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## 2. Continuous Distributions

### Basic Theory

As usual, suppose that we have a random experiment with probability measure $$\P$$ on an underlying sample space $$\Omega$$. A random variable $$X$$ taking values in $$S$$ is said to have a continuous distribution if

$\P(X = x) = 0 \text{ for all } x \in S$

The fact that $$X$$ takes any particular value with probability 0 might seem paradoxical at first, but conceptually it is the same as the fact that an interval of $$\R$$ can have positive length even though it is composed of points each of which has 0 length. Similarly, a region of $$\R^2$$ can have positive area even though it is composed of points (or curves) each of which has area 0.

If $$X$$ has a continuous distribtion then $$\P(X \in C) = 0$$ for any countable $$C \subseteq S$$.

Proof:

Since $$C$$ is countable, it follows from the additivity axiom of probability that

$\P(X \in C) = \sum_{x \in C} \P(X = x) = 0$

Thus, continuous distributions are in complete contrast with discrete distributions, for which all of the probability mass is concentrated on a discrete set. For a continuous distribution, the probability mass is continuously spread over $$S$$. Note also that $$S$$ itself cannot be countable. In the picture below, the light blue shading is intended to suggest a continuous distribution of probability.

#### Probability Density Functions

Suppose again that $$X$$ has a continuous distribution on $$S \subseteq \R^n$$. A real-valued function $$f$$ defined on $$S$$ is said to be a probability density function for $$X$$ if $$f$$ satisfies the following properties:

1. $$f(x) \ge 0$$ for all $$x \in S$$
2. $$\int_S f(x) dx = 1$$
3. $$\P(X \in A) = \int_A f(x) dx$$ for $$A \subseteq S$$

Property (c) in the definition is particularly important since it implies that the probability distribution of $$X$$ is completely determined by the probability density function. Conversely, any function that satisfies properties (a) and (b) is a probability density function, and then property (c) can be used to define a continuous probability distribution on $$S$$.

If $$n \gt 1$$, the integrals in properties (b) and (c) are multiple integrals over subsets of $$\R^n$$ with $$\bs{x} = (x_1, x_2, \ldots, x_n)$$ and $$d \bs{x} = dx_1 dx_2 \cdots dx_n$$. In fact, technically, $$f$$ is a probability density function relative to the standard $$n$$-dimensional measure, which we recall is given by

$\lambda_n(A) = \int_A \bs{1} d \bs{x}, \quad A \subseteq \R^n$

Note that $$\lambda_n(S)$$ must be positive (perhaps infinite). In particular,

1. If $$n = 1$$, $$S$$ must be a subset of $$\R$$ with positive length ($$S$$ is usually an interval).
2. If $$n = 2$$, $$S$$ must be a subset of $$\R^2$$ with positive area.
3. If $$n = 3$$, $$S$$ must be a subset of $$\R^3$$ with positive volume.

However, we recall that except for exposition, the low dimensional cases ($$n \in \{1, 2, 3\}$$) play no special role in probability. Interesting random experiments often involve several random variables (that is, a random vector). Finally, note that we can always extend $$f$$ to a probability density function on all of $$\R^n$$ by defining $$f(x) = 0$$ for $$x \notin S$$. This extension sometimes simplifies notation.

Just as in the discrete case, an element $$x \in S$$ that maximizes the probability density function $$f$$ is called a mode of the distribution. If there is only one mode, it is sometimes used as a measure of the center of the distribution.

Probability density functions of continuous distributions differ from their discrete counterparts in several important ways:

• A probability density function $$f$$ for a continuous distribution is not unique. Note that the values of $$f$$ on a finite (or even countable) set of points could be changed to other nonnegative values, and properties (a), (b), and (c) would still hold. The critical fact is that only integrals of $$f$$ are important.
• Note that it is possible to have $$f(x) \gt 1$$ for some (or even all) $$x \in S$$. Indeed, $$f$$ can be unbounded on $$S$$.
• Finally, note that $$f(x)$$ is not a probability; it is the probability density at $$x$$. That is, $$f(x) dx$$ is approximately the probability that $$X$$ is in a small region of size $$dx$$ about $$x$$.

#### Constructing Probability Density Functions

Suppose that $$g$$ is a nonnegative function on $$S \subseteq \R^n$$. Let

$c = \int_S g(x) dx$

If $$0 \lt c \lt \infty$$ then $$f(x) = \frac{1}{c} g(x)$$ for $$x \in S$$ defines a probability density function on $$S$$.

Note that $$f$$ is just a scaled version of $$g$$. Thus, the result in the last exercise can be used to construct probability density functions with desired properties (domain, shape, symmetry, and so on). The constant $$c$$ is sometimes called the normalizing constant.

#### Conditional Densities

Suppose that $$X$$ is a random variable taking values in $$S \subseteq \R^n$$ with a continuous distribution that has probability density function $$f$$. The probability density function of $$X$$, of course, is based on the underlying probability measure $$\P$$ on the sample space $$\Omega$$. This measure could be a conditional probability measure, conditioned on a given event $$E \subseteq \Omega$$ (with $$\P(E) \gt 0$$ of course). The usual notation is

$f(x \mid E), \quad x \in S$

Note, however, that except for notation, no new concepts are involved. The function above is a probability density function for a continuous distribution. That is, it satisfies properties (a) and (b) while property (c) becomes

$\int_A f(x \mid E) dx = \P(X \in A \mid E)$

All results that hold for probability density functions in general have analogies for conditional probability density functions.

Suppose that $$B \subseteq S$$ with $$\P(X \in B) \gt 0$$. The conditional probability density function of $$X$$ given $$X \in B$$ is

$f(x \mid X \in B) = \frac{f(x)}{\P(X \in B)}$
Proof:

For $$A \subseteq B$$,

$\int_A \frac{f(x)}{\P(X \in B)} \, dx = \frac{1}{\P(X \in B)} \int_A f(x) \, dx = \frac{\P(X \in A)}{\P(X \in B)} = \P(X \in A \mid X \in B)$

### Examples and Applications

#### The Exponential Distribution

Let $$f(t) = r e^{-r t}$$ for $$t \ge 0$$, where $$r \gt 0$$ is a parameter. Then

1. $$f$$ is a probability density function.
2. The graph of $$f$$ is decreasing and concave upward on $$[0, \infty)$$
3. The mode occurs at $$t = 0$$.

The distribution defined by the probability density function in the previous exercise is called the exponential distribution with rate parameter $$r$$. This distribution is frequently used to model random times, under certain assumptions. Specifically, in the Poisson model of random points in time, the times between successive arrivals have independent exponential distributions, and the parameter $$r$$ is the average rate of arrivals. The exponential distribution is studied in detail in the chapter on Poisson Processes.

The lifetime $$T$$ of a certain device (in 1000 hour units) has the exponential distribution with parameter $$r = \frac{1}{2}$$. Find

1. $$\P(T \gt 2)$$
2. $$\P(T \gt 3 \mid T \gt 1)$$
1. $$e^{-1} \approx 0.3679$$
2. $$e^{-1} \approx 0.3679$$

In the gamma experiment, set $$n =1$$ to get the exponential distribution. Vary the rate parameter $$r$$ and note the shape of the probability density function. For various values of $$r$$, run the simulation 1000 times and note the apparent convergence of the empirical density function to the true probability density function.

#### A Random Angle

In Bertrand's problem, a certain random angle $$\Theta$$ has probability density function $$f(\theta) = \sin(\theta)$$ for $$0 \le \theta \le \frac{\pi}{2}$$.

1. Show that $$f$$ is a probability density function.
2. Graph $$f$$ and identify the mode.
3. Find $$\P(\Theta \lt \frac{\pi}{4})$$.
1. mode $$\theta = \frac{\pi}{2}$$
2. $$1 - \frac{1}{\sqrt{2}} \approx 0.2929$$

Bertand's problem is named for Joseph Louis Bertrand and is studied in more detail in the chapter on Geometric Models.

In Bertrand's experiment, select the model with uniform distance. Run the simulation 1000 times and compute the empirical probability of the event $$\{\Theta \lt \frac{\pi}{4}\}$$. Compare with the true probability in the previous exercise.

#### Gamma Distributions

Let $$g_n(t) = e^{-t} \frac{t^n}{n!}$$ for $$t \ge 0$$ where $$n \in \N$$ is a parameter. Then

1. $$g_n$$ is a probability density function for each $$n$$.
2. $$g_n(t)$$ is strictly increasing for $$t \lt n$$ and strictly decreasing for $$t \gt n$$.
3. The distribution is unimodal with mode at $$t = n$$.
Proof:

For (a), use induction on $$n$$.

Remarkably, we showed in the last section on discrete distributions, that $$f_t(n) = g_n(t)$$ is a probability density function on $$\N$$ for each $$t \ge 0$$ (it's the Poisson distribution with parameter $$t$$). The distribution defined by the probability density function $$g_n$$ is an example of a gamma distribution; $$n + 1$$ is called the shape parameter. The family of gamma distributions is studied in more generality in the chapter on Poisson Processes. Note that the special case $$n = 0$$ gives the exponential distribution with rate parameter 1.

Suppose that the lifetime of a device $$T$$ (in 1000 hour units) has the gamma distribution in Exercise 9 with $$n = 2$$. Find $$\P(T \gt 3)$$.

$$\frac{17}{2} e^{-3} \approx 0.4232$$

In the gamma experiment, set the rate parameter $$r = 1$$ and the shape parameter $$k = 3$$ to get the distribution in the previous exercise. Run the experiment 1000 times and compute the empirical probability of the event $$\{T \gt 3\}$$. Compare with the theoretical probability in the previous exercise.

#### Beta Distributions

Let $$g(x) = x (1 - x)$$ for $$0 \le x \le 1$$.

1. Sketch the graph of $$g$$.
2. Find the probability density function $$f$$ proportional to $$g$$.
3. Find the mode.
4. Find $$\P\left( \frac{1}{2} \lt X \lt \frac{3}{4} \right)$$ where $$X$$ is a random variable with the probability density function in (b).
1. $$f(x) = 6 \, x (1 - x)$$ for $$0 \le x \le 1$$
2. mode $$x = \frac{1}{2}$$
3. $$\frac{11}{32}$$

Let $$g(x) = x^2 (1 - x)$$ for $$0 \le x \le 1$$.

