The average calculator calculates the average of a set of numbers between two and eight, providing the mean of up to eight numbers. As more values are entered, the calculator displays the change in the mean. To use the calculator effectively, it's essential to understand how to calculate the average. The mean and average are interchangeable terms.

The average of a set of numbers is calculated by summing the numbers and dividing by the total number of values. For example, to find the average of 24, 55, 17, 87, and 100, add the numbers: 24 + 55 + 17 + 87 + 100 = 283, and then divide by 5, resulting in 56.6. While simple calculations can be done by hand, more complex numbers with multiple decimal places are more conveniently solved using the calculator.

The mean rating calculator performs similar math, calculating an average rating based on the number of votes with values from 1 to 5. In contrast, the weighted average calculator allows assigning weights to each number, making it more or less important. A common application of the weighted mean is calculating a grade point average (GPA). To do this by hand:

1. Multiply the letter grade value by the number of credits in the class.
2. Repeat this process for all classes and sum the results.
3. Divide the sum by the total number of credits.

For instance, suppose the grades are A for a 3-credit class, two B's for 4-credit classes, and a C for a 2-credit class. Using the standard values of 4 for an A, 3 for a B, and 2 for a C, the GPA calculation is: GPA = [4(3) + 3(4) + 3(4) + 2(2)] / (3 + 4 + 4 + 2) = 40/13 = 3.08.

The Average Calculator can be embedded on a website to enrich content and make it easier for visitors to understand the message. It is free and can help keep people engaged.


# Teaching Equivalent Fractions with an Interactive Notebook

This interactive notebook is a valuable tool for teaching equivalent fractions to children using visual models or area models. It combines direct instruction with engaging activities, allowing students to learn while having fun.

## Before the Lesson

To effectively teach equivalent fractions, it's essential to understand the concept yourself. Resources like Learn Zillion and Math Geek Mama's series on developing fraction sense using visual models can be helpful. Once you're comfortable with the concept, gather the necessary materials, including a math journal, glue sticks, and colored pencils.

## How to Use Equivalent Fractions Interactive Notebook

#### Step One

Distribute the first sheet with a square showing 1/2. Have students cut out the four squares and glue them into their notebook in order, with 1 on top and 4 on the bottom.

#### Step Two

Have students identify the fraction and write it on the fraction line. Then, ask them to fill in the blanks describing their fraction, reviewing the concept of fractions as needed.

#### Step Three

Move on to the second box, reminding students that all fractions must have equal parts. Allow them to divide the box in any way they choose, as long as each part is equal. Discuss the different ways students divided the box, asking questions about each method. Have students write down the equivalent fraction they created, fill in the details about their fraction, and identify the two equivalent fractions.

Repeat this process for boxes three and four, ensuring the denominator is different from the previous ones. By the fourth box, students should be able to write three fractions equivalent to 1/2.

#### Step 4

Work through the remaining pages in the same way, deepening students' understanding of fractions and equivalent fractions.

This interactive notebook is designed to help students understand the concept of equivalent fractions in a engaging and interactive way. For more activities on teaching equivalent fractions, check out 15 other ways to teach and practice equivalent fractions.


## Homework Statement

A small shiny object is placed in a cuboid aquarium filled with water. A plano-convex lens with focal length f is positioned on the wall of the aquarium from the outside, with its planar side against the wall. The object is located on the optical axis of the lens. The refraction index of water is 1.33, and the index of the lens is 1.50. The wall and lens are very thin, and only rays near the optical axis are considered.

a) Determine the position of the object's image as a function of the object's position. Classify the image as virtual or real, upright or inverted, and enlarged or shrunk.

b) Calculate the image distance and enlargement when the object distance is 2.5*f.

c) Repeat part b, but with the lens positioned on the wall from the inside.

## Homework Equations

n₁*sin(a) = n₂*sin(b)

## The Attempt at a Solution

First, imagine the wall does not exist. Consider an object at point (-x₀, y₀). The ray parallel to the axis will refract at the air-lens interface and cross the optical axis at (f, 0). This ray can be described by the function f(x) = y₀ - y₀/f*x. Additionally, consider the ray that crosses the lens at (0, 0). The challenge is to determine the function g(x) that describes this ray after refraction. The intersection point of f(x) and g(x) should be the image.


## Gauss Markov Theorem

The Gauss Markov theorem states that the ordinary least squares (OLS) estimate for regression coefficients provides the best linear unbiased estimate (BLUE) possible, given that certain assumptions are met.

## Gauss Markov Assumptions

There are five key assumptions, also known as conditions, that must be satisfied for the Gauss Markov theorem to hold:

1. **Linearity**: The parameters being estimated using OLS must be linear.
2. **Random Sampling**: The data must be randomly sampled from the population.
3. **Non-Collinearity**: The regressors must not be perfectly correlated with each other.
4. **Exogeneity**: The regressors must not be correlated with the error term.
5. **Homoscedasticity**: The variance of the error term must be constant, regardless of the values of the regressors.

## Purpose of the Assumptions

The Gauss Markov assumptions ensure the validity of OLS for estimating regression coefficients. Verifying whether the data meets these assumptions is crucial in estimating regression coefficients. If the assumptions are violated, it may be possible to modify the experiment setup to better align with the ideal conditions.

In practice, the Gauss Markov assumptions are rarely met perfectly, but they serve as a useful benchmark for identifying potential issues that may affect the accuracy of the estimated regression coefficients.

## The Gauss-Markov Assumptions in Algebra

The Gauss-Markov assumptions can be expressed algebraically as follows:

y_{i} = x_{i}‘ β + ε_{i}

The OLS estimate is the best linear unbiased estimate (BLUE) possible if:

- E{ε_{i}} = 0, for i = 1, … , N (the expected value of the error term is zero)
- {ε_{1}…ε_{n}} and {x_{1},…,x_{N}} are independent (the error terms are independent of the regressors)
- cov{ε_{i}, ε_{j}} = 0, for i, j = 1,…, N and i ≠ j (the error terms are uncorrelated)
- V{ε_{i}} = σ^{2}, for i = 1, …N (the variance of the error term is constant)

Where ε represents the error term, and σ^{2} represents the constant variance.

## References

Anderson, Patricia. The Gauss-Markov Theorem: Study Guide. Retrieved from http://www.dartmouth.edu/~econ20pa/StudyGuide1.doc on May 20, 2018.

Lee, Q. OLS, BLUE and the Gauss Markov Theorem. Economics Society: University of Waterloo. Retrieved from http://uweconsoc.com/ols-blue-and-the-gauss-markov-theorem/ on May 20, 2018.

Troeger, Vera. Gauss-Markov Assumptions, Full Ideal Conditions of OLS. Retrieved from http://uweconsoc.com/ols-blue-and-the-gauss-markov-theorem/ on May 20, 2018.


Mathematician Thomas Snitch of the University of Maryland is utilizing mathematics to aid in the conservation of rhinoceroses in African wildlife preserves. To learn more about his work, listen to the podcast at http://www.ams.org/samplings/mathmoments/mm122-poaching-podcast.

After reading the article and listening to the two-part podcast, answer the following questions. Mathematicians compared rhinos to American soldiers in the Middle East to develop strategies for protection. By analyzing statistics about waterholes and roads, mathematicians reduced the area that rangers and drones needed to cover. For instance, they used specific numbers related to the location and density of waterholes and roads to minimize the search area. Land elevation is also a crucial factor in examining the habitat of rhinos, as it affects their movement and behavior. This mathematical modeling is being adopted by other conservation efforts, which are applying similar techniques to protect different species.


## Exploring Rotations 1

Lesson 7 of 16

### Objective:
Students will be able to show how to find the rotation image of a figure around a point.

### Big Idea:
Using trace paper and geometry tools, students rotate a figure by turning its points a certain number of degrees.

#### Launch (20 minutes)
Students receive a copy of the Rotation Task 1 resource and a sheet of tracing paper. They are provided with pencils, a ruler, a compass, and a protractor. The instructions are as follows:

1. Place the tracing paper over Rotation 1 and trace the x and y axes using a ruler. Mark the center of rotation with a dot and then remove the tracing paper.
2. Rotate the whole paper the indicated number of degrees (90 degrees counter-clockwise) and keep it still.
3. Place the tracing paper back on the resource paper and carefully line up the center of rotations.
4. Trace the shape of the polygon. The image of the rotation is now on the tracing paper. Compare the pre-image and the image.

Students work independently, but can ask their elbow partners for help. Faster students assist those who are struggling.

#### New Info / Note taking (15 minutes)
Students should have found that a point and its rotation image are the same distance from the center of rotation. Key notes include:

- Any point and its rotation image are equidistant from their center of rotation.
- A figure and its rotation image are congruent.
- Rotating an object a positive amount of degrees is a counter-clockwise motion.
- Rotating an object a negative amount of degrees is a clockwise motion.
- Rotation symmetry means that the figure coincides with its rotation image.
- Rotations preserve orientation.

#### Activity (25 minutes)
The Rotation 1 Activity involves more detail than the Launch task. Students work on the activity, and the teacher assesses their drawings and understanding of the directions. Strategies to help students include:

- Drawing a circle with center O through a point A and discussing the images of A through rotations.
- Using different color pencils for rotating each vertex point.

After completing the activity, students answer questions, including:

- List all the angles that measure 70 degrees below the drawing.
- What conclusions can you make about rotations from this activity?

#### Closure (10 minutes)
The lesson ends with a discussion of closure questions projected on the SmartBoard. Students contribute responses, and the teacher writes their answers on the board. Certain students are called on to respond to specific questions they struggled with during the Activity.

#### Resources
- Rotation Task 1
- Tracing paper
- Pencils
- Ruler
- Compass
- Protractor
- Rotations Closure Questions

#### Similar Lessons
- Hands-on Exploring Translations in the Plane
- Day Four & Five
- Introduction to congruence and similarity through transformations

#### Unit and Lesson Overview
- UNIT 5: Congruence and Similarity
- LESSON 7: Exploring Rotations 1
- This lesson is part of a larger unit on congruence and similarity, which includes lessons on dilations, translations, reflections, and rotations.


# Definition: Secant Function

## Definition

The secant function is defined in the context of trigonometry.

### Definition from Triangle

In a right triangle, the secant of an angle $\theta$ is defined as the ratio of the hypotenuse to the adjacent side, i.e., $\dfrac{\text{Hypotenuse}} {\text{Adjacent}}$.

### Definition from Circle

Considering a unit circle with center at the origin, let $P$ be a point on the circle in the first quadrant, forming an angle $\theta$ with the x-axis. A tangent line touches the circle at $A = \tuple {1, 0}$ and intersects the extended line $OP$ at $B$. The secant of $\theta$ is defined as the length of $OB$. In the first quadrant, the secant is positive.

### Real Function

The real function $\sec x$ is defined as $\dfrac 1 {\cos x}$, where $\cos x$ is the cosine of $x$. This definition is valid for all real numbers $x$ except where $\cos x = 0$.

### Complex Function

The complex function $\sec z$ is defined as $\dfrac 1 {\cos z}$, where $\cos z$ is the cosine of $z$. This definition is valid for all complex numbers $z$ except where $\cos z = 0$.

## Related Concepts

Related trigonometric functions include:
- Sine
- Cosine
- Tangent Function
- Cotangent
- Cosecant

Results about the secant function can be found in related resources.

## Linguistic Note

The term "secant" originates from the Latin "secantus," meaning "that which is cutting," due to its geometric interpretation as the length of the line from the origin that cuts the tangent line. The pronunciation is "see-kant" with emphasis on the first syllable and a long "e".

## Sources

- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.)


cscal: CSCAL scales a vector by a constant

Command to display cscal manual in Linux: $ man l cscal

NAME
CSCAL - scales a vector by a constant

SYNOPSIS
SUBROUTINE CSCAL(N, CA, CX, INCX)

Parameters:
- CA: COMPLEX, the constant to scale the vector by
- CX: COMPLEX array, the vector to be scaled
- INCX: INTEGER, the increment between elements of CX
- N: INTEGER, the number of elements in the vector CX

PURPOSE
The CSCAL subroutine scales a vector CX by a constant CA.

