#### Discover more from Infinitely More

The project to prove the irrationality of root two in rhyming verse began two years ago when Hannah Hoffman inquired about it on Twitter. The challenge was accepted, and a day of thinking, writing, and revising with Barbara Gail Montero and daughter Hypatia in Oxford resulted in a set of lyrics providing the proof in rhyme and meter. The goal was to highlight the logic of the proof and tell the story of Hippasus, credited with the discovery, who was drowned at sea as punishment from the Gods.

### Ode to Hippasus

*Lyrics by Joel David Hamkins, Barbara Gail Montero & their daughter Hypatia*
*Music by Hannah Hoffman*

The diagonal of a square is incommensurable with its side, an astounding fact the Pythagoreans hid. Hippasus rebelled and spoke the truth with irrefutable proof. It's absurd to suppose that the root of two is rational, namely, *p* over *q*. Squaring both sides shows that twice *q* squared is the square of *p*. Since *p* squared is even, *p* is also even. If *p* is 2*k*, then 2*q* squared equals 4*k* squared, revealing *q*'s even fate when halved. Thus, root two as a fraction, *p* over *q*, must have a numerator and denominator with factors of two, and to lowest terms, it can't be reduced. Root two is irrational, as Hippasus deduced. As retribution for revealing this irrationality, Hippasus was drowned in the sea, but his proof lives on.

This is the first of several mathematical/philosophical music videos undertaken with Hannah Hoffman, to be posted on Infinitely More. A paper was written with Conway on proofs of the irrationality of sqrt(2), classifying 7 "essentially different" ones based on how they generalized. Recently, an 8th proof was discovered, generalizing to all algebraic irrationals. A 9th proof using ultraproducts also exists, generalizing to all roots of irreducible polynomials that are unsolvable mod p for infinitely many p, but is less useful due to the need for the Chebotarev density theorem.

What is the 8th proof of the irrationality of sqrt(2) that generalizes to all algebraic irrationals?
A) Using geometric progressions
B) Using ultraproducts
C) Using algebraic number theory
D) Using a new, unknown method

Note: The answer is not provided, as it is a question for the reader to consider.


Intermediate Algebra
Tutorial 37: Radicals

Learning Objectives
After completing this tutorial, you should be able to:
 Find the principal nth root of an expression.
 Find the nth root of an expression raised to the nth power.

Introduction
The root of an expression is the reverse of raising it to a power. For example, the square root of an expression is another expression that, when squared, gives the original expression. This concept applies to all roots.

Key Concepts
When looking for the nth radical or nth root, you want the expression that, when raised to the nth power, gives the radicand (what is inside the radical sign). If there is no index number, n, it is understood to be a 2 or square root.

Examples
1. Simplify √9. We are looking for a number that, when squared, gives 9. Since 3² = 9, the answer is 3.
2. Simplify ∛(1/8). We are looking for a number that, when cubed, gives 1/8. Since (1/2)³ = 1/8, the answer is 1/2.
3. Simplify ∜81. We are looking for a number that, when raised to the fourth power, gives 81. Since 3⁴ = 81, the answer is 3.
4. Simplify √(x²). We are looking for an expression that, when squared, gives x². Since (√x)² = x², the answer is √x.
5. Simplify ∛(x³). We are looking for an expression that, when cubed, gives x³. Since (∛x)³ = x³, the answer is ∛x.
6. Simplify ∜(x⁴). We are looking for an expression that, when raised to the fourth power, gives x⁴. Since (∜x)⁴ = x⁴, the answer is ∜x.
7. Simplify √(x⁴). We are looking for an expression that, when squared, gives x⁴. Since (x²)² = x⁴, the answer is x².
8. Simplify ∛(x⁶). We are looking for an expression that, when cubed, gives x⁶. Since (x²)³ = x⁶, the answer is x².

Rule
If n is an even positive integer, then |x|ⁿ = xⁿ.
If n is an odd positive integer, then xⁿ = xⁿ.

Examples with Variables
9. Simplify √(y²). Since the root number and exponent are equal and even, we need to put an absolute value around y: |y|.
10. Simplify ∛(y³). Since the root number and exponent are equal and odd, the answer is y.
11. Simplify ∜(a²b²). Since the root number and exponent are equal and even, we need to put an absolute value around ab: |ab|.

Practice Problems
Simplify the following expressions, assuming variables represent positive real numbers:
1a. √(x²)
1b. ∛(x³)
1c. ∜(x⁴)
1d. √(x⁴)
2a. ∛(x⁶)
2b. ∜(x⁸)


The frequency of a photon with an energy of 3.26 x 10^-19 J can be calculated using the formula E = h * f, where h is Planck's constant (6.626 x 10^-34 J*s) and f is the frequency of the photon. Rearranging the formula gives f = E / h. Plugging in the given energy value and the Planck constant, the frequency of the photon is approximately 4.92 x 10^14 Hz.

The energy of a photon is directly proportional to its frequency, with the constant of proportionality being Planck's Constant, h = 6.63 x 10^-34 joule-second. The formula E = hf can be used to calculate the energy of a photon, where E is the energy of the photon and f is the frequency of the photon.

Alternatively, the energy of a photon can be calculated using the formula E = hc/λ, where c is the speed of light (3.00 x 10^8 m/s) and λ is the wavelength of the photon. The speed of light equation, c = λf, can be used to calculate the frequency of the photon, which can then be substituted into the energy equation to find the energy of the photon.