1. Sketch the graph of $$g$$.
2. Find the probability density function $$f$$ proportional to $$g$$.
3. Find the mode.
4. Find $$\P\left( \frac{1}{2} \lt X \lt 1 \right)$$ where $$X$$ is a random variable with the probability density function in (b).
1. $$f(x) = 12 \, x^2 (1 - x)$$ for $$0 \le x \le 1$$
2. mode $$x = \frac{2}{3}$$
3. $$\frac{11}{16}$$

Let $$g(x) = \frac{1}{\sqrt{x (1 - x)}}$$ for $$0 \lt x \lt 1$$.

1. Sketch the graph of $$g$$.
2. Find the probability density function $$f$$ proportional to $$g$$.
3. Find $$\P\left( 0 \lt X \lt \frac{1}{4} \right)$$ where $$X$$ is a random variable with the probability density function in (b).
1. In the integral, first use the simple substitution $$u = \sqrt{x}$$ and then recognize the new integral as an arcsine integral.
2. $$f(x) = \frac{1}{\pi \sqrt{x (1 - x)}}$$ for $$0 \lt x \lt 1$$
3. $$\frac{1}{3}$$

The distributions defined in the last three exercises are examples of beta distributions. The particular beta distribution in Exercise 14 is also known as the arcsine distribution. Beta distributions are studied in detail in the chapter on Special Distributions.

In the special distribution simulator, select the beta distribution. For the following parameter values, note the shape and location of the probability density function. Run the simulation 1000 times and note the apparent convergence of the empirical density function to the true probability density function.

1. $$a = 2$$, $$b = 2$$. This gives the beta distribution in Exercise 12.
2. $$a = 3$$, $$b = 2$$. This gives the beta distribution in Exercise 13.
3. $$a = \frac{1}{2}$$, $$b = \frac{1}{2}$$. This gives the arcsine distribution in Exercise 14.

#### The Pareto Distribution

Let $$g(x) = \frac{1}{x^b}$$ for $$1 \le x \lt \infty$$, where $$b \gt 0$$ is a parameter.

1. Sketch the graph of $$g$$.
2. Show that for $$0 \lt b \le 1$$, there is no probability density function proportional to $$g$$.
3. Show that for $$b \gt 1$$, the normalizing constant is $$\frac{1}{b - 1}$$.

The distribution defined in the last exercise is known as the Pareto distribution, named for Vilfredo Pareto. The parameter $$a = b - 1$$, so that $$a \gt 0$$, is known as the shape parameter. Thus, the Pareto distribution with shape parameter $$a$$ has probability density function $$f(x) = \frac{a}{x^{a+1}}$$ for $$1 \le x \lt \infty$$. The Pareto distribution is studied in detail in the chapter on Special Distributions.

Suppose that the income $$X$$ (in appropriate units) of a person randomly selected from a population has the Pareto distribution with shape parameter $$a = 3$$. Find $$\P(X \gt 2)$$.

$$\frac{1}{4}$$

In the special distribution simulator, select the Pareto with $$a = 3$$. Run the simulation 1000 times and compute the empirical probability of the event $$\{X \gt 2\}$$. Compare with the theoretical probability in the last exercise.

#### The Cauchy Distribution

Let $$g(x) = \frac{1}{x^2 + 1}$$ for $$x \in \R$$.

1. Sketch the graph of $$g$$.
2. Show that the normalizing constant is $$\pi$$.
3. Find $$\P(-1 \lt X \lt 1)$$ where $$X$$ has the probability density function proportional to $$g$$.
1. In the integral, note that the antiderivative is the arctangent function.
2. $$\frac{1}{2}$$

The distribution constructed in the previous exercise is known as the Cauchy distribution, named after Augustin Cauchy (it might also be called the arctangent distribution). The Cauchy distribution is studied in detail in the chapter on Special Distributions. The graph of $$g$$ is known as the witch of Agnesi, in honor of Maria Agnesi.

In the special distribution simulator, select the student $$t$$ distribution. Set $$n = 1$$ to get the Cauchy distribution. Run the simulation 1000 times and note how well the empirical density function fits the true probability density function.

#### The Standard Normal Distribution

Let $$g(z) = e^{-z^2/2}$$ for $$z \in \R$$.

1. Sketch the graph of $$g$$.
2. Show that the normalizing constant is $$\sqrt{2 \pi}$$.
Proof:

In (b), let $$c = \int_{-\infty}^\infty g(x) \, dx$$ denote the normalizing constant. Then

$c^2 = \int_{-\infty}^\infty e^{-x^2/2} \, dx \int_{-\infty}^\infty e^{-y^2/2} \, dy = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2) / 2} \, dx \, dy$

We now change to polar coordinates: $$x = r \cos(\theta)$$, $$y = r \sin(\theta)$$ where $$r \in [0, \infty)$$ and $$\theta \in [0, 2 \pi)$$. Then $$x^2 + y^2 = r^2$$ and $$dx \, dy = r \, dr \, d\theta$$. Hence

$c^2 = \int_0^{2 \pi} \int_0^\infty r e^{-r^2 / 2} \, dr \, d\theta$

If we use the simple substitution $$u = r^2$$, the inner integral is $$\int_0^\infty e^{-u} du = 1$$. Then the outer integral is $$\int_0^{2\pi} 1 \, d\theta = 2 \pi$$.

The distribution defined in the last exercise is the standard normal distribution, perhaps the most important distribution in probability. Normal distributions are widely used to model physical measurements that are subject to small, random errors. The family of normal distributions is studied in detail in the chapter on Special Distributions.

In the special distribution simulator, select the normal distribution (the default parameters give the standard normal distribution). Run the simulation 1000 times and note how well the empirical density function fits the true probability density function.

#### The Extreme Value Distribution

Let $$f(x) = e^{-x} e^{-e^{-x}}$$ for $$x \in \R$$.

1. Show that $$f$$ is a probability density function.
2. Sketch the graph of $$f$$ and identify the mode. Note in particular the asymmetry of the graph.
3. Find $$\P(X \gt 0)$$, where $$X$$ has probability density function $$f$$.
1. Mode $$x = 0$$
2. $$1 - e^{-1} \approx 0.6321$$

The distribution in the last exercise is the type 1 extreme value distribution, also known as the Gumbel distribution in honor of Emil Gumbel. Extreme value distributions are studied in detail in the chapter on Special Distributions.

In the special distribution simulator, select the extreme value distribution. Note the shape and location of the probability density function. Run the simulation 1000 times and note how well the empirical density function fits the true probability density function.

#### The Logistic Distribution

Let $$f(x) = \frac{e^x}{(1 + e^x)^2}$$ for $$x \in \R$$.

1. Show that $$f$$ is a probability density function.
2. Sketch the graph of $$f$$ and identify the mode. Note in particular the symmetry of the graph.
3. Find $$\P(X \gt 1)$$, where $$X$$ has probability density function $$f$$.
1. Mode $$x = 0$$
2. $$\frac{1}{1 + e} \approx 0.2689$$

The distribution in the last exercise is the logistic distribution. Logistic distributions are studied in detail in the chapter on Special Distributions.

In the special distribution simulator, select the logistic distribution. Note the shape and location of the probability density function. Run the simulation 1000 times and note how well the empirical density function fits the true probability density function.

#### Weibull Distributions

Let $$f(t) = 2 t e^{-t^2}$$ for.

1. Show that $$f$$ is a probability density function.
2. Sketch the graph of $$f$$ and identify the mode.
3. Find $$\P(T \gt \frac{1}{2})$$, where $$T$$ has probability density function $$f$$.
1. Mode $$t = \frac{1}{\sqrt{2}}$$
2. $$e^{-1/4} \approx 0.7788$$

Let $$f(t) = 3 t^2 e^{-t^3}$$ for $$t \ge 0$$.

1. Show that $$f$$ is a probability density function.
2. Sketch the graph of $$f$$ and identify the mode..
3. Find $$\P(T \gt \frac{1}{2})$$, where $$T$$ has probability density function $$f$$.
1. Mode $$t = \sqrt[3]{\frac{2}{3}}$$
2. $$e^{-1/8} \approx 0.8825$$

The distributions in the last two exercises are examples of Weibull distributions, name for Walodi Weibull. Weibull distributions are studied in more generality in the chapter on Special Distributions. They are often used to model random failure times of devices (in appropriately scaled units).

In the special distribution simulator, select the Weibull distribution. For each of the following values of the shape parameter $$k$$, note the shape and location of the probability density function. Run the simulation 1000 times and note how well the empirical density function fits the true probability density function.

1. $$k = 2$$. This gives the Weibull distribution in Exercise 27.
2. $$k = 3$$. This gives the Weibull distribution in Exercise 28.

Let $$f(x) = -\ln(x)$$ for $$0 \lt x \le 1$$.

1. Sketch the graph of $$f$$.
2. Show that $$f$$ is a probability density function.
3. Find $$\P(\frac{1}{3} \le X \le \frac{1}{2})$$ where $$X$$ has the probability density function in (b).
1. $$\frac{1}{2} \ln(2) - \frac{1}{3} \ln(3) + \frac{1}{6} \approx 0.147$$

Let $$f(x) = e^{-x} (1 - e^{-x})$$ for $$x \ge 0$$.

1. Sketch the graph of $$g$$ and find the probability density function $$f$$ proportional to $$g$$
2. Identify the mode.
3. Find $$\P(X \ge 1)$$ where $$X$$ has the probability density function in (b).
1. $$f(x) = 2 e^{-x} (1 - e^{-x})$$ for $$0 \le x \lt \infty$$
2. Mode $$x = \ln(x)$$
3. $$2 e^{-1} - e^{-2} \approx 0.6004$$

Let $$f(x, y) = x + y$$ for $$0 \le x \le 1$$, $$0 \le y \le 1$$.