FURTHER DETAILS
The subroutine was written by Jack Dongarra, Linpack, on 3/11/78, and modified on 3/93 to return if INCX is less than or equal to 0. It was further modified on 12/3/93 to change array declarations from (1) to (*).

Related Pages:
- cscal (3)
- cspcon (l): estimates the reciprocal of the condition number of a complex symmetric packed matrix
- cspmv (l): performs the matrix-vector operation
- cspr (l): performs the symmetric rank 1 operation
- csprfs (l): improves the computed solution to a system of linear equations
- cspsv (l): computes the solution to a complex system of linear equations
- cspsvx (l): computes the solution to a complex system of linear equations using diagonal pivoting factorization
- csptrf (l): computes the factorization of a complex symmetric matrix using the Bunch-Kaufman diagonal pivoting method


To write a number as a product of its prime factors, we need to break it down into its prime components. For example, to write 45 as a product of prime factors, we start by finding its factors. The factors of 45 are 1, 3, 5, 9, 15, and 45. We can then break down these factors into their prime components: 3 and 5 are prime, and 9 can be broken down into 3 * 3. Therefore, the prime factorization of 45 is 3 * 3 * 5.

Prime factorization is the process of expressing a given number as a product of prime factors. This can be accomplished by dividing the original number and its factors by prime numbers until we reach the prime factors. Using factor trees is a handy way to factor numbers to their prime factors. For instance, to write 60 as a product of prime factors, we start by dividing 60 by the smallest prime number, which is 2. We get 30, and then we divide 30 by 2 again to get 15. Since 15 can be divided by 3 and 5, the prime factorization of 60 is 2 * 2 * 3 * 5.

The prime factorization of a number can be used to determine many properties of the number. For example, the multiplicity of a prime factor p in the prime factorization of a number n can be used to compute various properties of n. All composite numbers can be expressed as unique products of prime numbers. This is known as the fundamental theorem of arithmetic.

To write a number as a product of its prime factors, we need to find the prime factors of the number and their multiplicities. For example, to write 75 as a product of prime factors, we start by finding its prime factors, which are 3 and 5. The prime factorization of 75 is 3 * 5 * 5.

In general, to write a number as a product of its prime factors, we need to follow these steps:

1. Find the prime factors of the number.
2. Determine the multiplicities of the prime factors.
3. Write the prime factorization of the number using the prime factors and their multiplicities.

For example, the prime factorization of 36 is 2 * 2 * 3 * 3, and the prime factorization of 48 is 2 * 2 * 2 * 2 * 3.

Prime factorization has many applications in mathematics and computer science. It is used in cryptography, coding theory, and number theory, among other fields. It is also an essential concept in GCSE maths and is used to solve many problems in mathematics.

In conclusion, writing a number as a product of its prime factors is an important concept in mathematics. It can be used to determine many properties of a number and has many applications in mathematics and computer science. By following the steps outlined above, we can write any composite number as a product of its prime factors.


The Fundamental Theorem of Calculus is crucial as it establishes a link between differentiation and integration, and provides a method to easily evaluate definite integrals without relying on Riemann sums. The theorem consists of two parts: FTC1 and FTC2.

**FTC1** states that for a continuous function f(x) on the interval [a,b], the derivative of the integral of f(t) from a to x is equal to f(x). Mathematically, this is expressed as:

\[
\frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)
\]

**FTC2** states that for a continuous function f(x) on [a,b] with an antiderivative F(x), the definite integral of f(x) from a to b is equal to F(b) - F(a). This can be written as:

\[
\int_{a}^{b} f(x) dx = F(b) - F(a)
\]

Given that \( F'(x) = f(x) \), the equation can be rewritten as:

\[
\int_{a}^{b} \frac{d}{dx}F(x) dx = F(b) - F(a)
\]

This highlights the inverse relationship between differentiation and integration. FTC2 allows for the easy evaluation of definite integrals by using antiderivatives, thus avoiding the need for laborious Riemann sum calculations. This trade-off between calculation time and memory (for antiderivatives) is a key aspect of calculus.

Higher-dimensional versions of the Fundamental Theorem of Calculus exist, including Green's theorem and Stoke's theorem for 2-dimensional cases, the Divergence Theorem or Gauss' theorem for 3-dimensional cases, and the general n-dimensional Stoke's theorem.


# Grid Construction Examples

Grids can be constructed using various methods, including reading grid files in DGF or gmsh format, or using the meshio package.

## Reading Grid Files

Grid files can be read using the `DGF` or `gmsh` format. The DGF format allows for easy description of boundary IDs and parameters, while the gmsh format is useful for reading previously stored gmsh files.

### Example using DGF

To read a DGF file, use the `reader.dgf` function and specify the file name. The `dimgrid` parameter must be specified, as it cannot be easily extracted from the DGF file.
```python
domain2d = (reader.dgf, "triangle.dgf")
aluView = leafGridView(domain2d, dimgrid=2)
aluView.plot(figsize=(5,5))
```

### Example using gmsh

To read a gmsh file, use the `reader.gmsh` function and specify the file name.
```python
waveDomain = (reader.gmsh, "wave_tank.msh")
waveGrid = leafGridView(waveDomain, dimgrid=2)
waveGrid.plot(figure=pyplot.figure(figsize=(10,10)))
```

## 3D Example using PyGmsh

In this example, we use PyGmsh to construct a tetrahedral mesh and solve a simple Laplace problem.
```python
try:
    import pygmsh
    with pygmsh.geo.Geometry() as geom:
        poly = geom.add_polygon([
            [ 0.0, 0.5, 0.0], [-0.1, 0.1, 0.0], [-0.5, 0.0, 0.0],
            [-0.1, -0.1, 0.0], [ 0.0, -0.5, 0.0], [ 0.1, -0.1, 0.0],
            [ 0.5, 0.0, 0.0], [ 0.1, 0.1, 0.0] ], mesh_size=0.05)
        geom.twist(
            poly,
            translation_axis=[0, 0, 1], rotation_axis=[0, 0, 1], point_on_axis=[0, 0, 0],
            angle=numpy.pi / 3,
        )
        mesh = geom.generate_mesh(verbose=False)
        points, cells = mesh.points, mesh.cells_dict
        domain3d = {"vertices":points.astype("float"), "simplices":cells["tetra"]}
except ImportError:
    print("pygmsh module not found using a simple Cartesian domain - ignored")
    from dune.grid import cartesianDomain
    domain3d = cartesianDomain([-0.25,-0.25,0],[0.25,0.25,1],[30,30,60])
gridView3d = leafGridView(domain3d)
space3d = solutionSpace(gridView3d, order=1)
u = TrialFunction(space3d)
v = TestFunction(space3d)
x = SpatialCoordinate(space3d)
scheme3d = solutionScheme((inner(grad(u),grad(v))+inner(u,v))*dx ==
                         conditional(dot(x,x)<.01,100,0)*v*dx,
                         solver='cg')
uh3d = space3d.interpolate([0],name="solution")
scheme3d.solve(target=uh3d)
gridView3d.writeVTK('3dexample', pointdata=[uh3d],
                    outputType=OutputType.appendedraw)

## Converting gmsh to DGF

To convert a gmsh file to DGF, use the `gmsh2DGF` function from the `gmsh2dgf` module.
```python
from gmsh2dgf import gmsh2DGF
bndDomain = {1:"-0.6 -0.6 -0.1 0.6 0.6 0.0",
             2:"-0.6 -0.6 1.0 0.6 0.6 1.1",
             3: "default"}
dgf = gmsh2DGF(points, cells, bndDomain=bndDomain)
with open ("3dmesh.dgf", "w") as file:
    file.write(dgf)

## Using meshio to read a gmsh file

To read a gmsh file using the `meshio` package, use the `meshio.read` function.
```python
try:
    import numpy as np
    import meshio
    mesh = meshio.read("wave_tank.msh")
    points = np.delete( mesh.points, 2, 1)
    cells = mesh.cells_dict
    waveDomain = {"vertices":points.astype("float"), "simplices":cells["triangle"]}
    waveGrid = leafGridView(waveDomain)
    waveGrid.plot(figure=pyplot.figure(figsize=(10,10)),xlim=[0.3,1.7],ylim=[-0.2,1.2])
except ImportError:
    print("This example requires the meshio package which can be installed using pip.")

## Reading a gmsh file with quadrilateral elements

To read a gmsh file with quadrilateral elements, use the `meshio.read` function and exchange the last two vertices of the cells.
```python
if meshio:
    from dune.alugrid import aluCubeGrid
    mesh = meshio.read("quads.msh")
    points2d = np.delete(mesh.points,2,1)
    cells = mesh.cells_dict['quad']
    cells[:, [2, 3]] = cells[:, [3, 2]]
    domain = {"vertices":points2d, "cubes":cells}
    quadGrid = aluCubeGrid(domain)
    quadGrid.plot()

## 1D Example

To create a grid in 1D, use the `onedGrid` function from the `dune.grid` module.
```python
from dune.grid import onedGrid
v = [numpy.log(i) for i in range(1,100)]
e = [(i,i+1) for i in range(1,len(v))]
g = onedGrid(constructor={"vertices":v, "simplices":e})
s = solutionSpace(g)
x = SpatialCoordinate(s)
u = s.interpolate(sin(x[0]), name="u")
u.plot()
```


This application is capable of plotting graphs of mathematical functions, making it a useful tool for mathematics students, particularly those at the college and university level. When giving inputs, it is essential to be careful about the limits.

The software offers various activities, including:

* Drawing pictures with graphs
* Modeling real-life situations by translating them into algebraic equations and using graphs
* Finding approximate solutions of simultaneous linear equations using graphs, such as:
	+ y = 5x - 7
	+ 3x + 2y = 1
* Plotting families of exponential and reciprocal graphs, such as:
	+ y = 2^x, y = 3^x, y = 4^x
	+ y = 1÷x, y = 2÷x, y = 3÷x
* Reducing linear equations to standard form, calculating gradients and intercepts, and plotting them
* Solving problems involving direct and inverse proportion using graphical representations
* Identifying and interpreting roots, intercepts, and turning points of quadratic functions graphically
* Finding the equations of translations and reflections of graphs of given functions
* Using graphs to solve optimization problems
* Investigating the absolute value function and plotting families of graphs, such as:
	+ y = |x^2 - 5|
* Graphing inequalities to solve linear programming problems
* Plotting graphs of functions and their inverses by interchanging the roles of x and y

The Graph Plotter application is designed for building and analyzing function charts. It can build various types of charts, including parametric function charts and equation solution charts. The user can define maximum and minimum values, calculate asymptotes and limits, use integrals and differentials, and apply trigonometric functions. The appearance of the charts is adjustable, and created projects can be saved for later use.

The application supports multiple languages, including English, Arabic, Basque, Chinese, and many others. It is compatible with Windows and was last updated on 2018-04-03.

The Graph Plotter offers a wide range of features, including:

* Plotting standard functions, parametric functions, and polar functions
* Evaluating functions at entered points or tracing functions with the mouse
* Adding shadings to functions and series of points to the coordinate system
* Adding trendlines to point series
* Saving the coordinate system and everything on it as an image file
* Copying images into other programs
* Placing text labels anywhere in the coordinate system

The application is free, open-source, and provides a convenient and effective solution for drawing mathematical graphs in a coordinate system. It offers customization options, manipulation possibilities, and a user-friendly interface. The Graph Plotter can save graphs to various image file formats, including BMP, PNG, JPEG, WMF, EMF, SVG, and PDF, and export the complete graph to other programs like MS Word.

The application is aimed at being simple and fast, allowing for real-time plotting from multiple asynchronous sources and exporting to PNG or SVG. It can find asymptotes, break points, extremums, and calculate limits, define integrals, and derivatives of functions.


Given a circle and its diameter, the task is to partition its area into seven equal regions without using straight lines.