Related concepts include the proportionality of photon energy to frequency and the use of Planck's Constant as the constant of proportionality. The energy of a photon with a given frequency can be calculated using the formula E = hf, and the frequency of a photon with a given energy can be calculated using the formula f = E / h.

Note: The correct answer is 4.92 x 10^14 Hz.


# Graph Isomorphisms

The definition of an isomorphism between two graphs $G$ and $H$ is as follows:

Definition: For two graphs $G = (V(G), E(G))$ and $H = (V(H), E(H))$, $G$ is an isomorphism of $H$ if there exists a bijection $f: V(G) \rightarrow V(H)$ such that $\left\{ {x, y}\right\} \in E(G)$ if and only if $\left\{ {f(x), f(y)}\right\} \in E(H)$. This is denoted as $G \cong H$.

In simpler terms, an isomorphism exists if there is a function $f$ that maps every pair of vertices $\{x, y \}$ forming an edge in $E(G)$ to a pair of vertices $\{ f(x), f(y) \}$ forming an edge in $E(H)$, and vice versa for non-edges.

To determine if two graphs are isomorphic, follow these steps:

1. Check if the number of vertices in $G$ is equal to the number of vertices in $H$.
2. Verify if the number of edges in $G$ is equal to the number of edges in $H$.
3. Compare the degree sequences of $G$ and $H$ to ensure they are the same.

Consider two graphs $G$ and $H$ with the same number of vertices and edges (7 edges). The degree sequences of $G$ and $H$ are $(2, 2, 3, 3, 4)$, indicating a potential isomorphism.

Since $\deg (a) = \deg (v) = 4$, a bijection $f$ can be defined as:

$f(a) = v$
$f(b) = x$
$f(c) = w$
$f(d) = y$
$f(e) = z$

Verify the correspondence of edges between $G$ and $H$ using the bijection $f$:

| Edge in G | Correspondence in H |
| --- | --- |
| $\{a, b\}$ | $\{v, x\}$ |
| $\{a, c\}$ | $\{v, w\}$ |
| $\{a, d\}$ | $\{v, y\}$ |
| $\{a, e\}$ | $\{v, z\}$ |
| $\{b, c\}$ | $\{x, w\}$ |
| $\{b, d\}$ | $\{x, y\}$ |
| $\{c, e\}$ | $\{w, z\}$ |

Since there is a bijection $f: V(G) \to V(H)$ and the number of edges in $G$ and $H$ are equal, $G \cong H$.


To solve the equation 2a + 3 = 11, we need to isolate the variable 'a'.

First, subtract 3 from both sides: 2a = 11 - 3, which simplifies to 2a = 8.

Then, divide both sides by 2: a = 8 / 2, resulting in a = 4.

In another equation, 2a + 13 = 5a + 1, we solve for 'a' by subtracting 2a from both sides: 13 = 3a + 1.

Subtract 1 from both sides: 12 = 3a.

Divide both sides by 3: a = 12 / 3, which simplifies to a = 4.

Given the equation 2A/3 = 8 + 4A, to solve for A, first multiply both sides by 3 to eliminate the fraction: 2A = 24 + 12A.

Then, subtract 2A from both sides: 0 = 24 + 10A.

Subtract 24 from both sides: -24 = 10A.

Divide both sides by 10: A = -24 / 10, which simplifies to A = -12/5.

For the equation -2a + 5 = 1, solve for 'a' by subtracting 5 from both sides: -2a = -4.

Divide both sides by -2: a = 2.

In the equation 2a + 5a + 9a = 16a, combine like terms: 16a = 16a, which is always true for any value of 'a'.

The equation a + a = 1 simplifies to 2a = 1, and solving for 'a' gives a = 1/2.

The expression 3 + 2a does not provide enough information to solve for 'a' without an equation.

Given 3a * 2a = 24, this simplifies to 6a^2 = 24.

Divide both sides by 6: a^2 = 4.

Taking the square root of both sides gives a = ±2.

For the equation 5a - 20 + 2a = 6a - 11, combine like terms: 7a - 20 = 6a - 11.

Add 20 to both sides: 7a = 6a + 9.

Subtract 6a from both sides: a = 9.

The equation 2ab - 3b^2 - 3b + 2a does not have a solution as it is not a complete equation without additional information.

Given 2a + 2b = 160 and b = a + 40, substitute b in the first equation: 2a + 2(a + 40) = 160.

This simplifies to 2a + 2a + 80 = 160, which further simplifies to 4a + 80 = 160.

Subtract 80 from both sides: 4a = 80.

Divide both sides by 4: a = 20.

Then, b = a + 40 gives b = 20 + 40 = 60.

So, the solution is a = 20, b = 60.

For the equation b = 2a + 11 and a + b = 5, substitute b in the second equation: a + (2a + 11) = 5.

This simplifies to 3a + 11 = 5.

Subtract 11 from both sides: 3a = -6.

Divide both sides by 3: a = -2.

Substitute a = -2 into b = 2a + 11: b = 2(-2) + 11 = -4 + 11 = 7.

Checking: a + b = -2 + 7 = 5, which confirms the solution.

Given a = 3a + 4, subtract 3a from both sides: -2a = 4.

Divide both sides by -2: a = -2.

The equation a = 1 is a direct statement of 'a'.

The formula A = h/2*(a + b) can be rearranged to find 'a': 2A = h*(a + b), then 2A/h = a + b, and finally a = 2A/h - b.

