To find the area of a kite, you need to know the length of its diagonals or the length of its sides and the angle between them. The area of a kite can be calculated using the formula: Area = (d1 * d2) / 2, where d1 and d2 are the lengths of the diagonals. Alternatively, you can use the formula: Area = a * b * sin(θ), where a and b are the lengths of the sides and θ is the angle between them.

The area of a kite can also be found by dividing it into two pairs of congruent triangles. Each triangle's area is equal to half the product of its base and height. Since the triangles are congruent, the area of the kite is equal to the sum of the areas of the two pairs of triangles.

A quadrilateral, such as a kite, has four sides and two diagonals. If the diagonals are perpendicular, the quadrilateral is a kite or a rhombus. The area of a kite can be calculated using the lengths of its diagonals or the lengths of its sides and the angle between them.

The Pythagorean Theorem can be used to find the length of the diagonals of a kite, given the lengths of its sides. The theorem states that a^2 + b^2 = c^2, where a and b are the lengths of the sides and c is the length of the hypotenuse.

To calculate the area of a kite, follow these steps:

1. Find the length of the diagonals or the length of the sides and the angle between them.
2. Use the formula: Area = (d1 * d2) / 2 or Area = a * b * sin(θ).
3. Divide the kite into two pairs of congruent triangles and calculate the area of each triangle.
4. Sum the areas of the two pairs of triangles to find the area of the kite.

For example, if the area of a kite is 126 cm^2 and one diagonal is 21 cm long, you can use the formula to find the length of the other diagonal. If the diagonals are 12 inches and 15 inches, the area of the kite is 360 square units.

Note that kites with a 90-degree angle will be square, while kites with an angle greater than 90 degrees will be less stable and may pitch forward during gusts. Kites with an angle less than 90 degrees will have little lift. Understanding the design and geometry of kites is essential for calculating their area and predicting their behavior in different wind conditions.


# Triangular Number Links

## Introduction
Mary knows that the 5th triangular number is 15 because it needs 15 counters to make the triangle. However, she doesn't know which of the given expressions is equal to the nth triangular number, T(n).

## Expressions for T(n)
The given expressions are:
(1) T(n) = 1 + 2 + 3 + 4 + . . . + (n – 3) + (n – 2) + (n – 1) + n;
(2) T(n) = ½n(n + 1);
(3) T(n) = 1 + 3 + 5 + … + (2n – 5) + (2n – 3) + (2n – 1);
(4) T(n) = n^{2} – (n – 1)^{2} + (n – 2)^{2} – (n – 3)^{2} + . . . + 4^{2} - 3^{2} + 2^{2} – 1^{2};
(5) T(n) = [(n + 1)^{2} + n^{2} + (n – 1)^{2} + ... + 1^{2}] – [(n^{2} + 2) + ((n – 1)^{2} + 2) + ((n – 2)^{2} + 2) + … + (1^{2} + 2) + 2];
(6) T(n) = T(n -1) + n and T(1) = 1.

## Solution
To determine which expressions are equal to T(n), we need to analyze each one.
Expression (1) can be proven using the backward/forward method. By adding the terms in reverse order, we get:
T(n) = n + (n – 1) + (n - 2) + (n – 3) + . . . . + 4 + 3 + 2 + 1.
Adding the two lines together gives:
2 T(n) = (n+1) + (n+1) + (n+1) + (n+1) + . . . . + (n+1) + (n+1) + (n+1) + (n+1).
Simplifying, we get:
2 T(n) = n x (n + 1).
And finally:
T(n) = ½n(n + 1).

Expression (2) is equal to T(n) as shown above.

Expression (3) is not equal to T(n). By substituting n = 2, we get:
1 + 3 = 4, but the second triangular number is 1 + 2 = 3.

Expression (4) is equal to T(n) for even values of n. Using the difference of two squares, we can simplify the expression:
n^{2} – (n – 1)^{2} = [n – (n – 1)][n + (n – 1)] = n + (n – 1).
However, this expression only holds for even values of n.

Expression (5) is not equal to T(n). By substituting n = 1, we get:
[2^{2} + 1^{2}] – [3 + 2] = 0, which is not equal to T(1) = 1.

