RS Aggarwal Solutions for Class 6 is a comprehensive chapter-wise solution for Mathematics subjects, available for free download in PDF format. This book, created by experts, follows CBSE guidelines, making it easier for students to grasp fundamental concepts and score well. The solutions provide step-by-step explanations, enabling students to understand and apply mathematical concepts effectively.

**Syllabus**

The syllabus is divided into 24 chapters, each focusing on a specific area of mathematics.

**Chapter 1: Number Systems**
A number system is a set of real numbers that can be represented on a number line. Real numbers include all rational and irrational numbers, such as 1,000,000, which is equal to 10,00,000.

**Chapter 2: Factors and Multiples**
Factors are the exact divisors of a number, while multiples are the products of a number and an integer. For example, the factors of 8 are 1, 2, 4, and 8, and its multiples are also 1, 2, 4, and 8.

**Chapter 3: Whole Numbers**
Whole numbers are a set of numbers that can be written without a fractional component, represented as "W". Examples include 0, 3, 5, and -135. All natural numbers are also whole numbers.

**Chapter 4: Integers**
Integers are whole numbers that can be positive, negative, or zero, without a fractional component or decimal. Examples include 1, 2, 3, 0, -3, -4, and -5.

**Chapter 5: Fractions**
A fraction represents a numerical value of a whole in parts. For example, if an apple is cut into 4 parts and a child takes 1 part, the fraction is 1/4.

**Chapter 6: Simplification**
Simplification involves simplifying complicated expressions into a simple or direct answer. The BODMAS formula is used to solve expressions, such as 23 + 7 x 3 = 23 + 21 = 44.

**Chapter 7: Decimals**
Decimals are fractional components represented in the form of decimals (points). For example, 1/6 can be represented as 0.1 in decimal notation.

**Chapter 8: Algebraic Expressions**
Algebraic expressions are sets of variables, mathematical operations, and constants. Variables have different values, and mathematical operations like addition, multiplication, subtraction, and division can be performed on them.

**Chapter 9: Linear Equations in One Variable**
Linear equations in one variable are equations that can be represented in the form ax + b = 0.

**Chapter 10: Ratio, Proportion, and Unitary Method**
A ratio is the comparison of two numbers by division. The unitary method involves finding the value of a unit and then the values of other units. For example, if the cost of 4 cans of juice is 200, the cost of 6 cans can be calculated using the unitary method.

**Chapter 11: Line Segment, Ray, and Line**
A line is drawn by two points, while a ray is a mixture of a line and a segment with an infinite terminating end.

**Chapter 12: Parallel Lines**
Lines are said to be parallel when two lines are in the same plane and do not intersect with each other.

**Chapter 13: Angles and Their Measurements**
The measurement between two arms is called an angle. There are six types of angles: acute, obtuse, right, reflex, full, and straight.

**Chapter 14: Constructions**
Constructions involve drawing different geometrical structures, such as drawing a ray AB and keeping the protector with its central point at C.

**Chapter 15: Polygons**
Polygons are made up of straight lines and connected by lines. Examples include triangles, pentagons, quadrilaterals, and hexagons.

**Chapter 16: Triangles**
Triangles are classified into groups based on their sides and angles.

**Chapter 17: Quadrilaterals**
A polygon with four sides and four vertices is called a quadrilateral. Examples include rectangles and squares.

**Chapter 18: Circles**
A circle is a closed curve that is not a polygon. Examples include balls, bangles, and coins.

**Chapter 19: Three-Dimensional Shapes**
Three-dimensional shapes are shapes that can be picked up, moved around, and touched. Examples include cubes, prisms, spheres, cones, and cylinders.

**Chapter 20: Two-Dimensional Reflection Symmetry**
Reflection symmetry is the symmetry concerning reflection. In two-dimensional reflection symmetry, a line or an axis of symmetry is formed.

**Chapter 21: Concept of Perimeter and Area**
The perimeter is the measurement of the length of the outline or boundary of an object or shape. The area is the space occupied by the object or shape, measured in square inches.

**Chapter 22: Data Handling**
Data handling involves collecting numbers to get information. This collection can be performed in various ways, including tally marks and pictographs.

**Chapter 23: Pictograph**
A pictograph is used to represent information in the form of objects or pictures. It is the study of numerical data in the form of pictographs.

**Chapter 24: Bar Graph**
A bar graph is a type of graphical representation of its length proportional to its value. It can be determined in a horizontal or vertical method, such as a graphical representation of the population scale of different years.

Regular practice of these chapters will make them easier to understand. RS Aggarwal provides total solutions to problems, helping students solve questions faster. Making notes of formulas and expressions for each chapter will aid in revision and understanding of mathematical concepts.


# The Monty Hall Problem in Python

The Monty Hall problem is a probability puzzle that can be challenging to understand. The problem states that a contestant is presented with three doors, behind one of which is a car, and the other two have goats. The contestant chooses a door, but before it is opened, the game show host opens one of the other two doors, revealing a goat. The contestant is then given the option to switch their choice to the remaining unopened door.

To test the problem, a Python script was created to simulate the Monty Hall problem. The script generates a set of three doors, randomly selects a winning door, and then chooses a door. After the door is chosen, another door is removed, revealing a goat. The script then determines whether switching doors would have resulted in a win or a loss.

The script uses the following functions:
- `generate_doors()`: generates a set of three doors, with one randomly selected as the winning door.
- `get_alternative_door()`: returns the index of the alternative door once one has been chosen and one removed.
- `is_door_winner()`: returns True or False if the selected door is a winning door.
- `monty_cycle()`: simulates the Monty Hall problem and returns True if swapping doors would have won.

The script runs 100 simulations and tracks the results. The output of the script shows the total number of runs, the number of runs that won by swapping, and the number of runs that won by staying. The success rates for swapping and staying are also calculated.

Example output:
```
Total Runs: 100
Runs that won by swapping: 67
Runs that won by staying: 33
Swap success: 67.0 %
Stay success: 33.0 %
```
This output confirms that switching doors gives the contestant a higher chance of winning, with a success rate of approximately 67%. The script provides a clear and concise way to understand and test the Monty Hall problem.

Key concepts:
- The Monty Hall problem is a probability puzzle that involves choosing a door and then switching or staying with the original choice.
- The problem can be simulated using a Python script to generate random doors and track the results.
- Switching doors gives the contestant a higher chance of winning, with a success rate of approximately 2/3.

Important numbers:
- 3: the number of doors in the Monty Hall problem.
- 100: the number of simulations run in the script.
- 67: the number of runs that won by swapping doors.
- 33: the number of runs that won by staying with the original door.
- 67.0%: the success rate for swapping doors.
- 33.0%: the success rate for staying with the original door.


# Mean Calculator

The mean calculator is a tool used to measure the central tendency of a large or small data set. It calculates the average of the data, which can be separated by commas.

## What is the Mean?

The mean is the average value of a given data set, determined by adding all numbers together and then dividing by the total number of values. It is also known as the arithmetic mean. The mathematical notation for the mean is **x̅**, and its formula is:

**x̅ = Σx/n**

Where:
- **Σx** = Sum of all values in the data set
- **n** = Total number of values

## How to Calculate the Mean

To calculate the mean, follow these steps:

1. Write down the total number of values and the data set.
2. Add all the values together and divide by the total number of values.

### Example 1:

Find the mean of the data set **13, 24, 12, 35, 20, 15**.

**Solution:**

1. **n = 6**, data set = **{13, 24, 12, 35, 20, 15}**
2. **x̅ = (13 + 24 + 12 + 35 + 20 + 15) / 6**
= **119 / 6**
= **19.83**

### Example 2:

Determine the mean of the data set **1.9, 1.7, 1.3, 15, 10**.

**Solution:**

1. Arrange the data set from least to greatest: **{1.3, 1.7, 1.9, 10, 15}**, **n = 5**
2. **x̅ = (1.3 + 1.7 + 1.9 + 10 + 15) / 5**
= **29.9 / 5**
= **5.98**


# Image Processing using Matlab/Octave — Part 1

This multipart series covers image processing using Matlab/Octave, including basics of Matlab/Octave, using different filters on images, and converting images into different color models.

## What is Image Processing?

Digital image processing refers to the processing of digital images by a digital computer. It is a subfield of digital signal processing. The process works as follows: the world around us is captured by a camera, processed by a computer, and the output is a processed image.

## What is an Image?

An image is a 2-D matrix (grayscale) or 3-D matrix (colored) represented by a mathematical function f(x,y), where x and y are the two coordinates horizontally and vertically. A matrix is a square arrangement of numbers with rows and columns. In an image, the numbers filling the matrix are intensities.

## Matrix and Pixel

The smallest addressable element in an image is called a pixel. The intensity is the representation of the value of the pixel in an image. A MxN image has M rows, N columns, and MN pixels. The intensity value lies in the range 0-255, where 0 represents black and 255 represents white for a grayscale image.

## Grayscale and Colored Images

A grayscale image has one channel, while a colored image has three channels (RGB). In a grayscale image, a single pixel is represented by 256 unique values (0-255) or 2^8 unique values, with 8 bits per pixel (bpp). A colored image has 24 bits per pixel (8+8+8), representing 16.7 million colors.

## RGB Color Model

The RGB color model is an additive model where red, green, and blue are added together to reproduce a broad array of colors.

## Image Size

The size of an image is calculated as rows*columns*bpp. For example, a 100x100 grayscale image has a size of 100*100*8 = 80,000 bits, which is approximately 78.12 bytes.

## Storing Images

Images are stored in a 3-D format (rows, columns, channels). Sometimes, in machine learning, the number of channels is put before rows, like (channels, rows, columns). There is also a binary image type, represented by only two values: 0 (black) and 1 (white).

## Working with Octave/Matlab

We will be working with Octave/Matlab. Both have similar functionality, but Octave is free, while Matlab is paid. Basic operations such as +, -, /, *, %, <, >, etc. work the same as in most programming languages.

## Matrix Operations

A 2-D matrix is defined by adding semicolons. A 3-D matrix is defined by adding depth to a 2-D matrix. Matrix operations include addition, subtraction, multiplication, and element-wise multiplication.

## Accessing Elements in a Matrix

Indexing in Matlab/Octave starts with 1, not 0. Elements can be accessed by specifying the row and column number or by using a colon to specify a range. Specific rows or columns can be accessed by defining a matrix of the desired elements.

## Basic Operations and Functions

Basic operations and functions in Octave/Matlab include:

*   **eye()** and **diag()** functions for creating diagonal matrices
*   **./** for element-wise division
*   **.** for element-wise calculations

By understanding these concepts, you will be well-prepared to work with image processing in Matlab/Octave. Practice these concepts to improve your skills.


Pythagoras, a 6th century BC philosopher and mathematician, is renowned for the Pythagorean Theorem, which states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b), or a^{2}+b^{2}=c^{2}. This relationship is fundamental in geometry and trigonometry.

Sets of integers (a, b, c) that satisfy the Pythagorean Theorem are known as Pythagorean Triples. Familiar examples of these triples include (3, 4, 5) and (5, 12, 13). Pythagorean Triples can be categorized into two types: trivial and non-trivial. Trivial triples are multiples of smaller triples. For instance, (6, 8, 10) is a trivial triple because it is a multiple of (3, 4, 5), as demonstrated by the equation:

6^{2} + 8^{2} = 10^{2}

This can be further simplified to show the multiplicative relationship:

(2*3)^{2} + (2*4)^{2} = (2*5)^{2}

Which simplifies to:

2^{2} * 3^{2} + 2^{2} * 4^{2} = 2^{2} * 5^{2}

And further to:

2^{2}(3^{2} + 4^{2}) = 2^{2}(5^{2})

Ultimately confirming the relationship:

3^{2} + 4^{2} = 5^{2}

The study and generation of Pythagorean Triples can be facilitated by computational tools. A JavaScript applet can generate these triples in increasing order, allowing users to explore both trivial and non-trivial triples. This applet provides a useful resource for students and mathematicians alike, enabling the discovery of Pythagorean Triples of significant size.

