Chapter 3 Practice Problems

Original Korean learning note

Translation of the complete original learning note. Original calculations and claims are preserved; separate editorial clarifications follow.

As I said in #1, I will now straighten out the concepts.

Heat ($Q$)

What is heat~? When asked that question… I… do not really know.

But I think we should accept heat simply as “energy in transit.”

Because,, when we rub our hands, heat appears in them.

The energy of rubbing must have turned into thermal energy, right???

Besides rubbing our hands, generating heat involves work,

or is it because we can do work with heat??? I should just shut my mouth…. T_T T_T T_T

I found a welcome passage in the book, so I will copy it down and move on.

“Because heat has meaning only when it is ‘in transit,’

there is no instrument that can read the amount of heat contained in a body, and it cannot be read.”

(Later, we will learn that a body contains energy.

Therefore, in principle, it is possible to measure the amount of energy contained in a body.)

I welcomed that passage by taking it to mean something like, “We did not know what heat was in the first place??”

This is what the book says. When I read this part, I felt even more confused,

but around Chapter 4 I understood a ti~~ny bit of what it meant.

For now, I think I should just move on.

Heat capacity ($C$)

Since we defined heat earlier as “energy in transit,” apparently the commonly used term “heat capacity” is slightly illogical.

“Energy capacity” would be a more appropriate expression, but historically people have used the other term a lot (even while knowing it is wrong), so we will just keep using it too.

And that heat capacity $C$ is

$$ C=\frac{dQ}{dT} $$

What it means is: “How much heat $dQ$ is needed to raise the temperature of a body by $dT$?”

And let us take along the advice that we should really look at “energy.”

However, this leaves out information about “the amount of the body,”

so

we can classify heat capacity according to whether it is per unit mass, per unit volume, and so on.

$$ \begin{aligned} C_M&=\frac C M\ [\mathrm J/\mathrm K\cdot\mathrm{kg}]\\ C_{n_m}&=\frac C M\ [\mathrm J/\mathrm K\cdot\mathrm{mol}]\\ C_V&=\frac C M\ [\mathrm J/\mathrm K\cdot\mathrm m^3] \end{aligned} $$

And and and and and there is this too.

We can also classify heat capacity as constant-volume or constant-pressure heat capacity!

Constant-volume heat capacity:

$$ C_V=\left(\frac{\partial Q}{\partial T}\right)_V $$

Differentiate partially with respect to $T$, holding $V$ constant.

Constant-pressure heat capacity:

$$ C_p=\left(\frac{\partial Q}{\partial T}\right)_p $$

Differentiate partially with respect to $T$, holding $p$ constant.

I will not deal with that in detail here.

For now, just think, “Oh, such a thing exists~,” and move on. We will cover it in great detail later. No worries, nope nope nope.

Chapter 3: Probability

Before starting quantum mechanics, I said a little about the necessary probability and statistics in Chapter 1 of the quantum book,

and here it is exactly the same. So I am just going to skip over it.

But this is something we learned in high school, and you have probably forgotten it, so I will summarize it before moving on. (I had forgotten it,)

Binomial distribution

I will prove it later, when solving the problems. For now I will just throw the results out there.

The discrete probability distribution for $k$ successes from $n$ Bernoulli trials (for example, coin tosses) is $P(n,k)$!

The mean of this distribution is $\langle k\rangle=np$.

The variance is $\sigma(k)^2=np(1-p)$.

There is also something I did not know, but I will treat it as general knowledge and move on.

The $k$th moments about the mean

1. First moment about the mean

$$ \left\langle(x-\langle x\rangle)^1\right\rangle=\text{mean deviation} $$

2. Second moment about the mean

$$ \left\langle(x-\langle x\rangle)^2\right\rangle=\text{mean squared deviation (= variance)} $$

3. Third moment about the mean—skewness

$$ \left\langle(x-\langle x\rangle)^3\right\rangle=\text{mean cubed deviation} $$

4. Fourth moment about the mean—kurtosis

$$ \left\langle(x-\langle x\rangle)^4\right\rangle=\text{mean fourth-power deviation} $$

Ex 1.3

Consider a system consisting of ten atoms.