Expression (6) is equal to T(n). This is a recurrence relation, which can be proven using mathematical induction. Starting with T(1) = 1, we can recursively find T(n) by adding n to T(n-1).

## Extension
There are many other ways to show that these results hold or don't hold. The extension problem asks to find multiple ways to show that expressions (1), (2), and (4) are triangular numbers.


### CAT Online Coaching - Permutation and Combination, Fixing the Errors II

This post provides the corrected solutions to questions 4, 5, and 6.

4. In how many ways can we select 2 cards from a card pack such that both belong to the same suit?

The correct approach is to select one of the four suits first and then select 2 cards from 13 in that suit. The number of ways of doing this = 4 * ^{13}C_{2} = 52 * 6.

5. When a coin is tossed 10 times, what is the probability of getting more heads than tails?

The probability can be calculated as ( ^{10}C_{6} + ^{10}C_{7} + ^{10}C_{8} + ^{10}C_{9} + ^{10}C_{10}) / 2^{10}. Alternatively, it can be thought of as ½ (1 - ^{10}C_{5}) / 2^{10}, removing the scenario where the number of heads and tails are equal.

6. In how many ways can we arrange 3 boys and 3 girls on a straight line such that no two boys stand next to each other?

There are 4 possible outlines: BGBGBG, GBGBGB, BGBGGB, and BGGBGB. The total number of possibilities = 3! * 3! * 4 = 144.

Labels: CAT 2015, CAT Coaching Online, CAT Combinatorics, CAT Online Preparation, CAT Permutation Combination


# Bohr's Model of Atom

Neils Bohr, a Danish physicist, proposed a model of the atom in 1913, addressing the limitations of Rutherford's Model. The key aspects of Bohr's Model are:

- Electrons revolve around the nucleus in fixed orbits, known as 'stationary orbits'.
- Each stationary orbit is associated with a fixed amount of energy, and electrons do not radiate energy as long as they remain in their fixed orbit.

The circular path around the nucleus is referred to as an orbit, energy level, or shell, represented by letters such as K, L, M, N, and so on. The orbits are denoted by numbers 1, 2, 3, and so on, with the corresponding letter designations:
- 1st orbit: K
- 2nd orbit: L
- 3rd orbit: M
- and so on.

## Distribution of Electrons in an Orbit or Shell

The distribution of electrons in an orbit can be calculated using the formula 2n^2, where 'n' is the number of the orbit.

### Number of Electrons in Each Shell

- K-shell (1st orbit): 2n^2 = 2 * 1^2 = 2 electrons
- L-shell (2nd orbit): 2n^2 = 2 * 2^2 = 8 electrons
- M-shell (3rd orbit): 2n^2 = 2 * 3^2 = 18 electrons
- N-shell (4th orbit): 2n^2 = 2 * 4^2 = 32 electrons
- O-shell (5th orbit): 2n^2 = 2 * 5^2 = 50 electrons
- P-shell (6th orbit): 2n^2 = 2 * 6^2 = 72 electrons

In general, the maximum number of electrons in any shell can be calculated using the formula 2n^2.

## Key Concepts

- A higher value of 'n' indicates an orbit farther from the nucleus, with a higher energy level.
- Orbits closer to the nucleus have lower energy levels.
- Electrons do not lose energy when moving in a definite orbit.
- When an atom gains energy from an external source, an electron jumps to a higher energy level.
- When an electron loses energy, it jumps down to a lower energy level.

## Quiz

1. Who proposed the model of the atom in 1913?
a) Neils Bohr
b) Rutherford
c) Other

2. How many electrons are in the K-shell?
a) 1
b) 2
c) 3
d) 4

3. How many electrons are in the L-shell?
a) 4
b) 6
c) 8
d) 10

4. How many electrons are in the M-shell?
a) 10
b) 12
c) 18
d) 20

5. What is the formula for the number of electrons in a shell?
a) n^2
b) 2n
c) 2n^2
d) n^3

6. How many electrons are in the N-shell?
a) 20
b) 24
c) 32
d) 36

7. How many electrons are in the O-shell?
a) 36
b) 40
c) 50
d) 60

8. What happens to an electron when it gains energy?
a) It jumps to a lower orbit
b) It jumps to a higher orbit
c) It remains in the same orbit
d) It is ejected from the atom

9. What happens to an electron when it loses energy?
a) It jumps to a higher orbit
b) It jumps to a lower orbit
c) It remains in the same orbit
d) It is ejected from the atom

10. Which shell is the first energy level?
a) K-shell
b) L-shell
c) M-shell
d) N-shell


To determine if 413 can be divided by 7, we can use two divisibility tricks. First, separate the last number from the rest, giving us 41 and 3.