1. Show that $$f$$ is a probability density function.
2. Find $$\P(Y \ge X)$$ where $$(X, Y)$$ has the probability density function in (a).
3. Find the conditional density of $$(X, Y)$$ given $$\{X \lt \frac{1}{2}, Y \lt \frac{1}{2}\}$$.
1. $$\frac{1}{2}$$
2. $$f(x, y \mid X \lt \frac{1}{2}, Y \lt \frac{1}{2}) = 8 (x + y)$$ for $$0 \lt x \lt \frac{1}{2}$$, $$0 \lt y \lt \frac{1}{2}$$

Let $$g(x, y) = x + y$$ for $$0 \le x \le y \le 1$$.

1. Find the probability density function $$f$$ that is proportional to $$g$$.
2. Find $$\P(Y \ge 2 X)$$ where $$(X, Y)$$ has the probability density function in (a).
1. $$f(x,y) = 2(x + y)$$, $$0 \le x \le y \le 1$$
2. $$\frac{5}{12}$$

Let $$g(x, y) = x^2 y$$ for $$0 \le x \le 1$$, $$0 \le y \le 1$$.

1. Find the probability density function $$f$$ that is proportional to $$g$$.
2. Find $$\P(Y \ge X)$$ where $$(X, Y)$$ has the probability density function in (a).
1. $$f(x,y) = 6 x^2 y$$ for $$0 \le x \le 1$$, $$0 \le y \le 1$$
2. $$\frac{2}{5}$$

Let $$g(x, y) = x^2 y$$ for $$0 \le x \le y \le 1$$.

1. Find the probability density function $$f$$ that is proportional to $$g$$.
2. Find $$P(Y \ge 2 X)$$ where $$(X, Y)$$ has the probability density function in (a).
1. $$f(x,y) = 15 x^2 y$$ for $$0 \le x \le y \le 1$$
2. $$\frac{1}{8}$$

Let $$g(x, y, z) = x + 2 y + 3 z$$ for $$0 \le x \le 1$$, $$0 \le y \le 1$$, $$0 \le z \le 1$$.

1. Find the probability density function $$f$$ that is proportional to $$g$$.
2. Find $$\P(X \le Y \le Z)$$ where $$(X, Y, Z)$$ has the probability density function in (a).
1. $$f(x, y, z) = \frac{1}{3}(x + 2 y + 3 z)$$ for $$0 \le x \le 1$$, $$0 \le y \le 1$$, $$0 \le z \le 1$$
2. $$\frac{7}{36}$$

Let $$g(x, y) = e^{-x} e^{-y}$$ for $$0 \le x \le y \lt \infty$$.

1. Find the probability density function $$f$$ that is proportional to $$g$$.
2. Find $$\P(X + Y \lt 1)$$ where $$(X, Y)$$ has the probability density function in (a).
1. $$f(x,y) = 2 e^{-x} e^{-y}$$, $$0 \lt x \lt y \lt \infty$$
2. $$1 - 2 e^{-1} \approx 0.2642$$

#### Continuous Uniform Distributions

In this subsection, we will study an important class of continuous distributions. First, recall again that the standard measure of size on $$\R^n$$ is

$\lambda_n(A) = \int_A \bs{1} d \bs{x}, \quad A \subseteq \R^n$

In particular,

• $$\lambda_1(A)$$ is the length of $$A$$ for $$A \subseteq \R$$
• $$\lambda_2(A)$$ is the area of $$A$$ for $$A \subset \R^2$$
• $$\lambda_3(A)$$ is the volume of $$A$$ for $$A \subseteq \R^3$$

Suppose that $$S \subseteq \R^n$$ with $$0 \lt \lambda_n(S) \lt \infty$$. Then

1. $$f(x) = 1 / \lambda_n(S)$$ for $$x \in S$$ defines a probability density function on $$S$$.
2. If $$X$$ has the probability density function in (a) then $$\P(X \in A) = \lambda_n(A) / \lambda_n(S)$$ for $$A \subseteq S$$.

A random variable $$X$$ with the probability density function in Exercise 38 is said to have the continuous uniform distribution on $$S$$. From part (b), note that the probability assigned to a subset $$A$$ of $$S$$ is proportional to the standard measure of $$A$$. Note also that in both the discrete and continuous cases, a random variable $$X$$ is uniformly distributed on a set $$S$$ if and only if the probability density function is constant on $$S$$. Uniform distributions play a fundamental role in various Geometric Models.

The most important special case is the uniform distribution on an interval $$[a, b]$$ where $$a, b \in \R$$, and $$a \lt b$$. In this case, the probability density function is $f(x) = \frac{1}{b - 1}, \quad a \le x \le b$ This distribution models a point chosen at random from the interval. In particular, the uniform distribution on $$[0, 1]$$ is known as the standard uniform distribution, and is very important because of its simplicity and the fact that it can be transformed into a variety of other probability distributions on $$\R$$. Almost all computer languages have procedures for simulating independent, standard uniform variables.

Give the probability density function of $$(X, Y)$$ and find $$\P(X \gt 0, Y \gt 0)$$ in each of the following cases:

1. $$(X, Y)$$ is uniformly distributed on the square $$S = [-6, 6]^2$$.
2. $$(X, Y)$$ is uniformly distributed on the triangle $$S = \{(x, y): -6 \le y \le x \le 6\}$$.
3. $$(X, Y)$$ is uniformly distributed on the circle $$S = \{(x, y): x^2 + y^2 \le 36\}$$.
1. $$f(x,y) = \frac{1}{36}$$ for $$-6 \le x \le 6$$, $$-6 \le y \le 6$$. $$\P(X \gt 0, Y \gt 0) = \frac{1}{4}$$
2. $$f(x,y) = \frac{1}{18}$$ for $$-6 \le x \le y \le 6$$. $$\P(X \gt 0, Y \gt 0) = \frac{1}{4}$$
3. $$f(x,y) = \frac{1}{36 \pi}$$ for $$x^2 + y^2 \le 36$$. $$\P(X \gt 0, Y \gt 0) = \frac{1}{4}$$.

In the bivariate uniform experiment, select each of the following domains and then run the simulation 1000 times. Watch the points in the scatter plot. Compute the empirical probability of the event $$\{X \gt 0, Y \gt 0\}$$ and compare with the true probability in the previous exercise.

1. Square
2. Triangle
3. Circle

Suppose that $$(X, Y, Z)$$ is uniformly distributed on the cube $$S = [0, 1]^3$$. Find $$\P(X \lt Y \lt Z)$$.

1. Compute the probability using the probability density function.
2. Compute the probability using a combinatorial argument.

$$\P(X \lt Y \lt Z) = \frac{1}{6}$$. In (b), argue that each of the 6 orderings of $$(X, Y, Z)$$ should be equally likely.

The time $$T$$ (in minutes) required to perform a certain job is uniformly distributed over the interval $$[15, 60]$$.

1. Find the probability that the job requires more than 30 minutes
2. Given that the job is not finished after 30 minutes, find the probability that the job will require more than 15 additional minutes.
1. $$\frac{2}{3}$$
2. $$\frac{1}{6}$$

Suppose that $$S \subseteq \R^n$$ and that $$0 \lt \lambda_n(S) \lt \infty$$ and that $$R \subseteq S$$ with $$\lambda_n(R) \gt 0$$. If $$X$$ is uniformly distributed on $$S$$, then the conditional distribution of $$X$$ given $$X \in R$$ is uniformly distributed on $$R$$.

Proof:

For $$A \subseteq R$$,

$\P(X \in A \mid X \in R) = \frac{\P(X \in A, X \in R)}{\P(X \in R)} = \frac{\P(X \in A)}{\P(X \in R)} = \frac{\lambda_n(A) / \lambda_n(S)}{\lambda_n(R) / \lambda_n(S)} = \frac{\lambda_n(A)}{\lambda_n(R)}$

#### Simulation

The last exercise has important implications for simulations. Suppose that $$R \subseteq \R^n$$ satisfies $$\lambda_n(R) \gt 0$$ as before, but suppose also that $$R$$ is bounded, so that $$R \subseteq S$$ where $$S \subseteq \R^n$$ is a Cartesian product of $$n$$ bounded intervals. It turns out to be quite easy to simulate a random variable $$X$$ that is uniformly distributed on the product set $$S$$. More generally, it is easy to simulate a sequence of independent random variables $$(X_1, X_2, \ldots)$$ each of which is uniformly distributed on $$S$$. Now let $$N = \min\{k \in \N_+: X_k \in R\}$$, the first time that one of the random variables lands in $$R$$. Note that $$N$$ has the geometric distribution on $$\N_+$$ with success parameter $$p = \lambda_n(R) / \lambda_n(S)$$. Let $$Y = X_N$$ so that $$Y$$ is the first term of the sequence that falls in $$R$$. We know from our work on independence and conditional probability, that the distribution of $$Y$$ is the same as the conditional distribution of $$X$$ given $$X \in R$$, which by the previous exercise, is uniformly distributed on $$R$$. Thus, we have derived an algorithm for simulating a random variable that is uniformly distributed on an irregularly shaped region $$R$$ (assuming that we have an algorithm for recognizing when a point $$x \in \R^n$$ falls in $$R$$). This method of simulation is known as the rejection method, and as we will see in subsequent sections, is more important that might first appear.

#### Data Analysis Exercises

If $$D$$ is a data set from a variable $$X$$ with a continuous distribution, then an empirical density function can be computed by partitioning the data range into subsets of small size, and then computing the probability density of points in each subset. Empirical probability density functions are studied in more detail in the chapter on Random Samples.

For the cicada data, $$BW$$ denotes body weight (in grams), $$BL$$ body length (in millimeters), and $$G$$ gender (0 for female and 1 for male). Construct an empirical density function for each of the following and display each as a bar graph:

1. $$BW$$
2. $$BL$$
3. $$BW$$ given $$G = 0$$
1.  BW Density $$(0, 0.1]$$ $$(0.1, 0.2]$$ $$(0.2, 0.3]$$ $$(0.3, 0.4]$$ 0.8654 5.8654 3.0769 0.1923
2.  BL Density $$(15, 29]$$ $$(20, 25]$$ $$(25, 30]$$ $$(30, 35]$$ 0.0058 0.1577 0.0346 0.0019
3.  BW Density given $$G = 0$$ $$(0, 0.1]$$ $$(0.1, 0.2]$$ $$(0.2, 0.3]$$ $$(0.3, 0.4]$$ 0.3390 4.4068 5.0847 0.1695

For the cicada data, $$WL$$ denotes wing length and $$WW$$ wing width (both in millimeters). Construct an empirical density function for $$(WL, WW)$$.

