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Chapter 5: Problem 15

a) Applying Lagrange's theorem to the function \(\frac{1}{x^{\alpha}}\), where \(\alpha>0\), show that the inequality $$ \frac{1}{n^{1+\alpha}}<\frac{1}{\alpha}\left(\frac{1}{(n-1)^{\alpha}}-\frac{1}{n^{\alpha}}\right) $$ holds for \(n \in \mathbb{N}\) and \(\alpha>0\) b) Use the result of a) to show that the series \(\sum_{n=1}^{\infty} \frac{1}{n^{\sigma}}\) converges for \(\sigma>1\).

Short Answer

Expert verified
Question: Prove the following inequality for \(n \geq 2\) and \(\sigma > 1\): $$ \frac{1}{n^{\sigma}} < \frac{1}{\sigma-1}\left(\frac{1}{(n-1)^{\sigma-1}} - \frac{1}{n^{\sigma-1}}\right) $$ Then, use the result to show the convergence of the series: $$ \sum_{n=1}^{\infty} \frac{1}{n^{\sigma}} $$

Step by step solution

01

Understand Lagrange's theorem

Lagrange's theorem states that for a function \(f(x)\) continuous on an interval \([a, b]\) and differentiable on \((a, b)\), there exists a point \(c \in (a, b)\) such that $$ f'(c) = \frac{f(b) - f(a)}{b - a} $$ We will apply this theorem to the function \(\frac{1}{x^\alpha}\), where \(\alpha > 0\).
02

Determine the derivative of the function

First, we must find the derivative of the function \(f(x) = \frac{1}{x^\alpha}\). Using the power rule, the derivative is given by: $$ f'(x) = -\alpha x^{-\alpha - 1} $$
03

Apply Lagrange's theorem

Now, we will apply Lagrange's theorem to \(f(x)\) on the interval \([n - 1, n]\), where \(n \in \mathbb{N}\). This means we need to find a point \(c \in (n-1, n)\) such that: $$ -\alpha c^{-\alpha-1} = \frac{\frac{1}{n^{\alpha}} - \frac{1}{(n-1)^{\alpha}}}{n - (n - 1)} $$ Simplifying and rearranging, we get: $$ \frac{1}{c^{\alpha+1}} = \frac{1}{\alpha}\left(\frac{1}{(n-1)^{\alpha}}-\frac{1}{n^{\alpha}}\right) $$
04

Use the inequality of the interval

Since \(c\) is in the interval \((n - 1, n)\), we have \(n - 1 < c < n\). Taking the reciprocal of all three sides (and flipping the inequality signs) gives: $$ n < c^{-1} < n - 1 $$ Now, we'll raise all three sides to the power of \((\alpha + 1)\) and reverse the inequality signs: $$ n^{\alpha+1} > c^{-\alpha-1} > (n-1)^{\alpha+1} $$
05

Use the inequality to prove the result

Recall that: $$ \frac{1}{c^{\alpha+1}} = \frac{1}{\alpha}\left(\frac{1}{(n-1)^{\alpha}}-\frac{1}{n^{\alpha}}\right) $$ Using the inequality we derived, we have: $$ n^{\alpha+1} > \frac{1}{\alpha}\left(\frac{1}{(n-1)^{\alpha}}-\frac{1}{n^{\alpha}}\right) > (n-1)^{\alpha+1} $$ Dividing all three sides by \(n^{\alpha+1}\), we obtain the required inequality: $$ \frac{1}{n^{1+\alpha}}<\frac{1}{\alpha}\left(\frac{1}{(n-1)^{\alpha}}-\frac{1}{n^{\alpha}}\right) $$
06