- Double the 3 and subtract it from 41: 41 - 6 = 35, which is a multiple of 7, indicating 413 can be divided evenly by 7.
- Multiply the 3 by 5 and add it to 41: 41 + 15 = 56, also a multiple of 7, leading to the same conclusion.

413 is a composite number with a prime factorization of 413 = 7 x 59. The exponents in the prime factorization are both 1. Adding 1 to each exponent and multiplying them gives (1 + 1)(1 + 1) = 2 x 2 = 4, meaning 413 has exactly 4 factors.

The factors of 413 are: 1, 7, 59, 413. The factor pairs are 1 x 413 and 7 x 59. Since 413 has no square factors, its square root cannot be simplified, with √413 ≈ 20.3224.

Additionally, 413 cannot be expressed as a^2 + b^2 because both its prime factors are of the form 4m+3. It also cannot be of the form a^2 + ab + b^2 to be the area of a triangle on the triangular lattice since it is of the form 3n+2, making it unsuitable as the basis of a tessellation. Furthermore, 413 cannot be the hypotenuse in a Pythagorean triple.


Cupcakes and Algebra (Follow Up)

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## ML Aggarwal Class 6 Solutions for ICSE Maths Chapter 6 Fractions Ex 6.1

### Question 1: Division as Fractions
Write the following divisions as fractions:
(i) 3 ÷ 7 = \(\frac{3}{7}\)
(ii) 11 ÷ 78 = \(\frac{11}{78}\)
(iii) 113 ÷ 128 = \(\frac{113}{128}\)

### Question 2: Fractions in Words
Write the following fractions in words:
(i) \(\frac{2}{7}\) = two-sevenths
(ii) \(\frac{3}{10}\) = three-tenths
(iii) \(\frac{15}{28}\) = fifteen-twenty-eighths

### Question 3: Fractions in Number Form
Write the following fractions in number form:
(i) one-sixth = \(\frac{1}{6}\)
(ii) three-eleventh = \(\frac{3}{11}\)
(iii) seven-fortieth = \(\frac{7}{40}\)
(iv) thirteen-one hundred twenty fifth = \(\frac{13}{125}\)

### Question 4: Fraction of Shaded Part
Determine the fraction of each figure that is shaded.

### Question 5: Shade Parts of Figures
Shade the parts of the figures according to the given fractions.

### Question 6: Error Identification
Identify the error in stating that the shaded region is \(\frac{1}{4}\) of the whole region in the given figure.

### Question 7: Writing Fractions
Write the fraction with:
(i) numerator = 5 and denominator = 13: \(\frac{5}{13}\)
(ii) denominator = 23 and numerator = 17: \(\frac{17}{23}\)

### Question 8: Fraction of Dresses Stitched
Shabana has stitched 21 out of 35 dresses. The fraction of dresses she has stitched is \(\frac{21}{35}\) = \(\frac{3}{5}\).

### Question 9: Fraction of a Day
8 hours is \(\frac{8}{24}\) = \(\frac{1}{3}\) of a day.

### Question 10: Fraction of an Hour
45 minutes is \(\frac{45}{60}\) = \(\frac{3}{4}\) of an hour.

### Question 11: Natural and Prime Numbers
There are 11 natural numbers from 87 to 97. The prime numbers in this range are 89, 97. The fraction of prime numbers is \(\frac{2}{11}\).


# Machine Learning

## Causal Impact Analysis

The `causalimpact` library in Python is used to determine the causal relationship between an event and a response variable in time series data. It analyzes the differences between expected and observed time series data.