These atoms exist in one of two states: energy 0 units or energy 1 unit.

This “unit” of energy is called an energy quantum.

1. In the quantum state with energy 10, and

2. In the quantum state with energy 4,

how many possible quantum arrangements can the system have?

1. No questions, no arguing.

$$ 1 $$

2. No questions, no picking it apart.

$$ {}_{10}C_4=\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2}=210 $$

Ex 1.4

Calculate the size of $10^{23}!$.

$$ \begin{aligned} \ln n!&\approx n\ln n-n\\ \ln10^{23}!&\approx10^{23}\ln10^{23}-10^{23}\\ &=23\,10^{23}\ln10-10^{23}\\ &=10^{23}(23\ln10-1)\\ &=51.9\times10^{23}\approx52.0\times10^{23} \end{aligned} $$$$ \begin{aligned} \ln10^{23}!&=52.0\times10^{23}\\ 10^{23}!&=e^{52.0\times10^{23}} \end{aligned} $$

Let us say $e^{52.0\times10^{23}}=10^y$.

$$ \begin{aligned} y\ln10&=52.0\times10^{23}\\ y&=\frac1{\ln10}\,52.0\times10^{23}=2.25\times10^{24}\\ \therefore\quad10^{23}!&=10^y=10^{2.25\times10^{24}}\\ &=10^{2250000000000000000} \end{aligned} $$

Ex 2.1

A 1 kW electric heater is switched on for ten minutes. How much heat is generated?

W (watt): electrical energy consumed in one second; $1\,\mathrm W=1\,\mathrm{J/s}$.

An electric heater consuming 1 kW of power consumes 1000 J of energy per second.

Ten minutes is 600 seconds.

Over 600 seconds, it will have consumed a total of 6000000 J.

“Heat” is energy…. Heat $Q=600\,\mathrm{kJ}$.

Ex 2.2

At room temperature, the heat capacity of 0.125 kg of water is 523 J/K.

Calculate the heat capacity of water a) per unit mass and b) per unit volume.

At room temperature, the heat capacity of 0.125 kg of water is $C=523\,\mathrm{J/K}$.

Per unit mass:

$$ C_M=523\,\mathrm{J/K}\,\frac1{0.125\,\mathrm{kg}}=4184\,\mathrm J/\mathrm K\cdot\mathrm{kg}=4.184\,\mathrm J/\mathrm K\cdot\mathrm g $$

Heat capacity per unit mass is specifically called specific heat capacity.~

Per unit volume → Water with a volume of $\mathrm m^3$ has a mass of 1000 kg, you see.

So having 0.125 kg means having water with a volume of $0.000123\,\mathrm m^3$.

Heat capacity per unit volume:

$$ =523\,\mathrm{J/K}\,\frac1{0.000125\,\mathrm m^3}=4184000\,\mathrm J/\mathrm K\cdot\mathrm m^3 $$

Prob 3.3

There is a discrete probability distribution known as the Poisson distribution.

Let $x$ be a discrete random variable taking values $0,1,2,\ldots$.

If the probability $P(x)$ is given as follows, we say that $x$ follows a Poisson distribution.

$$ P(x)=\frac{e^{-m}\cdot m^x}{x!} $$

a) Show that $P(x)$ is normalized, with

$$ \sum_{x=0}^\infty P(x)=1 $$ $$ \sum_{x=0}^\infty\frac{e^{-m}\cdot m^x}{x!}=e^{-m}\sum_{x=0}^\infty\frac{m^x}{x!} $$

→ Expand it, go go!