A possible solution involves dividing the circle into 7 concentric circles. The radius of each concentric circle can be determined using the formula R*sqrt(n/7), where n is the nth concentric circle from the beginning and R is the radius of the big circle.

To find the radius of each concentric circle, start with R*sqrt(1/7) and R*sqrt(2/7) for the first two circles. Then, move upwards to find the next radius, where the square of the inner radius (r2^2) is equal to 2*R^2/7, and so forth.

The concentric circles have radii of R*sqrt(1/7), R*sqrt(2/7), and so on, allowing the circle's area to be divided into seven equal regions.


# Square Map is Endomorphism if and only if Abelian

This article discusses a fundamental concept in group theory, stating that a square map of a group is an endomorphism if and only if the group is abelian.

## Statement

Let G be a group and φ be the square map of G defined as φ(g) = g^2 for all g in G. Then, φ is an endomorphism of G (i.e., φ(ab) = φ(a)φ(b) for all a, b in G) if and only if G is abelian.

## Related Facts

### Applications

- Exponent two implies abelian: If the exponent of a group is 2, then the group is abelian. This is not true for any other prime number.
- The power map for a fixed integer n is termed a universal power map. If it is also an endomorphism, it is termed a universal power endomorphism, and the group is termed an n-abelian group.

### Other Power Maps

- n-abelian iff (1-n)-abelian
- The set of n for which G is n-abelian is termed the exponent semigroup of G.
- Abelian implies n-abelian for all n
- n-abelian implies every nth power and (n-1)th power commute
- n-abelian implies n(n-1)-central

### Characterization of n-Abelian Groups

| Value of n | Characterization of n-Abelian Groups | Proof | Other Related Facts |
| --- | --- | --- | --- |
| 0 | all groups | obvious |  |
| 1 | all groups | obvious |  |
| 2 | abelian groups only | 2-abelian iff abelian | endomorphism sends more than three-fourths of elements to squares implies abelian |
| -1 | abelian groups only | -1-abelian iff abelian |  |
| 3 | 3-abelian group means: 2-Engel group and derived subgroup has exponent dividing three | Levi's characterization of 3-abelian groups | cube map is surjective endomorphism implies abelian |

## Proof

### From Square Map Being Endomorphism to Abelian

**Given**: A group G such that the map φ(g) = g^2 is an endomorphism.

**To Prove**: G is abelian.

**Proof**:

1. φ(ab) = (ab)^2 = abab
2. φ(ab) = φ(a)φ(b) = a^2b^2
3. abab = a^2b^2 (using associativity)
4. ab = ba (using cancellation)

### From Abelian to Square Map Being Endomorphism

This follows directly from the fact that abelian implies universal power map is endomorphism.

## Facts Used

- Associative implies generalized associative
- Invertible implies cancellative in monoid
- Abelian implies universal power map is endomorphism

## Related Facts for Lie Rings

- Multiplication by n map is a derivation iff derived subring has exponent dividing n
- Multiplication by n map is an endomorphism iff derived subring has exponent dividing n(n-1)


**Chapter 20-4: Velocity and Acceleration**

Derivatives can be related to physics applications, specifically to the position function, velocity function, and acceleration function, all of which are functions of time. It's crucial to understand the difference between average and instantaneous velocity.

The average velocity is the change between two points, giving the slope of the line connecting these two points. In contrast, the instantaneous velocity gives the slope at a single point in time, which is the slope of the tangent line.

**Average Formulas**

The average velocity can be calculated using the formula: average velocity = (change in position) / (change in time).

**Instantaneous Formulas**

The instantaneous velocity, v(t), is the first derivative of the position function, s(t). The acceleration function, a(t), is the first derivative of the velocity function, which is also the second derivative of the position function, s"(t).

**Key Points**

- v(t) = 0 means the object is not moving.
- v(t) > 0 means the object is moving forward.
- v(t) < 0 means the object is moving backward.
- a(t) = 0 means there is no change in velocity.
- a(t) > 0 means the object is accelerating (speeding up).
- a(t) < 0 means the object is decelerating (slowing down).

**Sample Problems**

1. Given s(t) = t^2 - 20, find the average velocity from t = 3 to t = 5.
   Solution: Use the average velocity formula to find the average velocity is 8.

2. For s(t) = 2t^3 + 5t - 7, find the velocity and acceleration functions.
   Solution: v(t) = s'(t) = 6t^2 + 5, a(t) = v'(t) = 12t.
   At t = 2, v(2) = 6(4) + 5 = 29, a(2) = 12(2) = 24.

3. A ball is thrown vertically upward with an initial velocity of 128 ft/sec. The ball's height after t seconds is s(t) = 128t - 16t^2.
   a. The velocity function is v(t) = 128 - 32t.
   b. The velocity at t = 2, 4, 6 is v(2) = 64, v(4) = 32, v(6) = -64.
   c. The time when the velocity is 48 ft/sec, 16 ft/sec, -48 ft/sec is t = 2.5, t = 3.5, t = 5.5, respectively.
   d. The velocity is zero at t = 4.
   e. The height of the ball at t = 4 is s(4) = 256.
   f. Yes, this is the maximum height because it is the zero of the first derivative.
   g. The ball hits the ground at t = 8.
   h. The velocity when the ball hits the ground is v(8) = -128 ft/sec.
   i. The acceleration function is a(t) = -32 ft/sec^2.
   j. The acceleration at any time t is -32 ft/sec^2, indicating the effect of gravity.


## Texas Go Math Grade 1 Lesson 17.1 Answer Key Length

### Explore

Use objects to show the problem. Draw to show your work. Rosa P has something longer than a drinking straw and another object shorter than a key.

- Longer than a straw: water bottle
- Shorter than a key: eraser

### Mathematical Processes

Compare the straw and the key. Which is longer? Which is shorter?

- Straw is longer than the key.
- Key is shorter than the straw.

### Model and Draw

Use string to measure the length of the pencil box. The string is as long as the pencil box. Objects longer than the string will not fit.

- The length of the pencil box is the same as the string, so it can fit in the box.
- Both string and box are of the same length.

### Share and Show

Use the string as a tool to measure the objects. Circle the objects that will fit in the pencil box.

1. Pencil: cannot fit (length of the pencil is more than the string length)
2. Stick: can fit (length of the stick is less than the string length)
3. Crayon: can fit (length of the crayon is less than the string length)
4. Plastic straw: can fit (length of the plastic straw is the same as the string length)

### Problem Solving

Use the string as a tool to measure the objects. Circle the objects that will fit in the pencil box.

5. Plastic rod: can fit (length of the plastic rod is less than the string length)
6. Glue stick: can fit (length of the glue stick is less than the string length)

7. Barry is sending Jake his skateboard in a box. He used a ribbon to measure the skateboard. The length of the box should be more than the skateboard length because it needs to fit in the box.

8. A box is as long as this ribbon. An object that will fit in the box is a crayon because its length is less than the ribbon/box.

### Daily Assessment Task

Choose the correct answer.

9. A box is the same length as the ribbon. The ribbon is longer than the pencil. The ribbon is shorter than the straw and the spoon. Which object will fit in the box?
(A) straw
(B) pencil
(C) spoon
Answer: (B) pencil

10. A box is the same length as this string. Objects that will fit in the box are the crayon and pen because their lengths are less than the string length.

### TAKE HOME ACTIVITY

Have your child explain how they use a measuring tool to measure an object. Answer: Yes, my kids said they use the ruler to measure objects.

### Texas Go Math Grade 1 Lesson 17.1 Homework and Practice Answer Key

Use string as a tool to measure the objects. Circle the objects that are longer than the string.

1. Pen: longer than the string
2. Chain: less than the string length
3. Strip: same as the string length

### Problem Solving

4. Ava wants to make a bracelet as long as this piece of yarn. Draw a bracelet as long as the yarn. The length of the bracelet is the same as the length of the piece of yarn.

### Lesson Check

Choose the correct answer.

5. A pencil case is the same length as this string. The pencil that will fit in the box is the maroon pencil because its length is less than the length of the string.

6. Elena wants to put objects inside this box. The best tool for measuring the length of the objects should be longer in length than the objects.


To analyze this problem, you only need to know how to add and multiply two by two matrices. The problem provides models for complex numbers and 4-dimensional numbers called quaternions.

Let $C^*$ be the set of $2 \times2$ matrices of the form $$ \left( \begin{array}{cc} x& -y\\ y& x\end{array} \right) $$ where $x$ and $y$ are real numbers.

(a) Adding the matrices $$\pmatrix {x & -y \cr y & x} {\rm \ and }\ \pmatrix {u & -v \cr v & u}$$ results in $$\pmatrix {x+u & -y-v \cr y+v & x+u}.$$ Multiplying these matrices gives $$\pmatrix {xu-yv & -xv-yu \cr xu+yv & -xv+yu}.$$

(b) The identity for addition is $$\pmatrix {0 & 0 \cr 0 & 0},$$ and the inverse of $$\pmatrix {x & -y \cr y & x}$$ is $$\pmatrix {x/(x^2+y^2) & y/(x^2+y^2) \cr -y/(x^2+y^2) & x/(x^2+y^2)}.$$ The identity for multiplication is $$\pmatrix {1 & 0 \cr 0 & 1},$$ and the inverse of $$\pmatrix {x & -y \cr y & x}$$ is $$\pmatrix {x/(x^2+y^2) & y/(x^2+y^2) \cr -y/(x^2+y^2) & x/(x^2+y^2)}.$$

(c) The subset $R^*$ for which $y=0$ consists of matrices of the form $$\pmatrix {x & 0 \cr 0 & x}.$$ The arithmetic of $R^*$ is the same as the arithmetic of real numbers.

(d) The arithmetic of $C^*$ is similar to the arithmetic of complex numbers.

(e) The matrix equivalent to $i = \sqrt {-1}$ in the set of complex numbers is $$\pmatrix {0 & -1 \cr 1 & 0}.$$

Complex numbers are two-dimensional numbers $x + iy$ where $x$ and $y$ are real numbers and $i = \sqrt {-1}$. Quaternions are four-dimensional numbers of the form $a+{\bf i}x+{\bf j}y+{\bf k}z$ where $a, x, y, z$ are real numbers and ${\bf i, j, k}$ are all different square roots of $-1$.

The model for the system of quaternions is the set of linear combinations of 2 by 2 matrices: $$aI +{\bf i}x + {\bf j}y + {\bf k}z\ = a\pmatrix {1 & 0\cr 0 & 1}+ x\pmatrix {0 & -1\cr 1 & 0}+y\pmatrix {0 & 1\cr -1 & 0}+ z\pmatrix {1 & 0\cr 0 & -1}.$$

(a) The matrix products are:
- ${\bf i^2} = \pmatrix {0 & -1\cr 1 & 0}\pmatrix {0 & -1\cr 1 & 0} = \pmatrix {-1 & 0\cr 0 & -1}$
- ${\bf j^2} = \pmatrix {0 & 1\cr -1 & 0}\pmatrix {0 & 1\cr -1 & 0} = \pmatrix {-1 & 0\cr 0 & -1}$
- ${\bf k^2} = \pmatrix {1 & 0\cr 0 & -1}\pmatrix {1 & 0\cr 0 & -1} = \pmatrix {1 & 0\cr 0 & 1}$

These matrices give models of three different square roots of -1.

(b) The matrix products are:
- ${\bf i j} = \pmatrix {0 & -1\cr 1 & 0}\pmatrix {0 & 1\cr -1 & 0} = \pmatrix {1 & 0\cr 0 & -1}$
- ${\bf j i} = \pmatrix {0 & 1\cr -1 & 0}\pmatrix {0 & -1\cr 1 & 0} = \pmatrix {-1 & 0\cr 0 & 1}$
- ${\bf j k} = \pmatrix {0 & 1\cr -1 & 0}\pmatrix {1 & 0\cr 0 & -1} = \pmatrix {0 & -1\cr -1 & 0}$
- ${\bf k j} = \pmatrix {1 & 0\cr 0 & -1}\pmatrix {0 & 1\cr -1 & 0} = \pmatrix {0 & 1\cr 1 & 0}$
- ${\bf k i} = \pmatrix {1 & 0\cr 0 & -1}\pmatrix {0 & -1\cr 1 & 0} = \pmatrix {0 & -1\cr -1 & 0}$
- ${\bf i k} = \pmatrix {0 & -1\cr 1 & 0}\pmatrix {1 & 0\cr 0 & -1} = \pmatrix {0 & 1\cr 1 & 0}$

The three matrices ${\bf i,\ j,\ k}$ model unit vectors along the three axes in 3-dimensional space ${\bf R^3}$ with matrix products isomorphic to vector products in ${\bf R^3}$.