### Degenerate Continuous Distributions

Unlike the discrete case, the existence of a probability density function for a continuous distribution is an assumption that we are making. A random variable can have a continuous distribution on a subset $$S \subseteq \R^n$$ but with no probability density function; the distribution is sometimes said to be degenerate. In this subsection, we explore the common ways in which such distributions can occur.

#### Reducing the Dimension

First, suppose that $$\bs{X}$$ is a random variable taking values in $$S \subseteq \R^n$$ with $$\lambda_n(S) = 0$$. It is possible for $$\bs{X}$$ to have a continuous distribution, but not for $$\bs{X}$$ to have a probability density function relative to $$\lambda_n$$. In particular, property (c) in the definition could not hold, since $$\int_A f(\bs{x}) d \bs{x} = 0$$ for any function $$f$$ and any $$A \subseteq S$$. However, in many cases, $$\bs{X}$$ may be defined in terms of continuous random variables on lower dimensional spaces that do have probability density functions.

For example, suppose that $$\bs{U}$$ is a random variable with a continuous distribution on $$T \subseteq \R^k$$ where $$k \lt n$$, and that $$\bs{X} = h(\bs{U})$$ for some continuous function $$h : T \to S$$. Any event defined in terms of $$\bs{X}$$ can be changed into an event defined in terms of $$\bs{U}$$. The following exercise illustrates this situation.

Suppose that $$\Theta$$ is uniformly distributed on the interval $$[0, 2 \pi)$$. Let $$X = \cos(\Theta)$$, $$Y = \sin(\Theta)$$.

1. Show that $$(X, Y)$$ has a continuous distribution on the circle $$C = \{(x, y): x^2 + y^2 = 1\}$$.
2. Show that $$(X, Y)$$ does not have a probability density function on $$C$$ (with respect to the area measure $$\lambda_2$$ on $$\R^2$$).
3. Find $$\P(Y \gt X)$$.
1. $$\frac{1}{2}$$

#### Mixed Components

Another situation occurs when a random vector $$\bs{X}$$ in $$\R^n$$ (with $$n \gt 1$$) has some components with discrete distributions and others with continuous distributions. Such distributions with mixed components are studied in more detail in the section on mixed distributions; however, the following exercise gives an illustration.

Suppose that $$X$$ is uniformly distributed on the set $$\{0, 1, 2\}$$, $$Y$$ is uniformly distributed on the interval $$[0, 2]$$, and that $$X$$ and $$Y$$ are independent.

1. Show that $$(X, Y)$$ has a continuous distribution on the product set $$S = \{0, 1, 2\} \times [0, 2]$$.
2. Show that $$(X, Y)$$ does not have a probability density function on $$S$$ (with respect to the area measure $$\lambda_2$$ on $$\R^2$$).
3. Find $$\P(Y \gt X)$$.
1. $$\frac{1}{2}$$

#### Singular Continuous Distributions

Finally, it is also possible to have a continuous distribution on $$S \subseteq \R^n$$ with $$\lambda_n(S) \gt 0$$, yet still with no probability density function. Such distributions are said to be singular, and are rare in applied probability. However, it is not difficult to construct such a distribution. Let $$(X_1, X_2, \ldots)$$ be a sequence of Bernoulli trials with success parameter $$p \in (0, 1)$$. We will indicate the dependence of the probability measure $$\P$$ on the parameter $$p$$ with a subscript. Thus, we have a sequence of independent indicator variables with

$\P_p(X_i = 1) = p, \quad \P_p(X_i = 0) = 1 - p$

We interpret $$X_i$$ as the $$i$$th binary digit (bit) of a random variable $$X$$ taking values in $$[0, 1]$$. That is,

$X = \sum_{i=1}^\infty \frac{X_i}{2^i}$

Conversely, recall that every number $$x \in [0, 1]$$ can be written in binary form:

$x = \sum_{i=1}^\infty \frac{x_i}{2^i} \text{ where } x_i \in \{0, 1\} \text{ for each } i \in \N_+$

This representation is unique except when $$x$$ is a binary rational of the form $$x = \frac{k}{2^n}$$ for some $$k \in \{1, 3, \ldots 2^n - 1\}$$. In this case, there are two representations, one in which the bits are eventually 0 and one in which the bits are eventually 1. By convention, we will use the first representation. Note, however, that the set of binary rationals is countable.

$$X$$ has a continuous distribution. That is, $$\P_p(X = x) = 0$$ for $$x \in [0, 1]$$.

Proof:

If $$x \in [0, 1]$$ is not a binary rational, then

$\P_p(X = x) = \P_p(X_i = x_i \text{ for all } i \in \N_+) = \lim_{n \to \infty} \P_p(X_i = x_i \text{ for } i = 1, \; 2 \ldots, \; n) = \lim_{n \to \infty} p^y (1 - p)^{n - y} \text{ where } y = \sum_{i=1}^n x_i$

Let $$q = \max\{p, 1 - p\}$$. Then $$p^y (1 - p)^{n - y} \le q^n \to 0$$ as $$n \to \infty$$. Hence, $$\P_p(X = x) = 0$$. If $$x \in [0, 1]$$ is a binary rational, then there are two bit strings that represent $$x$$, say $$(x_1, x_2, \ldots)$$ (with bits eventually 0) and $$(y_1, y_2, \ldots)$$ (with bits eventually 1). Hence $$\P_p(X = x) = \P_p(X_i = x_i \text{ for all } i \in \N_+) + \P_p(X_i = y_i \text{ for all } i \in \N_+)$$. But both of these probabilities are 0 by the same argument as before.

Next, we define the set of numbers for which the limiting relative frequency of 1's is $$p$$. Let

$C_p = \left\{ x \in [0, 1]: \frac{1}{n} \sum_{i = 1}^n x_i \to p \text{ as } n \to \infty \right\}$

The sets $$\{C_p: p \in [0, 1]\}$$ satisfy the following properties:

1. $$C_p \cap C_q = \emptyset$$ for $$p \ne q$$.
2. $$\P_p(X \in C_p) = 1$$
Proof:

Part (a) is trivial, since limits are unique. Part (b), follows from the strong law of large numbers which we will study later in the chapter on random samples. The basic idea is simple: in a sequence of Bernoulli trials with success probability $$p$$, the long-term relative frequence of successes is $$p$$.

It follows that the distributions of $$X$$, as $$p$$ varies from 0 to 1, are mutually singular; that is, as $$p$$ varies, $$X$$ takes values in mutually disjoint sets.

If $$p = \frac{1}{2}$$ then $$X$$ is uniformly distributed on $$[0, 1]$$.

Proof:

Let $$F(x) = \P_p(X \le x) = \P_p(X \lt x)$$ for $$x \in [0, 1]$$. The function $$F$$ is the distribution function of $$X$$. Distribution functions are studied in more detail later in this chapter. If $$x \in [0, 1]$$ is not a binary rational, then $$X \lt x$$ if and only if there exists $$n \in \N_+$$ such that $$X_i = x_i$$ for $$i \in \{1, 2, \ldots, n - 1\}$$ and $$X_n = 0$$ while $$x_n = 1$$. Hence

$\P_{1/2}(X \lt x) = \sum_{n=1}^\infty \frac{x_n}{2^n} = x$

It turns out, as we will see, that the distribution function of a continuous distribution is continuous, so $$F(x) = x$$ for all $$x \in [0, 1]$$. This means that $$X$$ has the uniform distribution on $$[0, 1]$$.

Note that the uniform distribution on $$[0, 1]$$ is the same as the standard measure $$\lambda_1$$ on $$[0, 1]$$.

If $$p \ne \frac{1}{2}$$, the distribution of $$X$$ does not have a probability density function relative to the standard measure on $$[0, 1]$$.

Proof:

Suppose that $$X$$ does have a probability density function $$f$$. Then for $$A \subseteq [0, 1]$$, $$\P_p(X \in A) = \int_A f(x) \, dx$$, so in particular

$1 = \P_p(X \in C_p) = \int_{C_p} f(x) \, dx$

But since $$p \ne 1/2$$, $$\lambda_1(C_p) = \P_{1/2}(X \in C_p) = 0$$ which implies that $$\int_{C_p} f(x) \, dx = 0$$, so we have a contradiction.

For an applied example of the ideas in this subsection, see Bold Play in the game of Red and Black.


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# Measurement of higher-order harmonic azimuthal anisotropy in PbPb collisions at a nucleon-nucleon center-of-mass energy of 2.76 TeV

Abstract : Measurements are presented by the CMS Collaboration at the Large Hadron Collider (LHC) of the higher-order harmonic coefficients that describe the azimuthal anisotropy of charged particles emitted in sqrt(s[NN]) = 2.76 TeV PbPb collisions. Expressed in terms of the Fourier components of the azimuthal distribution, the n = 3-6 harmonic coefficients are presented for charged particles as a function of their transverse momentum (0.3
Document type :
Journal articles
Domain :

http://hal.in2p3.fr/in2p3-00879540
Contributor : Dominique Girod <>
Submitted on : Monday, November 4, 2013 - 11:12:48 AM
Last modification on : Friday, June 5, 2020 - 10:52:07 AM

### Citation

S. Chatrchyan, M. Besancon, F. Couderc, M. Dejardin, D. Denegri, et al.. Measurement of higher-order harmonic azimuthal anisotropy in PbPb collisions at a nucleon-nucleon center-of-mass energy of 2.76 TeV. Physical Review C, American Physical Society, 2014, 89, pp.044906. ⟨10.1103/PhysRevC.89.044906⟩. ⟨in2p3-00879540⟩

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text:
## 18-Dec-2013: Convert to sparse file utility (win32)

Just wrote utility intended for converting files into sparse ones on Windows NTFS file system. Sparse files are those in which long zero blocks are not stored on hard disk, but replaced to information about them (metadata) instead. These files are very useful for saving space on storing half-empty ISO files, half-downloaded torrent files, virtual machine disk images.