$$ \begin{aligned} e^{-m}\sum_{x=0}^\infty\frac{m^x}{x!}&=e^{-m}\left(1+\frac{m^1}{1!}+\frac{m^2}{2!}+\frac{m^3}{3!}+\cdots\right)\\ &=e^{-m}\cdot e^m=1 \end{aligned} $$

Taylor expansion:

$$ e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots $$

b) Show that the mean of the probability distribution is

$$ \langle x\rangle=\sum xP(x)=m $$ $$ \begin{aligned} \langle x\rangle&=\sum xP(x)=\sum x\frac{e^{-m}\cdot m^x}{x!}\\ &=e^{-m}\sum x\frac{m^x}{x!}\\ &=e^{-m}\left(0+\frac1{1!}m+\frac2{2!}m^2+\frac3{3!}m^3+\cdots\right) \end{aligned} $$

Cancel-to-the-terms, go go!

$$ =e^{-m}\left(0+m+\frac1{1!}m^2+\frac1{2!}m^3+\cdots\right) $$

Take out one $m$ from each term.

$$ \begin{aligned} &=e^{-m}m\left(0+1+\frac1{1!}m^1+\frac1{2!}m^2+\cdots\right)\\ &=e^{-m}me^m=m\\ \therefore\quad\langle x\rangle&=m \end{aligned} $$

c) The Poisson distribution is useful for describing very rare events that are independent and whose rate of occurrence does not change in our region of interest. Examples include the number of newborns with disabilities recorded each year, annual traffic accidents at a particular intersection, typos on a page, and Geiger-counter activations per minute.

In fact, the first recorded measurement of the Poisson distribution, motivated by Poisson himself,

concerned a very rare accident such as someone in the Prussian army being kicked to death by a horse.

The numbers of soldiers kicked to death by horses in the Prussian army were recorded over twenty years, from 1875 to 1894, in ten corps, as follows.

First, let us find the probabilities.

Let us calculate the expectation with these.

$$ \begin{aligned} &0\cdot P(0)+1\cdot P(1)+2P(2)+3P(3)+4P(4)\\ &=0\cdot\frac{109}{200}+1\cdot\frac{65}{200}+2\cdot\frac{22}{200}+3\cdot\frac3{200}+4\cdot\frac1{200}\\ &=\frac{65+44+9+4}{200}\\ &=\frac{122}{200}\\ &=0.61 \end{aligned} $$

$\langle x\rangle=0.61$: we already found out that this is the $m$ in the Poisson distribution, you know!?!?!

$$ P(x)=\frac{e^{-0.61}\cdot(0.61)^x}{x!} $$

We can say that, then.

Let us check!

Let us check.

$$ P(0)=\frac{e^{-0.61}\cdot1}{0!}=e^{-0.61}=0.543350\ldots $$

(Oh, similar to 0.545!)

$$ P(1)=\frac{e^{-0.61}(0.61)^1}{1!}=e^{-0.61}\times(0.61)=0.331444\ldots $$

(Wow… similar to 0.325… Goosebumps.)

$$ P(2)=\frac{e^{-0.61}(0.61)^2}{2!}=\frac12(P(1))(0.61)=0.10109\ldots $$

(Damn! This is amazingly cool—similar to 0.11!)

$$ P(3)=\frac{e^{-0.61}(0.61)^3}{3!}=\frac16(P(2)\cdot2)(0.61)=0.2055\ldots $$

(Damn, the gap really is starting to widen… 0.15.)

$$ P(4)=\frac{e^{-0.61}(0.61)^4}{4!}=0.00313\ldots $$

(The gap from 0.05 is widening again…)

Prob 3.4

This problem concerns a continuous probability distribution known as the exponential distribution.

Let $x$ be

$$ x\ge0 $$

a continuous random variable taking arbitrary values satisfying this condition.