Credits for the development and modification of the Pythagorean Triple Calculator include:
- The original Java Applet by Nick Exner (1998)
- Modifications by Michael McKelvey (2001)
- A JavaScript and HTML5 remake by Evan Ramos (2014)

Additional resources for calculator programs compatible with the TI-83 and TI-89, written by James P. Dildine in 1999, are available for those interested in further exploration of Pythagorean Triples using graphing calculators.


Graphs of Sine and Cosine

Graph the function f(x) = cos(x) + 3. The addition of 3 moves the parent function up 3 units, relocating the sinusoidal axis to 3.

Graph the function f(x) = cos(x + 11) - 9. The addition of 11 inside the parenthesis shifts the graph 11 units to the left. The subtraction of 9 outside the parenthesis moves the graph down 9 units, positioning the sinusoidal axis at -9.

Graph the function f(x) = 3cos(1/3(x + 1)) + 4. To find the period, divide 2π by the B value (1/3), resulting in a period of 6π. The coefficient 3 increases the amplitude to 3. The addition of 1 inside the parenthesis shifts the graph 1 unit to the left, and the addition of 4 outside the parenthesis moves the graph up 4 units, placing the sinusoidal axis at 4.

Now, apply this knowledge to a real-world problem: A man dives off a 12 ft diving board, and his trajectory is modeled by f(x) = 8cos(2x) + 12. Graph his trajectory into and out of the water. If the pool is 6 ft deep, did the man hit the bottom?

For more examples of graphing sine and cosine, visit http://www.youtube.com/watch?v=s_NI50p-pcg.


# Normal Line and Two Flat Mirrors at Right Angles

When two plane mirrors are at right angles to each other, a ray of light reflects from one mirror into the other and back out again. The outgoing reflected ray is always parallel to the incident ray. This can be demonstrated using the law of reflection and geometry.

To prove this, we start by drawing normals at the points where the ray touches the mirrors and extend them to meet each other. By adding the relevant angles, we observe that angles A and B are equal to `90^@ - (minus) theta`. Regardless of the angle `theta`, the incident ray and reflected ray will always be parallel.

This principle can be illustrated with different angles, as shown in the following examples:

**Example 1**
Angle A equals angle B and equals `90^@ - (minus) theta`, demonstrating the parallel relationship between the incident and reflected rays.

**Example 2**
Similarly, angle A equals angle B and equals `90^@ - (minus) theta`, further confirming that the incident and reflected rays are parallel, regardless of the angle `theta`.


## Description

The text consists of multiple problems related to probability and statistics.

1. Chapter 12, problems 17 and 20, and Chapter 20, problems 11 and 15, are mentioned but not elaborated upon.

2. A random vector (X, Y, Z) is given, where Y and Z are marginally Bernoulli random variables. The conditional probability of Y given X is expressed as P(Y = 1|X) = e^(α+XTβ) / (1 + e^(α+XTβ)). The conditional probability of Z given X and Y is P(Z = 1|X, Y) = π(X)φ(Y). It is shown that P(Y = 1|X, Z = 1) = e^(α∗+XTβ) / (1 + e^(α∗+XTβ)), and an expression for α∗ is provided.

3. A properly normalized density function f(x) on the real line and its natural exponential family f(x|θ) = f(x)e^(θx-ϕ(θ)) are considered.
(a) For f(x|θ) to be a well-defined density function, certain conditions must be met.
(b) The choice of ϕ(θ) that makes f(x|θ) a properly normalized density function is such that ∫f(x|θ)dx = 1.
(c) With the choice of ϕ in (b), it is shown that ϕ'(θ) = ∫xf(x|θ)dx and ϕ''(θ) = ∫(x - ϕ'(θ))^2 f(x|θ)dx, where ϕ' and ϕ'' are the first and second derivatives of ϕ. If ϕ'(θ) is a monotone strictly increasing function, the variance is determined by its mean for natural exponential families.

4. A dataset and a generalized linear model logit^-1{P(Y = 1|X)} = β1X are given. The maximum likelihood estimate of β1 is reported, and it is noted that the estimate may not make sense due to the small sample size. A plot of the log-likelihood as a function of β1 is mentioned to explain this phenomenon. A sufficient and necessary condition for the MLE of logistic regression to exist is provided.

The problem serves as a warning for future work, highlighting the importance of checking the identifiability of parameters in a model, especially when the model is complex or the sample size is small.


To find the derivative of f(x) = kx + b using the definition of a derivative, we start with the formula:

f'(x) = lim(as h goes to 0) ((k(x+h) + b) - (k(x) + b))/h

Expanding the equation gives:

f'(x) = lim(as h goes to 0) ((kx + kh + b) - (kx + b))/h

Simplifying further:

f'(x) = lim(as h goes to 0) (kx + kh + b - kx - b)/h

This simplifies to:

f'(x) = lim(as h goes to 0) (kh)/h

Since h is approaching 0, we can cancel out the h terms:

f'(x) = k

This result shows that the derivative of f(x) = kx + b is f'(x) = k.

Note: The key steps involve expanding the given function, simplifying the expression, and applying the limit as h approaches 0 to find the derivative.


A national organization has been working with utilities throughout the nation to find sites for large wind machines that generate electricity. Wind speeds must average more than 22 miles per hour (mph) for a site to be acceptable. Recently, the organization conducted wind speed tests at a particular site. Based on a sample of n=33 wind speed recordings (taken at random intervals), the wind speed at the site averaged 22.8 mph, with a standard deviation of 4.3 mph. To determine whether the site meets the organization's requirements, perform the following hypothesis test at the 1% significance level and state your conclusion.
Ho: μ = 22
HA: μ > 22

The correct conclusion is:
a) At α = .01, there is sufficient evidence to conclude the true mean wind speed at the site exceeds 22 mph.
b) At α = .01, there is not sufficient evidence to conclude the true mean wind speed at the site exceeds 22 mph.
c) We are 99% confident that the site meets the organization's requirements.
d) We are 99% confident that the site does not meet the organization's requirements

Given the sample mean (22.8 mph) is greater than the required mean (22 mph), and considering the sample size (n=33) and standard deviation (4.3 mph), the hypothesis test will likely provide evidence to support the alternative hypothesis (HA: μ > 22) at the 1% significance level. Therefore, the correct answer is:
a) At α = .01, there is sufficient evidence to conclude the true mean wind speed at the site exceeds 22 mph.


Three circles with radii 1, 2, and 3 ft are externally tangent to each other. The task is to find the area of the sector of the circle with a radius of 1 ft that is cut off by the line segments joining its center to the centers of the other two circles.

Connecting the centers of the three circles forms a triangle. The length of the sides of this triangle is the sum of the radii of the tangent circles. This results in a right triangle, which can be proven using the Pythagorean Theorem.

The right angle of the triangle is opposite the hypotenuse, which is 5 ft. Consequently, the angle of the sector for the circle with a radius of 1 ft is also a right angle, or 90º. If the triangle were not a right triangle, the Cosine Law would be used to find the angle opposite the hypotenuse.

The area of the sector of a circle can be calculated using the formula: Area = (1/2) * r^2 * θ, where r is the radius of the sector and θ is the angle of the sector in radians. The angle θ in radians is calculated as θ = 90º * (π / 180º) = π/2 radians.

Therefore, the area of the sector is Area = (1/2) * (1 ft)^2 * (π/2) = (1/2) * 1 * (π/2) = π/4 square feet.


To find the optimal window size for a sender, we consider the formula N = 1 + 2a, where N is the sender window size and a = Tp / Tt. Given that the Round-Trip Time (RTT) between two points, x and y, is 60 ms (or 60 * 10^-3 seconds) and the bandwidth of the link between X and Y is 512 KBps (or 512 * 10^3 Bytes/seconds), we can calculate the total data sent from x to y.

Since it's a full-duplex channel, the total data sent is equal to 2 * Tp * Bandwidth, which is also equal to RTT * bandwidth. Therefore, the total data sent is (60 * 10^-3) * (512 * 10^3) = 60 * 512 = 30720 Bytes.

Given a packet size of 64 bytes, the number of windows required is 30720 / 64 = 480. The optimal window size, N, is then 1 + 2a = 1 + 480 = 481.

In the case of optimal window size, maximum throughput should be achieved, meaning link utilization is 100%. The window size can also be defined as the number of frames sent in 1 RTT, which is in terms of packets. Therefore, the optimal window size is 1 greater than the number of frames sent in 1 RTT.

Key points to note:
- Optimal window size formula: N = 1 + 2a
- RTT between x and y: 60 ms
- Bandwidth of the link: 512 KBps
- Total data sent: 30720 Bytes
- Packet size: 64 bytes
- Number of windows required: 480
- Optimal window size: 481

This information is crucial for understanding how to achieve maximum throughput and 100% link utilization in a full-duplex channel.


**NTSGrad Spring 2018/Abstracts**

## Jan 23
### Solly Parenti
#### Rankin-Selberg L-functions
What do you get when you cross an Eisenstein series with a cuspform? An L-function! Since there's no modular forms course this semester, I will try to squeeze in an entire semester's course on modular forms during the first part of this talk, and then I'll explain the Rankin-Selberg method of establishing analytic continuation of certain L-functions.

## Jan 30
### Wanlin Li
#### Intersection Theory on Modular Curves
My talk is based on the paper by François Charles with title "FROBENIUS DISTRIBUTION FOR PAIRS OF ELLIPTIC CURVES AND EXCEPTIONAL ISOGENIES". I will talk about the main theorem and give some intuition and heuristic behind it. I will also give a sketch of the proof.

## Feb 6
### Dongxi Ye
#### Modular Forms, Borcherds Lifting and Gross-Zagier Type CM Value Formulas
During the course of past decades, modular forms and Borcherds lifting have been playing an increasingly central role in number theory. In this talk, I will partially justify these by discussing some recent progress on some topics in number theory, such as representations by quadratic forms and Gross-Zagier type CM value formulas.

## Feb 20
### Ewan Dalby
#### The Cuspidal Rational Torsion Subgroup of J_0(p)
I will define the cuspidal rational torsion subgroup for the Jacobian of the modular curve J_0(N) and try to convince you that in the case of J_0(p) it is cyclic of order (p-1)/gcd(p-1,12).

## Feb 27
### Brandon Alberts
#### A Brief Introduction to Iwasawa Theory
A bare bones introduction to the subject of Iwasawa theory, its main results, and some of the tools used to prove them. This talk will serve as both a small taste of the subject and a prep talk for the upcoming Arizona Winter School.

## Mar 13
### Solly Parenti
#### Do You Even Lift?
Theta series are generating functions of the number of ways integers can be represented by quadratic forms. Using theta series, we will construct the theta lift as a way to transfer modular(ish) forms between groups.

## Mar 20
### Soumya Sankar
#### Finite Hypergeometric Functions: An Introduction
Finite Hypergeometric functions are finite field analogues of classical hypergeometric functions that come up in analysis. I will define these and talk about some ways in which they are useful in studying important number theoretic questions.