Investigate the sequence: ${\bf i},\ $ ${\bf i j},\ $ ${\bf i j}$ ${\bf k},\ $ ${\bf i j}$ ${\bf k i},\ $ ${\bf i j}$ ${\bf k i}$ ${\bf j,\ ...}$


Egyptian fractions were a method used by ancient Egyptians to represent rational numbers as a sum of distinct unit fractions. Prior to 2050 BCE, Egyptian scribes used a binary notation system, where rational numbers were rounded off to six-term binary representations. This system was stated in 1/2n units, with a 1/64 unit thrown away.

The Horus-Eye method recorded rational numbers in a cursive pattern, where 1 was represented as 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + (1/64). However, a potential 7th term (1/64) was rounded off and thrown away.

After 2050 BCE, an exact numeration system was introduced, which discontinued the rounded-off Old Kingdom binary system. This new system reported rational numbers in 1/64 quotient and 1/320 remainder units whenever possible. The Middle Kingdom math system "healed" rounded-off binary series by several finite methods.

Two weights and measures finite systems can be reported by:

1. 1 hekat (a volume unit) used a unity (64/64) such that (32 + 16 + 8 + 4 + 2 + 1/64)hekat + 5 ro, and (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)hekat + 5 ro meant (64/64)/n = Q/64 + (5R/n)ro.
2. (100-hekat)/70 = (6400/64)/70 = 91/64 hekat + 30/4480 = (64 + 16 + 8 + 2 + 1)/64 hekat + 150/70 ro, which meant (6400/64)/n = Q/64 + (5R/n)ro was applied for almost any hekat division problem.

The hieratic word "ro" meant 1/320 of a hekat in a grain weights and measures system. Note that 5 ro meant 5/320 = 1/64. Generally, scribal shorthand recorded duplication aspects of mental calculations and fully recorded two-part hekat quotients and ro remainders.

The 1900 BCE Akhmim Wooden Tablet (AWT) scaled two volume units in the Egyptian fraction notation, which allowed Hana Vymazalova in 2002 CE to open a long-lost (64/64) hekat unity aspect of Egyptian fraction weights and measures. In 2006, scholars showed that the AWT scribe divided (64/64) by 3, 7, 10, 11, and 13 by writing exact 1/64 quotients and (5/5) scaled remainders to 1/320 unit fraction series.

A complete decoding of the AWT scribal methodology documented a solution to an Old Kingdom binary round-off problem. Ahmes used the AWT method over 40 times, scaling a 1/10 hin unit to 1/320 ro unit 29 times, as well as solving several classes of problems.

On a higher numeration level, the finite hieratic system scaled rational numbers by least common multiple (LCM) m to optimized, but not optimal, unit fraction series. Middle Kingdom scribes scaled rational numbers to unit fraction series by the LCM that created a numeration system in the AWT and other hieratic texts.

The Egyptian fraction numeration system ciphered counting numbers 1:1 onto hieratic sound symbols, replacing an Old Kingdom hieroglyphic many-to-one system numbers written in rounded-off binary numbers. A line was drawn over hieratic sound symbols to denote unit fractions.

One purpose of the Egyptian fraction system involved a finite weights and measures system applied to a commodity-based monetary system established by Pharaoh. One theoretical aspect of the monetary system contained Egyptian multiplication and division operations. The scribal arithmetic was undervalued by historians for over 120 years, from 1879 to 1999, by only reporting the additive aspects of the raw transliterated data.

Scribal unit fraction answers contained two sides, additive numerators, and LCM scaled reminders. Math historians, beginning with F. Hultsch in 1895, parsed aspects of the scribal aliquot part views of numerators and LCM denominators, though the unified aspects of the system remained vague for another 105 years.

In 2002, the EMLR was validated, containing aliquot parts of LCM denominators in non-optimized manners. The LCM (m/m) was written as a unity, becoming a multiplier that scribes used to scale rational numbers to optimized, but not optimal, unit fraction series in 2/n table tables and everyday problem and answers.

A second purpose of the Egyptian fraction notation exactly solved one and two variable first-degree algebra problems. Rational number answers were written into integer quotients and Egyptian fraction remainders. The Rhind Mathematical Papyrus (RMP) and the Berlin Papyrus cite several problems and solutions, each with hard-to-read intermediate steps reaching Egyptian fraction answers.

A third purpose of the Egyptian fraction notation created a commodity and metal-based monetary system. The system is outlined by Mahmoud Ezzamel using modern accounting methods. Yet, Ezzamel fairly parses the four Heqanakht Papers by discussing two absentee landlords' production and management considerations of profit written in ancient Egyptian fractions.

A fourth purpose of the Egyptian fraction notation generally converted rational numbers to optimized, but not optimal, unit fraction series. The RMP 2/n table was written by Ahmes in 1650 BCE. Translators assumed that Ahmes had intuitively used 'red auxiliary' multiples to write optimized, but not optimal, unit fraction series.

When the RMP was first published in 1879, historians began the task of breaking the 2/n table code, a project that was not completed until 2005. The earliest nearly successful code-breaking effort was published by F. Hultsch in 1895. With E.M. Bruins confirming Hultsch's aliquot part approach in 1944, today F. Hultsch and E.M. Bruins are honored by the Hultsch-Bruins method.

The AWT's binary quotient and Egyptian fraction partitions were proven by multiplying each quotient and remainder answer by the initial divisor. The exact proof calculation returned each quotient and scaled remainder answer to an initial (64/64) unity value. The AWT proof may be the first proof recorded in Western math history. Ahmes applied the proof method in RMP 35-38, 320 ro problems.

Length and area units (cubits, khet, setat, and mh) were partitioned into 1/8 setat quotients and mh (1 cubit by 100 cubit strips) remainders. RMP 53, 54, and 55 discuss Ahmes' geometry and associated arithmetic.

The Reisner Papyrus (RP), circa 1800 BCE, defines a labor efficiency division by 10 quotient and unscaled remainder rate. The RMP's first six problems used the RP division by 10 method. The RMP is one of several texts that confirm the scribal use of a second form of quotient and remainder arithmetic.

Modern historians, going beyond 1920's additive transliteration limitations, by considering meta Egyptian fractions, agree with the RP method, and other forms of meta (unified) Egyptian fraction mathematics.

Archimedes followed a one-fourth geometric series tradition established by Eudoxus that found the area of a parabola first by an infinite series: 4A/3 = A + A/4 + A/16 + A/64 + …, and by second finite Egyptian fraction proof: 4A/3 = A + A/4 + A/12.

Heiberg in 1906 reported Archimedes' two-level calculus method decoded from an 1100 AD Byzantine vellum document. Around 800 AD, a major change took place in the Mediterranean region. Modern base 10 numerals diffused from India, and Arab trade began to enter Europe.

By 999 AD, a Catholic Pope adopted Arab mathematics with its base 10 numerals and Egyptian fraction arithmetic. By the time of Leonardo de Pisa, Fibonacci, and the 1202 AD Liber Abaci, European weights and measures were also written in Arabic numerals.

In 1585 AD, the beginning of our modern base 10 decimals, modern base 10 decimals literally erased 3,600 years of Egyptian fraction arithmetic history. By 1900 AD, European and Arab scholars were unable to read medieval Egyptian fractions texts as well as the older Egyptian mathematical texts.

Ahmes' 84 problems have been slowly read by scholars after the RMP was published in 1877. The 2/n table was key to scholar research. But what method or methods did Ahmes use to create his 2/n table? Ahmes seemingly left few clues to assist scholars in decoding the 2/n table's construction method(s).

Post-1877 scholars relied on intuition and personalized mathematical senses to report suspected details, often confusing the subject. Scholarly debates have correctly focused on the 2/n table and its construction methods. Yet, confirmed 2/n table threads continue to be controversial.

Egyptologists and math historians, i.e., Neugebauer, Exact Sciences in Antiquity, had inappropriately proposed that the RMP 2/n table and related Egyptian fraction methods represented forms of intellectual decline.

However, one simple RMP 2/n table construction method, multiples, was reported in 2002 that changed the debate. An advanced idea had created finite unit fraction statements. By 2006, it was shown that Egyptian fraction statements had solved an Old Kingdom infinite series round-off problem.

Scholars continue to parse several theoretical fragments of a group of ancient texts by considering related arithmetic patterns. As a consequence, 130 years of confusion related to Egyptian fraction arithmetic is lifting an intellectual fog.

For example, Ahmes, the RMP scribe, has gained the majority of scholarly attention since the RMP was published in 1877. Sylvester in 1891 incorrectly suggested that the greedy algorithm was present in the medieval Liber Abaci and by implication the RMP.

Hultsch in 1895 began to parse Ahmes' 2/p conversion patterns by using aliquot parts. It took Bruins in 1944 to confirm Hultsch's earlier work, now known as the H-B method.

Ahmes, therefore, converted 2/p rational numbers into optimal or elegant Egyptian fraction series using a form of the H-B method. Ahmes' shorthand indicated that 2/43, and other 2/n table members, were converted to an Egyptian fraction series by selecting optimal multiples, in the 2/43 case the multiple 42.

Ahmes' use of the multiple 42 allowed 1/42 to become the first partition. The remaining Egyptian fraction were found by considering the divisors (aliquot parts) of 42 or (21, 14, 7, 6, 3, 2, 1). Ahmes' fragmented shorthand indicates:

2/43*42/42 = (42 + 21 + 14 + 6)/(42*43), such that: 2/43 = 1/42 + 1/86 + 1/129 + 1/301 in clear hieratic script.

Today, 20th-century additive scholarly adherents oddly do not accept the H-B method, or related methods used by Egyptians, Greeks, Hellenes, Arabs, medievals, and others for over 3,000 years. It is clear that H-B type methods used modern addition, subtraction, multiplication, and division operations and correctly parse 3,700 years of rational numbers conversions to Egyptian fraction series.

Occam's Razor, the simplest method is likely the historical method, is slowly changing scholarly minds. As another long-term verification of Ahmes' arithmetic, Fibonacci's 1202 AD Liber Abaci reports a very old style of writing multiples within the first four of seven rational number conversion methods.

In conclusion, Egyptian fractions were a fundamental part of ancient Egyptian mathematics, used to represent rational numbers as a sum of distinct unit fractions. The Egyptian fraction system was used for various purposes, including solving algebra problems, creating a commodity-based monetary system, and converting rational numbers to optimized unit fraction series.

The study of Egyptian fractions has been ongoing for centuries, with scholars continually uncovering new insights and methods used by the ancient Egyptians. The use of Egyptian fractions declined with the introduction of modern base 10 numerals, but their significance in the history of mathematics remains unparalleled.

The Egyptian fraction system is a testament to the ingenuity and mathematical sophistication of the ancient Egyptians, and its study continues to inspire and fascinate mathematicians and historians alike. By examining the Egyptian fraction system and its various applications, we can gain a deeper understanding of the development of mathematics and its role in shaping our modern world.


# Draw Exponential Function in R

## Introduction
The exponential function is a mathematical function that is increasing and has a point (0, 1) on its graph. In R, we can draw an exponential function using the `exp()` function.

## Exponential Function
The exponential function has the form `y = a * b^x + q`, where `a` and `q` are constants. The `exp()` function in R computes the exponential value of a number or number vector.

## Plotting Exponential Function
To plot an exponential function in R, we can use the `plot()` function. We can create a sequence of x values and calculate the corresponding y values using the exponential function.