If the probability between

$$ x $$

and

$$ x+dx $$

is given as follows, we call this the exponential distribution.

$$ P(x)dx=Ae^{-x/\lambda}dx $$

($\lambda,A$ are constants.)

a) Find the value of $A$ that makes $P(x)$ a probability distribution normalized as

$$ \int_0^\infty p(x)dx=1 $$ $$ \begin{aligned} \int_0^\infty Ae^{-x/\lambda}\,dx&=A\left[-\lambda e^{-\frac x\lambda}\right]_0^\infty=A\lambda=1\\ \therefore\quad A&=\frac1\lambda\\ \therefore\quad P(x)&=\frac1\lambda e^{-\frac x\lambda} \end{aligned} $$

b) Show that the mean of this probability distribution is $\langle x\rangle=\lambda$.

$$ \langle x\rangle=\int_0^\infty xP(x)dx=\int_0^\infty x\frac1\lambda e^{-\frac x\lambda}dx $$

Let

$$ \begin{aligned} \frac x\lambda&=t\\ \frac1\lambda dx&=dt\\ dx&=\lambda dt \end{aligned} $$$$ \begin{aligned} &=\lambda\int_0^\infty te^{-t}dt=\lambda\Gamma(2)=\lambda\\ \therefore\quad\langle x\rangle&=\lambda \end{aligned} $$

'

c) Find the variance and standard deviation.

$$ \langle x^2\rangle=\int_0^\infty x^2P(x)dx=\int_0^\infty\frac{x^2}{\lambda}e^{-\frac x\lambda}dx $$

Let

$$ \frac x\lambda=t\quad//\quad\frac1\lambda dx=dt\quad//\quad x^2=\lambda^2t^2 $$$$ =\int_0^\infty\lambda^2t^2e^{-t}dt=\lambda^2\Gamma(3)=2\lambda^2 $$

Variance:

$$ \sigma^2=\langle x^2\rangle-\langle x\rangle^2=2\lambda^2-\lambda^2=\lambda^2 $$

Standard deviation:

$$ \sigma=\lambda $$

Prob 3.5

$\theta$ is a continuous random variable, uniformly distributed between $0$ and $\pi$.

Write an expression for $P(\theta)$. And write the values of the following averages.

Since it is uniformly distributed between $0$ and $\pi$, $P(\theta)=1/\pi$.