## Apr 3
### Brandon Alberts
#### Certain Unramified Metabelian Extensions Using Lemmermeyer Factorizations
We use conditions on the discriminant of an abelian extension K/ℚ to classify unramified extensions L/K normal over ℚ where the (nontrivial) commutator subgroup of Gal(L/ℚ) is contained in its center. This generalizes a result due to Lemmermeyer stating that the quadratic field of discriminant d, ℚ(√d), has an unramified extension M/ℚ(√d) normal over ℚ with Gal(M/ℚ(√d)) = H_8 (the quaternion group) if and only if the discriminant factors d = d_1 d_2 d_3 into a product of three coprime discriminants, at most one of which is negative, satisfying (d_i d_j/p_k) = 1 for each choice of {i, j, k} = {1, 2, 3} and prime p_k | d_k.

## Apr 10
### Niudun Wang
#### Nodal Domains of Maass Forms
Hecke-Maass cusp forms on modular surfaces produce nodal lines that divide the surface into disjoint nodal domains. I will briefly talk about this process and estimate the number of nodal domains as the eigenvalues vary.

## Apr 17
### Qiao He
#### An Introduction to Automorphic Representations
Automorphic representation is a powerful tool to study L-functions. For me, Tate's marvelous thesis is the real beginning of the whole theory. So I will start with Tate's thesis, which is really the automorphic representation of GL_1. Then I will talk about how to generalize Tate's idea to higher dimensions and explain some ideas behind Langlands program. If there is still time left, I will also mention the trace formula and use it to prove the classical Poisson summation formula.

## Apr 23
### Iván Ongay Valverde
#### Definability of Frobenius Orbits and a Result on Rational Distance Sets
In this talk I will present a paper by Héctor Pastén. We will talk about the meaning of definability in a ring and how having a formula that identifies Frobenius orbits can help you show an analogous case of Hilbert's tenth problem (the one asking for an algorithm that tells you if a diophantine equation is solvable or not). Finally, if time permits, we will do an application that solves the existence of a dense set in the plane with rational distances, assuming some form of the ABC conjecture. This last question was proposed by Erdös and Ulam.

## Apr 24
### Brandon Boggess
#### Moving from Local to Global
What do problems over local fields tell us about global problems?

## May 1
### Qiao He
#### An Introduction to Automorphic Representations Part II
Last time I talked about Tate's thesis, which is actually the theory of automorphic representation of GL_1. This time I will continue. First, I will give the definition of automorphic representation, and use Hecke characters and modular forms to motivate the definition. Then I will explain some classical results about automorphic representation, and discuss how automorphic representations are related to L-functions.

## May 8
### Sun Woo Park
#### Parametrization of Elliptic Curves by Shimura Curves
Let f be a weight-2 newform on Γ_0(N). Given a fixed isogeny class of semistable elliptic curves over ℚ, for some N there exists a distinguished element A of the isogeny class such that A is the strong modular curve attached to f. In fact, A is a quotient of J_0(N) by an abelian variety, from which we can obtain a covering map π: X_0(N) → A. Based on Ribet and Takahashi’s paper, I will discuss the properties of the covering map as well as its generalization to Shimura curves.


Decision trees are a type of supervised machine learning algorithm used for categorization or making predictions based on previous sets of questions or choices. They are a form of supervised learning where data is trained and tested on a separate set of data with desired categorization. Decision trees can be used for both classification and regression problems.

A decision tree works like a tree with branches, with a root node at the base and decision nodes representing choices or decisions to be made. These decision nodes are also called leaf nodes, which represent the result of the decisions. Each section of a decision tree is called a "branch." For example, if the question is "Are you a diabetic?" the leaf nodes can be 'yes' or 'no'.

Key terminologies in decision trees include:
- Root node: the base of the decision tree
- Splitting: the process of dividing a node into sub-nodes
- Decision node: when a sub-node is further split into additional sub-nodes
- Leaf node: when a sub-node depicts the possible outcomes and cannot be further split
- Pruning: the process by which sub-nodes of a decision tree are removed
- Branches: the subsection of the decision tree consisting of multiple nodes

Loss functions quantify the losses generated by errors in estimating parameters of a statistical model or using a predictive model. The ultimate goal of machine learning algorithms is to decrease this loss and statistical risk. Loss functions can be different for regression and classification models.

In decision trees, the loss function works on the concept of purity in the split. The primary methods for calculating impurity are Gini and entropy. Cleaner splits result in lower scores. The formula for entropy is calculated on a set of observations, where 'pi' refers to the fraction of observations belonging to a particular value, and 'c' is the number of possible values of the response variable.

Gini impurity can be calculated by subtracting the sum of the squared probabilities of each class from 1. Gini impurity is biased towards bigger partitions and is easy to implement, while information gain is biased towards smaller partitions with multiple values.

Decision trees can be implemented using Python's Scikit-learn library, with the DecisionTreeClassifier class. The criterion parameter can be set to 'gini' or 'entropy' to use these metrics.

In conclusion, decision trees are a bureaucratic approach to classification and regression problems. The use of values can be done to split questions at each choice or branch, ending up in multiple leaf nodes. A small tree can be made easily but may have high variance, while a tall tree with many splits generates better classifications but may be overfitting. Using the right criterion or loss function is crucial to ensure the tree is not overfitted or underfit. Entropy or Gini metrics can be used to split branches accordingly. If one decision tree is not enough, multiple decision trees can be used, which is called a random forest.


A circle and a quadrilateral have the same total number of degrees, which is 360 degrees. A quadrilateral is made up of four straight sides, and since a 4-sided quadrilateral can be divided into two triangles, the total degrees in a quadrilateral is 2 * 180 = 360 degrees. Similarly, a complete circle is made up of 360 degrees.

To find the angle made by the hands of a clock at 4 o'clock, divide the clock into 12 equal parts. The ratio of the time to the total number of parts is 4/12. Cross-multiply this ratio with the ratio of the unknown degrees to the total number of degrees in a circle (x/360) to get 4 * 360 = 12x. Solving for x gives 120 degrees.

A quadrilateral can be made up of various shapes, including rectangles, and its internal angles always total 360 degrees. If one angle is 90 degrees, the remaining angles must total 270 degrees, making at least one of them obtuse.

In contrast, a circle is made up of a continuous curved line, equivalent to 360 degrees. The circle of illumination on the Earth includes 180 degrees of latitude, which remains constant throughout the year, regardless of the Earth's tilt. This is evident during the equinoxes, where illumination ranges from 90 degrees south to 90 degrees north latitude, totaling 180 degrees of latitude.


**1. An accurate clock shows 8 o'clock in the morning. Through how many degrees will the hour hand rotate when the clock shows 2 o'clock in the afternoon?**

**A. 154°**
**B. 180°**
**C. 170°**
**D. 160°**
**Answer: Option B**
**Explanation:** The angle traced by the hour hand in 12 hours = 360°. From 8 to 2, there are 6 hours. Angle traced by the hour hand in 6 hours = (6/12) * 360° = 180°.

**2. A clock is started at noon. By 10 minutes past 5, the hour hand has turned through**
**A. 155°**
**B. 145°**
**C. 152°**
**D. 140°**
**Answer: Option A**
**Explanation:** The angle traced by the hour hand in 12 hours = 360°. Time duration from noon to 10 minutes past 5 = 5 hours 10 minutes. Hence, the angle traced by the hour hand from noon to 10 minutes past 5 = (5/12) * 360° + (10/60) * 30° = 150° + 5° = 155°.

**3. At what time between 7 and 8 o'clock will the hands of a clock be in the same straight line but not together?**
**A. 5 minutes past 7**
**B. 21.82 minutes past 7**
**C. 22 minutes past 7**
**D. 21.818 minutes past 7**
**Answer: Option D**
**Explanation:** The two hands of a clock will be in the same straight line but not together between H and (H+1) o'clock at (22 + 10/11) minutes past H. Here, H = 7. Hands of the clock will point in opposite directions at 21.818 minutes past 7.

**4. At what time between 5.30 and 6 will the hands of a clock be at right angles?**
**A. 44 minutes past 5**
**B. 21.82 minutes past 5**
**C. 21.818 minutes past 5**
**D. 43 minutes past 5**
**Answer: Option C**
**Explanation:** The two hands of a clock will be at right angles between H and (H+1) o' clock at (22 + 10/11) minutes past H 'o clock and (55 - 10/11) minutes past H 'o clock. Let's take H = 5. Hence, the two hands will be at right angles between 5 and 6 at 21.818 minutes past 5.

**5. At what angle are the hands of a clock inclined at 15 minutes past 5?**
**A. 67.5°**
**B. 75°**
**C. 72.5°**
**D. 82.5°**
**Answer: Option A**
**Explanation:** The angle between hands of a clock = |30*H - (11/2)*M|, where H is the hour and M is the minutes. Here, H = 5, M = 15. Hence, the angle = |30*5 - (11/2)*15| = |150 - 82.5| = 67.5°.

**6. At 3:40, the hour hand and the minute hand of a clock form an angle of**
**A. 135°**
**B. 130°**
**C. 120°**
**D. 125°**
**Answer: Option B**
**Explanation:** The angle between hands of a clock = |30*H - (11/2)*M|, where H is the hour and M is the minutes. Here, H = 3, M = 40. Hence, the angle = |30*3 - (11/2)*40| = |90 - 220| = 130°.


## Explanation

A fraction is a value expressed by the quotient of two whole numbers, signifying a part of the whole. To calculate 4/5, divide four units by five, resulting in 0.8 units. The decimal form is created by dividing the numerator by the denominator. The fraction 4/5 is equivalent to fractions like 8/10, 16/20, and the decimal 0.8.

## Key Concepts

A fraction is a division of two numbers, usually integers, known as the numerator and the denominator. Two fractions are equivalent if they generate the same result. There are infinitely many equivalent fractions of a number, obtained by multiplying the numerator and the denominator by the same number.

## Converting Fractions to Decimals

To change a fraction to a decimal, divide the numerator by the denominator. For example, to convert 4/5 to a decimal, divide four by five, resulting in 0.8.

## Converting Decimals to Fractions

To change a decimal to a fraction, write the decimal as a fraction of one, then multiply the numerator and denominator by ten until an integer is obtained in the numerator. Finally, shorten the fraction if it is not in its simplest form.

## Fractional Operations

To solve fractional fractions, multiply the numerators and denominators together. For example, 4/5 * 4 = 16/5 = 3 1/5 = 3.2. The result can be expressed in words as "three and one-fifth" or "sixteenth".

## Principles of Expression with Fractions

Fractions are expressed using a forward slash to divide the numerator by the denominator. Mixed numbers have a space between the whole number and the fraction. The colon (:) is used as the fraction division operator. Decimals are entered with a decimal point and can be automatically converted to fractions.

## Conclusion

To solve fraction problems like 4/5, remember to multiply the numerators and denominators together. This will result in the correct answer, which can be expressed as a fraction, decimal, or mixed number.


Motion is when an object changes its position. It can be represented mathematically through calculations of speed, velocity, and acceleration. Velocity has both speed and direction, and is calculated as velocity = distance + direction / time. Displacement is the distance and direction between the starting and ending position of an object.

For example, a car moved 60 km East and 90 km West. To find the total distance, we add the distances traveled: 60 km + 90 km = 150 km. To find the displacement, we subtract the distance traveled in one direction from the distance traveled in the other direction: 90 km - 60 km = 30 km West.

Similarly, a bike rider biked 50 km North and 30 km South. The total distance is 50 km + 30 km = 80 km, and the displacement is 50 km - 30 km = 20 km North.

Velocity is calculated as distance + direction / time. For instance, if a person heads 60 meters southwest from their room to the track in 20 seconds, their velocity is v = 60 m SW / 20 sec = 3 m/sec SW.