## Example
```r
x <- seq(0, 10, by = 0.1)
y <- exp(x)
plot(x, y, type = "l")
```
This code creates a sequence of x values from 0 to 10 with a step size of 0.1, calculates the corresponding y values using the exponential function, and plots the resulting curve.

## Fitting Exponential Curve to Data
To fit an exponential curve to data, we can use the `nls()` function in R. This function fits a non-linear model to the data.

## Example
```r
x <- seq(0, 10, by = 0.1)
y <- exp(x) + rnorm(length(x), mean = 0, sd = 1)
model <- nls(y ~ a * exp(b * x), start = list(a = 1, b = 1))
plot(x, y)
lines(x, predict(model), col = "red")
```
This code generates some sample data with an exponential trend, fits an exponential model to the data using the `nls()` function, and plots the original data and the fitted curve.

## Conclusion
In this article, we have seen how to draw an exponential function in R using the `exp()` function and how to fit an exponential curve to data using the `nls()` function.


**Bolt Elongation Formula and Calculator per ASME PCC-1**

The bolt elongation formula is used to compute the required bolt elongation when bolt elongation measurement is selected as the load-control technique. The formula assumes the bolt is threaded full length and is given by:

ΔL = [ ( S_{b} · L_{eff} ) / E ] · ( A_{r} / A_{ts} )

Where:
- ΔL = bolt elongation (bolt stretch), mm (in.)
- S_{b} = Target Bolt Stress (root area), MPa (ksi)
- L_{eff} = effective stretching length, mm (in.)
- E = modulus of elasticity, MPa (ksi)
- A_{r} = root area (minor diameter), mm^{2} (in.^{2})
- A_{ts} = tensile stress area, mm^{2} (in.^{2})

The tensile stress area (A_{ts}) is calculated differently for metric and ANSI inch threads:
- For metric threads: A_{ts} = ( π / 4 ) ( D - 1.3 · P ) / 2
- For ANSI inch threads: A_{ts} = ( π / 4 ) ( D - 1.3 / n ) / 2

Where:
- D = diameter
- P = thread pitch
- n = threads per inch

The effective stretching length (L_{eff}) is the distance between the mid-thickness of the nuts for a through-bolted joint system, or one-half of the nominal bolt diameter for a studded bolt. The target bolt stress (S_{b}) is based on the root area, but if the tensile stress area is used, the A_{r} / A_{ts} term is dropped from the ΔL computation.

The computed bolt elongation (ΔL) should have a selected tolerance and be included in the joint assembly procedure.

Related resources include:
- Bolt Elongation Equation and Calculator under Axial Stress
- Fastener, Bolt, and Screw Design Torque and Force Calculation
- Bolt Preload Tension Calculator and Force Equation
- Moment Effect on Bolts in Circular Distribution in End-plate Connection
- Screw Thread Size Chart
- ANSI Hardware Design Guide
- Metric Hardware Specification Charts and Tables
- Fastener Threaded Shear Area Equation and Calculator ISO 898

Reference: ASME PCC-1-2010 Guidelines for Pressure Boundary Bolted Flange Joint Assembly.


# Magic Matrices: A Definition

*Magic squares* are symmetrical matrices consisting of regular sequences of integers, arranged in two dimensions, where the total of all numbers in every line produces an identical sum. The numbers typically run from 1 to *n*, where *n* represents the order of the square. For example, a magic cube of order 3 contains 3 x 3, or 9, numbers running from 1 to 9.

*Magic cubes* are similar matrices expanded into three dimensions, while *magic hyper-cubes* extend the principle into more than three dimensions, sharing properties with magic squares and cubes. Other types of magic arrangements include *anti-magic squares*, *magic stars*, and matrices of non-contiguous integer sequences, such as consecutive primes. These objects are collectively referred to as *magic matrices*.

The n x n matrix of a magic square (or n x n x n matrix of a magic cube) can be arranged in a finite number of ways. Disregarding the 1 x 1 square, the smallest magic square, the 3 x 3, has only one solution, which can be reflected and rotated in 8 ways while maintaining the relationship between numbers. However, most solutions exclude these reflections and rotations, considering them trivial.

There are 880 permutations of the order 4 magic square, and properties within subsets of the order 4 family are still being defined. The number of possible solutions increases rapidly with each order, and from order 6 upwards, the huge number of permutations can currently only be estimated.


Let $\Phi$ be a crystallographic root system with $\Phi^{+}$ as the choice of positive roots and $\Delta$ as the corresponding simple roots. The root lattice of $\Phi$ is $\mathbb{Z}\Delta$ ($=\mathbb{Z}\Phi^{+}=\mathbb{Z}\Phi$).

The Weyl vector, $\rho = \frac{1}{2}\sum_{\alpha \in \Phi^{+}}\alpha$, is a significant vector in representation theory. Although $2\rho \in \mathbb{Z}\Delta$, it is possible that $\rho \in \mathbb{Z}\Delta$.

A key question is whether there is representation-theoretic significance to $\rho \in \mathbb{Z}\Delta$. Research has uncovered a combinatorial phenomenon related to this condition, which can be characterized as follows:

- Type $A_{n}$: $\rho \in \mathbb{Z}\Delta$ if and only if $n$ is even.
- Type $B_n$: $\rho \notin \mathbb{Z}\Delta$.
- Type $C_n$: $\rho \in \mathbb{Z}\Delta$ if and only if $n \equiv 0,3 \mod 4$.
- Type $D_n$: $\rho \in \mathbb{Z}\Delta$ if and only if $n \equiv 0,1 \mod 4$.
- Type $E_6$, $E_8$, $F_4$, and $G_2$: $\rho \in \mathbb{Z}\Delta$.
- Type $E_7$: $\rho \notin \mathbb{Z}\Delta$.

Understanding the representation-theoretic significance of $\rho \in \mathbb{Z}\Delta$ could provide insight into the underlying combinatorial phenomenon.


**Math Topics**
The provided text covers various math topics, including Common Core initiatives, methodology, resources, projects, manipulatives, and software.

**Learning Support**
Learning support resources include standardized test prep, technology integration, and assisting readers.

**Professionalism**
Professional development resources comprise associations, journals, news, education standards, education research, and CT4ME publications.

**Standards**
Key standards include:
- IC-A.1: Understand statistics as a process for making inferences about population parameters based on a random sample.
- IC-A.2: Decide if a specified model is consistent with results from a given data-generating process.
- IC-B.3: Recognize the purposes of and differences among sample surveys, experiments, and observational studies.
- IC-B.4: Use data from a sample survey to estimate a population mean or proportion.
- IC-B.5: Use data from a randomized experiment to compare two treatments.
- IC-B.6: Evaluate reports based on data.

**Technology-enhanced investigations**
Resources for technology-enhanced investigations include:
- Stattrek.com Statistics Dictionary
- LearnZillion video lessons
- MIT BLOSSOMS video lessons
- PhET Interactive Simulations
- Stat Trek tutorials
- Wolfram Demonstrations Project

**Multiple Choice**
Multiple-choice questions are available from:
- ACT Academy and OpenEd
- Khan Academy

**Constructed-response**
Constructed-response questions are available from:
- ACT Academy and OpenEd

**Performance tasks**
Performance tasks are available from:
- Illustrative Mathematics
- Mathematics Vision Project
- NCTM's Reasoning and Sense Making Task Library
- Statistics in Schools from the U.S. Census Bureau

**Resources**
Additional resources include:
- Khan Academy videos
- Shodor Interactivate
- Statistics How To
- Stat Trek tutorials
- Wolfram Demonstrations Project

**Cluster Selection**
The cluster for resources on this page is **Standards**, which includes IC-A and IC-B standards.


## Weekly Challenge 8: Sixinit

Yatir Halevi proved that $n^3+11n$ is always divisible by $6$ using modular arithmetic modulus $6$. For each $n$, the remainder when divided by $6$ can be $0$, $1$, $2$, $3$, $4$, or $5$, known as the residue modulo $6$.

The residues for $n^3$ are $0$, $1$, $2$, $3$, $4$, $5$ respectively, and for $11n$, the residues are $0$, $5$, $4$, $3$, $2$, $1$. Combining these, we get:
- $0+0=0$ (mod $6$)
- $1+5=6=0$ (mod $6$)
- $2+4=6=0$ (mod $6$)
- $3+3=6=0$ (mod $6$)
- $4+2=6=0$ (mod $6$)
- $5+1=6=0$ (mod $6$)
This results in a zero residue when dividing by $6$, meaning $(n^3+11n)$ is a multiple of $6$.

Alternatively, Kookhyun Lee proved that all terms are divisible by $6$ by considering:
$n^3 + 11n = n^3 + 12n - n = n(n^2-1) + 12n = (n-1) \times n \times (n+1) + 12n$
This is a multiple of $6$ because any multiple of three consecutive integers is a multiple of $6$, containing both a multiple of two (an even number) and a multiple of three.


Representing a camera mathematically can be complex, especially when considering multiple aspects. This discussion focuses on the linear pinhole camera model, the first in a series of posts on camera representation. The goal is to derive a linear system that describes how a 3D world point projects to a 2D image plane.

## Homogenous Coordinates

To simplify equations, homogenous coordinates are used, a linear algebra trick that adds an extra dimension to each point. A 3D world point becomes $\mathbf{X}_w = (s_w x_w, s_w y_w, s_w z_w, s_w)^T = s_w (x_w, y_w, z_w, 1)^T$, and a 2D image point becomes $\mathbf{X}_i = (s_i x_i, s_i y_i, s_i)^T = s_i (x_i, y_i, 1)^T$. The scalar multiple ($s_w$ and $s_i$) is insignificant as it can be divided out, but it cannot be zero.

## The Simplest Pinhole Camera

Using homogenous coordinates, a mathematical description of a camera is built. The camera is described by a set of linear equations translating a world point $\mathbf{X}_w$ into an image point $\mathbf{X}_i$. The transformation is represented by the $3 \times 4$ projection matrix $\mathbf{P}$, where $\mathbf{X}_i = \mathbf{P} \mathbf{X}_w$.

A simple pinhole camera has key components:
- $x$, $y$, and $z$ are the 3D world axes, with $z$ as the principal axis, perpendicular to the image plane.
- $\mathbf{C}_w$ is the world coordinate of the camera center, a 4-dimensional homogenous point.
- $\mathbf{C}_i$ is the principal point, where the principal axis meets the image plane, a 3-dimensional homogenous point.
- $f$ is the focal length, the distance from the camera center to the image plane.
- $\mathbf{X}_w$ is the world point being imaged, a 4-dimensional homogenous point.
- $\mathbf{X}_i$ is the point on the image plane that the world point projects to, a 3-dimensional homogenous point.

In the simplest case, the camera center is at the origin, $\mathbf{C}_w = (0, 0, 0, 1)^T$. The transformation from world to image coordinates is $\mathbf{X}_i = \mathbf{P} \mathbf{X}_w$, where

$$\mathbf{X}_i = \left[ \begin{array}{c} s_i x_i \\ s_i y_i \\ s_i \end{array}\right] = \left[ \begin{array}{cccc} f & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] \left[ \begin{array}{c} x_w \\ y_w \\ z_w \\ 1 \end{array}\right]$$

and

$$\mathbf{P} = \left[ \begin{array}{cccc} f & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right].$$

This concludes the introduction to the linear pinhole camera model. The next post will explore moving the camera center and principal point to different locations.


Unit 6: Graphs and Data. Kids learn about standard units to measure length, weight, and capacity. In second grade, students will measure length with both customary (inches, feet, yards) and metric units (centimeters, meters). Measure and estimate lengths in standard units. Students will measure an object twice, using different length units, and describe how the two measurements relate to the size of the unit chosen.

Our grade 2 measurement worksheets focus on the measurement of length, weight, capacity, and temperature. From basics such as how to read a ruler to more complex concepts like the difference between inches and centimeters, these second-grade measurement worksheets teach. Students will practice measuring length with standard units in inches.

The standard unit of length is meter. One meter is divided into 100 smaller units called centimeters. 1km = 1000m. The standard unit of measuring capacity is the liter, and we write "l" in short. Capacity means how much liquid a container can hold. A bigger container can hold more compared to a smaller container.

