/*! 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 Obtain equation \((9-15)\) fro \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 30

Obtain equation \((9-15)\) fro \(m(9-\mid 4)\). Make use of the following sums, correct when \(|x|<1\) : $$\begin{aligned}\sum_{n=0}^{\infty} x^{n} &=\frac{1}{1-x} \\\\\sum_{n=0}^{\infty} n x^{n} &=\frac{x}{(1-x)^{2}}\end{aligned}$$

Short Answer

Expert verified
-1

Step by step solution

01

Simplify the equation

First, simplify the given equation \(m(9-|4|)\). Here, the absolute value |4| is 4. Therefore, the equation becomes \(m \times 5\), which means \(m = \frac{(9-15)}{5}\).
02

Apply the first summation formula

Next, apply the first provided summation formula, \(\sum_{n=0}^{\infty} x^{n} =\frac{1}{1-x}\). This could represent \(m=\frac{1}{1-x}\) if \(x= -1\), because the outcome -5 of the series matches with the numerator of derived equation from Step 1.
03

Apply the second summation formula

Then, consider the second provided summation formula, \(\sum_{n=0}^{\infty} n x^{n} =\frac{x}{(1-x)^{2}}\). This equation could stand for \(x= -1\) again because the outcome -5 of the series matches with the numerator of the derived equation from Step 1, so \(m=\frac{-1}{(1-(-1))^2}\).
04

Compare and verify

Finally, verify both findings from Step 2 and Step 3, and it can be seen that the equations derived from both formulas are consistent with the simplified equation from Step 1: \(m=\frac{-1}{(1-(-1))^2} = \frac{1}{1-(-1)} = \frac{(9-15)}{5}\).

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 91影视!

Key Concepts

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

Geometric series
A geometric series is a sequence of terms where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, in the series \(1, x, x^2, x^3, \ldots\), each term is multiplied by \(x\), which is the common ratio.
Understanding the formula for the sum of an infinite geometric series is crucial. This formula is given by:
  • \(\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}\)
This formula allows us to find the sum of the series as long as the absolute value of the common ratio \(x\) is less than 1. This is important because it ensures that the terms get closer to zero, making the infinite sum finite.
We used this formula in our exercise to help represent part of the expression \(m\) where \(x\) potentially equals -1, though typically conditions like \(|x| < 1\) need to be checked carefully.
Series convergence
Series convergence refers to the concept of determining whether a series adds up to a finite number as more and more terms are included. When a series converges, its sequence of partial sums approaches a specific value.
The condition \(|x| < 1\) for convergence of a geometric series is vital. It ensures that each successive term gets smaller and aggregated locations of sums are fleshed out without bouncing off to infinity. If \(x\) lies outside this range, the terms will not shrink, and the series will either diverge or not exist in any meaningful way.
Consider the exercise where we need to examine the convergence conditions closely, particularly since playing with special scenarios such as \(x = -1\) can yield interesting insights into our calculated result.
Summation techniques
Summation techniques are methods used to handle and simplify series. They enable us to make complex calculations more tractable. One valuable technique involves using known formulas, like the geometric series sum, to simplify expressions.
For example, the exercise makes use of summation formulas
  • \(\sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}\)
  • \(\sum_{n=0}^{\infty} n x^{n} = \frac{x}{(1-x)^2}\)
These formulas allow us to find a generalized sum easily, providing solutions that might be cumbersome through traditional summation means.
In practical contexts like that in the exercise, these techniques allow one to transition from a complex arrangement of numbers to a manageable expression, reflecting both consistency and elegance in solutions.

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

Figure 9.8 cannot do justice to values at the very high. speed end of the plot. This exercise investigates how small it really gets. However, although integrating the Maxwell speed distribution over the full range of speeds from 0 to infinity can be carried out (the so-called Gaussian integrals of Appendix \(K\) ). over any restricted range, it is one of those integrals that. unfortunately. cannot be done in closed form. Using a computational aid of your choice, show that the fraction of molecules moving faster than \(2 v_{\text {ms }}\) is \(\sim 10^{-2}\); faster than \(6 v_{\text {ms }}\) is \(-10^{-23} ;\) and faster than \(10 v_{\mathrm{ms}}\) is \(\sim 10^{-64}\), where \(v_{\mathrm{m} 2}\) from Exercise \(41 .\) is \(\sqrt{3 k_{B} T / m}\). (Exercise 48 uses these values in an interesting application.)

You have six shelves, one above the other and all above the floor, and six volumes of an encyciopedia, A. B. C. D. \(E\), and \(F\). (a) L.ist all the ways you can arrange the volumes with five on the floor and one on the sixth/top shelf. One way might be \(\mid \mathrm{ABCDE},-,-,-,-,-, \mathrm{F}\\}\) (b) List all the ways you can arrange them with four on the floor and two on the third shelf. (c) Show thal there are many more ways, relative to parts \((\mathrm{a})\) and \((\mathrm{b})\), to strange the six volumes with two on the floor and two eachon the first and second shelves. (There are several ways to answer this, but even listing them all won't take forever it's fewer than \(100 .)\) (d) Suddenly, a fantastic change! All six volumes are volume \(X\) - \(\mathrm{it}\) 's impossible to tell them apar. For each of the three distributions described in parts (a), (b), and (c). how many different (distinguishable) ways are there now? (e) If the energy you expend to lift a volume from the floor is proportional to a shelf's height, how do the total energies of distributions (a), (b), and (c) compare? (I) Use these ideas to atgue that the relative probabili. ties of occupying the lowestenergy states should be higher for hosons than for classically distinguishable particles. g) Combine these ideas with a famous principle to atgue that the relative probabilities of occupying the lowest states should be lower for fermions than for classically distinguishable particies

Four distinguishable harmonic oscillators \(a, b, c,\) and \(d\) may exchange energy. The energies allowed particle \(a\) are \(E_{a}=n_{d} h \omega_{0} ;\) those allowed particle \(b\) are \(E_{b}=n_{b} h \omega_{0}\) and so \(\mathrm{cm}\). Consider an overall state (macrustate) in which the total energy is \(3 \hbar \omega_{0}\). One possible microstate would have particles \(\alpha\) b. and \(c\) ' in their \(n=0\) states and particle \(d\) in its \(n=3\) state: that is, \(\left(n_{u}, n_{b}, n_{c}, n_{d}\right)=(0,0,0,3)\) (a) List all possible microstates. (b) What is the probability that a given particle will be in its \(n=0\) state? (c) Answer par (b) for all other possible values of \(n\). (d) Plot the probability versus \(n\).

Determine the density of states \(D(E)\) for a \(2 D\) infinite well (ignoring spin) in which $$ E_{A_{x+} n_{2}}=\left(n_{x}^{2}+n_{y}^{2}\right) \frac{\pi^{2} \hbar^{2}}{2 m L^{2}} $$

Pruve that fur any sine function \(\sin (k x+\phi)\) of wavelength shonter than \(2 a\), where \(a\) is the atomic spacing. there is a sine function with a wavelength longer than \(2 a\) that has the same values at the points \(x=a .20,3 a\). and so on. Wotr: It is probably easier to work with wave number than with wavelength. We seck to show that for every wave number greater than \(\pi / a\) there is an equivalent une less than \(\pi / a\).)
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Particle Model of Matter

Read Explanation

Famous Physicists

Read Explanation

Quantum Physics

Read Explanation

Work Energy and Power

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
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.