For 2a + 3a + 4 = 5a + 4, combine like terms: 5a + 4 = 5a + 4, which is always true for any value of 'a'.


## Introduction

This post compares the performance of two methods to simulate samples from a pair of independent normal random variables: Box-Muller and Marsaglia-Bray. The Python functions for these methods are:

```python
import math
import numpy as np
from numpy import pi

def box_muller_sample(n=1):
    u1 = np.random.rand(n)
    u2 = np.random.rand(n)
    r = np.sqrt(-2*np.log(u1))
    x = np.cos(2*pi*u2)
    y = np.sin(2*pi*u2)
    z1 = r*x
    z2 = r*y
    return z1, z2

def marsaglia_sample(n=1):
    def marsaglia():
        while True:
            w1 = (np.random.rand() * 2) - 1
            w2 = (np.random.rand() * 2) - 1
            s = w1 * w1 + w2 * w2
            if s < 1:
                t = math.sqrt((-2) * math.log(s)/s)
                z1 = w1 * t
                z2 = w2 * t
                return z1, z2
    z1 = np.empty(n)
    z2 = np.empty(n)
    for i in range(n):
        z1[i], z2[i] = marsaglia()
    return z1, z2
```

To compare their performance, 1000 samples of size $n=100000$ are generated, and the average time for this task is calculated.

## Performance Comparison

The Box-Muller function simulates a sample of size $n=100000$ on an average of 0.002097 seconds. In contrast, the Marsaglia function simulates a sample of size $n=100000$ on an average of 0.148888 seconds. This indicates that the Box-Muller function is significantly faster than the Marsaglia function.

## First Improvement – Marsaglia 1.1

An improved version of the Marsaglia function using numpy vector functions is:

```python
def marsaglia_sample_vectorial(n=1):
    def marsaglia():
        while True:
            w1 = (np.random.rand() * 2) - 1
            w2 = (np.random.rand() * 2) - 1
            if w1 * w1 + w2 * w2 < 1:
                return w1, w2
    w1 = np.empty(n)
    w2 = np.empty(n)
    for i in range(n):
        w1[i], w2[i] = marsaglia()
    s = w1*w1 + w2*w2
    t = np.sqrt(-2*np.divide(np.log(s), s))
    z1 = w1*t
    z2 = w2*t
    return z1, z2
```

The vectorial version of the Marsaglia function simulates a sample of size $n=100000$ on an average of 0.120829 seconds, which is faster than the original Marsaglia function but still slower than the Box-Muller function.

## Understanding the Accept-Reject Method

The acceptance probability of the Marsaglia function is $p = \mathbb{P}(w_1^2 + w_2^2 < 1) = \frac{\pi}{4} \approx 0.7854$. To avoid the while loop, the minimum number of simulations $N$ required to guarantee at least $n$ accepted points with high probability needs to be determined.

## Finding the Right N

The probability of accepting at least $n$ points can be calculated using the binomial distribution or the normal approximation. A Python function to calculate this probability is:

```python
def calculate_probability(n, N):
    p = pi/4
    mu = N*p
    sigma = np.sqrt(N*p*(1-p))
    prob = 1 - norm.cdf(n, mu, sigma)
    return prob
```

## Second Improvement – Marsaglia 1.2

Using the finding that the probability of accepting at least $n$ points can be calculated, a new version of the Marsaglia function without the while loop is introduced:

```python
p = pi/4
aux = p*(1-p)

def marsaglia_optimal_sample(n=1):
    x = (3*math.sqrt(aux) + math.sqrt(9*aux + p*n))/p
    N = math.ceil(x*x)
    w1 = np.random.rand(N) * 2 - 1
    w2 = np.random.rand(N) * 2 - 1
    s = w1 * w1 + w2 * w2
    index = s<1
    w1 = w1[index][:n]
    w2 = w2[index][:n]
    s = s[index][:n]
    t = np.sqrt(-2*np.divide(np.log(s), s))
    z1 = w1*t
    z2 = w2*t
    return z1, z2
```

The new version of the Marsaglia function simulates a sample of size $n=100000$ on an average of 0.004235 seconds, which is closer to the performance of the Box-Muller function.


**Problem Description**
Li Ming is a student who wants to know his rank in class after every exam. His teacher provides him with the scores of all students and asks him to determine his own rank. Since there are many classmates, Li Ming needs help to find his rank.

**Input**
The first line contains an integer N (1 <= N <= 10000), representing the number of students in Li Ming's class. The next N lines contain the names of the students, each with no more than 30 letters. The (N+2)th line contains an integer M (1 <= M <= 50), representing the number of exams. Each exam consists of N lines, with each line containing a positive integer S (no more than 100) and a name P, which must match a name in the initial list.

**Output**
The output contains M lines, each giving Li Ming's rank after the corresponding exam. The rank is determined by the total scores, with Li Ming ranked higher in case of a tie.

**Sample Input**
3
Li Ming
A
B
2
49 Li Ming
48 A
80 B
85 B
83 Li Ming

**Sample Output**
1
2

**Note**: The problem can be solved using a map or other data structures, but the provided code uses the qsort function. The goal is to find Li Ming's rank after each exam by calculating the total scores.