I need it primarily for VMware WS disk images "compressing". I suppose, many other Oracle specialists use VMware machines with a lot of Oracle versions as well :-)

Compiled executable file: //yurichev.com/utils/cvt2sparse.exe

That is how it looks after converting:

This one-liner is to be run inside of *NIX virtual machine to write zeroes to unused parts of file system:

dd if=/dev/zero of=empty_file; rm empty_file

This open sourced site and this page in particular is hosted on GitHub. Patches, suggestions and comments are welcome.


text:
# ListPlot in a Table with different colours and labels

I am making a Table of ListPlots, in which I superimpose two ListPlots with Show.

I would like to make each nth element of a different colour and with a diffenret label on the legend.

The line in the code is this:

plots =  Table[
Show[ListPlot[cumulativeMaximaOut[[n]][[1]]],
ListPlot[cumulativeMaximaOut[[n]][[2]]]], {n, 1, Length[xm]}]

Is it possible to make each nth element automatically of a different colour (and label), or do I have to define a string of labels and of colours before hand (of finite length then)?

I tried this:

plots =  Table[
ListPlot[{Labeled[cumulativeMaximaOut[[n]][[1]], "sqrt"],
Labeled[cumulativeMaximaOut[[n]][[2]], "log"]}], {n, 1,
Length[xm]}]

and it says:

Requested order is too high; order has been reduced to {2}

--

Data for cumulativeMaximaOut is here: http://pastebin.com/cb9i9FeP,

xm = {2,3,4}

• This may have something to do with the internal structure of your data. You need to provide some representative data for your cumulative... list, and xm. – MarcoB Nov 24 '16 at 19:17
• Do I just write down what values the arrays have? Is there a better way of providing data here? – SuperCiocia Nov 24 '16 at 19:21
• You can use e.g. pastebin to share larger amounts of data: paste the data there and then add a link to your original question. – MarcoB Nov 24 '16 at 19:22

I created some synthetic data that is the value of Sin from 0 to 2π with a variable frequency.

data = Table[{θ, Sin[n θ]}, {n, 1, 6}, {θ, 0, 2 π, 2 π/500}];

ListPlot itself can handle the this with multiple colors.

ListPlot[data]

You need to do something to handle the legends

ListPlot[
data,
PlotStyle -> PointSize[0.01],
PlotLegends -> Flatten@Table[
{"Sin[" ~~ ToString[n] ~~ " x]",
"Sin[" ~~ ToString[n + 1] ~~ " x]"},
{n, 1, Length[data] - 1, 2}
]
]

There are many approaches you can use to create the legends depending upon the character of your data.

I am not entirely sure that I understand your final desired formatting. However, here is an attempt at differentiating the two data series present in each component of your data set:

ListPlot[#,
PlotStyle -> {PointSize[0.03], PointSize[0.005]},
PlotLegends -> {"sqrt", "log"},
PlotRange -> {{-300, 6800}, {-3, 49}},
Axes -> False, Frame -> True,
ImageSize -> Medium
] & /@ cumulativeMaximaOut

• Oh yeah sorry I guess I was wrong in the code. I actually meant to plot all the three graphs on top of each other, each with a different colour and label. Do you have any idea on how to do that? – SuperCiocia Nov 24 '16 at 22:10


text:
# Continuity of Function and Derivative at boundary.

1. Oct 18, 2009

### Prologue

I am reading both Griffiths and Gasiorowicz and I can't get either of them to tell me why the continuity of the derivative of the natural log of the amplitude

$$\frac{d(ln(u(x)))}{dx}=\frac{1}{u(x)}\frac{du(x)}{dx}$$

or put a different way

$$\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}$$

at a boundary a implies that

$$u(a^{-})=u(a^{+})$$

and

$$\frac{du(a^{-})}{dx}=\frac{du(a^{+})}{dx}$$

Why is this true?

It looks to me that there still is some ambiguity (as in the top equation is neccessary but not sufficient) because

$$\frac{1}{u(a^{-})}\frac{du(a^{-})}{dx}= \frac{1}{u(a^{+})}\frac{du(a^{+})}{dx}$$

can be reconfigured to

$$\frac{u(a^{+})}{u(a^{-})}=\frac{\frac{du(a^{+})}{dx}}{\frac{du(a^{-})}{dx}}$$

Which just means that the ratios have to be equal. But say for instance that $$u(a^{+})$$ is twice as big as $$u(a^{-})$$ so that the ratio is 2. Then all that means is the ratio of the slopes at a has to be 2. So, I don't see where they are limited to be identical.

Last edited: Oct 18, 2009
2. Oct 19, 2009

### Avodyne

It isn't true.

What is true is that continuity of both u and du/dx can be derived from the Schrodinger equation. And then, since the overall normalization is not fixed by the Schrodinger equation, it's equivalent to just consider continuity of the log derivative.

3. Oct 19, 2009

### Prologue

Thank you! I appreciate the reply. I am glad to hear that I wasn't missing something in the math. I understand that Schrodinger's equation forces the continuity of Psi and it's derivative. But there still is a gap in my understanding, I do not know what is meant by "since the overall normalization is not fixed by the Schrodinger equation" and then how the log derivative follows from that. Can you expand on that just a little more?

Is the "normalization" that you are referring to the normalization of |Psi|^2 over all space?

Last edited: Oct 19, 2009
4. Oct 20, 2009

### Avodyne

Oops, what I said isn't quite true either.

You need both conditions to get both the energy eigenvalue and the eigenfunction. But if we are interested in only the eigenvalue (or we want to get that first), then taking the ratio removes the unknown info about the eigenfunction.

And yes, by normalization I mean the integral of psi^2.

5. Oct 20, 2009

### Prologue

So the problem for me lies in this: It is our task to match two different functions and their derivatives at the boundary because Schrodinger's equation tells us it has to be that way. If we then just find two solutions in different regions that meet at a boundary then they must be continuous as well as their derivatives because the Schrodinger equation tells us it must be that way. But the book says that having the very first equation (composite equation) be true necessarily means the other two equations are true and I just don't see how that happens.

In other words, we need to find two solutions that fit into Schrodinger's constraints at the boundary. These solutions to start with have arbitrary constants all over the place so they can take on any finite value at the boundary. Then we apparently use the combined equation to simplify the finding out of these constants. Instead of doing two constraints we apply just one that apparently contains both. I don't understand how that combined equation can contain both for the reasons given in the first post.

Can you explain exactly how the normalization combined with this composite equation guarantees that we will find the constants that give us identical boundaries?

*Another view.*

Suppose we do the normalization integral, then we will get a sum of products/etc of constants that have to equal 1. The let's say we pick these constants so that they purposely solve the normalization equation, but also so that they do not satisfy the boundary rules that the Schrodinger equation insists on. Is it possible to do this? If not why are we guaranteed that we can't do this?

Last edited: Oct 20, 2009
6. Oct 20, 2009

### Avodyne

You don't see how that happens because the book is wrong.

Oops, I did it again! This is only correct in special situations, where, before matching, you know the wave function in each region up to overall normalization. This happens if the potential is even, and you therefore know that the energy eigenfunctions are even or odd.

But, more generally, you don't know this, and then you do indeed have to separately match the wave function and its derivative at each boundary.


text:
# Spacers for stacked PCBs

#### clangray

Joined Nov 4, 2018
227
I'm considering stacking a few very small boards with some M1.6 brass standoffs (the smallest standoffs I could find). To decrease the space between them even more, I could instead make a custom space with some M1.6 screws and hex nuts between the boards or even shorter like M1.4. However would the brass spacers be more suitable because they are "anti-short" and are more electrically friendly? Am I risking the circuit with the custom version?

#### MrChips

Joined Oct 2, 2009
27,686
Brass spacers are not "anti-short". Use nylon spacers instead.

#### clangray

Joined Nov 4, 2018
227
Brass spacers are not "anti-short". Use nylon spacers instead.
Thx for the reply. Il look into this.
But since I am here, could you tell me would the screws be problematic? If they are I need to abandon this route all together.

#### MrChips

Joined Oct 2, 2009
27,686
You can buy nylon screws, nuts, washers, and spacers just as if they were made of metal.

#### Boggart

Joined Jan 31, 2022
67
You can buy nylon screws, nuts, washers, and spacers just as if they were made of metal.

View attachment 282571
Yes, this. I use black nylon spacers with nylon bolts on some of the kits I sell, no worries about shorts between boards and the spacers can even sit on top of tracks without issue, so you don't have to worry about exclusion zones around the holes, although it is preferred to avoid spacers sitting on tracks if possible.

Also, nylon spacers are a lot cheaper than brass, a big plus when doing lots of boards, I buy them in 1000qtys for a $20 to$60, depending on length.

#### clangray

Joined Nov 4, 2018
227
Yes, this. I use black nylon spacers with nylon bolts on some of the kits I sell, no worries about shorts between boards and the spacers can even sit on top of tracks without issue, so you don't have to worry about exclusion zones around the holes, although it is preferred to avoid spacers sitting on tracks if possible.

Also, nylon spacers are a lot cheaper than brass, a big plus when doing lots of boards, I buy them in 1000qtys for a $20 to$60, depending on length.
Hi, this makes very good sense. However I can't find Nylon standoffs < M2 size. That's my reason to try to customize with the small screws M1.6, 1.4 or less. It makes a big impact on the profile of my board. Now if the possibility of shorts between the board and standoffs are reduced by nylon I can ditch the customized plan, increase the profile of the board to accommodate the nylon screws. Or I can go with brass that I found on EBay for M1.6.

#### Lo_volt

Joined Apr 3, 2014
280
Brass spacers should be fine if you don't connect the mounting holes to any nets. Isolate the holes from your circuit and the brass won't affect your circuit.

#### clangray

Joined Nov 4, 2018
227
What if instead I used very small screws, stainless steel/brass and anchored it by passing through both boards and arranged one end with two hex nuts to keep it from sliding. I've attached a rudimentary illustration to demonstrate how this would work. 0 Nets touching the screw; exclusion zones necessary?

Last edited:

#### ericgibbs

Joined Jan 29, 2010
16,792
hi

#### Attachments

• 10.7 KB Views: 12

#### clangray

Joined Nov 4, 2018
227
Hi Eric I don't understand this post. Also since its been up I have no new posted. Maybe coincidence.