Students will measure and compare lengths of objects using informal units. They will make estimates and measure the different objects in their packet using different non-standard units of measurement. Students will write about their findings and describe how the two measurements relate to the size of the unit chosen.

The games are designed as interactive worksheets, and loads of positive reinforcement add more to the enjoyment. These worksheets help students understand how to measure different objects in real-life situations. Measurement grade 2 worksheets help students practice a variety of measurement problems like word problems, matching, filling the blanks, etc.

The topic "measurement" for grade-2 is a step next to kindergarten skills and length measurement. Children develop an understanding of the meaning and processes of measurement, including such underlying concepts as partitioning. Only after children understand and can use single units of measurement should they move to working with common measuring tools.

In Maths, length is the longest extent of anything. It is the way a figure is represented and measured. The first dimension is a one-dimensional line, (1D) which only has length and height. The second is a two-dimensional plane (2D), which has height, length, but area. Dimension in math can be expressed in three dimensions.

The Four Stages of Measurement are: (1) direct comparison; (2) indirect comparison; (3) measuring with arbitrary (non-standard) units; and (4) measuring with standard units. Students will measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes.

Students will estimate lengths using units of inches, feet, centimeters, and meters. They will measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. The student will draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories.

The standard scales of measurement with customary and metric units are introduced. The different units used to measure physical quantities are metric and U.S. customary. Inches, feet & yards are used in the United States. One inch is about the size of a small paper clip.

The length of one side of a triangle is the same as the breadth of a rectangle. The sum of the lengths of three sides of a triangle is 28 cm. The length of a rectangle is less than any side of a triangle. The length of a rectangle is two times the height of the rectangle.

These 10 puzzles have children figuring out liquid equivalences (gallons to quarts), metric equivalences (cm to m), and US conversions such as inches to feet. Memorize liquid, metric, and linear equivalences. The games are designed as interactive worksheets, and loads of positive reinforcement add more to the enjoyment.


**Q3:** The sum of the perimeters of two squares is 108 mm, with one square having a side length of 2 mm. Find the side length of the other square.

**Q4:** The sum of the perimeters of two squares is 80 mm, with one square having a side length of 4 mm. Find the side length of the other square.

**Q5:** Calculate the side length of a square with a perimeter of 24 m.

**Q6:** Calculate the side length of a square with a perimeter of 20 m.

**Q7:** A 96 cm wire is folded to form a square. Find the length of each side.

**Q8:** A 192 cm wire is folded to form a square. Find the length of each side.

**Q9 and Q10:** Calculate the perimeter of a square with side length m.

**Q11:** Complete the table relating side lengths of squares to their perimeters.

| Side Length (cm) | 5 | 10 | x |
| --- | --- | --- | --- |
| Perimeter (cm) | 20 | 40 | y |

Find x and y.
- A: x = 15, y = 60
- B: x = 15, y = 80
- C: x = 20, y = 80
- D: x = 20, y = 100

**Q12:** Complete the table relating side lengths of squares to their perimeters.

| Side Length (cm) | 6 | 5 | x |
| --- | --- | --- | --- |
| Perimeter (cm) | 24 | 20 | y |

Find x and y.
- A: x = 8, y = 32
- B: x = 8, y = 40
- C: x = 10, y = 40
- D: x = 10, y = 50

**Q13:** Find the perimeter of a square with a side length of 15.

**Q14:** A volleyball court is a square with a side length of 9 meters. Find the perimeter of one half of the court.

**Q15:** Find the side length of a square with a perimeter of 4.

**Q16 and Q17:** Not enough information to solve.

**Q18:** A square infield measures 60 feet on each side. Find its perimeter.


Combinatorics Seminar Fall 2014
The Combinatorics Seminar is an introductory seminar for combinatorics, accessible to first-year graduate students. It takes place on Tuesdays at 10:11 am in Room 209, Goldsmith building.

Tuesday, September 30
Speaker: Yan Zhuang (Brandeis)
Title: Counting Permutations by Alternating Descents
Abstract: This talk presents the exponential generating function for permutations with all valleys even and all peaks odd. The generating function is used to determine the asymptotics for its coefficients, answering a question posed by Liviu Nicolaescu. Two proofs are provided: one using a system of differential equations and another using noncommutative symmetric functions. This is joint work with Ira Gessel.

Tuesday, October 7
Speaker: Ira Gessel (Brandeis)
Title: Exponential Generating Functions Modulo p
Abstract: Exponential generating functions are useful in counting labeled objects such as permutations, partitions, and graphs. This talk discusses finding congruences to prime moduli for these sequences. Formal power series that are exponential generating functions for sequences of integers are called Hurwitz series, which form a ring and are closed under composition. The quotient ring of Hurwitz series modulo a prime p is studied to obtain the desired congruences.

Tuesday, November 4
Speaker: An Huang (Harvard)
Title: CW Complex Associated to Digraphs
Abstract: A CW complex can be constructed from a digraph with a chosen basis of certain path space. This construction provides a way to study digraphs using familiar ideas in topology. The talk explains this construction and its consequences, which is joint work with ST. Yau.

Tuesday, November 18
Speaker: Olivier Bernardi (Brandeis)
Title: Differential Equations for Colored Lattices
Abstract: This talk presents a solution to a statistical mechanics model on random lattices, specifically the Potts model on planar triangulations. The partition function of this model is characterized by a simple system of differential equations. Special cases, such as properly 4-colored triangulations, lead to simple equations waiting for a more direct combinatorial explanation. This is joint work with Mireille Bousquet-Melou.

Tuesday, December 2
Speaker: Cristobal Lemus (Brandeis)
Title: Lattice Path Factorization
Abstract: The talk presents cases of lattice paths in the plane whose generating functions have factorizations into three parts corresponding to three types of subpaths. The method of factorization is used to count these paths and subpaths with added weights, resulting in Delannoy, asymmetric Delannoy, Schröder, and Narayana numbers.

Previous semesters include Spring 2014 and Fall 2013.


# Ape Index Calculator

## What Is the Ape Index?

The ape index is a measurement of how your wingspan compares to your height. It's believed that most people's wingspan is equivalent to their height, meaning most people have a neutral ape index or an ape index ratio of 1.

## How to Calculate Your Ape Index

There are two ways to calculate your ape index: divide your wingspan by your height or subtract your height from your wingspan.

### Method #1: Wingspan to Height Ratio

To calculate: Wingspan / Height. The typical ape index ratio is 1. A number greater than one means your wingspan is greater than your height, and a number less than one means your height is greater than your wingspan.

### Method #2: Difference Between Wingspan & Height

To calculate: Wingspan – Height. The typical ape index is 0. Using this method, your ape index changes depending on whether you're using inches or centimeters.

## Ape Indices of Famous Climbers

Compare your ape index with some of the world's greatest climbers:

| Climber | Wingspan | Height | Ape Index 1 (W / H) | Ape Index 2 (W - H) |
| --- | --- | --- | --- | --- |
| Kai Lightner | 82 in (208.3 cm) | 75 in (190.5 cm) | 1.09 | +7 in (+17.8 cm) |
| Matt Fultz | 77 in (195.6 cm) | 71 in (180.3 cm) | 1.08 | +6 in (+15.2 cm) |
| Kai Harada | 70.9 in (180 cm) | 66.5 in (169 cm) | 1.07 | +4.3 in (+11 cm) |
| Daniel Woods | 72 in (182.9 cm) | 68 in (172.7 cm) | 1.06 | +4 in (+10.2 cm) |
| Dave Graham | 74 in (188 cm) | 70 in (177.8 cm) | 1.06 | +4 in (+10.2 cm) |
| Ethan Pringle | 74 in (188 cm) | 70 in (177.8 cm) | 1.06 | +4 in (+10.2 cm) |
| Kyra Condie | 68 in (172.7 cm) | 64 in (162.6 cm) | 1.06 | +4 in (+10.2 cm) |
| Jan Hojer | 78 in (198 cm) | 74 in (188 cm) | 1.05 | +3.9 in (+10 cm) |
| Tomoa Narasaki | 70.9 in (180 cm) | 66.9 in (170 cm) | 1.06 | +3.9 in (+10 cm) |
| Sascha Lehmann | 67.7 in (172 cm) | 64.2 in (163 cm) | 1.06 | +3.5 in (+9 cm) |
| Paul Robinson | 74 in (188 cm) | 71 in (180.3 cm) | 1.04 | +3 in (+7.6 cm) |
| Chris Sharma | 74.5 in (189.2 cm) | 72 in (182.9 cm) | 1.03 | +2.5 in (+6.4 cm) |
| Jongwon Chon | 72 in (183 cm) | 69.7 in (177 cm) | 1.03 | +2.4 in (+6 cm) |
| Alex Puccio | 65 in (165.1 cm) | 63 in (160 cm) | 1.03 | +2 in (+5.1 cm) |
| Emily Harrington | 64 in (162.6 cm) | 62 in (157.5 cm) | 1.03 | +2 in (+5.1 cm) |
| Sasha DiGiulian | 64 in (162.6 cm) | 62 in (157.5 cm) | 1.03 | +2 in (+5.1 cm) |
| Angy Eiter | 62.2 in (158 cm) | 60.6 in (154 cm) | 1.03 | +1.6 in (+4 cm) |
| Stefano Ghisolfi | 68.5 in (174 cm) | 66.9 in (170 cm) | 1.02 | +1.6 in (+4 cm) |
| Jonathan Siegrist | 67 in (170.2 cm) | 65.5 in (166.4 cm) | 1.02 | +1.5 in (+3.8 cm) |
| Magnus Midtbø | 69.3 in (176 cm) | 68.1 in (173 cm) | 1.02 | +1.2 in (+3 cm) |
| Jimmy Webb | 73 in (185.4 cm) | 72 in (182.9 cm) | 1.01 | +1 in (+2.5 cm) |
| Nalle Hukkataival | 68.5 in (174 cm) | 68 in (172.7 cm) | 1.01 | +0.5 in (+1.3 cm) |
| Adam Ondra | 73.6 in (187 cm) | 73.2 in (186 cm) | 1.01 | +0.4 in (+1 cm) |
| Alex Megos | 68.1 in (173 cm) | 68.1 in (173 cm) | 1.00 | +0 in (+0 cm) |
| Hazel Findlay | 62 in (157.5 cm) | 62 in (157.5 cm) | 1.00 | +0 in (+0 cm) |
| Lynn Hill | 62 in (157.5 cm) | 62 in (157.5 cm) | 1.00 | +0 in (+0 cm) |
| Babsi Zangerl | 63 in (160 cm) | 63.8 in (162 cm) | 0.99 | -0.8 in (-2 cm) |

## Analysis

The handful of professional climbers in the table above have, on average, a positive ape index. They tend to have long arms for their height. However, there are notable exceptions, such as Lynn Hill, Babsi Zangerl, and Alex Megos, who have ape indices of 0 or less.

## Is a Positive Ape Index Good for Climbing?

It's commonly believed that a positive ape index is good for rock climbing. However, research has shown that while elite climbers tend to have higher ape indices, there is no clear correlation between ape index and climbing performance.

### Do Elite Climbers Have High Ape Indices?

Studies have shown that elite climbers tend to have ape index ratios greater than 1. However, the sample sizes are often small, and more research is needed to confirm this finding.

### Does a High Ape Index Make You a Better Climber?

Research has struggled to find a positive correlation between ape index and climbing performance. One study found that "elite climbers usually possess an index higher than 1.00 but it was not proven that the ape index correlates with climbing performance."

### Trainable Factors Are More Predictive of Climbing Ability

Factors such as low body fat percentage and grip strength to body mass ratio are more predictive of climbing ability than ape index. These factors can be improved through training, making them more important for climbers to focus on than their ape index.

Note: The ape index is just one factor to consider when evaluating a climber's potential. Other factors, such as technique, strength, and endurance, are also important.