The Roman Numeral system is a way of writing numbers using a unique set of symbols. The Roman numerals with their corresponding Hindu-Arabic numerals are as follows:

To write numbers in the Roman system, the following rules are used:
- When a Roman numeral is repeated more than once, its value is added each time to get the number.
- The same symbol is not repeated more than 3 times together.
- The symbols V, L, and D are never repeated.
- There is no symbol for zero in the Roman system.
- Symbols I, X, and V are used to write Roman numerals up to 39.

When a smaller Roman numeral is written to the right of a greater Roman numeral, the value of the greater numeral is added to the smaller numeral. Conversely, when a smaller Roman numeral is written to the left of a greater Roman numeral, it is always subtracted from the value of the greater numeral.

Key rules to note:
- Symbol I can be subtracted only once.
- The symbols V, L, and D are never subtracted.
- When a smaller Roman numeral is placed between two greater Roman numerals, its value is subtracted from the value of the greater numeral on its right.

Example 1: Write the following in Hindu-Arabic numerals.
(a) XII
(b) VII
(c) XV
(d) XIV
Solutions:
(a) XII = 10 + 1 + 1 = 12
(b) VII = 5 + 1 + 1 = 7
(c) XV = 10 + 5 = 15
(d) XIV = 10 + (5 – 1) = 14

Example 2: Write the following in Roman numerals.
(a) 37
(b) 16
(c) 28
(d) 34
Solutions:
(a) 37 = 30 + 7 = X + X + X + V + I + I = XXXVII
(b) 16 = 10 + 6 = X + V + I = XVI
(c) 28 = 20 + 8 = X + X + V + I + I + I = XXVIII
(d) 34 = 30 + 4 = X + X + X + IV = XXXIV


**Estimation** is used to find an approximate answer by solving a simpler problem. When an operation is involved, estimation can be done by rounding numbers before performing the operation.

To estimate the difference between two numbers, round both numbers to the nearest hundred and then subtract. For example, to estimate 6,391 - 605, round both numbers to the nearest hundred: 6,400 - 600 = 5,800. The actual difference is 5,786, making 5,800 a good estimate.

To estimate a quotient, round the dividend to the nearest thousand before dividing. For instance, to estimate 5,000 ÷ 5, round the dividend to the nearest thousand and then divide: 5,000 ÷ 5 = 1,000. This method provides a quick and reasonable estimate of the actual result.


Formula for Sum and Product of Roots

The sum and product of roots can be derived from a quadratic equation of the form ax^2 + bx + c = 0, where a, b, and c are constants.

Given a quadratic equation with a = 3, b = 2p, and c = -p, we can analyze the roots in two scenarios:

a) When the roots are equal, let the roots be α and α. The sum of roots is α + α = 2α, and the product of roots is α * α = α^2. Using the equation α^2 + (2p/3)α - (p/3) = 0, we can substitute to find relationships between α and p.

b) When one root is the reciprocal of the other, let the roots be α and 1/α. The sum of roots is α + 1/α, and the product of roots is α * (1/α) = 1. Using the product of roots, we can find the value of p. Since α * (1/α) = -c/a = -(-p)/3 = p/3, we get p/3 = 1, which implies p = 3.

It's essential to note that the roots of an equation do not always have to be denoted as α and β; other notations like β and 1/β can also be used. The key is to understand the relationship between the roots and the coefficients of the quadratic equation.


A comparison function is one of six primitive dyadic scalar functions that determine a Boolean result based on the numerical ordering of two arguments. These functions are subject to comparison tolerance and can be divided into two equalities (=, ≠) and four inequalities (<, ≤, ≥, >). The equalities depend only on whether the arguments are equal, while the inequalities depend on the ordering of the arguments.

The six comparison functions are:
- Less Than (<)
- Less Than or Equal (≤)
- Equal to (=)
- Greater Than or Equal (≥)
- Greater Than (>)
- Not Equal to (≠)

These functions have Boolean results on Boolean arguments, making them Boolean functions. The majority of nontrivial dyadic Boolean functions are comparison functions.

The result of a comparison function on real numbers is determined by their ordering, which falls into three possible cases: the left argument is smaller, the arguments are tolerantly equal, or the right argument is larger. Each comparison function can be thought of as a choice of one Boolean value for each of these three cases, resulting in eight total possibilities. However, two of these possibilities (all 0 and all 1) are trivial and do not depend on the ordering of the arguments.

The remaining possibilities correspond to the APL comparisons:
- Function <: 1 0 0
- Function ≤: 1 1 0
- Function=: 0 1 0
- Function ≥: 0 1 1
- Function >: 0 0 1
- Function ≠: 1 0 1

The traditional APL ordering of the comparison functions resembles a Gray code ordering, where adjacent values differ by only a change of one bit. However, it excludes the all 0 and all 1 combinations.


The Guild 100J guitar amp features a unique four-triode phase inverter. Each of the two triodes shown is actually a pair of triodes in parallel, connected plate-to-plate, grid-to-grid, and cathode-to-cathode. The 150kΩ plate load resistors carry the plate current of two triodes, so the equivalent value for one triode is double: 300kΩ. The 1.2kΩ and 68kΩ resistors carry the plate current of four triodes, so the equivalent value for one triode is quadruple: 4.8kΩ and 272kΩ, respectively.

The equivalent DC circuit for one triode can be analyzed to determine the load line. If the plate current is 0mA, the plate-to-cathode voltage is equal to the plate supply voltage, so one endpoint of the DC load line is at 360V, 0mA. If the plate-to-cathode voltage is 0V, then according to Ohm's Law, the plate current is 360V / (300kΩ + 4.8kΩ + 272kΩ) = 0.62mA. The other endpoint for the load line is therefore at 0V, 0.62mA.