#### ericgibbs

Joined Jan 29, 2010
16,792
Hi clang.
I added a missing Nut under the top PCB.
Also, I would extend the holes to the edge of the PCB, using a round needle file, you will find it easier to disassemble/reassemble as required
E


text:
# Here is our 2020 budget

Have you ever wondered how do we know how much \rm{CO_2}we are emitting globally? And where is this \rm{CO_2} going to? For sure I did. It turns out that a global organization called the Global Carbon Project set this goal to itself back in 2001. In their own words:

The scientific goal of the Global Carbon Project is to develop a complete picture of the global carbon cycle, including both its biophysical and human dimensions together with the interactions and feedback between them.

Since 2005 they have been providing a report on what they call the "Global Carbon Budget" almost every year. And they release the report under the Creative Common license! 🙂 So I have had a look at their last report: Global Carbon Budget 2020.

## Carbon Emissions

The carbon budget tracks the anthropogenic (read human made) emissions of \rm{CO_2}. They are divided into two categories:

• E_{\rm{fos}}: emissions resulting from the combustion and oxidation of fossil fuels from all energy and industrial processes (including cement production and carbonation). It includes emissions from a wide range of activities like transport, heating and cooling, industry, the production and carbonation of cement and other process emissions like the production of chemicals and fertilizers. Its measurement is essentially based on energy consumption data.
• E_{\rm{luc}}: emissions resulting from deliberate human activities on land, including land-use change. It is the net emissions (sum of emissions and removals) due to all anthropogenic activities such as deforestation, afforestation, logging, forest degradation, shifting cultivation. It is estimated by so-called bookkeeping models keeping track of human activities, such as turning forest to crops, around the world.

Both terms are measured in giga-tons of carbon per year (GtCyr^{-1}), a giga-ton being 10^{12}kilograms. The graph below, taken from the publication shows the global fossil emissions (a), together with breakdowns of the emissions per categories (b), for selected regions (c), and for selected regions per capita (d). We see that the global emissions from fossil fuels are still rising as of 2020 despite a slow decrease in the EU27 and the USA. The emissions per capita allow to put into perspective the impressive increase in emissions of China. Yes China is now and by far the country using the most fossil fuels, but looking per capita the emissions just recently reached the level of EU and are still twice below the USA.

## Carbon Sinks

Where are all these emissions going? The different reservoirs the emissions go to are called sinks. There is an active carbon cycle that circulates carbon from different sinks: the atmosphere, the ocean, the terrestrial biosphere and geologic reservoirs. Exchanges between the 3 firsts occur on time scale ranging from hours to millennia. Geologic reservoirs like oil fields are formed on geological time scales and are not considered in the carbon budget (although the action of emptying these reservoirs is taken into account by E_{\rm{fos}}). Lets have a look at the 3 main sinks:

• As you might have guessed a fair fraction is going to the atmosphere. The Global Monitoring Laboratory part of the National Oceanic and Atmospheric Administration of the USA measures the \rm{CO_2}concentration of the atmosphere in part per million (ppm) since 1958. Daily measurements are available on there website. At the time of writing these lines, the last measurement available was for June 19th at 414.94 ppm (It was 277ppm in 1750). The graph below, taken from their website, shows the evolution of the \rm{CO_2}concentration of the atmosphere. The upward trend results from human activity while the seasonality is due to the vegetation: growing plants absorb \rm{CO_2}from the atmosphere via photosynthesis, \rm{CO_2}that is then re-emitted as the plants (or their leaves) die. Because of the Northern hemisphere having more lands, and thus more vegetation coverage than the Southern hemisphere, photosynthesis removes more \rm{CO_2}from the atmosphere during boreal Summer than austral Summer. This "breathing cycle" of the Earth was beautifully put into a visualization by Nadieh Bremer. But lets get back to our carbon budget. From the direct measurements we can derive the growth rate of the atmospheric \rm{CO_2}concentration G_{\rm{atm}}, that is the amount of \rm{CO_2}that has been added to the atmosphere in a year.
• Although we don't put much \rm{CO_2}directly in the ocean (or do we?), the ocean is in contact with the atmosphere. When the concentration of \rm{CO_2}in the atmosphere grows, some of the \rm{CO_2}dissolves in the ocean (reaching a so-called phase equilibrium). The reverse is also true, meaning that if we were to remove some of the \rm{CO_2}from the atmosphere, some of the \rm{CO_2}dissolved into the ocean would evaporate into the atmosphere... The yearly amount of \rm{CO_2}dissolved into the ocean is noted S_{\rm{ocean}}. Measurements are not direct but based on models that are constrained by observations.
• The last sink is the terrestrial one. It results from an increase in plant growth due to a higher \rm{CO_2}atmospheric concentration and the lengthening of the growing season due to climate change. It is noted S_{\rm{land}}. It is different from E_{\rm{luc}}as it does not include direct human land usage. Its measurement is also model based.

## Global Carbon Budget

The global carbon budget equation equalizes the anthropogenic emissions to the sinks:

E_{\rm{fos}} + E_{\rm{luc}} = G_{\rm{atm}} + S_{\rm{ocean}} + S_{\rm{land}} + B_{\rm{im}}

You may have notice that a new term, B_{\rm{im}}, appeared. In reality global emissions are fully partitioned between the atmosphere, the ocean and the land. In practice, because all terms in the equation are measured with an uncertainty, the reported numbers do not add up to zero. The resulting sum is coined the budget imbalance B_{\rm{im}}. The graph below, taken from the Global Carbon Budget 2020 publication, shows the emissions and their distribution among the different sinks. It shows the cumulative changes in giga-tons of carbon (GtC) since 1850, as well as mean fluxes in giga-tons of carbon per year (GtCyr^{-1}) between 2010 and 2019.

We see that in both cases the budget imbalance is small, indicating that the overall budget is consistent. Since 2010, 85% of anthropogenic \rm{CO_2}emissions are due to fossil fuel usage while the remaining 15% belong to land usage. About 46% of the \rm{CO_2}is going to the atmosphere while around 31% is going into the terrestrial biosphere sink and 23% to the ocean sink.


text:
Electronic Identification (eIDAS)

### Article 37 — Electronic seals in public services

1. If a Member State requires an advanced electronic seal in order to use an online service offered by, or on behalf of, a public sector body, that Member State shall recognise advanced electronic seals, advanced electronic seals based on a qualified certificate for electronic seals and qualified electronic seals at least in the formats or using methods defined in the implementing acts referred to in paragraph 5.
2. If a Member State requires an advanced electronic seal based on a qualified certificate in order to use an online service offered by, or on behalf of, a public sector body, that Member State shall recognise advanced electronic seals based on a qualified certificate and qualified electronic seal at least in the formats or using methods defined in the implementing acts referred to in paragraph 5.
3. Member States shall not request for the cross-border use in an online service offered by a public sector body an electronic seal at a higher security level than the qualified electronic seal.
4. The Commission may, by means of implementing acts, establish reference numbers of standards for advanced electronic seals. Compliance with the requirements for advanced electronic seals referred to in paragraphs 1 and 2 of this Article and Article 36 shall be presumed when an advanced electronic seal meets those standards. Those implementing acts shall be adopted in accordance with the examination procedure referred to in Article 48(2).
5. By 18 September 2015, and taking into account existing practices, standards and legal acts of the Union, the Commission shall, by means of implementing acts, define reference formats of advanced electronic seals or reference methods where alternative formats are used. Those implementing acts shall be adopted in accordance with the examination procedure referred to in Article 48(2).


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Drossel found a mysterious box today. And the logical way to kill schrodinger's cat is to let it out of the box. And so Drossel set out to opening the box to find out its contents...

It turned out to be an ordinary cup with a stick of glue inside.

Kagami: Hold on a second. I've seen that somewhere before.
Tsukasa: ?

Kagami: I've DEFINITELY seen it somewhere before.

Kagami: What do you want?

Kagami: DON'T FLIP MY SKIRT!

written by astrobunny \\ bowl, drossel, kagami, nice, nice bowl, shima, shimabowl, shimapan, tsukasa


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Torus package on a torus

Let $$G$$ to be a nilpotent Lie group and $$H to be two closed subgroups such as
$$G / H to G / J$$ is a bundle of fibers with fibers $$J / H$$.

Assume that $$G / J = (S ^ 1) ^ m$$ and $$J / H = (S ^ 1) ^ m$$. Is the package trivial?


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+1.617.933.5480

# Q: Finance 3040

'Q2: As the winner of the Housecleaners sweepstakes (won at the beginning of this year), you are entitled to one of the following prizes:

A. $999,999 immediately. B.$100,000 per year forever.

C. $100,000 at the time of winning, and then$180,000 per year for the next 10 years starting from the end of this year.

In terms of present values, which prize should be chosen if r = 9%?'

Solution: Discount Rate = 9% Option A: Present Value = $999999 Option B:... Related Questions in Personal Finance • Q: You win a contest. The prize is cash, and you are offered... (Solved) September 15, 2014 You win a contest. The prize is cash, and you are offered several alternative payment plans. Which plan should you choose? Assume... Solution Preview : Solution: Assumption: 5% return and no inflation a)$ 30000 today b) $40000 in five years, PV of$ 40000 in 5 year = $40000/(1.05)^4 =$ 32908
• Q:
finance help!! please!! January 16, 2014

.00%, but you must make interest payments at the end of each quarter and then pay off the $10 ,000 principal amount at the end of the year . What... • Q: Business Finance Year 3 February 22, 2013 period increases. d. Both " b " and " c ". e. All of the above 5 You have just won a magazine sweepstakes and have a choice of three alternatives.... • Q: Please help me answer these finance questions. (Solved) March 31, 2013 I need someone to answer these homework questions for my Finance class. I need to study them for the test I have tomorrow night! Please help me! Solution Preview : PART-I 1. The primary operating goal of a publicly-owned firm interested in serving its stockholders should be to a. Maximize its expected total corporate income. b. Maximize its expected... • Q: Business Math January 18, 2014 has 360 days. 1. The firm will collect$1,133 in Q 2 2. The accounts receivable balance at the beginning of Quarter 4 will be \$767 3. The firm will...

