/*! 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 three nonzero ter... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the first three nonzero terms of the Maclaurin series for each function and the values of \(x\) for which the series converges absolutely. \(f(x)=(\sin x) \ln (1+x)\)

Short Answer

Expert verified
The first three nonzero terms are \( x^2, -\frac{x^3}{2}, -\frac{x^4}{6} \) and the series converges for \( |x| < 1 \).

Step by step solution

01

Recall the Maclaurin series

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \dots \) The series is a special case of the Taylor series evaluated at \( a=0 \).
02

Find the Maclaurin series for \( \sin(x) \) and \( \ln(1+x) \)

The Maclaurin series for \( \sin x \) is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} + \dots \). The Maclaurin series for \( \ln(1+x) \) is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \).
03

Determine product of series terms up to required order

To find the Maclaurin series of \( f(x) = \sin x \ln(1+x) \), multiply the first few terms of both series:\[\left(x - \frac{x^3}{6} + \frac{x^5}{120}\right) \times \left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)\]
04

Distribute and collect terms

Calculate the product term by term:- First nonzero term: \( x \cdot x = x^2 \)- Next term: \( x \cdot -\frac{x^2}{2} = -\frac{x^3}{2} \)- Next terms: \( -\frac{x^3}{6} \cdot x = -\frac{x^4}{6} \) and combine with other terms of same order to continue with next powers.Focus on collecting terms up to the cubic term.
05

Identify the first three nonzero terms

By distributing and collecting terms, the first three nonzero terms are identified as:- \( x^2 \)- \( -\frac{x^3}{2} \)- \( -\frac{x^4}{6} \)Adding these gives the first three nonzero terms of the series.
06

Determine the interval of absolute convergence

The series \( \sin(x) \) converges for all \( x \) and \( \ln(1+x) \) converges for \( |x| < 1 \). Hence the product \( f(x) \) converges absolutely for \( |x| < 1 \).

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Key Concepts

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

Convergence of Series
A series is a sum of terms that follows a specific rule or pattern. Converging series are the ones where the sum approaches a finite value as you add more terms. This behavior is vital in mathematics because it tells us whether a series gives us a meaningful result when extended infinitely. In the context of series convergence, it is important to focus on absolute convergence. A series is said to converge absolutely if the series formed by taking the absolute value of each term also converges.
For instance, in the given exercise, the function involves the product of two series: the Maclaurin series expansions of \( \sin(x) \) and \( \ln(1+x) \). While the \( \sin(x) \) series converges for all \( x \), the \( \ln(1+x) \) series only converges for \( |x| < 1 \).
By multiplying these series, it was determined that the product also converges absolutely within the interval \( |x| < 1 \). This insight is crucial in understanding which values of \( x \), the series sum is reliable and meaningful.
Taylor Series
The Taylor series is a way to represent functions as infinite sums of terms derived from their derivatives at a single point. The Maclaurin series is a special case of the Taylor series created when the expansion point is zero. This makes Maclaurin series especially useful for approximating functions near the origin.
Mathematically, a function \( f(x) \) can be expressed using its Taylor series as:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots \)
When \( a=0 \), this becomes the Maclaurin series:
  • \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \dots \)
In our exercise, the product of the Maclaurin series for \( \sin(x) \) and \( \ln(1+x) \) helps us find the series expansion for a more complex function, \( f(x) = \sin(x) \ln(1+x) \).
The function is represented around \( x=0 \), allowing for efficient approximation and calculation of the function values nearby zero.
Series Expansion
Series expansion is about breaking down complex functions into simpler components, expressed as a sum of terms. This process makes analysis and computation more manageable. The series expansion transforms functions into infinite polynomials.
For example, the Maclaurin series expansion for \( \sin(x) \) is:
  • \( x - \frac{x^3}{6} + \frac{x^5}{120} + \dots \)
And for \( \ln(1+x) \), it is:
  • \( x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \)
To find the Maclaurin series of \( f(x) = \sin(x) \ln(1+x) \), we multiply and collect terms between these two expansions. The resulting terms describe \( f(x) \) as a series expansion.
This multiplication and combination, outlined in the steps of the exercise, yield the first three non-zero terms of the resulting Maclaurin series: \( x^2 \), \( -\frac{x^3}{2} \), and \( -\frac{x^4}{6} \). These steps help compactly and efficiently express \( f(x) \), providing valuable approximations around zero.

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Most popular questions from this chapter

Suppose that \(a_{n}>0\) and \(b_{n}>0\) for \(n \geq N(N\) an integer). If \(\lim _{n \rightarrow \infty}\left(a_{n} / b_{n}\right)=\infty\) and \(\sum a_{n}\) converges, can anything be said about \(\sum b_{n}^{?}\) Give reasons for your answer.

Use series to evaluate the limits. \begin{equation} \lim _{y \rightarrow 0} \frac{y-\tan ^{-1} y}{y^{3}} \end{equation}

It is not yet known whether the series \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n^{3} \sin ^{2} n}\end{equation} converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps. \begin{equation} \begin{array}{l}{\text { a. Define the sequence of partial sums }}\end{array} \end{equation} \begin{equation} s_{k}=\sum_{n=1}^{k} \frac{1}{n^{3} \sin ^{2} n}. \end{equation} \begin{equation} \begin{array}{l}{\text { What happens when you try to find the limit of } s_{k} \text { as } k \rightarrow \infty ?} \\ {\text { Does your CAS find a closed form answer for this limit? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Plot the first } 100 \text { points }\left(k, s_{k}\right) \text { for the sequence of partial }} \\ \quad {\text { sums. Do they appear to converge? What would you estimate }} \\ \quad {\text { the limit to be? }} \\ {\text { c. Next plot the first } 200 \text { points }\left(k, s_{k}\right) . \text { Discuss the behavior in }} \\ \quad {\text { your own words. }} \\ {\text { d. Plot the first } 400 \text { points }\left(k, s_{k}\right) \text { . What happens when } k=355 \text { ? }} \\\ \quad {\text { Calculate the number } 355 / 113 . \text { Explain from you calculation }} \\ \quad {\text { what happened at } k=355 . \text { For what values of } k \text { would you }} \\ \quad {\text { guess this behavior might occur again? }} \end{array} \end{equation}

Approximation properties of Taylor polynomials Suppose that \(f(x)\) is differentiable on an interval centered at \(x=a\) and that \(g(x)=b_{0}+b_{1}(x-a)+\cdots+b_{n}(x-a)^{n}\) is a polynomial of degree \(n\) with constant coefficients \(b_{0}, \ldots, b_{n}\) . Let \(E(x)=\) \(f(x)-g(x) .\) Show that if we impose on \(g\) the conditions i) \(E(a)=0\) ii) $$\lim _{x \rightarrow a} \frac{E(x)}{(x-a)^{n}}=0$$ then $$\begin{array}{r}{g(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots} \\ {+\frac{f^{(n)}(a)}{n !}(x-a)^{n}}\end{array}$$ Thus, the Taylor polynomial \(P_{n}(x)\) is the only polynomial of degree less than or equal to \(n\) whose error is both zero at \(x=a\) and negligible when compared with \((x-a)^{n}.\)

Show that the sum of the first 2\(n\) terms of the series $$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots$$ is the same as the sum of the first \(n\) terms of the series $$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\frac{1}{5 \cdot 6}+\cdots$$ Do these series converge? What is the sum of the first \(2 n+1\) terms of the first series? If the series converge, what is their sum?

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