Using Ohm's Law, the current through the 4.8kΩ resistor can be calculated for different grid-to-cathode voltages: 1V / 4.8kΩ = 0.21mA, 1.5V / 4.8kΩ = 0.31mA, and 2V / 4.8kΩ = 0.42mA. These points define the blue line segments, which intersect the load line at the DC operating point: 1.7V grid-to-cathode, 155V plate-to-cathode, and 0.36mA plate current.

In response to an audio signal, the plate current increases for the inverted phase and decreases for the non-inverted phase, so the 95V across the tail ideally stays constant. The AC plate load is the 150kΩ plate load resistor in parallel with the 220kΩ grid-leak resistor: 89kΩ. This value is doubled to get the equivalent AC load for one tube: 178kΩ. The AC load line has this slope and passes through the DC operating point.

The maximum output voltage swing is from 20V to 319V, representing 135V to +164V relative to 155V at the DC operating point. The gain of the LTP is approximately half the gain of a common cathode voltage amplifier. Two triodes in parallel are the equivalent of one triode with double the transconductance and half the plate resistance. The amplification factor is the same.

For a traditional 2-triode LTP design, the plate supply voltage is 360V and the desired voltage across the tail is 95V. The equivalent plate supply voltage for a common cathode voltage amplifier is the voltage difference between the actual plate supply and the tail voltage: 265V. The cathode resistor value for one triode is 1.235kΩ, and the value for two triodes is half: 620Ω. The DC plate current is 0.85mA for one triode, double for the current through the tail resistor: 1.7mA.

The tail resistor value for a desired voltage drop of 95V is 95V / 1.7mA = 56kΩ. The green AC load line indicates that the maximum output swing is from 71V to 210V, representing 65 to +74 relative to 136V at the DC operating point. This is quite a bit less swing than for the 4-triode, parallel design.

A performance summary of the Guild 100J and traditional design is as follows:
- Output, peak-to-peak: 299V vs 139V
- Gain: 31dB vs 29dB
- Output impedance: 26kΩ vs 44kΩ
The 2dB increase in gain is not worth an extra tube. The pair of triodes can instead be configured as typical voltage amplifiers for a total loaded gain of 71dB. The decrease in output impedance impacts the dynamics of overdrive, because the Guild design can drive more grid current. The main difference in the two designs is the top line: a substantially greater output voltage swing.


Statistical data sets are collections of data maintained in an organized form, serving as the foundation for statistical analysis. The collection of data is the initial step in any statistical analysis, which is then examined using statistical tools. These data sets form the basis from which statistical inferences can be drawn.

The scope of information recorded in a statistical data set depends on the requirements of the experiment. For instance, a study on the relationship between height and age might only record these two parameters. However, a more comprehensive study may require additional data, such as height at birth, weight, nutritional background, and family history. Therefore, researchers must determine beforehand what data is necessary for their statistical data sets.

Certain characteristics are common to all statistical data sets. The order of the data is insignificant, allowing researchers to organize the data in a convenient manner. Creating a statistical data set is only the first step in research; the interpretation and validity of the inferences drawn from the data are of utmost importance.

Numerous statistical data sets are already available for various areas of study. For example, the international genealogical index contains extensive family history data, which can be utilized by researchers to study patterns and statistical data. This availability of data sets simplifies the researcher's task. A particular statistical data set can be used for multiple research studies, such as census data, which provides comprehensive demographic information that can be used by social scientists to study family structures, incomes, and other topics.

A statistical data set is not an end in itself but rather a starting point where data is stored. The collection and interpretation of data depend on the researcher studying the data. Statistical data sets are essential in social sciences, medical sciences, and physical sciences, serving as the foundation for most research in these fields.


To determine the common ratio and find the next four terms of the geometric sequence, we analyze the given terms: a^(-6), a^(-2), a^2. The common ratio is the factor by which each term is multiplied to obtain the next term. In this case, the common ratio is a^4, because when multiplying numbers with exponents having the same base, the exponents are added. For example, a^(-6) * a^4 = a^(-6+4) = a^(-2).

To find the next four terms, multiply a^2 by a^4 to get the next term, then multiply that result by a^4, and repeat this process. The calculation for the next term would be a^2 * a^4 = a^(2+4) = a^6. Continuing this pattern, the next three terms would be a^6 * a^4 = a^(6+4) = a^10, then a^10 * a^4 = a^(10+4) = a^14, and finally a^14 * a^4 = a^(14+4) = a^18.

Therefore, the common ratio is a^4, and the next four terms of the geometric sequence are a^6, a^10, a^14, and a^18.


#### Answer

Using the ASA method, we can prove that $\triangle ABC\cong\triangle DBC$ based on 2 congruent pairs of angles and one included pair of sides.

#### Work Step by Step

Since $\vec{CB}$ bisects $\angle ACD$, the resulting angles $\angle 3$ and $\angle 4$ are congruent. Additionally, we know that $\angle A\cong\angle D$ and $\overline{AC}\cong\overline{CD}$. With 2 angles and the included side of $\triangle ABC$ being congruent to 2 angles and the included side of $\triangle DBC$, according to the ASA method, $\triangle ABC\cong\triangle DBC$.


Your First Step in Data Analysis
Before undertaking data analysis, it's essential to understand the nature of the data you're working with. Data can be diverse, and it's crucial to recognize the type of data you have to develop an effective analysis. There are two primary types of data: Quantitative and Qualitative.