Speed is the distance an object travels in a certain amount of time, and is calculated as speed = distance / time. Average speed is the total distance divided by the total time, while constant speed is a speed that does not change.

Acceleration is a change in the direction or speed of an object over time, and can be caused by unbalanced forces. It is measured in units of m/s/s or km/hr/s. For example, if a car is moving at 70 km/hr due north and then changes direction to northeast while maintaining the same speed, its velocity changes but its speed remains the same.

Force is the cause of motion, and has four components: push, pull, size, and direction. It always comes in pairs, and is measured using a spring scale. One newton is the force required to accelerate a 1 kg mass at a rate of 1 m/s^2.

Net force is the sum of all forces acting on an object. If the forces are in the same direction, they are added together, while if they are in opposite directions, the smaller force is subtracted from the larger force. Balanced forces result in no motion, while unbalanced forces cause motion.

Newton's laws describe the relationship between force and motion. The first law, also known as the Law of Inertia, states that an object at rest will remain at rest, and an object in motion will remain in motion, unless acted upon by an unbalanced force. The second law states that an object acted on by an unbalanced force will accelerate in the direction of the force. The third law states that for every action, there is an equal and opposite reaction.

For example, if a 2000 kg truck and a 500 kg car are racing, and the same amount of force is applied to each of them, the car will accelerate faster than the truck. If a person is running at a speed of 5 m/s and then accelerates to 10 m/s in 2 seconds, their acceleration is 5 m/s^2.

In another example, if a train runs 4200 km in 21 hours, its speed is 4200 km / 21 hr = 200 km/hr. If two people head 60 meters southwest in 20 seconds, their velocity is v = 60 m SW / 20 sec = 3 m/sec SW.

The four components of force are push, pull, size, and direction. Speed is the distance an object travels in a certain amount of time, while acceleration is the rate at which the speed or direction of an object is changing.

To calculate the force needed to accelerate an object, we use the formula F = ma, where F is the force, m is the mass, and a is the acceleration. For example, to accelerate a 2000 kg car at a rate of 8 m/s^2, we need a force of F = 2000 kg * 8 m/s^2 = 16000 N.

In a multiple-choice question, which of the following best describes the difference between speed and acceleration? A) Acceleration is the distance an object travels within a specific unit of time; speed is the measure of the force necessary to change the acceleration of an object. B) Speed is the distance an object travels within a specific unit of time; acceleration is a measure of the force necessary to change the speed of an object. C) Speed is the distance an object travels within a specific unit of time; acceleration is the rate at which the speed or direction of an object is changing.

The correct answer is C) Speed is the distance an object travels within a specific unit of time; acceleration is the rate at which the speed or direction of an object is changing.

Another multiple-choice question: What is the formal name for Newton's 1st Law? A) The Law of Action/Reaction. B) The Law of Inertia. C) The Law of Force and Acceleration.

The correct answer is B) The Law of Inertia.

In another question, a book is sitting on a table at rest. The forces acting on the book include gravity pushing down and the table pushing up, and these forces are balanced.

Tom is teaching Jenny how to ice skate. Both skaters, who are about the same size, are on the ice and wearing their skates. To help Jenny get started, Tom gives her a gentle push from behind. Which of the following best describes what will happen next? A) Jenny will move backward; Tom will move forward. B) Jenny will move forward. C) Jenny will move forward; Tom will move backward.

The correct answer is B) Jenny will move forward.

The four components of force are A) Push, Pull, Speed, Direction. B) Push, Size, Speed, Direction. C) Push, Pull, Size, Direction.

The correct answer is C) Push, Pull, Size, Direction.

How much force is needed to accelerate a 2000 kg car at a rate of 8 m/s^2? A) 24,000 N. B) 4,000 N. C) 16,000 N.

The correct answer is C) 16,000 N.

If Sara's father presses on the brakes quickly, slowing the car down, what slows Sara down? A) The seat pushing forward on her. B) Her seat belt pulling back on her. C) Air resistance on the outside of the car.

The correct answer is B) Her seat belt pulling back on her.

Jodie sets two objects on a flat surface. Object A has a mass of 50g, while object B has a mass of 250g. Jodie pushes both objects with the same force and watches them accelerate. Which object will accelerate more? A) Object B accelerates more. B) Object A accelerates more. C) They accelerate the same amount.

The correct answer is B) Object A accelerates more.

A car is traveling south at a speed of 65 miles per hour and then begins traveling west but continues traveling at the same speed. Which of the following has changed? A) Speed. B) Mass. C) Velocity.

The correct answer is C) Velocity.

The definition of inertia is A) The tendency of a body at rest to stay at rest or of a body in motion to remain in motion. B) The tendency of an object in motion to decelerate. C) The rate at which an object accelerates.

The correct answer is A) The tendency of a body at rest to stay at rest or of a body in motion to remain in motion.

The theories of continental drift and sea-floor spreading combined to make the theory of A) Plate tectonics. B) Rejoined continents. C) Pangaea.

The correct answer is A) Plate tectonics.

The Atlantic Ocean is currently growing wider as its floor enlarges due to the process of sea-floor spreading. In sea-floor spreading, pieces of oceanic crust A) Move sideways against each other along a transform plate boundary. B) Move together along a convergent plate boundary. C) Move apart along a divergent plate boundary.

The correct answer is C) Move apart along a divergent plate boundary.

A tectonic plate applies a downward force onto the asthenosphere, which lies beneath the plate. According to Newton's 3rd Law, which can be described as the law of action-reaction, A) The asthenosphere applies a force of double magnitude upward onto the tectonic plate. B) The asthenosphere applies a force of equal magnitude upward onto the tectonic plate. C) The asthenosphere applies a force of equal magnitude downward as well.

The correct answer is B) The asthenosphere applies a force of equal magnitude upward onto the tectonic plate.

A subduction zone is an area where one tectonic plate sinks underneath another one. Volcanoes often form as a result. Which of the following plate collisions would most likely result in a subduction zone? A) A piece of continental crust and a piece of oceanic crust meeting at a convergent boundary. B) Two pieces of oceanic crust meeting at a divergent boundary. C) A piece of continental crust and a piece of oceanic crust meeting at a divergent boundary.

The correct answer is A) A piece of continental crust and a piece of oceanic crust meeting at a convergent boundary.

Which of the following provides the most evidence for the theory of plate tectonics? A) The spreading of the sea floor. B) The locations of the glaciers. C) The changing of the seasons.

The correct answer is A) The spreading of the sea floor.


## Calculator Program in Scala

In this post, we will explore a Scala program that implements basic calculator operations. The program consists of a `Calculator` class with four methods: `+` (addition), `-` (subtraction), `*` (multiplication), and `/` (division).

The `Calculator` class is defined as follows:
```scala
package com.clearurdoubt

class Calculator {
  def +(a: Int, b: Int): Int = a + b
  def -(a: Int, b: Int): Int = a - b
  def *(a: Int, b: Int): Long = a * b
  def /(a: Int, b: Int): Int = {
    require(b != 0, "denominator cannot be 0")
    a / b
  }
}
```
The `main` method in the `Calendar` object demonstrates the usage of the `Calculator` class:
```scala
object Calendar {
  def main(args: Array[String]) = {
    val calc = new Calculator()
    println("Addition: " + calc.+(10, 2))
    println("Subtraction: " + calc.-(10, 2))
    println("Multiplication: " + calc.*(10, 2))
    println("Division: " + calc./(10, 2))
  }
}
```
The program creates a `Calculator` object and performs basic arithmetic operations using the `+`, `-`, `*`, and `/` methods. Note that the `/` method includes a check to prevent division by zero.


### About Increasing, Decreasing, and Constant Intervals

A function is increasing on an interval if f(a) > f(b) for all a, b in that interval where a > b. Conversely, a function is decreasing on an interval if f(a) < f(b) for all a, b in that interval where a > b.

### Test Objectives
- Demonstrate the ability to find intervals where a function is increasing, decreasing, or constant.

### Problems and Solutions

1. **Instructions**: Find intervals where the function is increasing, decreasing, or constant.
   - **Solution**: Increasing on: (-∞, -5), Decreasing on: (-5, ∞)

2. **Instructions**: Find intervals where the function is increasing, decreasing, or constant.
   - **Solution**: Increasing on: (1, ∞), Decreasing on: (-∞, 1)

3. **Instructions**: Find intervals where the function is increasing, decreasing, or constant.
   - **Solution**: Increasing on: (-1, 1), Decreasing on: (-∞, -1) ∪ (1, ∞)

4. **Instructions**: Find intervals where the function is increasing, decreasing, or constant.
   - **Solution**: Increasing on: (-∞, ∞)

5. **Instructions**: Find intervals where the function is increasing, decreasing, or constant.
   - **Solution**: Increasing on: (-4, 0) ∪ (4, ∞), Decreasing on: (-∞, -4) ∪ (0, 4)


Today, we're going to solve a problem called "geniuses and chocolates." This problem exercises your knowledge of properties of probability laws.

The problem states that 60% of students are geniuses, 70% of students love chocolate, and 40% fall into both categories. We need to determine the probability that a randomly selected student is neither a genius nor a chocolate lover.

To solve this problem, we can use a Venn diagram, which is a tool that helps us understand how different sets relate to each other and how their corresponding probabilities relate to each other. We draw a rectangle to denote the sample space, and two intersecting circles to represent geniuses and chocolate lovers.

We can label the regions in the Venn diagram as X, Y, Z, and W. X represents students who are neither geniuses nor chocolate lovers, Y represents students who are geniuses but not chocolate lovers, Z represents students who are both geniuses and chocolate lovers, and W represents students who are chocolate lovers but not geniuses.

We know that the probability of a student being a genius is 0.6, the probability of a student being a chocolate lover is 0.7, and the probability of a student being both is 0.4. Using these values, we can set up equations to solve for the probabilities of Y, Z, and W.

We find that the probability of Y is 0.2, the probability of Z is 0.4, and the probability of W is 0.3. Finally, we can use the fact that X, Y, Z, and W form a partition of the sample space to solve for the probability of X.

We have that the probability of the sample space is equal to 1, which is also equal to the sum of the probabilities of X, Y, Z, and W. Plugging in the values we found earlier, we get that the probability of X is 0.1.

Therefore, the probability that a randomly selected student is neither a genius nor a chocolate lover is 0.1. This problem demonstrates the usefulness of Venn diagrams and partitions in solving probability problems.


Take any three-digit number, such as 123. Multiply it by 13 to get 1599, then by 11 to get 17589, and finally by 7 to get 123123. Notice that the original number 123 is repeated twice. This occurs because 13 x 11 x 7 equals 1001, so multiplying by these numbers is equivalent to multiplying by 1001. Any number multiplied by 1001 results in the number repeated twice.

Division by 1001 yields a similar pattern. For example, 123 ÷ 1001 = 0.122877122877... The first three digits are 122, which is 123 - 1, and the next three digits are 877, where 122 + 877 = 999. This pattern continues, but the proof is unknown.

To divide by 1001, subtract 1 from the first three digits and subtract the result from 999 for the next three digits. However, dividing by 1001 may not seem impressive since it is close to 1000. Instead, use the multiplication pattern to divide by 13, then 7, then 11.

Consider the number 33 divided by 91. To make it a division by 1001, multiply the denominator 91 by 11, and also multiply the numerator 33 by 11. This results in 363 ÷ 1001. Using the division pattern, 363 - 1 = 362 and 999 - 362 = 637, so the answer is 0.362637362637...

To divide by 91, multiply the number by 11, subtract 1, and subtract the result from 999. This principle can also be applied to divide by 77 (11 x 7) and 143 (13 x 11) by multiplying the top number by 13 or 7 instead of 11. The 1001 pattern and multiplication by 11 can be combined to create an impressive mental math trick for division by 91.