Question Status: Solved


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# Linear motion problems solved with differential equations?

A body of mass m, which is initially at rest, falls vertically from the height h under the influence of constant gravitational acceleration, g = 9.81 ms−2 .

Assuming that the body’s air resistance is proportional to the square of its velocity v and opposite to the direction of v, i.e. F = −kv^2 , with k a positive constant, determine the velocity vimpact with which the body hits the ground level (height zero).

I have created this relation in the system:

$F=mg-kv^2=ma=m\dot{v}=mv*\frac{\mathrm{dv}&space;}{\mathrm{d}&space;x}$

However I am very stuck where to move from here, any guidance would be greatly appreciated.

Thanks

i think your equation is given by $$mg-k\left(\frac{dx}{dt}\right)^2=m\frac{d^2x}{dt^2}$$ substituting $$u(t)=\frac{dx(t)}{dt}$$ then we get $$\frac{m\frac{du(t)}{dt}}{gm-ku(t)^2}=1$$ integrating both sides we obtain $$\frac{\sqrt{m}arctanh\left(\frac{\sqrt{k}u(t)}{\sqrt{gm}}\right)}{\sqrt{gk}}=t+C_1$$

• Once I have derived this how can I obtain V-impact? – QuestionAnswer13 Mar 2 '18 at 15:33

I think you have no problem to derive $$g-Av^2=\frac{dv}{dt}$$ where $A=k/m$.

Now seperating variables: $$dt=\frac{dv}{g-Av^2}$$

Integrate both sides: $$t+C=-\frac{\ln\left(\frac{|Av-\sqrt{Ag}|}{|Av+\sqrt{Ag}|}\right)}{2\sqrt{Ag}}$$(by wolfy)

After some algebra I think you are able to make $v$ the subject.

$C$ is determined my initial conditions, which, in this case, I think is the initial velocity.


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# Expanded Notation

In this expanded notation activity, learners write the expanded notation for a set of 10 numbers. Students are given an example, and the numbers are 2 or 3 digits.


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# zbMATH — the first resource for mathematics

Real algebraic curves and flexible curves on ruled surfaces of base $$\mathbb{C} \mathbb{P}^1$$. (Courbes algébriques réelles et courbes flexibles sur les surfaces réglées de base $$\mathbb{C} \mathbb{P}^1$$.) (French) Zbl 1016.14019
A first aim of this paper is to answer, in the case of real ruled surfaces $$X_l$$ of base $$\mathbb{C}\mathbb{P}^1$$, $$l \geq 2$$, a question of V. A. Rokhlin [Usp. Mat. Nauk 33, No. 5(203), 77-89 (1978; Zbl 0437.14013)]: The equivariant isotopy class does not suffice to distinguish the connected components of the space of smooth real algebraic curves of $$X_l$$. A second aim is to prove that there exists in these surfaces some real schemes realized by real flexible curves but not by smooth real algebraic curves.
These two results of real algebraic geometry are deduced from the following comparison theorem: When $$m = l+ 2k$$, $$k>0$$, the discriminants of the surface $$X_m$$ are deduced from those of the surface $$X_l$$ via weighted homotheties. All these results are obtained thanks to a study of a deformation of ruled surfaces.

##### MSC:
 14J26 Rational and ruled surfaces 14H10 Families, moduli of curves (algebraic) 14P25 Topology of real algebraic varieties
Full Text:


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# 7.1: Strain

As was discussed in Chapters 4 and 6, in a continuum, i.e. a system of particles so close to each other that the system discreteness may be neglected, the particle displacement $$\mathbf{q}$$ may be considered as a continuous function of space and time. In this chapter, we will consider only small deviations from the rigid-body approximation discussed in Chapter 4, i.e. small deformations. The deformation smallness allows one to consider the displacement vector $$\mathbf{q}$$ as a function of the initial (pre-deformation) position of the particle, $$\mathbf{r}$$, and time $$t$$ - just as was done in Chapter 6 for 1D waves.

The first task of the deformation theory is to exclude from consideration the types of motion considered in Chapter 4, namely the translation and rotation, unrelated to deformations. This means, first of all, that the variables describing deformations should not depend on the part of displacement distribution, that is independent of the position $$\mathbf{r}$$ (i.e. is common for the whole media), because that part corresponds to a translational shift rather than to a deformation (Figure 1a). Moreover, even certain nonuniform displacements do not contribute to deformation. For example, Eq. (4.9) (with $$d \mathbf{r}$$ replaced with $$d \mathbf{q}$$ to comply with our current notation) shows that a small displacement of the type $d \mathbf{q}_{\text {rotation }}=d \varphi \times \mathbf{r} \text {, }$ where $$d \varphi=\omega d t$$ is an infinitesimal vector common for the whole continuum, corresponds to its rotation about the direction of that vector, and has nothing to do with its deformation (Figure 1b).

This is why to develop an adequate quantitative characterization of deformation, we should start with finding suitable functions of the spatial distribution of displacements, $$\mathbf{q}(\mathbf{r})$$, that exist only due to deformations. One of such measures is the change of the distance $$d l=|d \mathbf{r}|$$ between two close points: $\left.(d l)^{2}\right|_{\text {after deformation }}-\left.(d l)^{2}\right|_{\text {before deformation }}=\sum_{j=1}^{3}\left(d r_{j}+d q_{j}\right)^{2}-\sum_{j=1}^{3}\left(d r_{j}\right)^{2}$ where $$d q_{j}$$ is the $$j^{\text {th }}$$ Cartesian component of the difference $$d \mathbf{q}$$ between the displacements $$\mathbf{q}$$ of these close points. If the deformation is small in the sense $$|d \mathbf{q}|<<|d \mathbf{r}|=d l$$, we may keep in Eq. (2) only the terms proportional to the first power of the infinitesimal vector $$d \mathbf{q}$$ : $\left.(d l)^{2}\right|_{\text {after deformation }}-\left.(d l)^{2}\right|_{\text {before deformation }}=\sum_{j=1}^{3}\left[2 d r_{j} d q_{j}+\left(d q_{j}\right)^{2}\right] \approx 2 \sum_{j=1}^{3} d r_{j} d q_{j} .$ Since $$q_{j}$$ is a function of three independent scalar arguments $$r_{j}$$, its full differential (at fixed time) may be represented as $d q_{j}=\sum_{j^{\prime}=1}^{3} \frac{\partial q_{j}}{\partial r_{j^{\prime}}} d r_{j^{\prime}}$ The coefficients $$\partial q_{j} / \partial r_{j}$$, may be considered as elements of a tensor $${ }^{1}$$ providing a linear relation between the vectors $$\mathrm{d} \mathbf{r}$$ and $$\mathrm{d} \mathbf{q}$$. Plugging Eq. (4) into Eq. (2), we get $\left.(d l)^{2}\right|_{\text {after deformation }}-(d l)_{\mid \text {before deformation }}^{2}=2 \sum_{j, j^{\prime}=1}^{3} \frac{\partial q_{j}}{\partial r_{j^{\prime}}} d r_{j} d r_{j^{\prime}} .$ The convenience of the tensor $$\partial q_{j} / \partial r_{j}$$, for characterizing deformations is that it automatically excludes the translation displacement (Figure 1a), which is independent of $$r_{j}$$. Its drawback is that its particular components are still affected by the rotation of the body (though the sum (5) is not). Indeed, according to the vector product definition, Eq. (1) may be represented in Cartesian coordinates as $\left.d q_{j}\right|_{\text {rotation }}=\left(d \varphi_{j^{\prime}} r_{j^{\prime \prime}}-d \varphi_{j^{\prime \prime}} r_{j^{\prime}}\right) \varepsilon_{j j j^{\prime \prime}},$ where $$\varepsilon_{i j i j}$$ is the Levi-Civita symbol. Differentiating Eq. (6) over a particular Cartesian coordinate of vector $$\mathbf{r}$$, and taking into account that this partial differentiation $$(\partial)$$ is independent of (and hence may be swapped with) the differentiation $$(d)$$ over the rotation angle $$\varphi$$, we get the amounts, $d\left(\frac{\partial q_{j}}{\partial r_{j^{\prime}}}\right)_{\text {rotation }}=-\varepsilon_{i j j^{\prime \prime}} d \varphi_{j^{\prime \prime}}, \quad d\left(\frac{\partial q_{j^{\prime}}}{\partial r_{j}}\right)_{\text {rotation }}=-\varepsilon_{j i j^{\prime \prime}} d \varphi_{j^{\prime \prime}}=\varepsilon_{j j j^{\prime \prime}} d \varphi_{j^{\prime \prime}},$ which may differ from 0 . However, notice that the sum of these two differentials equals zero for any $$d \varphi$$, which is possible only if $$^{2}$$ $\left(\frac{\partial q_{j^{\prime}}}{\partial r_{j}}+\frac{\partial q_{j}}{\partial r_{j^{\prime}}}\right)_{\text {rotation }}=0, \quad \text { for } j \neq j^{\prime} .$ This is why it is convenient to rewrite Eq. (5) in a mathematically equivalent form, $\left.(d l)^{2}\right|_{\text {affer deformation }}-(d l)^{2} \mid \text { |before deformation }=2 \sum_{j, j^{\prime}=1}^{3} s_{i j^{\prime}} d r_{j} d r_{j^{\prime}} \text {, }$ where $$s_{j j}$$ ’ are the elements of the so-called symmetrized strain tensor, defined as $s_{i j^{\prime}} \equiv \frac{1}{2}\left(\frac{\partial q_{j}}{\partial r_{j^{\prime}}}+\frac{\partial q_{j^{\prime}}}{\partial r_{j}}\right) \text {. }$ (Note that this modification does not affect the diagonal elements: $$s_{j j}=\partial q_{j} / \partial r_{j}$$.). The advantage of the symmetrized tensor $$(9 \mathrm{~b})$$ over the initial tensor with elements $$\partial q_{j} / \partial r_{j}$$, is that according to Eq. (8), at pure rotation, all elements of the symmetrized strain tensor vanish.