$$ \begin{aligned} \text{a)}\quad\langle\theta\rangle&=\int_0^\pi\theta P(\theta)d\theta=\int_0^\pi\theta\frac1\pi d\theta\\ &=\frac1\pi\left[\frac12\theta^2\right]_0^\pi=\frac1\pi\left(\frac12\pi^2\right)=\frac\pi2\\ \text{b)}\quad\left\langle\theta-\frac\pi2\right\rangle&=\int_0^\pi\left(\theta-\frac\pi2\right)P(\theta)d\theta\\ &=\frac1\pi\int_0^\pi\left(\theta-\frac\pi2\right)d\theta\\ &=\frac1\pi\left[\frac12\theta^2-\frac\pi2\theta\right]_0^\pi=0\\ \text{c)}\quad\langle\theta^2\rangle&=\int_0^\pi\theta^2P(\theta)d\theta=\frac1\pi\int_0^\pi\theta^2d\theta\\ &=\frac1\pi\left[\frac13\theta^3\right]_0^\pi=\frac{\pi^2}3\\ \text{d)}\quad\langle\theta^n\rangle&=\frac1\pi\int_0^\pi\theta^n d\theta\\ &=\frac1\pi\left[\frac1{n+1}\theta^{n+1}\right]_0^\pi=\frac{\pi^n}{n+1} \end{aligned} $$$$ \begin{aligned} \text{e)}\quad\langle\cos\theta\rangle&=\int_0^\pi\cos\theta P(\theta)d\theta=\frac1\pi\int_0^\pi\cos\theta d\theta\\ &=\frac1\pi[\sin\theta]_0^\pi=0\\ \text{f)}\quad\langle\sin\theta\rangle&=\int_0^\pi\sin\theta P(\theta)d\theta=\frac1\pi\int_0^\pi\sin\theta d\theta\\ &=\frac1\pi[-\cos\theta]_0^\pi=\frac1\pi\\ \text{g)}\quad\langle|\cos\theta|\rangle&=\int_0^\pi|\cos\theta|P(\theta)d\theta\\ &=\frac1\pi\left(\int_0^{\pi/2}\cos\theta d\theta+\int_{\pi/2}^\pi-\cos\theta d\theta\right)\\ &=\frac1\pi\left\{[\sin\theta]_0^{\pi/2}-[\sin\theta]_{\pi/2}^\pi\right\}\\ &=\frac1\pi\{(1-0)-(0-1)\}=\frac2\pi \end{aligned} $$$$ \begin{aligned} \text{h)}\quad\langle\cos^2\theta\rangle&=\int_0^\pi\cos^2\theta P(\theta)d\theta=\frac1\pi\int_0^\pi\cos^2\theta d\theta\\ &=\frac1\pi\int_0^\pi\frac{1+\cos2\theta}2d\theta\\ &=\frac1{2\pi}\left[\theta+\frac12\sin2\theta\right]_0^\pi=\frac1{2\pi}(\pi+0)=\frac12\\ \text{i)}\quad\langle\sin^2\theta\rangle&=\int_0^\pi\sin^2\theta P(\theta)d\theta=\frac1\pi\int_0^\pi\frac{1-\cos2\theta}2d\theta\\ &=\frac1{2\pi}\left[\theta-\frac12\sin2\theta\right]_0^\pi=\frac12\\ \text{j)}\quad\langle\sin^2\theta+\cos^2\theta\rangle&=\int_0^\pi1\cdot P(\theta)d\theta\\ &=\frac1\pi\int_0^\pi d\theta=\frac1\pi\cdot\pi=1 \end{aligned} $$

Prob 3.7

When $n\gg1$ but $np$ is small, show that the binomial distribution is approximated by the Poisson distribution.

(This means $p\ll1$, so success is very rare.)

Start!

The probability in the binomial distribution:

$$ P(n,k)={}_n C_kp^k(1-p)^{n-k} $$ $$ \begin{aligned} \langle\text{success}\rangle&=\sum_{k=0}^n k\cdot{}_n C_kp^k(1-p)^{n-k}\\ &=\sum_{k=1}^n k\cdot{}_n C_kp^k(1-p)^{n-k} \end{aligned} $$$$ \begin{aligned} k\cdot{}_n C_k&=k\frac{n!}{k!(n-k)!}\\ &=\frac{n!}{(k-1)!(n-k)!}\\ &={\color{red}\frac{n(n-1)!}{(k-1)!(n-k)!}}=n\cdot{\color{red}{}_{n-1}C_{k-1}} \end{aligned} $$ $$ k\cdot{}_n C_k=n\cdot{}_{n-1}C_{k-1} $$

Another proof:

The number of ways to choose $k$ people from $n$, then choose one captain from those $k$ → $k\cdot{}_n C_k$.

The number of ways to choose one captain from $n$ people, then choose $k-1$ people from the remaining $n-1$ → $n\cdot{}_{n-1}C_{k-1}$.

They count the same cases, do they not? So

$$ k\cdot{}_n C_k=n\cdot{}_{n-1}C_{k-1} $$

c) Find the variance and standard deviation.

$$ \langle x^2\rangle=\int_0^\infty x^2P(x)dx=\int_0^\infty\frac{x^2}{\lambda}e^{-\frac x\lambda}dx $$

Let

$$ \frac x\lambda=t\quad//\quad\frac1\lambda dx=dt\quad//\quad x^2=\lambda^2t^2 $$$$ \langle x^2\rangle=\int_0^\infty\lambda^2t^2e^{-t}dt=\lambda^2\Gamma(3)=2\lambda^2 $$