Types of Data
Quantitative Data is related to objective measures and can be further categorized into Continuous and Discrete data. Quantitative data can be measured and compared objectively, such as length, weight, sound, and number of people.

Quantitative Data
Quantitative Data can be divided into two subcategories:
- Continuous Data: This type of data can be infinitely divided, such as length, weight, and temperature.
- Discrete Data: This type of data cannot be infinitely divided, such as the number of people in a room or the number of objects.

Examples of Continuous and Discrete Data
Continuous Data examples include weight, length, velocity, sound, and temperature. Discrete Data examples include people, objects, and money.

Qualitative Data
Qualitative Data, on the other hand, is related to subjective descriptions or states. It serves to classify or categorize something subjective or describe the state of something. Qualitative data can be categorized into Binary, Nominal, and Ordinal data.

Qualitative Data Types
- Binary Data: Variables that can have only two possible values, such as yes/no or true/false.
- Nominal Data: Descriptive non-numerical data, such as colors, sensations, or feelings.
- Ordinal Data: Subjective classification or a certain order, such as ranking or evaluating similarity.

Examples of Qualitative Data
Binary Data examples include yes/no, true/false, and on/off. Nominal Data examples include colors, sensations, and flavors. Ordinal Data examples include similarity, likability, and subjective size.

Importance of Understanding Data Types
Understanding the type of data you're working with is crucial for developing an effective analysis. By recognizing the type of data, you can give it the importance it deserves and quantify it properly. This enables you to make informed decisions and develop a better analysis.

Tips for Handling Data
When handling data, keep the following tips in mind:
1. Handle continuous variables whenever possible.
2. Group non-quantifiable variables and study them independently.
3. Don't mix subjective and objective variables, as it can cause confusion and misunderstandings.

Real-World Example
Imagine studying how a variable evolves over a year. Instead of using dates, it's more efficient to number the days from 1 to 365. This approach allows for easier analysis and calculation of subperiods of time.

Summary
In summary, there are two primary types of data: Quantitative and Qualitative. Quantitative Data can be further categorized into Continuous and Discrete data, while Qualitative Data can be categorized into Binary, Nominal, and Ordinal data. Understanding the type of data you're working with is essential for developing an effective analysis and making informed decisions. By recognizing the type of data and quantifying it properly, you can develop a better analysis and achieve your goals.


Simplicial complexes are collections of finite sets where the power set of each member set is contained in the collection. The geometric notions attached to a simplicial complex include facets, adjacency, and chains of adjacent simplices.

A facet of a simplicial complex is a subset of codimension one. Two simplices are adjacent if they share a facet in common. A chain of adjacent simplices is a sequence of simplices where each pair of consecutive simplices is adjacent.

The locale SimplicialComplex defines a simplicial complex as a collection of finite sets with certain properties. It includes definitions for maximal simplices, adjacent sets, and chains of maximal simplices. A maximal simplex is a simplex that is not a subset of any other simplex in the complex. An adjacent set is a set of simplices that are adjacent to a given simplex. A chain of maximal simplices is a sequence of maximal simplices where each pair of consecutive simplices is adjacent.

The locale SimplicialComplexIsomorphism defines an isomorphism between two simplicial complexes. It includes lemmas about the properties of isomorphisms, such as preserving adjacency and maximal simplices.

The concept of a poset (partially ordered set) is also defined, and a simplicial complex can be constructed from a poset. The locale ComplexLikePoset defines a poset that can be used to construct a simplicial complex. It includes lemmas about the properties of this construction, such as the relationship between the poset and the simplicial complex.

Key definitions and lemmas include:

* Facet: a subset of codimension one
* Adjacent: two simplices that share a facet in common
* Chain of adjacent simplices: a sequence of simplices where each pair of consecutive simplices is adjacent
* Maximal simplex: a simplex that is not a subset of any other simplex in the complex
* Adjacent set: a set of simplices that are adjacent to a given simplex
* Chain of maximal simplices: a sequence of maximal simplices where each pair of consecutive simplices is adjacent
* Isomorphism: a bijective map between two simplicial complexes that preserves adjacency and maximal simplices
* Poset: a partially ordered set
* Simplicial complex constructed from a poset: a simplicial complex where each simplex is a set of pseudominimal elements in the poset.

Important numbers and concepts include:

* Codimension: the difference between the dimension of a simplex and the dimension of its facet
* Adjacency: the relationship between two simplices that share a facet in common
* Maximal simplex: a simplex that is not a subset of any other simplex in the complex
* Isomorphism: a bijective map between two simplicial complexes that preserves adjacency and maximal simplices
* Pseudominimal element: an element in a poset that is minimal with respect to a given subset.


#### Abstract

This paper develops a generalized matrix Riccati technique for the self-adjoint matrix Hamiltonian system. New oscillation and interval oscillation criteria are established using the standard integral averaging technique and positive functionals. These criteria extend and improve previous results.

#### 1. Introduction

The linear Hamiltonian system is considered, where A and B are real matrix-valued functions, A is Hermitian, and B is positive definite. A solution is nontrivial if it is not identically zero, and conjoined if its Wronski matrix is constant. Previous studies by Kumari and Umamaheswaram, Yang and Cheng, and Wang used the substitution to study oscillation.