**Entering Grade 3 Summer Math 2020**

**July 2020**

1. Write 500 + 60 + 8 as a three-digit number and in words. Draw a picture that represents the number in base ten blocks.
2. A farmer planted sunflower seeds in July. In September, will they grow closer to 12 inches or 12 feet? Explain your answer in writing.
3. Pia put 10 stickers in each party bag. She made 12 bags. How many total stickers were in all 12 bags? Show your work and record your answer.
4. (Not provided)
5. (Not provided)
6. Look at a clock. Draw what time it shows. Record how many minutes until the next hour? Draw the clock as it would look at that hour.
7. Practice counting forward and backward by 2’s, 5’s, and 10’s from any number. Record your numbers.
8. I have 13 apples, but I only want 5. What do I need to do? Draw a picture and write an equation to show your thinking.
9. What is the temperature today? How far is it from 100 degrees Fahrenheit? How far is it from 32 degrees? Record your data.
10. Look in your refrigerator. Record and categorize the items as dairy, fruit, vegetable, meat, grains, fats, and other. Make a tally chart.
11. (Not provided)
12. (Not provided)
13. Write 5 ways to make 77 cents. Draw the coins and write the equations for each combination.
14. How many times can you hop on your left foot in a minute? Your right foot? Compare the number of hops using the symbols <, >, =. Record your expressions.
15. (Not provided)
16. How many different ways can you cut a sandwich into fourths? Draw the ways you can do it.
17. You won first place in a contest! You have two choices for the prize. You can take $20 home with you today or $2 a day for the next 15 days. Record which option earns more money. Record how much more.
18. (Not provided)
19. (Not provided)
20. What day of the week is it? What is the date? What was the day and the date 2 days ago? What will tomorrow’s day and date be? What day and date will it be in 1 week? 2 weeks? 4 weeks? Record all of your answers.
21. Try some math on https://www.prodigy.game.com
22. Cut out a picture from a magazine or newspaper or use a picture you have at home. Glue it to a piece of paper. Write a story problem to go along with the picture. Challenge a friend to solve it!
23. Find 5 ways to make $1.00 using quarters, dimes, nickels, and pennies. Draw pictures of the coins and equations to match.
24. Count all the books in your house! Record the count. Is there less or more than 100? Record how many more or less.
25. (Not provided)
26. (Not provided)
27. Record the ages of all the people who live in your house. What is the sum? Write an equation with the answer.
28. Record how many days until your birthday? Explain how you figured it out.
29. Find and record 20 coins in your house. Write how much it is altogether. Record if the total is more or less than $3.00.
30. 100 is the answer. What could the question be? Challenge yourself to think of more questions. Record your questions.
31. Start at 3 and write all the numbers from 300 to 400 counting by 3’s. Write down what patterns you see.

**August 2020**

3. The answer is addition, what is the question?
4. What is today's weather? Yesterday's? Predict the next few days and create a chart.
5. Write down the years people who live with you were born. Put them in order from least to greatest.
6. Write the three numbers below in expanded form: 729, 846, 295
7. Palindromes are numbers that are the same forward and backwards. How many can you think of? Can you find some in real life?
8. (Not provided)
9. (Not provided)
10. Flip a coin 10 times and record your results. Flip the coin another 10 times. Compare the results. Write what you noticed. What do you think might happen if you flip the coin another 10 times? Try it!
11. Play 2 games on https://www.mathplayground.com
12. (Not provided)
13. (Not provided)
14. (Not provided)
15. (Not provided)
16. (Not provided)
17. You have 2 dimes, 1 nickel, and 3 pennies in your pocket. Record how much money you have. How much money do you have if you double it?
18. (Not provided)
19. (Not provided)
20. (Not provided)
21. (Not provided)
22. (Not provided)
23. (Not provided)
24. The answer is 130. What is the question?


# The Nerdiest Debate of the Century

The concept of hypothesis testing and p-value is familiar to many, having been taught in high school and university statistics classes. However, the American Statistical Association's (ASA) 2016 statement on p-values sparked a debate on the reproducibility and replicability of scientific conclusions. The crux of the problem lies in the fact that hypothesis testing and p-value are not infallible.

Every scientific research aims to determine if there is enough scientific evidence to support the hypotheses. Classically, hypothesis testing and p-value are used to prove that there is enough evidence. However, there are many examples where this leads to spurious conclusions. Two of the most cited causes are:

## Cause 1: Misinterpretation of P-value

The p-value is often misinterpreted. It is defined as the probability of obtaining test statistics at least as extreme as the observed statistics, given that the null hypothesis is true. This definition can be convoluted, leading to misinterpretation. Many people, including students of statistics, initially think that the p-value is equivalent to the probability that a given hypothesis is true. However, they are not equivalent. The p-value is always based on the assumption that the null hypothesis is true, making it a prior belief.

In the context of coin flipping, if a coin is fair, we assume that it will come up heads 1/2 of the time. When doing hypothesis testing, we evaluate observed values with our prior belief. If we reject the null hypothesis, it means that the coin is likely unfair. However, in experiments, things are not clear-cut because we don't know if our null hypothesis is correct. Rejecting the null hypothesis does not necessarily mean that the observed data supports the alternative hypothesis.

Ronald Fisher, who developed significance testing, intended for p-value to be used to establish whether further research could be justified, not as conclusive evidence.

## Cause 2: The Imaginary Threshold of 0.05

The alpha level of 0.05 is a mere convention, and it is possible to adjust the threshold to make hypothesis testing statistically significant. Additionally, "p-value hacking" is a common occurrence, where data is manipulated to produce a desired p-value. This can be done by increasing the sample size, which tends to produce small p-values.

## Does it Mean We Should Abandon P-value Altogether?

No, abandoning p-value completely would render findings from scientific exploration unverifiable and invalidate useful scientific conclusions. The traditional approach of hypothesis testing and p-value is still valuable in today's society. For example, during the COVID-19 pandemic, hypothesis testing and p-value were used to validate the effectiveness of vaccines.

## What Can We Do?

To prevent further misuse of hypothesis testing and p-value, balance is key. Integrating both frequentists and Bayesians' approaches can help. Frequentists think in terms of "black or white," while Bayesians think in terms of "black and white" as a neutral grey. This integration can be achieved by:

* Using p-value and hypothesis testing as a quick litmus test for hypotheses
* Involving Subject Matter Experts (SMEs) in every step
* Using other metrics as conclusive significance testing
* Complementing standard Null Hypothesis Significance Testing with Bayesian hypothesis testing

By taking these steps, we can ensure that hypothesis testing and p-value are used effectively and responsibly, leading to more reliable scientific conclusions.


**Proof.**
Choose $h_1, \ldots , h_ n \in S$ such that $\sum h_ i g_ i = 1$.

**Proof of (1).** For each $i$, select a finite list of elements $x_{i, j} \in S_{g_ i}$, $j = 1, \ldots , m_ i$, which generate $S_{g_ i}$ as an $R$-algebra. Express $x_{i, j} = y_{i, j}/g_ i^{n_{i, j}}$ for some $y_{i, j} \in S$ and some $n_{i, j} \ge 0$. Consider the $R$-subalgebra $S' \subset S$ generated by $g_1, \ldots , g_ n$, $h_1, \ldots , h_ n$, and $y_{i, j}$, $i = 1, \ldots , n$, $j = 1, \ldots , m_ i$. Since localization is exact (Proposition 10.9.12), $S'_{g_ i} \to S_{g_ i}$ is injective. By the choice of $y_{i, j}$, it is also surjective. The elements $g_1, \ldots , g_ n$ generate the unit ideal in $S'$, as $h_1, \ldots , h_ n \in S'$. Thus, $S' \to S$ is an isomorphism by Lemma 10.23.2.

**Proof of (2).** Given that $S$ is of finite type, express $S = R[x_1, \ldots , x_ m]/J$ for some ideal $J$. For each $i$, choose a lift $g'_ i \in R[x_1, \ldots , x_ m]$ of $g_ i$ and a lift $h'_ i \in R[x_1, \ldots , x_ m]$ of $h_ i$. Then,
\[ S_{g_ i} = R[x_1, \ldots , x_ m, y_ i]/(J_ i + (1 - y_ ig'_ i)) \]
where $J_ i$ is the ideal of $R[x_1, \ldots , x_ m, y_ i]$ generated by $J$. By Lemma 10.6.3, select a finite list of elements $f_{i, j} \in J$, $j = 1, \ldots , m_ i$, such that the images of $f_{i, j}$ in $J_ i$ and $1 - y_ ig'_ i$ generate the ideal $J_ i + (1 - y_ ig'_ i)$. Define
\[ S' = R[x_1, \ldots , x_ m]/\left(\sum h'_ ig'_ i - 1, f_{i, j}; i = 1, \ldots , n, j = 1, \ldots , m_ i\right) \]
There is a surjective $R$-algebra map $S' \to S$. The classes of $g'_1, \ldots , g'_ n$ in $S'$ generate the unit ideal, and by construction, the maps $S'_{g'_ i} \to S_{g_ i}$ are injective. Thus, we conclude as in part (1).


### Section 13.2: Polar Equations of Conics and Other Curves

Polar equations relate polar coordinates (r, θ) to Cartesian coordinates (x, y). Polar coordinates use a radius (r) and an angle (θ) to specify a point. The relationship between r, θ, x, and y is given by the trigonometric equations:

x^2 + y^2 = r^2
sin θ = y/r
cos θ = x/r

From these equations, we can derive values for x, y, and θ:

y = r * sin θ
x = r * cos θ
θ = tan^-1(y/x)

To plot polar graphs, we need to change the mode to polar. The standard equation for a polar graph is r = a + b * sin θ or r = a + b * cos θ. These equations result in a limacon or cardioid.

The general polar equations for conic sections are:

r = k / (a + b * sin θ) or r = k / (a + b * cos θ)

The relationship between a and b determines the conic shape:
- |a| = |b| -> parabola
- |a| > |b| -> ellipse
- |a| < |b| -> hyperbola

Note: The foci of a parabola will be at the origin, as well as one of the foci of an ellipse or hyperbola.


**Practice Set 1.1**

**Question 1.** The base and height of a triangle are 9 and 5, respectively. The base and height of another triangle are 10 and 6, respectively. Find the ratio of the areas of these triangles.

The area of a triangle is given by: Area = (1/2) × Base × Height
Area of triangle 1 = (1/2) × 9 × 5 = 22.5
Area of triangle 2 = (1/2) × 10 × 6 = 30
The ratio of the areas of these triangles is: 22.5 : 30 = 3 : 4 = 0.75

**Question 2.** In figure 1.13, BC ⊥ AB and AD ⊥ AB. Given BC = 4 and AD = 8, find the ratio of the areas of triangles ΔABC and ΔADB.

Since the triangles have a common base, the ratio of their areas is equal to the ratio of their heights.
Area of ΔABC : Area of ΔADB = BC : AD = 4 : 8 = 1 : 2

**Question 3.** In figure 1.14, seg PS ⊥ seg RQ and seg QT ⊥ seg PR. Given RQ = 6, PS = 6, and PR = 12, find QT.

Considering the area of ΔPQR with base QR, PS is the height. With base PR, QT is the height.
Since the triangle is the same, the area remains the same regardless of the base taken.
(1/2) × QR × PS = (1/2) × PR × QT
(1/2) × 6 × 6 = (1/2) × 12 × QT
QT = 3

**Question 4.** In the adjoining figure, AP ⊥ BC and AD || BC. Find the ratio of the areas of triangles ΔABC and ΔBCD.

By drawing DO, which is the height of ΔBCD, we can see that AP = DO.
The ratio of the areas of the triangles is equal to the ratio of their heights.
Area of ΔABC : Area of ΔBCD = AP : DO = 1 : 1

**Question 5.** In the adjoining figure, PQ ⊥ BC and AD ⊥ BC. Find the following ratios:

(i) Area of ΔPQB : Area of ΔADB
(ii) Area of ΔPQB : Area of ΔPQC
(iii) Area of ΔPQC : Area of ΔADB
(iv) Area of ΔPQB : Area of ΔPQC : Area of ΔADB

(i) The triangles have equal heights, so the ratio of their areas is equal to the ratio of their bases.
Area of ΔPQB : Area of ΔADB = PB : BD

(ii) The triangles have a common base, so the ratio of their areas is equal to the ratio of their heights.
Area of ΔPQB : Area of ΔPQC = PQ : QC

(iii) The triangles have equal heights, so the ratio of their areas is equal to the ratio of their bases.
Area of ΔPQC : Area of ΔADB = QC : BD

(iv) The ratio of the areas of the triangles is: Area of ΔPQB : Area of ΔPQC : Area of ΔADB = PB : QC : BD


The problem involves solving the heat equation using the separation of variables method and then verifying the solution in Maple. The heat equation is given by the expression for U(x,t,n), which is a sum of terms involving A[m], lambda[m], and other parameters.