Show the convergence of the series

Using the result from part a, we have: $$ \frac{1}{n^{\sigma}} < \frac{1}{\sigma-1}\left(\frac{1}{(n-1)^{\sigma-1}} - \frac{1}{n^{\sigma-1}}\right) $$ Now, consider the series: $$ \sum_{n=2}^{\infty} \frac{1}{n^{\sigma}} $$ This series is bounded by the integral: $$ \int_{1}^{\infty} \frac{1}{x^{\sigma}} dx = \frac{1}{\sigma - 1} $$ Using the result obtained in part a and the integral test for convergence, we can conclude that the series converges whenever \(\sigma > 1\). Addition of the first term (\(n=1\)) does not affect convergence due to its finiteness. Therefore, the series: $$ \sum_{n=1}^{\infty} \frac{1}{n^{\sigma}} $$ converges for all \(\sigma > 1\).

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Key Concepts

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

Mathematical Analysis
Mathematical analysis encompasses a broad range of topics in mathematics, dealing with limits and related theories such as differentiation, integration, measure, infinite series, and analytic functions. In the context of this problem, we apply the concepts of differentiation and series within analysis. Understanding these fundamental tools allows us to dissect and understand the behavior of functions and series over real numbers. Differentiation, in particular, is key as it helps us find slopes of tangents and, by extension, analyze changes in functions rigorously. Analyzing the behavior of a function through its derivative offers insights, such as identifying points where a function might increase or decrease.
Series Convergence
Convergence of a sequence or series is a fundamental concept in mathematical analysis. It refers to the situation where, as more terms are added, the series approaches a specific value, also known as the limit. For the series \( \sum_{n=1}^{\infty} \frac{1}{n^{\sigma}} \), the notion of convergence tells us under what conditions this sum approaches a fixed number rather than diverging to infinity.
  • This series is called a p-series, where specific rules apply regarding its convergence.
  • In general, a p-series \( \sum \frac{1}{n^p} \) converges if \( p > 1 \) and diverges if \( p \leq 1 \).
  • This simple rule about the convergence of p-series allows us to quickly determine whether a series like this one will have a definite limit.
Understanding the concept of convergence is crucial not only in theory but also in various applications ranging from engineering to physical sciences.
Integral Test
The integral test is a powerful tool to determine the convergence of infinite series. It connects continuous functions with discrete series. If a function \( f(x) \) is positive, continuous, and decreasing on \( [1, \infty) \) and \( f(n) = a_n \), where \( a_n \) is the term of a series, then the series \( \sum_{n=1}^{\infty} a_n \) and the integral \( \int_{1}^{\infty} f(x) \, dx \) either both converge or both diverge.
  • This method simplifies proving convergence because calculating an integral can often be easier than managing an infinite series.
  • For the series \( \sum_{n=1}^{\infty} \frac{1}{n^{\sigma}} \), using the integral test implies setting up the corresponding integral \( \int_{1}^{\infty} \frac{1}{x^{\sigma}} \, dx \) to investigate convergence.
  • Upon evaluation, if \( \sigma > 1 \), the integral converges, implying that the series does too.
  • This linkage helps us conclude definitively about the behavior of the series.
Power Rule
The power rule is a crucial derivative technique in calculus, facilitating the differentiation of functions of the form \( x^n \). The power rule states that if \( f(x) = x^n \), then its derivative \( f'(x) = nx^{n-1} \). This simple rule significantly aids in handling complex functions by breaking them into manageable pieces through differentiation.
  • In this exercise, the power rule was applied to differentiate \( \frac{1}{x^\alpha} \), giving us \( f'(x) = -\alpha x^{-(\alpha+1)} \).
  • By understanding how to differentiate such functions, we can better manipulate and analyze them.
  • The power rule not only helps in obtaining derivatives but also serves as a basis for more complex operations in calculus.
Mastering the power rule can simplify your approach to many problems that involve calculus, making it an indispensable tool in the mathematical toolkit.