Theorem A states that if there exists a function such that certain conditions are met, then the system is oscillatory. Theorem B uses a monotone subhomogeneous functional to establish oscillation. Theorem C uses a positive linear functional and a function to establish oscillation. Theorem D uses the standard integral averaging technique and a positive linear functional to establish oscillation.

#### 2. Main Results

A lemma is introduced, stating that if a positive linear functional exists, then a certain inequality holds. Theorem 2.2 states that if there exist a function, a matrix function, and a positive linear functional such that certain conditions are met, then the system is oscillatory. Theorem 2.3 states that if there exists a function such that certain conditions are met, then the system is oscillatory.

Example 2.4 illustrates the application of Theorem 2.3 to a linear Hamiltonian system. Remark 2.5 notes that Theorems 2.2 and 2.3 generalize and improve previous results. Remark 2.6 states that the theorems provide a wide range of possibilities for deriving explicit oscillation criteria.

#### 3. Interval Oscillation Criteria

Interval oscillation criteria are established, which are given by the behavior of the system on a sequence of subintervals. Lemmas 3.1 and 3.2 are introduced, stating that certain inequalities hold for nontrivial prepared solutions. Theorem 3.3 states that if there exist a matrix function and a positive linear functional such that certain conditions are met, then any nontrivial prepared solution has at least one zero in a given interval. Theorem 3.4 states that if certain conditions are met, then the system is oscillatory.

The results extend, improve, and complement previous theorems, and can be used to derive easily verifiable oscillation criteria. The choice of functions and can be used to obtain different explicit oscillation criteria.

The paper was supported by the National Natural Science Foundation of China, the National Ministry of Education, and the Natural Sciences Foundation of Shandong Province.


**NonUniformCartesianGrid**
==========================

### Class Definition

`plasmapy.plasma.grids.NonUniformCartesianGrid(*seeds, num=100, **kwargs)`

Bases: `AbstractGrid`

A Cartesian grid in which the `_make_mesh` method produces a non-uniformly spaced grid.

### Attributes

* `ax0`: First axis of the grid (only valid for uniform grids)
* `ax1`: Second axis of the grid (only valid for uniform grids)
* `ax2`: Third axis of the grid (only valid for uniform grids)
* `grid`: A single grid of vertex positions of shape `(N0, N1, N2, 3)` (only defined for grids with a defined `unit` property)
* `grid_resolution`: A scalar estimate of the grid resolution, calculated as the closest spacing between any two points
* `grids`: Three grids of vertex positions (in each coordinate), each having shape `(N0, N1, N2)`
* `is_uniform`: A boolean value reflecting whether or not the grid points are uniformly spaced
* `pts0`: Array of positions in dimension 1
* `pts1`: Array of positions in dimension 2
* `pts2`: Array of positions in dimension 3
* `quantities`: A list of the keys corresponding to the quantities currently defined on the grid
* `recognized_quantities`: A dictionary of standard key names representing particular physical quantities
* `shape`: Shape of the grid
* `si_scale_factors`: 3-element list containing unitless scale factors for converting the corresponding axis from its stored units to SI
* `unit`: The unit for the entire grid (only valid if all dimensions of the grid have the same units)
* `unit0`: Unit of dimension 1
* `unit1`: Unit of dimension 2
* `unit2`: Unit of dimension 3

### Methods

* `add_quantities(**kwargs)`: Adds a quantity to the dataset as a new DataArray
	+ Parameters: `**kwargs` (key, array pairs) - The key will be used as the dataset key, while the array holds the quantity
* `nearest_neighbor_interpolator(pos, *args, persistent=False)`: Interpolate values on the grid using a nearest-neighbor scheme with no higher-order weighting
	+ Parameters:
		- `pos` (ndarray or Quantity array, shape `(n,3)`) - An array of positions in space, where the second dimension corresponds to the three dimensions of the grid
		- `*args` (str) - Strings that correspond to DataArrays in the dataset
		- `persistent` (bool) - If `True`, the interpolator will assume the grid and its contents have not changed since the last interpolation
* `on_grid(pos)`: Given a list of positions, determines which are in the region bounded by the grid points
* `require_quantities(req_quantities, replace_with_zeros=False)`: Check to make sure that a list of required quantities are present
	+ Parameters:
		- `req_quantities` - List of required quantities
		- `replace_with_zeros` (bool) - If `True`, create missing quantities and fill them with an array of zeros

### Raises

* `ValueError`: If all grid dimensions do not have identical units


Key math categories and their corresponding answer counts are as follows:
- Pre-Algebra: 12.7k answers
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The platform contains a total of 86,155 questions, 92,154 answers, and 2,246 comments, with 23,902 registered users.


**June 2018 Question Paper**

**Instructions:**
1. Question 1 is compulsory.
2. Attempt any three questions from Q. 2 to Q. 6.

**Questions:**

1. Explain how interference in a wedge-shaped film is used to test the optical flatness of a given glass plate. (Chapter: Interference and Diffraction of Light)

2. Define diffraction grating and explain the advantage of increasing the number of lines in the grating. (Chapter: Interference and Diffraction of Light)

3. With a neat ray diagram, explain the concept of total internal reflection (TIR). (Chapter: Fibre Optics)

4. Differentiate between spontaneous and stimulated emission. (Chapter: Lasers)

5. Find the cylindrical coordinates of a point (3i + 4j + k). (Chapter: Electrodynamics)

6. In Newton's ring experiment, determine the order of the dark ring that will have double the diameter of the 40th dark ring. (Chapter: Interference and Diffraction of Light)

