/*! 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 30 Find the first four nonzero term... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the first four nonzero terms in the Maclaurin series for the functions. $$\frac{\ln (1+x)}{1-x}$$

Short Answer

Expert verified
The first four nonzero terms are \( x + \frac{x^2}{2} - \frac{x^3}{6} + \cdots \).

Step by step solution

01

Understand Maclaurin Series

The Maclaurin series is a Taylor series around 0. It is a power series representation of a function in the form of \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). For this problem, we aim to find the series of \( \frac{\ln(1+x)}{1-x} \).
02

Series for \(\ln (1+x)\)

Recall the known series expansion for \( \ln(1+x) \): \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \).
03

Series for \(\frac{1}{1-x}\)

Using the geometric series formula, the series expansion for \( \frac{1}{1-x} \) is: \( 1+x+x^2+x^3+ \cdots \).
04

Multiply the Series

Multiply the series for \( \ln(1+x) \) and \( \frac{1}{1-x} \) to combine them. Start with the first few terms: - \( (x - \frac{x^2}{2} + \frac{x^3}{3})(1 + x + x^2 + x^3) \). - Expand, starting with each term: - \( x \times 1 = x \) - \( x \times x = x^2 \) - \( x \times x^2 = x^3 \) - \( -\frac{x^2}{2} \times 1 = -\frac{x^2}{2} \) - \( -\frac{x^2}{2} \times x = -\frac{x^3}{2} \) - \( \frac{x^3}{3} \times 1 = \frac{x^3}{3} \) - Combine terms: This results in \( x + \frac{x^2}{2} + \left(\frac{1}{3} - \frac{1}{2}\right)x^3 = x + \frac{x^2}{2} - \frac{x^3}{6} \).
05

Verify First Four Nonzero Terms

Among these terms, note that the expansion begins with terms \( x, \frac{x^2}{2}, -\frac{x^3}{6} \). Thus, ensuring that we have found the first three nonzero terms only, the fourth nonzero term would come from sequential higher powers.

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Key Concepts

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

Taylor series
The Taylor series is an incredibly useful tool in mathematics. It allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. A Taylor series for a function \(f(x)\) around a point \(a\) is represented as:- \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \)One special case of the Taylor series is the Maclaurin series, which is centered at 0. This simplifies our formula to:- \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \)By using this series, we can approximate complex functions with polynomials. These polynomials are easier to work with, especially when calculating limits or approximating values for calculations.
It's important to note that the degree of the polynomial we use will determine the accuracy of our approximation.
power series
Power series are a crucial concept in calculus and mathematics in general. They provide an infinite polynomial representation of functions. A power series is generally expressed as:- \( a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots \)In the context of Taylor and Maclaurin series, power series help us break down functions into manageable terms. Each term of the power series involves a coefficient (often derived from derivatives of the function) and increasing powers of \(x\).
One key reason power series are helpful is that they have a radius of convergence, within which they perfectly represent the function. Outside of this radius, though, the series might not converge to the function.
Understanding power series is fundamental when dealing with more advanced series like Taylor and Maclaurin, as they form the backbone of these expansions.
geometric series
A geometric series is a simpler form and specific type of power series where each term is a constant multiple of the preceding term. In other words, a geometric series is represented by:- \( a + ar + ar^2 + ar^3 + \cdots \)Here, \(a\) is the first term, and \(r\) is the common ratio between the terms. The sum of an infinite geometric series, where \(|r| < 1\), is given by:- \( \frac{a}{1-r} \)For example, the series for \( \frac{1}{1-x} \) can be written using the formula for a geometric series starting with \(a = 1\) and \(r = x\), resulting in:- \( 1 + x + x^2 + x^3 + \cdots \)This expansion is very handy in a variety of problems, including finding Maclaurin series, because it gives us a simple infinite series representation of a function.
ln(1+x) expansion
The expansion of \(\ln(1+x)\) is a classic example of a power series, more specifically a Maclaurin series. The function \(\ln(1+x)\) can be expanded as:- \(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \)This series alternates signs and is derived by integrating the geometric series of \(\frac{1}{1-x}\). The convergence is valid for \(-1 < x \leq 1\). The key takeaway from this series is its representation as an infinite polynomial. This is particularly valuable in calculus for approximations and solving complex problems like finding the Maclaurin series for combinations of functions, such as \(\frac{\ln(1+x)}{1-x}\). By analyzing this, students can see the practical applications of series expansions.

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Most popular questions from this chapter

For a sequence \(\left\\{a_{n}\right\\}\) the terms of even index are denoted by \(a_{2 k}\) and the terms of odd index by \(a_{2 k+1} .\) Prove that if \(a_{2 k} \rightarrow L\) and \(a_{2 k+1} \rightarrow L,\) then \(a_{n} \rightarrow L\).

Show that if \(M\) is a number less than \(1,\) then the terms of \(\\{n /(n+1)\\}\) eventually exceed \(M .\) That is, if \(M<1\) there is an integer \(N\) such that \(n /(n+1)>M\) whenever \(n>N .\) since \(n /(n+1)<1\) for every \(n,\) this proves that 1 is a least upper bound for \(\\{n /(n+1)\\}\).

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. \(a_{n}=\int_{1}^{n} \frac{1}{x^{p}} d x, \quad p>1\)

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) \(a_{n}=n \sin \frac{1}{n}\)

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