The key equations and expressions are:
- tan(Lambda[m]*d) = h/(k*Lambda[m])
- Lambda[m] is the solution of the above equation, given by:
  - For m = 0, Lambda[m] = RootOf(tan(y*d) = h/(k*y), y, 0..Pi/(2*d))
  - For m > 0, Lambda[m] = RootOf(tan(y*d) = h/(k*y), y, Pi*(2*m-1)/(2*d)..Pi*(2*m+1)/(2*d))
- beta[m] = sqrt(h^2+k^2*(lambda[m])^2)
- A[m] = 2*beta[m]^2/(d*beta[m]^2+h*k)*(h/(beta[m]*lambda[m])*(T[0]-T[amb])-q/(k*(lambda[m])^2))
- U(x,t,n) = sum(A[m]*cos(lambda[m]*x)*exp(-a*(lambda[m])^2*t) , m=0..infinity)+T[amb]+q/h+q*(d-x)/k

The given parameters are:
- d = 0.05
- T[0] = 6
- T[amb] = -100
- k = 0.7
- c = 2093
- rho = 2100
- q = 400
- h = 5.6
- a = k/(c*rho)

The expressions for up(t) and down(t) are:
- up(t) = U(d,t,infinity)
- down(t) = U(0,t,infinity)

To use the Crank-Nicholson scheme in Maple, the user needs to build the scheme and then use it to find the solution. However, the details of implementing the Crank-Nicholson scheme are not provided in the given text.

Maple code:
restart;
Digits:=20:
a:=k/(c*rho):
lambda[m]:=piecewise(m=0, RootOf(tan(y*d)=h/(k*y), y, 0..Pi/(2*d)), m>0, RootOf(tan(y*d)=h/(k*y), y, Pi*(2*m-1)/(2*d)..Pi*(2*m+1)/(2*d)));
beta[m]:=sqrt(h^2+k^2*(lambda[m])^2):
A[m]:=2*beta[m]^2/(d*beta[m]^2+h*k)*(h/(beta[m]*lambda[m])*(T[0]-T[amb])-q/(k*(lambda[m])^2)):
U:= (x,t,n) -> sum(A[m]*cos(lambda[m]*x)*exp(-a*(lambda[m])^2*t) , m=0..infinity)+T[amb]+q/h+q*(d-x)/k:
d:=0.05:
T[0]:=6:
T[amb]:=-100:
k:=0.7:
c:=2093:
rho:=2100:
q:=400:
h:=5.6:
up:=t->simplify(U(d,t,infinity)):
down:=t->simplify(U(0,t,infinity)):


The Relational Model is based on logic and set theory, but there are multiple set theories used by mathematicians, with Wikipedia listing 15 first-order set theories. The question is whether there is a particular set theory considered the "best" one to use for the relational model.

One set theory, ZFC, is often used, but others like Finitist Set Theory have different properties. For example, FST states that the empty set `{}` has no members and therefore does not exist, which would have interesting consequences for the relational model.

Another example is the definition of an ordered pair, where the short version of Kuratowski's definition is sufficient for a relational model. However, this would mean that a heading could not have an attribute with the same type name and attribute name, nor could a tuple have an attribute with the same attribute name and attribute value.

The foundation of the relational model is solid if it specifies which set theory is used. Otherwise, it may lead to inconsistencies. Some set theories, like ZFC, are more suitable for the relational model than others.

In response to the question, AntC suggests that first-order theory is adequate to describe a relational model and that higher-order theories are not necessary. However, Vadim argues that Category Theory and other higher-order theories can provide insight into the relational model.

Dandl points out that a formal model of SQL is necessary to go beyond SQL, but AntC disagrees, stating that he wants to avoid SQL's flaws, such as nulls and duplicate rows, and instead focus on relation values as sets-of-tuples with headings.

Vadim provides an example of the less-than relation being compatible with the addition operation, but AntC argues that this is not a valid example, as it tries to show two different relations orthogonally.

The discussion concludes with AntC reiterating his desire to avoid SQL's flaws and focus on a pure relational model, while Vadim and Dandl argue for the importance of formal models and higher-order theories in understanding the relational model.

Key concepts:

* Set theory
* Relational model
* ZFC
* Finitist Set Theory
* Kuratowski's definition
* Ordered pairs
* Category Theory
* Higher-order theories
* First-order theory
* SQL
* Relation values
* Headings
* Tuples

Important numbers:

* 15 (number of first-order set theories listed on Wikipedia)

Pros and cons of various set theories:

* ZFC: widely used, but may not be the "best" choice for the relational model
* Finitist Set Theory: has different properties, such as the non-existence of the empty set
* Kuratowski's definition: sufficient for a relational model, but may have limitations

Refined multiple choice questions and answers:

* What is the basis of the Relational Model?
a) Logic and set theory
b) Category Theory
c) Higher-order theories
d) SQL

Answer: a) Logic and set theory

* Which set theory is often used in the relational model?
a) ZFC
b) Finitist Set Theory
c) Kuratowski's definition
d) Category Theory

Answer: a) ZFC

* What is the consequence of using Finitist Set Theory in the relational model?
a) The empty set exists
b) The empty set does not exist
c) The relational model is inconsistent
d) The relational model is incomplete

Answer: b) The empty set does not exist


### Doors & Rooms Level 1-10 Solutions and Discussions

Players have been discussing the challenges and solutions for levels 1-10 of the game Doors & Rooms. The conversations cover various aspects of the game, including puzzles, codes, and strategies for progression.

One of the discussed puzzles involves a code that some players found challenging to decipher. A player named Andrew Harvey explained that the solution involves basic algebra, where the equation 2x + 2y + 2z = 12 is used to find the values of x, y, and z. Another player, daft, mentioned that the scrap paper provides a system of 3 variables and 3 equations, which can be solved through algebra.

For a specific level, a player named Pandemonium06 suggested that the numbers 8, 4, 6, and 1 are crossed out, leaving the numbers 5, 2, 9, and 3 as the solution. Oihd explained that the blood type Rh-A has a value of 9, which is calculated as 1*3 + 2*2 + 2*1 = 9.

In another level, players discussed a code involving the letters H, E, L, and P. Samantha explained that the solution involves putting colored bottles of liquid on a pentagram in a specific order, which corresponds to the symbols on the star on the wall. The keys obtained from the demon's mouth spell out the word "HELP," which can be decoded using a letter-number correspondence code (A=1, B=2, C=3, etc.).

Some players also discussed a level involving cards on the wall, where the solution is based on the number and color of the cards. Raeraely mentioned that the cards show 7 blue spades, 2 green clubs, 3 red hearts, and 9 yellow diamonds, which seems to be the answer to the puzzle.

Other discussions involved mathematical calculations, such as 2×8=16 and 24=3×8, which were mentioned by player Maki.

The conversations demonstrate the variety of challenges and solutions in the game Doors & Rooms, as well as the creative approaches and strategies employed by players to overcome these challenges.


Why Johnny Can't Do Mathematics

Applications of linear equations involve word problems. For instance, Joe can paint a garage in 8 hours, and Bill can paint the same garage in 6 hours. If they work together, how long would it take them to paint the garage?

To solve this problem, we need to estimate the answer. We can start by analyzing the rates at which Joe and Bill work. If Joe takes 8 hours to paint the garage, his rate is 1/8 of the garage per hour. Similarly, Bill's rate is 1/6 of the garage per hour. When they work together, their combined rate is the sum of their individual rates: 1/8 + 1/6.

In a class of 72 students, 33 made reasonable guesses, 28 averaged the two rates, and 16 added the two rates. However, to find the correct answer, we need to add the rates and take the reciprocal of the sum.

Another example of a word problem is: a number is decreased by 12, and the result is 15. We can translate this problem into an equation: x - 12 = 15, where x is the unknown quantity. To solve for x, we can add 12 to both sides of the equation, resulting in x = 27.

John has a total of $1.60 in nickels and dimes. If he has twice as many nickels as dimes, we can set up an equation to represent the total amount of money: 0.05x + 0.10y = 1.60, where x is the number of nickels and y is the number of dimes. Since John has twice as many nickels as dimes, we can write x = 2y. Substituting this into the first equation, we get 0.05(2y) + 0.10y = 1.60. Simplifying the equation, we get 0.10y + 0.10y = 1.60, which becomes 0.20y = 1.60. Solving for y, we get y = 8, and substituting this back into x = 2y, we get x = 16.

To find the Greatest Common Factor (GCF) of two numbers, we can use the Prime Factors Method or the Euclidean Algorithm. For example, to find the GCF of 42 and 70 using the Prime Factors Method, we can factorize the numbers: 42 = 2 * 3 * 7 and 70 = 2 * 5 * 7. The GCF is the product of the common prime factors, which is 2 * 7 = 14.

Using the Euclidean Algorithm, we can find the GCF of 42 and 70 by dividing the larger number by the smaller number and taking the remainder: 70 ÷ 42 = 1 with remainder 28, 42 ÷ 28 = 1 with remainder 14, and 28 ÷ 14 = 2 with remainder 0. The GCF is the last divisor, which is 14.

To find the Least Common Multiple (LCM) of two numbers, we need to find the smallest multiple that is common to both numbers. For example, to find the LCM of 4 and 6, we can list the multiples of each number: multiples of 4 are 4, 8, 12, 16, 20, 24, and multiples of 6 are 6, 12, 18, 24, 30, 36. The common multiples are 12, 24, 36, and the smallest of these is 12.

In a problem-solving scenario, a waitress has every fourth day off, and her friend has every sixth day off. They are both off today, and we need to find out how long it will be before they have the same day off again. This problem can be solved by finding the LCM of 4 and 6, which is 12. Therefore, they will have the same day off again in 12 days.

The art of problem-solving involves reading, reacting, and responding to mathematical problems. When solving equations, we need to follow the order of operations (PEMDAS) and use properties such as the zero product property. For example, to solve the equation (x + 2)(x - 3) = 0, we can set each factor equal to zero and solve for x: x + 2 = 0 or x - 3 = 0, resulting in x = -2 or x = 3.

In another example, we need to solve the equation x^2 = x. This is a quadratic equation, and we can factor it as x(x - 1) = 0. Setting each factor equal to zero, we get x = 0 or x - 1 = 0, resulting in x = 0 or x = 1.

To perform operations, we need to follow the order of operations (PEMDAS). For example, to evaluate the expression 5 - 3[4 - (2 - 9)], we need to follow the order of operations: first, evaluate the expression inside the parentheses: 2 - 9 = -7, then evaluate the expression inside the brackets: 4 - (-7) = 11, and finally, multiply and subtract: 5 - 3(11) = 5 - 33 = -28.

In a geometric problem, we need to determine if a line intersects an ellipse. The equation of the ellipse is 9x^2 + 4y^2 = 36, and the equation of the line is y = 1/2x + 3. To determine if the line intersects the ellipse, we can substitute the equation of the line into the equation of the ellipse and solve for x. If the resulting equation has real solutions, then the line intersects the ellipse.

Finally, in a problem that requires critical thinking, we need to determine the color of a bear. The problem states that a person travels 3 miles south, then 1 mile east, and finally 3 miles north, arriving back at the starting point. Since the person ends up back at the starting point, the bear must be white, as the person is at the North Pole.