7. Draw the block diagram of a cathode ray tube and briefly explain its parts. (Chapter: Charge Particle in Electric and Magnetic Fields)

8. Derive the conditions of maxima and minima due to interference of light reflected from a thin film of uniform thickness. (Chapter: Interference and Diffraction of Light)

9. Derive the formula for the numerical aperture of a step-index fiber and explain its physical significance. Given that the N.A of an optical fiber is 0.5 and the core refractive index is 1.54, find the refractive index of the cladding. (Chapter: Fibre Optics)

10. Discuss Fraunhoffer diffraction at a single slit and obtain the condition for minima.

11. In a plane transmission grating, the angle of diffraction for the second-order principal maxima for a wavelength of 5 x 10^-5 is 35°. Calculate the number of lines/cm for this diffraction grating. (Chapter: Interference and Diffraction of Light)

12. Explain the difference between photography and holography, and describe the holography technique to obtain a 3D image of an object. (Chapter: Lasers)

13. Find the divergence of the vector field F = x^2yz i + xz j. (Chapter: Electrodynamics)

14. Explain how A.C voltage and its frequency are measured using a CRO. (Chapter: Charge Particle in Electric and Magnetic Fields)

15. A wedged-shaped air film having an angle of 40 seconds is illuminated by monochromatic light, and fringes are observed vertically through a microscope. The distance measured between consecutive fringes is 0.12 cm. Calculate the wavelength of the light used. (Chapter: Interference and Diffraction of Light)

16. Explain Newton's ring experiment and show that the diameters of nth dark rings are proportional to the square root of natural numbers. (Chapter: Interference and Diffraction of Light)

17. Write Maxwell's equation and explain its physical significance. (Chapter: Electrodynamics)

18. Explain the construction and working of an atomic force microscope. (Chapter: Nanoscience and Nanotechnology)

19. Explain the different types of carbon nanotubes and their applications. (Chapter: Nanoscience and Nanotechnology)

20. Explain the construction and working of an Nd:YAG LASER. (Chapter: Lasers)

21. Write a note on electrostatic focusing. (Chapter: Charge Particle in Electric and Magnetic Fields)


Using vectors, prove that the diagonals of a rhombus are perpendicular. Join O to F. Take O as origin. Let be the position vectors of A, B, C respectively. The length of longer diagonal of the parallelogram on 5 a + 2 b and a − 3 b.

Let two adjacent sides of the parallelogram be the vectors A and B (as shown in the figure). Then we have the two diagonals are A + B and A − B. In vector notation, AC = AB + BC = a + b BD = BC + CD = b - a AC.

(b-a) = (b+a). (b-a) = b^2 - a^2 = 0 (as a =b).

Using dot or scalar product, prove that the diagonals of a rhombus are perpendicular to each other. Let u and v be nonzero vectors, and let denote the angle between them.

Show that the diagonals of a rhombus are perpendicular. The sides of a Rhombus are equal. From this is easy to say that the diagonals are perpendicular.

Prove by vector method that the Internal bisectors of the angles of a triangle are concurrent.

Use vectors to prove that the diagonals of a rhombus are perpendicular.


Bill has a deck of 12 playing cards with 2 suits of 6 cards each, having different values from 1 to 6. He shuffles the deck, turns over 4 cards, and looks for pairs of cards with the same value. The goal is to find the chance that Bill finds at least one pair of cards with the same value.

The original solution attempts to calculate this probability but contains errors. The correct approach involves using the complement rule: P(at least 1) = 1 - P(None). To calculate P(None), we need to find the number of ways to select 4 distinct cards with no pairs.

The correct calculation involves finding the total number of ways to choose 4 cards from 12, which is 12C4. Then, we calculate the number of ways to choose 4 distinct cards with no pairs, considering the different suits and values.

The provided multiple-choice answers are:
8/33
62/165
17/33
103/165
25/33

The original solution incorrectly calculates the probability as 18/33. However, this approach is flawed, and the correct solution will yield a different result.

To find the correct answer, we need to calculate P(None) and subtract it from 1. This involves finding the number of ways to choose 4 distinct cards with no pairs and dividing it by the total number of ways to choose 4 cards from 12.

Let's re-evaluate the problem and calculate the correct probability. The total number of ways to choose 4 cards from 12 is 12C4 = 495. To find the number of ways to choose 4 distinct cards with no pairs, we consider the different suits and values.

There are 6 values, and we need to choose 4 distinct values. This can be done in 6C4 = 15 ways. For each value, there are 2 suits, so we have 2^4 = 16 ways to choose the suits. Therefore, the number of ways to choose 4 distinct cards with no pairs is 15 * 16 = 240.

Now, we can calculate P(None) = 240/495. Finally, we apply the complement rule: P(at least 1) = 1 - P(None) = 1 - 240/495 = 255/495 = 17/33.

The correct answer is 17/33.


This course covers mathematical topics in trigonometry, focusing on the study of triangle angles and lengths, with applications extending beyond triangles. The course aims to prepare students for a first-semester single-variable calculus course. Upon completion, students will be able to:

1. Evaluate trigonometric functions using the unit circle and right triangle approaches.
2. Solve trigonometric equations.
3. Verify trigonometric identities.
4. Prove and apply basic trigonometric identities.
5. Manipulate trigonometric expressions using standard identities.
6. Solve right triangles.
7. Apply the Law of Sines and the Law of Cosines to solve problems.