## The Web as a Graph

The goal is to assign an "importance score" $x_i \in [0,1]$ to each web page $v_i$ in a way that captures the idea of which websites are likely to be important. To achieve this, we need a mathematical description of the internet's structure, which can be represented using graph theory.

**Definition:** A *web* $W$ is a directed graph $(V, E, \varphi)$ with web pages $v_i \in V$, hyperlinks $e_i \in E$, and $\varphi : E \to V \times V$ providing incidence structure. A web has $n$ pages, and incoming links are called *backlinks*. The number of incoming directed edges is the *in-degree* of a vertex, denoted $\textup{in}(v_i)$, while the number of outgoing edges is the *out-degree*, denoted $\textup{out}(v_i)$.

Our analysis of importance within a web relies on the axiom that content creators usually link to websites they find important. The importance score of a web page should be related to the density of incoming links. However, simply counting the in-degree of each vertex is not descriptive enough, as it assigns equal value to every link.

## Crawl, then Walk

A more sophisticated ranking would be to compute a page's importance score as the sum of the scores of pages that link to it, weighted by the out-degree of the corresponding vertex. This can be represented by the equation:

$$\displaystyle x_i = \sum\limits_{j \in S_i} \dfrac{x_j}{\textup{out}(v_j)}$$

where $S_i$ is the set of indices of pages with links to $v_i$. This equation represents a voting system where votes correspond to links within a web.

## Example and Linear Algebra

Consider a web $W$ with four pages. Writing out the equation for each page $v_i$, we get a system of four equations with four unknowns. This system can be rewritten in matrix form as $A \mathbf{x} = \mathbf{x}$, where $A$ is the *link matrix* for $W$. The problem of ranking pages in this web reduces to finding an eigenvector for $A$ corresponding to the eigenvalue 1.

## Assume the Scary Thoughts Away

The link matrix $A$ for a web $W$ with no dangling nodes has 1 as an eigenvalue. This can be proven by showing that $A$ is column-stochastic, meaning that its columns sum to 1. However, it is not the case that these rankings are unique up to scaling. The dimension of the eigenspace corresponding to the eigenvalue 1 can be greater than 1, leading to ambiguity in the ranking.

## Conclusion

We have solved a restricted version of the ranking problem, providing sensible rankings for webs with no dangling nodes and a one-dimensional eigenspace corresponding to the eigenvalue 1. However, this algorithm is not yet suitable for the real world, where the internet has many dangling nodes and disconnected subwebs. Further refinement is needed to create a finished PageRank algorithm.


If $f(x)=(π−2x)_{2}1−sinx$ when $x ≠ π/2$ and $f(π/2) = λ$, then $f(x)$ will be a continuous function at $x = π/2$ if
$\lim_{x→π/2}f(x)=f(π/2)$.
Let's evaluate $\lim_{x→π/2}f(x)$.
Suppose $(π/2 - x) = t$, then
$\lim_{t→0}[(π - 2(π/2 - t))_{2}1 - sin(π/2 - t)] = \lim_{t→0}[(2t)_{2}1 - cos(t)]$.
Using the trigonometric identity $sin^2(x) + cos^2(x) = 1$, we can rewrite $1 - cos(t)$ as $2sin^2(t/2)$.
Thus, $\lim_{t→0}[4t^2 \cdot 2sin^2(t)] = \lim_{t→0}[8t^2 \cdot sin^2(t)]$.
Since $\lim_{t→0} \frac{sin(t)}{t} = 1$, we have $\lim_{t→0} \frac{sin^2(t)}{t^2} = 1$.
Therefore, $\lim_{t→0}[8t^2 \cdot sin^2(t)] = \lim_{t→0}[8t^2 \cdot \frac{sin^2(t)}{t^2} \cdot t^2] = 8 \cdot 1 \cdot 0 = 0$ is incorrect and
$\lim_{t→0}[8t^2 \cdot sin^2(t)] = \lim_{t→0}[8 \cdot (t \cdot sin(t))^2] = 8 \cdot 0^2 = 0$ is also incorrect.
However,
$\lim_{t→0}[(2t)_{2}1 - cos(t)] = \lim_{t→0}[4t^2 \cdot 2sin^2(t/2)] = \lim_{t→0}[8t^2 \cdot sin^2(t/2)]$.
Using the identity $sin^2(x) = \frac{1 - cos(2x)}{2}$,
$\lim_{t→0}[8t^2 \cdot \frac{1 - cos(t)}{2}] = \lim_{t→0}[4t^2 \cdot (1 - cos(t))] = \lim_{t→0}[4t^2 \cdot 2sin^2(t/2)]$.
Since $\lim_{t→0} \frac{sin(t)}{t} = 1$,
$\lim_{t→0} \frac{sin(t/2)}{t/2} = 1$ and $\lim_{t→0} \frac{sin^2(t/2)}{(t/2)^2} = 1$.
Thus, $\lim_{t→0}[4t^2 \cdot 2 \cdot \frac{sin^2(t/2)}{(t/2)^2} \cdot (\frac{t}{2})^2] = \lim_{t→0}[4t^2 \cdot 2 \cdot 1 \cdot (\frac{t}{2})^2] = \lim_{t→0}[4t^2 \cdot \frac{t^2}{2}] = \lim_{t→0}[2t^4] = 0$ is incorrect.
However,
$\lim_{x→π/2}f(x) = \lim_{t→0}[(2t)_{2}1 - cos(t)] = \lim_{t→0}[4t^2 \cdot 2sin^2(t/2)] = \lim_{t→0}[8t^2 \cdot \frac{1 - cos(t)}{2}]$.
Since $\lim_{t→0} \frac{1 - cos(t)}{t^2} = \frac{1}{2}$,
$\lim_{t→0}[8t^2 \cdot \frac{1 - cos(t)}{2}] = \lim_{t→0}[4t^2 \cdot \frac{1 - cos(t)}{t^2} \cdot t^2] = 4 \cdot \frac{1}{2} \cdot 0 = 0$ is incorrect.
However,
$\lim_{t→0}[4t^2 \cdot \frac{1 - cos(t)}{t^2}] = \lim_{t→0}[4 \cdot \frac{1 - cos(t)}{t^2} \cdot t^2] = 4 \cdot \frac{1}{2} = 2$.
Thus,
$\lim_{x→π/2}f(x) = \lim_{t→0}[4t^2 \cdot \frac{1 - cos(t)}{t^2}] = 2$.
For $f(x)$ to be continuous at $x = π/2$,
$\lim_{x→π/2}f(x) = f(π/2)$.
Therefore, $λ = f(π/2) = \lim_{x→π/2}f(x) = \frac{1}{2}$.
Hence, $f(x)$ will be continuous at $x = π/2$ if $λ = \frac{1}{2}$.
The correct answer is $\frac{1}{2}$.


### How do Microorganisms Swim?

#### Overview

As a microorganism, such as a bacterium or sperm cell, the world is experienced differently due to its small size (1µm tall) and the unique physics at this scale. Unlike humans, who are familiar with Newton's second law ($F = ma$), microorganisms live in an inertialess world where forces balance out constantly, making the acceleration term negligible.

Many microorganisms swim, and their movement can be understood through the Navier-Stokes equation, which governs fluid motion. However, the acceleration term in this equation is not a concern for microorganisms, simplifying the governing equations. This allows for the use of techniques learned over the last three years to analyze how microorganisms swim.

#### Project Structure

The project can be structured as follows:

- Introduction to Stokes flow, including its derivation as a limit of the Navier-Stokes equations
- Solutions to Stokes flow using Green's functions
- Exploring methods to solve the Stokes equations for microorganism shapes
- Coding a simple microorganism model using Python (for those interested)

#### Prerequisites

This applied math project involves computational and analytical work, requiring some familiarity with Python, but not expertise in numerical analysis. While not essential, taking Fluid Mechanics III and Mathematical Biology III can be useful for understanding Green's functions. The governing equations and physics intuition will be derived and learned together.

#### Resources

- "Life at low Reynolds number" (1976)
- "Small particles in a viscous fluid" course notes (2014)
- "A Physical Introduction to Suspension Dynamics" textbook (2012)
- Stokesian Dynamics online game (to understand the physics)


# Algebra 1

**Session 1 Review: (Chapters 1 & 2)**
- Perform arithmetic operations on integers and fractions.
- Apply PEMDAS (Order of Operations).
- Understand necessary vocabulary.
- Solve two-step and multi-step equations and inequalities.
**Quiz**

**Session 2: Functions (Chapters 3 & 4)**
- Define a function and identify whether data represents a function.
- Graph linear and quadratic functions.
- Evaluate functions from graphs.
- Identify inputs and outputs.
- Introduce domain and range.
**Quiz**

**Session 3: Linear Functions (Chapters 3 & 4)**
- Identify and calculate linear functions, including slope (growth rate) and y-intercept (initial value).
- Graph linear functions using rise/run.
- Calculate slope.
**Quiz**

**Session 4: Linear Equations (Chapters 3 & 4)**
- Use Slope-Intercept Form and Point-Slope Form.
- Graph linear functions.
- Identify parallel and perpendicular functions.
**Quiz**

**Session 5: System of Equations (Chapter 5)**
- Solve systems by graphing, substitution, and elimination.
- Identify the point of intersection.
**Quiz and Homework**

**Session 6: Exponential Functions and Sequences (Chapter 6)**
- Apply properties of exponents.
- Use radicals and rational exponents.
- Solve exponential equations.
**Quiz**

**Session 7-8: Quadratic Functions and Factoring (Chapters 7, 8, & 9)**
- Formulate quadratic functions in intercept and vertex form.
- Factor quadratic functions.
- Graph quadratic functions.
**Quiz**


## Step 1: Analyzing the Problem Statements
The provided text contains multiple problem statements related to linear algebra, including vector spaces, spanning sets, basis, linear transformations, and matrices. Each problem requires specific solutions or proofs.

## Step 2: Identifying Key Concepts and Removing Irrelevant Information
Key concepts include vector spaces, spanning sets, basis, linear independence, linear transformations, and matrix operations. Irrelevant information, such as redundant explanations or unnecessary examples, will be removed to refine the text.

## Step 3: Reorganizing Content for Clarity and Conciseness
The refined text will be organized by topic, with clear headings and concise explanations. Problem statements will be preserved, and solutions will be provided in a step-by-step format where applicable.

## Step 4: Preserving Original Length and Refining Multiple Choice Questions
The refined text will maintain the original length as much as possible, focusing on clarity and conciseness. Multiple choice questions will be refined to ensure they are clear and relevant to the topic.

## Step 5: Outputting the Refined Text
Given the nature of the task, the output will focus on presenting the refined problem statements and solutions in a clear and organized manner.

Let $W$ be the set of $3\times 3$ skew-symmetric matrices. To show that $W$ is a subspace of the vector space $V$ of all $3\times 3$ matrices, we must verify that $W$ is closed under addition and scalar multiplication.

A set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^3$. The question of whether $S_2=\{\mathbf{v}_1\}$ can be a spanning set for $V$ depends on the linear independence of the vectors in $S_1$.

Consider a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ that is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then the set $S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}$ is still a spanning set for $V$ because it contains all the vectors necessary to span $V$.

Given $A=\begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 &3 & 0 & 5 \\ 1 & 1 & 6 & 3 \end{bmatrix}$, to find a basis for the nullspace of $A$, we solve the equation $A\mathbf{x} = \mathbf{0}$.

To find a basis for the row space of $A$, we consider the rows of $A$ as vectors and find a linearly independent subset of these vectors.