Therefore,

$$ \begin{aligned} \langle\text{success}\rangle&=\sum_{k=1}^n n\cdot{}_{n-1}C_{k-1}\cdot p^k(1-p)^{n-k}\\ &=\sum_{k=1}^n np\cdot{}_{n-1}C_{k-1}\cdot p^{k-1}(1-p)^{n-k}\\ &=np\sum_{k=1}^n{}_{n-1}C_{k-1}\cdot p^{k-1}(1-p)^{n-k} \end{aligned} $$$$ \begin{aligned} n-1&=N\\ k-1&=K\\ k=1&\longrightarrow K=0 \end{aligned} $$$$ \begin{aligned} &=np\sum_{K=0}^N{}_N C_K\cdot p^K(1-p)^{N-K}\\ &=np\bigl(p+(1-p)\bigr)^N=np \end{aligned} $$

Damn, why did I prove the mean of the binomial distribution?

The real start!

The probability of $k$ successes:

$$ p(k)={}_n C_kp^k(1-p)^{n-k} $$

Set $p=m/n$, and let us take the limit later.

$$ \begin{aligned} &={}_n C_kp^k(1-p)^{n-k}\\ &={}_n C_k\left(\frac mn\right)^k\left(1-\frac mn\right)^{n-k}\\ &=\frac{n!}{k!(n-k)!}\frac{m^k}{n^k}\left(1-\frac mn\right)^{n-k}\\ &=\frac{(n-k+1)\cdots(n-1)n}{k!}\frac{m^k}{n^k}\left(1-\frac mn\right)^{n-k} \end{aligned} $$

Now we are going to apply $\lim_{n\to\infty}$ to this:

$$ (n-k+1)\cdots(n-1)n $$

There are $k$ lots of $n$ here!!!! So its highest power of $n$ is $n^k$.

$$ \lim_{n\to\infty}\frac{{\color{red}(n-k+1)\cdots(n-1)n}}{k!}\frac{m^k}{{\color{red}n^k}}\left(1-\frac mn\right)^{n-k} $$

The red bits fly away.

$$ \begin{aligned} &=\lim_{n\to\infty}\frac{m^k}{k!}\left(1-\frac mn\right)^{n-k}\\ &=\frac{m^k}{k!}\lim_{n\to\infty}\left(1-\frac mn\right)^{n-k}\\ &=\frac{m^k}{k!}\lim_{n\to\infty}\left(1-\frac mn\right)^n\left(1-\frac mn\right)^{-k}\\ &=\frac{m^k}{k!}\lim_{n\to\infty}\frac{\left(1-\frac mn\right)^n}{\left(1-{\color{red}\frac mn}\right)^k} \end{aligned} $$

The denominator tends to 1.

$$ =\frac{m^k}{k!}\lim_{n\to\infty}\left(1-\frac mn\right)^n $$

Do you remember advanced calculus in high school?

$$ \lim_{n\to\infty}\left(1+\frac1n\right)^n=e $$

Let us make use of that definition.