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Most popular questions from this chapter

a) Show that the integral $$ \int x^{m}\left(a+b x^{n}\right)^{p} \mathrm{~d} x $$ whose differential is a binomial, where \(m, n\), and \(p\) are rational numbers, can be reduced to the integral $$ \int(a+b t)^{p} t^{q} \mathrm{~d} t $$ where \(p\) and \(q\) are rational numbers. b) The integral (5.194) can be expressed in terms of elementary functions if one of the three numbers \(p, q\), and \(p+q\) is an integer. (Chebyshev showed that there were no other cases in which the integral (5.194) could be expressed in elementary functions.)

Show that if \(f \in C^{(\infty)}[-1,1]\) and \(f^{(n)}(0)=0\) for \(n=0,1,2, \ldots\), and there exists a number \(C\) such that sup \(_{-1 \leq x \leq 1}\left|f^{(n)}(x)\right| \leq n ! C\) for \(n \in \mathbb{N}\), then \(f \equiv 0\) on \([-1,1]\)

Show that a) any polynomial \(P(x)\) admits a representation in the form \(c_{0}+c_{1}\left(x-x_{0}\right)+\) \(\cdots+c_{n}\left(x-x_{0}\right)^{n}\) b) there exists a unique polynomial of degree \(n\) for which \(f(x)-P(x)=o((x-\) \(\left.x_{0}\right)^{n}\) ) as \(E \ni x \rightarrow x_{0} .\) Here \(f\) is a function defined on a set \(E\) and \(x_{0}\) is a limit point of \(E\) \({ }^{14} \mathrm{Ch}\). Hermite (1822-1901) - French mathematician who studied problems of analysis; in particular, he proved that e is transcendental.

Write the formulas for approximate computation of the following values: a) \(\sin \left(\frac{\pi}{6}+\alpha\right)\) for values of \(\alpha\) near 0 ; b) \(\sin \left(30^{\circ}+\alpha^{\circ}\right)\) for values of \(\alpha^{\circ}\) near 0; c) \(\cos \left(\frac{\pi}{4}+\alpha\right)\) for values of \(\alpha\) near 0 ; d) \(\cos \left(45^{\circ}+\alpha^{\circ}\right)\) for values of \(\alpha^{\circ}\) near 0 .

Efficiency in rocket propulsion. a) Let \(Q\) be the chemical energy of a unit mass of rocket fuel and \(\omega\) the outflow speed of the fuel. Then \(\frac{1}{2} \omega^{2}\) is the kinetic energy of a unit mass of fuel when ejected. The coefficient \(\alpha\) in the equation \(\frac{1}{2} \omega^{2}=\alpha Q\) is the efficiency of the processes of burning and outflow of the fuel. For engines of solid fuel (smokeless powder) \(\omega=2 \mathrm{~km} / \mathrm{s}\) and \(Q=1000 \mathrm{kcal} / \mathrm{kg}\), and for engines of liquid fuel (gasoline with oxygen) \(\omega=3 \mathrm{~km} / \mathrm{s}\) and \(Q=2500 \mathrm{kcal} / \mathrm{kg} .\) Determine the efficiency \(\alpha\) for these cases. b) The efficiency of a rocket is defined as the ratio of its final kinetic energy \(m_{\mathrm{R}} \frac{v^{2}}{2}\) to the chemical energy of the fuel burned \(m_{\mathrm{F}} Q .\) Using formula \((5.139)\), obtain a formula for the efficiency of a rocket in terms of \(m_{\mathrm{R}}, m_{\mathrm{F}}, Q\), and \(\alpha\) (see part a)). c) Evaluate the efficiency of an automobile with a liquid-fuel jet engine, if the automobile is accelerated to the usual city speed limit of \(60 \mathrm{~km} / \mathrm{h}\). d) Evaluate the efficiency of a liquid-fuel rocket carrying a satellite into low orbit around the earth. e) Determine the final speed for which rocket propulsion using liquid fuel is maximally efficient. f) Which ratio of masses \(m_{\mathrm{F}} / m_{\mathrm{R}}\) yields the highest possible efficiency for any kind of fuel?
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