For the range of $A$, a basis consisting of column vectors of $A$ can be found by identifying the linearly independent columns of $A$.

Let $A$ be an $n\times n$ nonsingular matrix, and let $\mathbf{v}, \mathbf{w}$ be linearly independent vectors in $\R^n$. To prove that $A\mathbf{v}$ and $A\mathbf{w}$ are linearly independent, we assume $c_1A\mathbf{v} + c_2A\mathbf{w} = \mathbf{0}$ and show that $c_1 = c_2 = 0$.

Consider vectors $\mathbf{a}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}, \mathbf{b}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \mathbf{c}=\begin{bmatrix} 2 \\ 3 \end{bmatrix}, \mathbf{d}=\begin{bmatrix} 3 \\ 2 \end{bmatrix}, \mathbf{e}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \mathbf{f}=\begin{bmatrix} 5 \\ 6 \end{bmatrix}$ in $\R^2$. For each set $S$, we determine whether $\Span(S)=\R^2$ and provide an algebraic description and geometric shape of $\Span(S)$ if $\Span(S)\neq \R^2$.

Let $A$ be a $3\times 3$ matrix with $A\mathbf{v}=-\mathbf{v}$ and $A\mathbf{w}=2\mathbf{w}$ for $\mathbf{v}=\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$ and $\mathbf{w}=\begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$. We find $A^5\begin{bmatrix} -1 \\ 8 \\ -9 \end{bmatrix}$ using the properties of linear transformations and matrix exponentiation.

The vector space $\mathrm{P}_n$ of polynomials of degree at most $n$ has a standard basis $B = \{1, x, x^2, \cdots, x^n\}$. The map $T : \mathrm{P}_n \rightarrow \mathrm{P}_{n+1}$ defined by $T(f)(x) = xf(x)$ is a linear transformation. We prove this and find its range and nullspace.

For an $n \times m$ matrix $M$ composed of column vectors $\mathbf{b}_1, \cdots, \mathbf{b}_m$, a vector $\mathbf{v} \in \R^n$ can be written as a linear combination of the column vectors if and only if there is a vector $\mathbf{x}$ that solves $M\mathbf{x} = \mathbf{v}$.

In $P_2$, the vector space of all polynomials of degree $2$ or less, let $S=\{p_1(x), p_2(x), p_3(x)\}$ where $p_1(x)=x^2+1, p_2(x)=6x^2+x+2, p_3(x)=3x^2+x$. Using the basis $B=\{x^2, x, 1\}$, we prove that $S$ is a basis for $P_2$ and find the coordinate vector of $p(x)=x^2+2x+3$ with respect to the basis $S$.

Given a vector space $V$ over a scalar field $K$ and vectors $\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k$ in $V$, the subset $W=\{a_1\mathbf{v}_1+a_2\mathbf{v}_2+\cdots+ a_k\mathbf{v}_k \mid a_1, a_2, \dots, a_k \in K \text{ and } a_1+a_2+\cdots+a_k=0\}$ consists of linear combinations of $\mathbf{v}_1, \dots, \mathbf{v}_k$ where the sum of the coefficients is zero.


The following are examples of 6th Grade Math Word Problems that deal with ratio and proportions, solved using tape diagrams, block diagrams, or bar models (Singapore Math).

**Example 1:**
Mark and Fred had money in the ratio 6:1. Mark gave half of his money to Fred. Find the ratio of the amount of money Mark had left to the amount of money Fred had in the end.

**Example 2:**
Carol puts green and red unit cubes in a box with a ratio of 2:1. She adds 12 more red cubes, and the ratio becomes 4:5.
a) How many green cubes are there in the box?
b) How many red cubes does Carol have in the end?

To solve this problem, we can use the following steps:
1 unit = 12 ÷ 3 = 4 cubes
4 units = 4 × 4 = 16 cubes
There are 16 green cubes in the box.
5 units = 5 × 4 = 20 cubes
Carol has 20 red cubes in the end.

**How to Solve Ratio Word Problems:**

1. A board was cut into two pieces with lengths in the ratio 2:5. The longest piece was 85 inches. How long was the shortest piece?
2. At a track meet, 5/8 of the students were boys. There were 48 more boys. How many students were there altogether?
3. Theresa read two and one-fourth times as many pages on Saturday as on Sunday. She read 120 pages more on Saturday. How many pages did she read altogether?

**How to Solve More Difficult Ratio Word Problems:**

1. The ratio of pigs to cows to sheep on the farm is 2:4:7. There are 65 more sheep than pigs. How many cows are there on the farm?
2. Archimedes, Hypatia, and Zeno shared a sum of money. Archimedes received 1/6 of the money. Hypatia and Zeno shared the rest in the ratio 2:3. Zeno got $7 more than Hypatia. How much money did they share altogether?
3. One-third of Elmer's age is the same as two-fifths of Sam's age.
a) Draw a bar model representing the situation.
b) Elmer is 5 years older than Sam. How old is Sam?

**How to Use Proportions to Solve Ratio Problems:**

1. At Al's Catering Service, the ratio of cups of apples to cups of oranges in the recipe is 8:5.
a) If they use 15 cups of oranges, how many cups of apples should they use?
b) If they only use 2 cups of apples, how many cups of oranges should they use?

Related Topics: More Math Word Problems, Algebra Word Problems, More Singapore Math Word Problems.


**Math Standards**

The math standards covered include representing and solving problems involving multiplication and division.

**Represent and Solve Problems Involving Multiplication and Division**

* **MGSE3.OA.1**: Interpret products of whole numbers. For example, interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each. Describe a context where a total number of objects can be expressed as 5 x 7.
* **MGSE3.OA.2**: Interpret whole number quotients of whole numbers. For example, interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each. Describe a context where a number of shares or a number of groups can be expressed as 56 ÷ 8.
* **MGSE3.OA.3**: Use multiplication and division within 100 to solve word problems involving equal groups, arrays, and measurement quantities. Use drawings and equations with a symbol for the unknown number to represent the problem.
* **MGSE3.OA.4**: Determine the unknown whole number in a multiplication or division equation relating three whole numbers using the inverse relationship of multiplication and division. For example, determine the unknown number that makes the equation true in each of the equations: 8 × ? = 48, 5 = ? ÷ 3, 6 × 6 = ?.

**Understand Properties of Multiplication and the Relationship Between Multiplication and Division**

* **MGSE3.OA.5**: Apply properties of operations as strategies to multiply and divide. Examples include:
	+ Commutative property of multiplication: If 6 × 4 = 24, then 4 × 6 = 24.
	+ Associative property of multiplication: 3 × 5 × 2 can be found by 3 × 5 = 15, then 15 × 2 = 30, or by 5 × 2 = 10, then 3 × 10 = 30.
	+ Distributive property: Knowing that 8 × 5 = 40 and 8 × 2 = 16, one can find 8 × 7 as 8 × (5 + 2) = (8 × 5) + (8 × 2) = 40 + 16 = 56.
* **MGSE3.OA.6**: Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8. Conversations should also include connections between division and subtraction.
* **MGSE3.OA.7**: Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division. For example, knowing that 8 × 5 = 40, one knows 40 ÷ 5 = 8.

**Important Reminder**
Always review addition, subtraction, and multiplication facts with your child.


**Lesson 15: Writing Systems of Equations**

**15.1: How Many Solutions?**

This activity connects algebraic representations of systems of equations to the number of solutions. Students recognize that the number of solutions can be determined without solving the system, using methods such as slope, y-intercept, or equation manipulation.

**Systems of Equations:**

1. \(\begin{cases} y=\frac{4}{3}x+4 \\ y = \frac{4}{3}x-1 \end{cases}\)
2. \(\begin{cases} y=4x-5 \\ y = 2x+7 \end{cases}\)
3. \(\begin{cases} 2x+3y = 8 \\ 4x+6y = 17 \end{cases}\)
4. \(\begin{cases} y= 5x-15 \\ y= 5(x-3) \end{cases}\)

**Number of Solutions:**

* No solutions
* One solution
* Infinitely many solutions

**15.2: Situations and Systems**

Students set up systems of equations for real-world scenarios and interpret the solutions in context.

**Scenarios:**

1. Lin's family bike ride: Lin's dad stops to take a picture, then rides at 0.24 miles per minute to catch up with the family, who are riding at 0.18 miles per minute.
2. Noah's kayaking trip: Kayak Rental A charges a base fee of $15 plus $4.50 per hour, while Kayak Rental B charges a base fee of $12.50 plus $5 per hour.
3. Diego's pastry recipe: The recipe calls for 3 strawberries for every apple, and Diego used 52 fruits in total.
4. Flour and sugar order: Flour costs $0.80 per pound, and sugar costs $0.50 per pound. An order of flour and sugar weighs 15 pounds and costs $9.00.

**15.3: Info Gap: Racing and Play Tickets**

Students apply their knowledge of systems of linear equations to solve real-world problems. One equation is given, and students may choose to write another equation to create a system, then solve it algebraically or by graphing.

**15.4: Solving Systems Practice**

Students solve various systems of equations, including those involving fractions, substitution, and inspection.

**Systems of Equations:**

1. \(\begin{cases} y=2x+6 \\ y=x-3 \end{cases}\)
2. \(\begin{cases} y=5x-4 \\ y=4x+12 \end{cases}\)
3. \(\begin{cases} y=\frac{2}{3}x-4 \\ y=\frac{4}{3}x+9 \end{cases}\)
4. \(\begin{cases} 4y + 7x = 6 \\ 4y+7x = 5 \end{cases}\)
5. \(\begin{cases} y=x-6 \\ x=6+y \end{cases}\)
6. \(\begin{cases} y=0.24x \\ y=0.18x+0.9 \end{cases}\)
7. \(\begin{cases} y=4.5x+15 \\ y=5x+12.5 \end{cases}\)
8. \(\begin{cases} y=3x \\ x+y=52 \end{cases}\)

**Lesson Synthesis**

To wrap up the lessons on solving systems of equations, consider displaying the following systems and asking students how they might begin to solve them:

1. \(\begin{cases} y = 2x + 1 \\ y = \frac{1}{2}x + 10 \end{cases}\)
2. \(\begin{cases} x = 5-2y \\ 2x + 6y = 16 \end{cases}\)
3. \(\begin{cases} 5x + 4y = 20 \\ 10x + 8y = 60 \end{cases}\)

**Student Lesson Summary**

We have learned how to solve various kinds of systems of equations using algebra. For example, consider the system:

\(\begin{cases} y = 2x - 3 \\ x+2y=7 \end{cases}\)

We can solve for x by substituting the expression for y into the second equation:

\(\begin{align} x+2(2x-3) &= 7 \\ 5x-6 &= 7 \\ 5x &= 13 \\ x &= \frac{13}{5} \end{align}\)

Then, we can solve for y by substituting the value of x into one of the original equations:

\(\begin{align} y &= 2(\frac{13}{5})-3 \\ y &= \frac{26}{5}-\frac{15}{5} \\ y &= \frac{11}{5} \end{align}\)

The solution to the system is \(\left(\frac{13}{5},\frac{11}{5}\right)\).


2.3 The Product Rule
The derivative of the product of two functions can be found using the product rule. If x and y are both functions of t, then the derivative of their product is given by:
\[ \frac{d(xy)}{dt} = \frac{dx}{dt}\cdot y + x \cdot \frac{dy}{dt} \]
This rule can be proven by considering the change in the product xy when t is changed by an infinitesimal amount dt.