$$ =\frac{m^k}{k!}e^{-m}\\ \therefore\quad p(k)=\frac{m^ke^{-m}}{k!} $$

Editorial clarifications separate from the original note

1. Heat is energy transferred because of a temperature difference; work is another mode of energy transfer. A body has internal energy, not a stored amount of heat. Rubbing hands transfers mechanical work and increases internal energy through dissipation; that increase is not itself a definition of heat. With work done by the system taken positive, the first law is $dU=\delta Q-\delta W$. Heat and work depend on the process, so $\delta Q$ denotes an inexact differential; the source’s $dQ$ and partial derivatives of $Q$ are retained above as historical notation. Heat capacity is a valid standard term, and its value requires a specified process. For a simple compressible closed system with only pressure-volume work under the usual equilibrium conditions, $C_V=(\partial U/\partial T)_V$ and $C_p=(\partial H/\partial T)_p$. Internal-energy differences can be measured; an absolute value requires a reference convention.
2. Total heat capacity has units J/K. Mass-specific, molar, and volume-specific capacities divide it by mass, amount in moles, and volume respectively, with units $\mathrm{J/(kg\,K)}$, $\mathrm{J/(mol\,K)}$, and $\mathrm{J/(m^3\,K)}$. The source’s repeated denominator $M$ and dot notation in the three unit expressions are not correct general definitions. Its $C_V$ also means two different things in different passages: per-volume capacity and constant-volume capacity. These must be distinguished. In Ex 2.2, taking water density as $1000\,\mathrm{kg/m^3}$ gives $V=0.000125\,\mathrm m^3$; the source’s prose value $0.000123$ is a typo. The resulting capacities are $4184\,\mathrm{J/(kg\,K)}$ and $4.184\times10^6\,\mathrm{J/(m^3\,K)}$ under that density approximation.
3. In Ex 2.1, $1000\,\mathrm{J/s}\times600\,\mathrm s=600000\,\mathrm J=600\,\mathrm{kJ}$. The source’s 6000000 J is ten times too large. The electrical energy input becomes this amount of heat delivered only under the assumed conversion and energy-accounting conditions; without those assumptions, the energy input, change in stored internal energy, and heat transferred need to be distinguished.
4. The first central moment is zero. The third and fourth central moments are not themselves the dimensionless skewness and kurtosis: for nonzero standard deviation $\sigma$, these are $\mu_3/\sigma^3$ and $\mu_4/\sigma^4$; excess kurtosis subtracts 3. For the two-state atom example, the counts 1 and 210 assume ten distinguishable atom positions with one state per energy level.
5. Ex 1.4 estimates the magnitude of $(10^{23})!$ using leading logarithmic Stirling approximation. Equalities following rounded quantities are approximations. The final written decimal exponent 2250000000000000000 does not equal $2.25\times10^{24}$; its omitted zeros must not be treated as a valid expansion. The unparenthesized source notation $10^{23}!$ means $(10^{23})!$ here, and $\ln n!$ means $\ln(n!)$.
6. In the horse-kick table, $3/200=0.015$ and $1/200=0.005$, rather than the source’s 0.15 and 0.05. The expectation 0.61 is correctly computed from the counts/fractions. At $m=0.61$, $P(3)\approx0.02055$, not the source’s 0.2055; comparison must use the corrected empirical decimals. Similarity of a few observed frequencies does not itself prove independent events or a constant event rate. The historical attribution to Poisson himself also requires qualification: the familiar published horse-kick analysis is associated with Ladislaus von Bortkiewicz’s 1898 work, using the earlier Prussian data.
7. A Poisson distribution uses $m\ge0$ and nonnegative integer $x$. A homogeneous Poisson-process interpretation additionally assumes the appropriate independence and constant-rate conditions. The exponential density here requires $\lambda>0$ and has support $x\ge0$ (zero outside that support); the source’s lowercase $p(x)$ in normalization denotes the same density as $P(x)$. Its mean $\lambda$, variance $\lambda^2$, and standard deviation $\lambda$ then follow.
8. In Prob 3.5(f), $\langle\sin\theta\rangle=2/\pi$, because $[-\cos\theta]_0^\pi=2$. The source’s $1/\pi$ is incorrect. The general moment formula $\langle\theta^n\rangle=\pi^n/(n+1)$ is valid for real $n>-1$ (and thus for the usual nonnegative integer moments).
9. The exponential second-moment calculation repeated in Prob 3.7 is an unrelated source insertion; it is preserved rather than silently deleted. The final binomial-to-Poisson limit holds with fixed nonnegative integer $k$, fixed finite $m=np$, and $n\to\infty$ with $p=m/n$. Merely saying “large $n$” and “small $np$” does not specify the limiting argument. The combinatorial identity used in the intermediate binomial-mean proof applies for $1\le k\le n$.

These bounded clarifications are new editorial material, not translated source prose.