-x) \). By subtracting these expansions, we approximate \( \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \). This requires calculating these series terms up to a specified number of derivatives to match the desired accuracy.
Geometric Series
A Geometric Series is a type of series that has a constant ratio between successive terms, and it converges for certain values of the variable. It plays a significant role in approximating functions, especially in handling rational expressions. The basic form of a geometric series is:
  • For a series \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio, the sum is given by: \[ \frac{a}{1-r} \] but only for \( |r| < 1 \).
In the exercise, we use the geometric series to expand the expression \( \frac{x}{1+ax^2} \). By considering \( \frac{1}{1+ax^2} \) as \( 1-x^2 \), we utilize the geometric series approach to provide an approximate polynomial form of the function.
Error Minimization
Error Minimization is crucial when finding approximations, as it ensures that the approximation is as close as possible to the true value. In mathematical approximations, especially with series, the error is often represented by the terms that are omitted or differently expressed.
  • We aim to minimize the error by choosing constants that equate significant error terms to zero.
  • This approach provides a function that best fits the original function's behavior up to a desired level of accuracy.
In the problem, we are dealing with an approximation error expressed as a polynomial series: \(\frac{x}{1+ax^2} - \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) = -\left(\frac{1}{3}+a\right)x^3 - \left(\frac{1}{5}-a^2\right)x^5 + \cdots \)To minimize the error, the coefficient of the first error term, \(-\left(\frac{1}{3}+a\right)x^3\), is set to zero, solving \( a \) to ensure the function is as close as possible to the true logarithmic expression for a wider range of \( x \).
Rational Expression
Rational Expressions involve ratios of polynomials, which are frequently encountered in various mathematical problems. Manipulating and approximating these expressions require a thorough understanding of series and expansions. In the context of the exercise:
  • The original expression being approximated is \( \frac{x}{1+ax^2} \), a rational function involving both a numerator and a denominator that depend on \( x \).
  • Approximations often involve expanding these functions into series, as seen with the use of geometric series.
The use of rational expression approximations is crucial in simplifying complex functions to polynomials. This facilitates easier calculations and deeper insights into the behavior of functions over specific domains. In practice, these approximations allow us to draw comparisons and compute approximations without extensive computational resources. By comparing these expressions, an informed choice of the constant \( a \) is crucial to obtain the best possible polynomial that matches the target function.

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 \(\left(10^{-2}(0.3251) \times 10^{-5}(0.2011)\right)\) and \(\left(10^{-1}(0.2168) \div 10^{2}(0.3211)\right.\) ) using four-digit floating-point arithmetic.

The impedance \(Z\) ohms of a circuit containing a resistance \(R\) ohms, inductance \(L\) henries and capacity C farads, when the frequency of the oscillation is \(n\) per second, is given by $$ Z=\sqrt{\left(R^{2}+\left(2 \pi n L-\frac{1}{2 \pi n C}\right)^{2}\right)} $$ (a) Make \(L\) the subject of this formula. (b) If \(n=50, R=15\) and \(C=10^{-4}\) show that there are two values of \(L\) which make \(Z=20\) but only one value of \(L\) which will make \(Z=100 .\) Find the values of \(Z\) in each case to two decimal places.

Newton's recurrence formula for determining the root of a certain equation is $$ x_{n+1}=\frac{x_{n}^{2}-1}{2 x_{n}-3} $$Taking \(x_{0}=3\) as your initial approximation, obtain the root correct to \(4 \mathrm{sf}\). By setting \(x_{n+1}=x_{n}=\alpha\) show that the fixed points of the iteration are given by the equation \(\alpha^{2}-3 \alpha+1=0\)

(a) If \(n=n_{1}+n_{2}+n_{3}\) show that $$ \left(\begin{array}{l} n \\ n_{1} \end{array}\right)\left(\begin{array}{c} n_{2}+n_{3} \\ n_{2} \end{array}\right)=\frac{n !}{n_{1} ! n_{2} ! n_{3} !} $$ (This represents the number of ways in which \(n\) objects may be divided into three groups containing respectively \(n_{1}, n_{2}\) and \(n_{3}\) objects.)(b) Expand the following expressions (i) \(\left(1-\frac{x}{2}\right)^{5}\) (ii) \((3-2 x)^{6}\)

Find the equation of the straight line (a) with gradient \(\frac{3}{2}\) passing through the point \((2,1)\) (b) with gradient \(-2\) passing through the point \((-2,3)\) (c) passing through the points \((1,2)\) and \((3,7)\); (d) passing through the points \((5,0)\) and \((0,3)\); (e) parallel to the line \(3 y-x=5\), passing through \((1,1)\) (f) perpendicular to the line \(3 y-x=5\), passing through \((1,1)\).

See all solutions

Recommended explanations on Physics 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.