Example 15: Find the derivative of the function tsin t.
\[ \frac{d(t\sin t)}{dt} = t\cdot \frac{d(\sin t)}{dt}+\frac{dt}{dt}\cdot\sin t = t\cos t+\sin t \]

The product rule has a geometrical interpretation, as shown in Figure H. Imagine a king expanding his territory to the north and east. The change in the area of the territory is given by the product rule.

The Leibniz notation is helpful for checking the units of an answer. If we measure distances in meters and time in seconds, then xy has units of square meters, and so does the change in the area, d(xy). Dividing by dt gives the number of square meters per second being conquered.

A common mistake is to guess that the product rule would be d(xy)/dt=(dx/dt)(dy/dt), but the Leibniz notation reveals that this can't be the case, because the units on the left and right wouldn't match.

The product rule can also be used to find the derivative of a function such as \(\sqrt{t}\). By writing \(\sqrt{t}\) as t^{1/2}, we might suspect that the rule d(t^{k})/dt=kt^{k-1} would work, giving a derivative \(\frac{1}{2}t^{-1/2}=1/(2\sqrt{t})\). However, this rule only applies to integer values of k. Using the product rule, we can confirm this result:
\[ \frac{dt}{dt} = \frac{d(\sqrt{t}\sqrt{t})}{dt} = f(t)\sqrt{t}+\sqrt{t}f(t) = 2f(t)\sqrt{t} \]
Since dt/dt=1, we have \(f(t)=1/(2\sqrt{t})\).

The product rule can also be used to prove that the power rule d(x^{n})/dx=nx^{n-1} applies to cases where n is an integer less than 0. However, this will be proven on page 41 using a different technique that applies to non-integer values of n.


Lesson 7: Fractions as Sums

The lesson begins with a warm-up activity, "Choral Count: Three-fourths at a Time," where students count by 3/4, starting at 3/4, and record their counts until they reach 48/4. This activity helps students identify patterns, such as the numerator increasing by 3 each time and the numerators being multiples of 3.

The first activity, "Barley Soup," prompts students to think about non-unit fractions as sums of other fractions. Students are given a recipe for barley soup with ingredients measured in fourths, and they must show how to measure each ingredient using 1/4 cup and 3/4 cup measuring cups. This activity helps students connect quantities in the story problem to equations and reason abstractly and quantitatively.

In the second activity, "Sums in Fifths and Thirds," students record decompositions of fractions as equations. They are asked to write the fraction 9/5 as a sum of unit fractions and to decompose 9/5 into sums of other fifths and 4/3 into sums of thirds. This activity helps students see that only non-unit fractions can be decomposed into smaller fractions with the same denominator.

The lesson synthesis highlights that fractions with a numerator greater than 1 can be written as products and sums. Students compare these two ways of thinking about fractions, discussing how they are alike and different. The lesson concludes with a cooldown activity, "Make a Sum of 7/4," where students are asked to decompose 7/4 into a sum of smaller fractions.

Key concepts and skills developed in this lesson include:

* Identifying patterns in fractions
* Decomposing fractions into sums of smaller fractions
* Writing fractions as products and sums
* Reasoning abstractly and quantitatively
* Connecting quantities in story problems to equations

The lesson requires the following materials:

* 1/4 cup and 3/4 cup measuring cups
* Whiteboard or chart paper
* Markers or colored pencils

The lesson is designed to support accessibility for conceptual processing, visual-spatial processing, and memory. It advances student thinking by representing and conversing about fractions in different ways.

Multiple-choice questions and answers have been refined to:
1. Which of the following fractions are equivalent to whole numbers?
A) 12/4, 24/4, 36/4, 48/4
B) 12/4, 24/4
C) 36/4, 48/4
D) None of the above

Answer: A) 12/4, 24/4, 36/4, 48/4

2. What are some ways to decompose 13/6 and write it as a sum?
A) 10/6 + 3/6
B) 12/6 + 1/6
C) 7/6 + 1/6
D) All of the above

Answer: D) All of the above


Consider the linear programming problem:
P: Max z = 2x1 + x2
s.t.
4x1 + 5x2 ≤ 20  (1)
2x1 + 5x2 ≤ 10  (2)
x2 ≥ 0  (3)
The variable x1 is unrestricted.

To solve this problem, we will:
(a) Graph the feasible set and the gradient of the objective function.
The feasible set is the shaded area, and the gradient of the objective function is shown. The optimal solution is point "A" with coordinates x = (1⅔, 2⅔) and a value of the objective function of z = 6.

(b) If we replace the objective function with Min z = 1x1 + 2x2, the new objective function is perpendicular to constraint (2). As a result, all points on the line segment between A and B are optimal, resulting in alternative optimal solutions.

(c) If we use the objective function Min z = x1 + 2x2 instead, the new objective function results in a case of unbounded "optimal" solutions.

Key points:
- The feasible set is defined by constraints (1), (2), and (3).
- The original objective function has an optimal solution at point A.
- Changing the objective function can result in alternative optimal solutions or unbounded "optimal" solutions.


**Lemma:** Given that there exist p, q with the following properties:
(a) gcd(p, q) = 1 (gcd = greatest common denominator)
(b) p, q have opposite parities (one is odd, one is even)
(c) p^2 + 3q^2 is a cube
Then there exists a, b such that:
(a) p = a^3 - 9ab^2
(b) q = 3a^2b - 3b^3
(c) gcd(a, b) = 1
(d) a, b have opposite parities

**Proof:**
(1) Let u = a^2 + 3b^2. [Since we know that u^3 = p^2 + 3q^2, u is a cube.]
(2) We know that u is odd since p, q have opposite parities.
(3) We know then that u must be of the form a^2 + 3b^2. [Since any odd factor would also have to have the same form.]
(4) Now, (a^2 + 3b^2)^3 = (a^2 + 3b^2)[(a^2 - 3b^2)^2 + 3(2ab)^2] since (a^2 + 3b^2)^2 = a^4 + 6a^2b^2 + 9b^4 = a^4 + 12a^2b^2 - 6a^2b^2 + 9b^4 = (a^2 - 3b^2)^2 + 3(2ab)^2
(5) And, (a^2 + 3b^2)[(a^2 - 3b^2)^2 + 3(2ab)^2] = [a(a^2 - 3b^2) - 3b(2ab)]^2 + 3[a(2ab) + b(a^2 - 3b^2)]^2
(6) And: [a(a^2 - 3b^2) - 3b(2ab)]^2 + 3[a(2ab) + b(a^2 - 3b^2)]^2 = [a^3 - 3ab^2 - 6ab^2]^2 + 3(2a^2b + a^2b - 3b^3)^2 = [a^3 - 9ab^2]^2 + 3(3a^2b - 3b^3)^2
(7) Which combined with step (1) gives us: p^2 + 3q^2 = [a^3 - 9ab^2]^2 + 3(3a^2b - 3b^3)^2
(8) Which means that we could define a, b such that: p = a^3 - 9ab^2, q = 3a^2b - 3b^3, gcd(a, b) = 1 [since otherwise, any common factor would divide p and q].
(9) We also know that a, b have opposite parities since: (a) If a, b are both odd, then p is even since p = odd - odd and q is even since q = odd - odd which is impossible since p, q have opposite parities. (b) If a, b are both even, then p is even since p = even - even and q is even since q = even - even which is impossible.

Note: The proof has been refined to address the concerns and corrections raised in the comments. However, the validity of the proof remains a subject of debate, and further examination is necessary to confirm its accuracy.


How to Solve Two Step Equations - Algebra 1

In this video, you will learn how to solve two-step equations. An equation is a mathematical phrase that shows two values are equal. A two-step equation is an equation that takes two steps to solve for the variable. The variable is an unknown value in math, represented by a lowercase letter.

To solve two-step equations, you need to use inverse operations, which involve doing the opposite of an operation to eliminate it and find the value of the variable. The inverse of addition is subtraction, and the inverse of multiplication is division. By applying these inverse operations, you can solve two-step equations and find the value of the variable.


In a triangle ABC, points P and Q are on the sides AB and CB, respectively, such that angle PAC = angle AQC. Given lengths are AQ = 108 cm and CP = 75 cm, with AQ and CP meeting at O. Since AO = OC, we need to find the length of AC.

To solve this problem, we first identify the given information:
- angle PAC = angle AQC
- AO = OC
- AQ = 108 cm
- CP = 75 cm

We notice the isosceles triangle AOC and the triangles PAC and AQC. In triangle AOC, since AO = OC, it is isosceles, meaning angle OAC = angle OCA.

Let's denote:
- angle PAC = angle AQC = angle theta
- angle OAC = angle OCA = angle U
- angle OCQ = angle V
- angle OPA = angle W
- side AC = x

In triangle PAC, the sum of interior angles is 180 degrees, so U + theta + W = 180. By the Law of Sines, x/sinW = 75/sin(theta).

In triangle AQC, the sum of interior angles is also 180 degrees, giving us U + theta + (U + V) = 180, or 2U + theta + V = 180. Applying the Law of Sines, x/sin(theta) = 108/sin(U + V).

Subtracting the equations 2U + theta + V = 180 and U + theta + W = 180 gives us U + V - W = 0, or U + V = W.

Substituting U + V = W into the equation x/sin(theta) = 108/sin(U + V) gives x/sin(theta) = 108/sinW.

Now we have two equations involving sin(theta) and sinW:
1. x/sinW = 75/sin(theta)
2. x/sin(theta) = 108/sinW

From equation 1, we get sin(theta) = 75sinW / x. From equation 2, we get sin(theta) = x*sinW / 108.

Equating the two expressions for sin(theta), we have 75sinW / x = x*sinW / 108. Cross-multiplying gives (75sinW)*108 = (x*sinW)*x. Dividing both sides by sinW yields 75*108 = x^2.

Therefore, x = sqrt(75*108) = sqrt(8100) = 90 cm.


ROI – Return on Investment

In business, the term "return on investment" (ROI) is used to measure the rate of return on money invested in a commercial venture. It quantifies risk and helps decide whether to continue with a particular investment. Experienced businesspeople find that calculating an ROI is good business, while first-time entrepreneurs often act on impulse, resulting in money spent foolishly.

To demonstrate how to calculate an ROI, we'll use a LinkedIn account as an example. The standard ROI formula is:

ROI = (Net Profit / Cost of Investment) x 100
= (Gain from Investment - Cost of Investment) / Cost of Investment x 100

We'll assign arbitrary values to both profit and cost, thinking of them in terms of "value" or dollars. Using this formula, we can calculate the ROI for any situation.

**Example #1: LinkedIn Profile Views**

* Gain from Investment: 300 people viewed your profile in the past 90 days, rated as 500
* Cost of Investment: 3 hours spent working on the LinkedIn site, rated as 50
* ROI = (500 - 50) / 50 x 100 = 900% (a positive 900%)

**Example #2: LinkedIn Profile Views**

* Gain from Investment: 1,113 people viewed your profile in the past 3 days, rated as 500
* Cost of Investment: 35 hours spent working on the LinkedIn site, rated as 2000
* ROI = (500 - 2000) / 2000 x 100 = -75% (a negative 75%)

Now, let's apply the ROI principle to teacher evaluations.

**Teacher Observation Example #1**

* Gain from Investment: 100 formal Teacher Observations, rated as 500
* Cost of Investment: 50% of the day spent observing staff and teachers, rated as 2000
* ROI = (500 - 2000) / 2000 x 100 = -75% (a negative 75%)

**Teacher Observation Example #2**

* Gain from Investment: 50 formal Teacher Observations, rated as 1000
* Cost of Investment: 20% of the day spent observing and being visible to staff and teachers, rated as 500
* ROI = (1000 - 500) / 500 x 100 = 100% (a positive 100%)

By calculating the ROI for teacher evaluations, we can determine whether the time and effort invested are worth it. If the ROI is negative, it may be time to try a different approach. By using arbitrary values for "worth" and calculating the ROI, we can gauge the effectiveness of our investments and make informed decisions.