Now let us discuss the physical meaning of this tensor. As was already mentioned in Sec. 4.2, any symmetric tensor may be diagonalized by an appropriate selection of the reference frame axes. In such principal axes, $$s_{j j}=s_{j j} \delta_{j j}$$, so that Eq. (4) takes a simple form: $d q_{j}=\frac{\partial q_{j}}{\partial r_{j}} d r_{j}=s_{j j} d r_{j} .$ We may use this expression to calculate the change of each side of an elementary cuboid (parallelepiped) with sides $$d q_{j}$$ parallel to the principal axes: $\left.d r_{j}\right|_{\text {after deformation }}-\left.d r_{j}\right|_{\text {before deformation }} \equiv d q_{j}=s_{j j} d r_{j},$ and of cuboid’s volume $$d V=d r_{1} d r_{2} d r_{3}$$ : $\left.d V\right|_{\text {after deformation }}-\left.d V\right|_{\text {before deformation }}=\prod_{j=1}^{3}\left(d r_{j}+s_{j j} d r_{j}\right)-\prod_{j=1}^{3} d r_{j}=d V\left[\prod_{j=1}^{3}\left(1+s_{j j}\right)-1\right]$ Since all our analysis is only valid in the linear approximation in small $$s_{j j}$$, Eq. (12) is reduced to $\left.d V\right|_{\text {after deformation }}-\left.d V\right|_{\text {before deformation }} \approx d V \sum_{j=1}^{3} s_{j j} \equiv d V \operatorname{Tr}(\mathrm{s}),$ where $$\operatorname{Tr}(\text { trace })^{3}$$ of any matrix (in particular, any tensor) is the sum of its diagonal elements; in our current case $$^{4}$$ $\operatorname{Tr}(\mathrm{s}) \equiv \sum_{j=1}^{3} s_{j j}$ So, the diagonal components of the tensor characterize the medium’s compression/extension; then what is the meaning of its off-diagonal components? It may be illustrated on the simplest example of purely shear deformation, shown in Figure 2 (the geometry is assumed to be uniform along the $$z$$-axis normal to the plane of the drawing). In this case, all displacements (assumed small) have just one Cartesian component, in Figure 2 along the $$x$$-axis: $$\mathbf{q}=\mathbf{n}_{x} \alpha y$$ (with $$\alpha<1$$ ), so that the only nonzero component of the initial strain tensor $$\partial q_{j} / \partial r_{j}$$, is $$\partial q_{x} / \partial y=\alpha$$, and the symmetrized tensor ( $$9 \mathrm{~b}$$ ) is $\mathrm{s}=\left(\begin{array}{ccc} 0 & \alpha / 2 & 0 \\ \alpha / 2 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right) .$ Evidently, the change of volume, given by Eq, (13), vanishes in this case. Thus, off-diagonal elements of the tensor s characterize shear deformations.

To conclude this section, let me note that Eq. (9) is only valid in Cartesian coordinates. For the solution of some important problems with the axial or spherical symmetry, it is frequently convenient to express six different components of the symmetric strain tensor via three components of the displacement vector $$\mathbf{q}$$ in either cylindrical or spherical coordinates. A straightforward differentiation of the definitions of these curvilinear coordinates, similar to that used to derive the well-known expressions for spatial derivatives, $${ }^{5}$$ yields, in particular, the following formulas for the diagonal elements of the tensor:

(i) in the cylindrical coordinates: $s_{\rho \rho}=\frac{\partial q_{\rho}}{\partial \rho}, \quad s_{\varphi \varphi}=\frac{1}{\rho}\left(q_{r}+\frac{\partial q_{\varphi}}{\partial \varphi}\right), \quad s_{z z}=\frac{\partial q_{z}}{\partial z} .$ (ii) in the spherical coordinates: $s_{r r}=\frac{\partial q_{r}}{\partial r}, \quad s_{\theta \theta}=\frac{1}{r}\left(q_{r}+\frac{\partial q_{\theta}}{\partial \theta}\right), \quad s_{\varphi \varphi}=\frac{1}{r}\left(q_{r}+q_{\theta} \frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta} \frac{\partial q_{\varphi}}{\partial \varphi}\right) .$ These expressions, which will be used below for the solution of some problems for symmetrical geometries, may be a bit counter-intuitive. Indeed, Eq. (16) shows that even for a purely radial, axiallysymmetric deformation, $$\mathbf{q}=\mathbf{n}_{\rho} q(\rho)$$, the angular component of the strain tensor does not vanish: $$s_{\varphi \varphi}=$$ $$q / \rho$$. (According to Eq. (17), in the spherical coordinates, both angular components of the tensor exhibit the same property.) Note, however, that this relation describes a very simple geometric effect: the change of the lateral distance $$\rho d \varphi<<\rho$$ between two close points with the same distance from the symmetry axis, at a small change of $$\rho$$, that keeps the angle $$d \varphi$$ between the directions towards these two points constant.

$${ }^{1}$$ Since both $$d \mathbf{q}$$ and $$d \mathbf{r}$$ are legitimate physical vectors (whose Cartesian components are properly transformed as the transfer between reference frames), the $$3 \times 3$$ matrix with elements $$\partial q_{j} / \partial r_{j}$$, is indeed a legitimate physical tensor - see the discussion in Sec. 4.2.

$${ }^{2}$$ As a result, the full sum (5), which includes three partial sums (8), is not affected by rotation - as we already know.

$${ }^{3}$$ The traditional European notation for Tr is Sp (from the German Spur meaning "trace" or "track").

$${ }^{4}$$ Actually, the tensor theory shows that the trace does not depend on the particular choice of the coordinate axes.

$${ }^{5}$$ See, e.g., MA Eqs. (10.1)-(10.12).

This page titled 7.1: Strain is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Konstantin K. Likharev via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.


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Differences between revisions 13 and 14
 ⇤ ← Revision 13 as of 2007-01-23 03:08:45 → Size: 633 Editor: TimothyClemans Comment: ← Revision 14 as of 2007-01-23 03:13:28 → ⇥ Size: 640 Editor: TimothyClemans Comment: Deletions are marked like this. Additions are marked like this. Line 3: Line 3: * $2^n \pm 1$ for $n ≤ 100$ [http://www.euronet.nl/users/bota/medium-p.htm] * $2^n \pm 1$ for $1200 < n < 10000$ [http://www.euronet.nl/users/bota/medium-p.htm]
1. a^n \pm 1 for a = 2, 3, 5, 6, 7, 10, 11, 12 and large exponents n [http://homes.cerias.purdue.edu/~ssw/cun/index.html]

2. a^n \pm 1 for a ≤ 13 and a not a perfect number [http://wwwmaths.anu.edu.au/~brent/factors.html]

3. 2^n \pm 1 for 1200 < n < 10000 [http://www.euronet.nl/users/bota/medium-p.htm]

4. 10^n \pm 1 for n ≤ 100 [http://www.swox.com/gmp/repunit.html]

5. p^p \pm 1 where p is a prime number and p < 180. [http://homes.cerias.purdue.edu/~ssw/bell/r1]

6. 2^{2^n} + 1 (Fermat numbers) [http://www.prothsearch.net/fermat.html]

7. 2^{3^n} \pm 1 [http://www.alpertron.com.ar/MODFERM.HTM]

factorization_of_integers_of_special_forms (last edited 2008-11-14 13:41:51 by localhost)


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# Lesson 03-03: Program Design Tips: Part I¶

Learning Target: I can document my code with appropriate comments.
Learning Target: I can use good naming conventions with variables.
Learning Target: I can know when to use variables instead of hard-coding numbers.

Comments can be used in a number of ways. The most important thing to remember is that comments are used to enhance readability of your program to other programmers, not to non-programmers. This means that ideally, especially with python, since it is such an easy-to-read language, people should be able to know what your program is doing simply by reading your code. Good variables names and good code structure will help them understand.

However, it also helps to put little comments here and there to explain an algorithm, especially if it’s tricky or hard to understand on its own. If you’re ever not sure whether you should be commenting something - just comment it. It’s a lot easier to read code with a lot of explanation than it is to read code with no explanation.

## Variable Naming Conventions¶

Although there is an entire guide on python naming conventions, we’re going to keep it simple and give you only a few guidelines to follow in order to improve your code readability.

1. Variable names should be descriptive of what they are, but not too long in length.

2. Variable names should be all lowercase, with words separated by underscores.

• For example, sum_of_prices is preferrable to sumofprices or SumOfPrices.

Following these guidelines will improve your code readability, which is useful when others are reviewing your code, or when you look back on your code after a while - when you use single letters or nondescriptive names in your code, you will eventually forget what they mean.

## Variables vs Hard-Coding Numbers¶

Sometimes you will have programs that have a constant value. For example, let’s say that you’re writing a program that checks your age to see if you are allowed to drive (assuming 17 is the minimum).

On line 2, when you put age >= 17, we are making the assumption that 17 will never change. In this context, we are hard-coding the number 17 into the program. This means that it cannot be changed at all without diving into the source code. Let’s say that this number is used again and again throughout the program. It’s better to put the value in a variable, and use the variable instead. Like so:

This way, if the minimum age to drive were to ever change, we would only need to change one variable instead of all of them. Additionally, a user could be given the opportunity to change the value of age_min - no need to change the source code.

It is generally good practice to avoid hard-coding constant values, especially if those values are repeated throughout the program.

Next Section - Lesson 03-04: The for Loop


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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 added 29 characters in body

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ uniformly and independently, the abobe above probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

2 added 17 characters in body

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ chosen at random we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?

1

# Probability of all combinations of k numbers among n being coprime

A simple argument shows that if we choose $n$ positive integers at random, the probability of their greatest common divisor being 1 is $1/ \zeta (n)$ (in the sense that if we choose the numbers among $\lbrace 1, 2, 3, \dots , N \rbrace$ the abobe probability tends to some number $p(n)$ as $N \rightarrow \infty$).

In the above sense, what is the probability $p(n,k)$ that among n integers $x_1, x_2, \dots, x_n$ we have $\gcd (x_{i_1}, x_{i_2}, \dots, x_{i_k})=1$ for all possible combinations of the $x_{i_j}$ (with $s \neq t$ if for $u \neq v$, $x_{i_u}=x_s$ and $x_{i_v}=x_t$) ?