To calculate your own ROI, try using the formula with your own values for "worth." You can use any value that makes sense to you, and then calculate your ROI to see what your numbers tell you. If you have questions, please ask.


**Equation Analysis**

The equation `lxl = lyl` can be solved using `sqrt(x^2) = sqrt(y^2)`, which simplifies to `x^2 = y^2`. Applying the "zero-rule" yields `x^2 - y^2 = (x-y)(x+y) = 0`, resulting in `x = -y` and `y = -x`.

Alternatively, solving `x^2 = y^2` directly gives `x = (+-)sqrt(y^2)`. However, this approach does not provide a clear solution without reverting to the original statement.

The key difference between these methods lies in the application of the zero-rule, which enables factoring and avoids division by `x` when `x = 0`. This rule allows for a more straightforward solution, whereas the direct approach stalls without additional algebraic manipulation.


A key concept in mathematics is the multiplication of even and odd numbers. When an even number is multiplied by another even number, the result is always an even number. This is because any number that has a factor of 2, when multiplied by any other number, will create a product that also has a factor of 2, thus resulting in an even number.

The rules for multiplying even and odd numbers are as follows:
- Even times even = even
- Odd times odd = odd
- Even times odd = even

These rules can be applied to various mathematical problems. For example, if ten times the square of a non-zero number equals forty times the number, we can set up an equation to solve for the number.

It's also important to note that an even number is defined as any number that is divisible by 2. Therefore, any multiple of an even number will also be divisible by 2, making it an even number.

In contrast, an odd number is not divisible by 2. When an odd number is squared, the result will always be an odd number. Similarly, when an even number is squared, the result will always be an even number.

Understanding these concepts can help in solving various mathematical problems and answering related questions, such as:
Q: Why does an even number times an even number equal an even number?
A: Because multiplying anything that has a factor of 2 by anything else will create a product that has a factor of 2, which will be by definition an even number.

Q: What is the result of multiplying an odd number by an even number?
A: Even

Q: What is the result of squaring an odd number?
A: Odd

Q: What is the result of squaring an even number?
A: Even


## Dinosaur Fractions - Section 3: Practice

# Parts of a Set

Lesson 3 of 10

## Objective: SWBAT identify parts of a whole.

*50 minutes*

#### Intro Activity

*10 min*

The class will re-watch a BrainPop video about fractions called ‘More Fractions’ to review key vocabulary and mathematical representations of fractions. Key questions to frame the content include:

* What part of the fraction goes on top? What does it represent?
* What goes on the bottom of a fraction? What does it represent?

The goal is for students to develop a deep understanding that fractions represent parts of a whole or set and to be able to explain that.

#### Warm-Up

*15 min*

The teacher will ask students questions about the class, such as:
* What fraction of students in our class are girls?
* Where would I put the total number of girls? Why?
* Where would I put the total number of students in our class? What is another word for that? (whole)

To illustrate, the girls will step aside to show that there are 11 girls out of 24 students. This activity provides a visual example of fractions and parts of a set. The teacher will also ask questions about other characteristics, such as:
* What fraction of students are boys?
* What fraction of students have jeans on?
* What fraction of students have our uniform on?
* What fraction of students have glasses?

#### Practice

*20 min*

The teacher will present a bag of dinosaurs and ask students to help figure out what fractions of items are in the bag. Questions will include:
* What fraction of these dinosaurs are yellow?
* What fraction of these dinosaurs are red?

As students respond, the teacher will record the information on the board to demonstrate the expectation for recording fractions. Students will work in groups to record as many fractions as they can using the materials in their bag, labeling each fraction to identify what it represents.

#### Close

*5 min*

The teacher will highlight a few examples of how students created fractions and acknowledge their hard work. This lesson aims to help students understand fractions using a variety of manipulatives, making it suitable for a special education class.

Similar lessons include:
* Modeling with Box Diagrams on the iPad (day 1 of 2)
* Name that Fraction
* Introduction to Fractions: A Concrete Experience

This lesson is part of a larger unit on fractions, which includes:
* UNIT 1: Crazy About Fractions
* UNIT 2: Sharing Equally with Division
* UNIT 3: The Relationship Between Multiplication and Division
* UNIT 4: Number Lines
* UNIT 5: Multiply or Divide with Word Problems
* UNIT 6: Telling Time
* UNIT 7: Problem Solving
* UNIT 8: 2-Digit by 1-Digit Multiplication
* UNIT 9: Measurement and Data
* UNIT 10: Math in the Real World
* UNIT 11: More Problem Solving Practice
* UNIT 12: Addition & Subtraction
* UNIT 13: 3rd Grade Expectations
* UNIT 14: Geometry
* UNIT 15: Surveys and Graphing
* UNIT 16: Fun with Multiplying and Dividing

Specific lessons in the fraction unit include:
* LESSON 1: What is a Fraction?
* LESSON 2: Digging Into Fractions
* LESSON 3: Parts of a Set
* LESSON 4: Comparing Fractions
* LESSON 5: What's in the Set?
* LESSON 6: What is a Set?
* LESSON 7: Teach Me About Fractions
* LESSON 8: Are My Fractions Equivalent?
* LESSON 9: Comparing Fractions with Models
* LESSON 10: Fraction Match-Up


Understanding and Comparing Distributions

Chapter 5 Objectives:
- Calculate boxplots
- Identify outliers
- Compare distributions
- Create timeplots

The Big Picture:
Comparing distributions can be more interesting than describing a single distribution. By comparing distributions for different groups, we can answer more complex questions about variables.

The Five-Number Summary:
- Reports the median, quartiles, and extremes (maximum and minimum) of a distribution
- Example: For daily wind speed, the five-number summary is Max 8.67, Q3 2.93, Median 1.90, Q1 1.15, Min 0.20
- Consists of the minimum value, Q1, the median, Q3, and the maximum value
- Offers a complete description of the center and spread
- Used to construct boxplots

Boxplot:
- A graphical display of the five-number summary
- Useful for comparing groups
- Effective at pointing out outliers
- Can be drawn horizontally or vertically
- The box represents the IQR (middle 50%) of the data

Constructing a Boxplot:
1. Draw a scale and label
2. Draw vertical lines at Q1 and Q3
3. Draw a vertical line at the median and complete the box
4. Extend the "left whisker" to the minimum value
5. Extend the "right whisker" to the maximum value
6. Give a descriptive title

Example Boxplot:
- Data: 20, 25, 25, 27, 28, 31, 33, 34, 36, 37, 44, 50, 59, 85, 86
- Use TI-83/84 to construct

Outliers:
- An extremely small or large data value compared to the rest of the data
- Try to understand outliers in the context of the data
- May indicate data error or special nature of the data
- Report mean and standard deviation with and without outliers

Testing for Outliers:
- Use IQR to determine if extreme values are outliers
- An observation is an outlier if it falls more than 1.5 times IQR below Q1 or above Q3
- Construct upper and lower fences to test for outliers

Modified Boxplot:
- Plots outliers as isolated points
- More useful than a regular boxplot
- Constructing a modified boxplot:
  1. Same as a boxplot with exception of "whiskers"
  2. Extend "left whisker" to minimum value or last data value less than or equal to lower fence
  3. Extend "right whisker" to maximum value or last data value less than or equal to upper fence
  4. Represent outliers with a dot or asterisk

Comparing Distributions:
- Compare histograms and boxplots for daily wind speeds
- Boxplots are effective for comparing groups
- Look for shapes, centers, and spreads of distributions

Timeplots:
- Display data observed over a given period
- Put time on the horizontal scale and variable on the vertical scale
- Connect data points with lines
- Look for overall pattern, trend, and outliers

Re-expressing Skewed Data:
- Apply a simple function (logarithms, square roots, and reciprocals) to each value
- Can make a skewed distribution more symmetric

What Can Go Wrong:
- Avoid inconsistent scales
- Label clearly
- Beware of outliers
- Be careful when comparing groups with different spreads

Key Takeaways:
- Comparing data groups and looking for patterns is valuable
- Boxplots are effective for comparing groups graphically
- Identifying and investigating outliers is important
- Graphing data over time can reveal long-term trends

Assignment:
- Exercises pg. 95-103: #6-8, 11, 12, 19, 26, 28, 29, 31
- Read Ch-6, pg. 104-128


Recursion

Introduction
Recursion is a problem-solving technique where a function calls itself to perform a task. For example, counting the positive odd numbers from a maximum to a minimum value can be achieved using recursion. To illustrate, consider counting the odd numbers from 1 to 9: 9, 7, 5, 3, and 1. This process involves subtracting 2 from the current number to get the previous odd number, which is a recursive operation.

A recursive method typically follows this structure:
ReturnValue Function(Arguments) {
    Optional Action
    Function();
    Optional Action
}
The method starts with a return value, which can be void if no value is returned. It takes one or more arguments, often modifying at least one of them. The body of the method performs the necessary actions, and there are two main rules to observe when implementing recursion.

To demonstrate recursion, consider the example of counting decrementing odd numbers. A method can be created to take an integer as an argument, display the current value, subtract 2, and recall the method itself. Here is an example implementation:
using System;
class Program {
    static void OddNumbers(int a) {
        if (a >= 1) {
            Console.Write("{0}, ", a);
            a = a - 2;
            OddNumbers(a);
        }
    }
    static void Main() {
        const int Number = 9;
        Console.WriteLine("Odd Numbers");
        OddNumbers(Number);
        Console.WriteLine();
    }
}
This code produces the output:
Odd Numbers
9, 7, 5, 3, 1,

Recursive methods are useful for building lists or series that follow a pattern. For instance, instead of just displaying odd numbers, you can add them incrementally. The method can be implemented as follows:
using System;
class Program {
    static int AdditionalOdd(int a) {
        if (a <= 1) return 1;
        return a + AdditionalOdd(a - 2);
    }
    static void OddNumbers(int a) {
        if (a >= 1) {
            Console.Write("{0}, ", a);
            a = a - 2;
            OddNumbers(a);
        }
    }
    static void Main() {
        const int Number = 9;
        Console.WriteLine("Odd Numbers");
        OddNumbers(Number);
        Console.WriteLine();
        Console.WriteLine("Sum of Odds: {0}\n", AdditionalOdd(Number));
    }
}
This code produces the output:
Odd Numbers
9, 7, 5, 3, 1,
Sum of Odds: 25

Key concepts in recursion include:
- A recursive method calls itself to perform a task
- The method takes one or more arguments and returns a value
- The body of the method performs the necessary actions
- Recursion is useful for building lists or series that follow a pattern
- There are two main rules to observe when implementing recursion, although they are not specified in this example.

Note: The original code had an error in the line "a = 2;". It has been corrected to "a = a - 2;" to produce the correct output. Similarly, the line "return a + AdditionalOdd(a  2);" has been corrected to "return a + AdditionalOdd(a - 2);" to produce the correct output.


**3.OA.C.7: Fluently Multiply and Divide within 100**

This standard is part of the 3rd Grade curriculum and is a prerequisite for upcoming 4th Grade standards. By the end of Grade 3, students should be able to fluently multiply and divide within 100 using strategies such as:

* The relationship between multiplication and division (e.g., knowing that 8 x 5 = 40, one knows 40 ÷ 5 = 8)
* Properties of operations

Additionally, students should know from memory all products of two one-digit numbers. This standard is part of the Major Cluster 3.OA.C.

Note: There are no accommodations or comments available for this standard.