To use `causalimpact`, first install the `pycausalimpact` library using `!pip install pycausalimpact`. Then, import the necessary libraries, including `numpy`, `pandas`, and `causalimpact`.

```python
import numpy as np
import pandas as pd
from causalimpact import CausalImpact
```

Generate a random sample of time series data using `ArmaProcess` from `statsmodels`. For example:

```python
np.random.seed(0)
ar = np.r_[1, 0.9]
ma = np.array([1])
arma_process = ArmaProcess(ar, ma)
X = 50 + arma_process.generate_sample(nsample=1000)
y = 1.6 * X + np.random.normal(size=1000)
y[800:] += 10
```

Create a `DataFrame` with the time series data and define the pre-period and post-period.

```python
data = pd.DataFrame({"y": y, "X": X}, columns=["y", "X"])
pre_period = [0, 799]
post_period = [800, 999]
```

Create a `CausalImpact` object and print the summary of the causal impact analysis.

```python
ci = CausalImpact(data, pre_period, post_period)
print(ci.summary())
print(ci.summary(output="report"))
ci.plot()
```

The output will show the average cumulative actual and predicted values, as well as the absolute and relative effects of the intervention.

## Preventing Data Leakage with Pipeline and GridSearchCV

When using `GridSearchCV` to tune hyperparameters, scaling the data before using `GridSearchCV` can lead to data leakage. To prevent this, assemble both the scaler and machine learning models in a pipeline and use it as the estimator for `GridSearchCV`.

```python
from sklearn.model_selection import train_test_split, GridSearchCV
from sklearn.preprocessing import StandardScaler
from sklearn.pipeline import make_pipeline
from sklearn.svm import SVC
from sklearn.datasets import load_iris

df = load_iris()
X = df.data
y = df.target
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)

make_pipe = make_pipeline(StandardScaler(), SVC())
grid_params = {"svc__C": [0.1, 1, 10, 100, 1000], "svc__gamma": [0.1, 1, 10, 100]}
grid = GridSearchCV(make_pipe, grid_params, cv=5)
grid.fit(X_train, y_train)
y_pred = grid.predict(X_test)
```

## Decomposing High-Dimensional Data with PCA

To decompose high-dimensional data into two or three dimensions, use the `PCA` class from the `yellowbrick` library.

```python
!pip install yellowbrick
from yellowbrick.datasets import load_credit
from yellowbrick.features import PCA

X, y = load_credit()
classes = ["account in defaut", "current with bills"]

visualizer = PCA(scale=True, classes=classes)
visualizer.fit_transform(X, y)
visualizer.show()
```

## Calculating RMSE with Sklearn

To calculate the root mean squared error (RMSE) using `sklearn`, pass `squared=False` to the `mean_squared_error` method.

```python
from sklearn.metrics import mean_squared_error
y_actual = [1, 2, 3]
y_predicted = [1.5, 2.5, 3.5]
rmse = mean_squared_error(y_actual, y_predicted, squared=False)
print(rmse)
```


# I Linear Programming

## Introduction to Linear Programming

Linear Programming (LP) is a technique used for the economic allocation of scarce resources to competing activities. It involves converting verbal descriptions and numerical data into mathematical expressions to represent relationships among decision factors, objectives, and restrictions on resource use. LP is used to optimize a given criterion, such as performance, return on investment, profit, cost, utility, time, or distance.

## History of Linear Programming

George B. Dantzig developed LP while working with the US Air Force during World War II, primarily to solve military logistics problems. Today, LP is extensively used in various fields, including business, hospitals, airlines, agriculture, military operations, oil refining, education, energy planning, pollution control, transportation planning, and research and development.

## Structure of Linear Programming

A general LP model consists of three components:

1. **Decision Variables (Activities)**: These are the alternatives or courses of action to be evaluated to arrive at the optimal value of the objective function. Decision variables are usually denoted by x1, x2, …, xn and represent the extent to which each activity is performed.
2. **The Objective Function**: This is a mathematical representation of the objective in terms of a measurable quantity, such as profit, cost, revenue, or distance. The objective function is represented as: Optimise (Maximise or Minimise) Z = c1x1 + c2x2 + … + cnxn.
3. **The Constraints**: These are limitations on the use of resources, such as labor, machine, raw material, space, or money, that limit the degree to which the objective can be achieved. Constraints must be expressed as linear equalities or inequalities in terms of decision variables.

## Application Areas of Linear Programming

LP has various application areas, including:

* Agricultural Applications: farm economics, farm management, and agricultural planning.
* Military Applications: selecting air weapon systems, transportation problems, and community defense.
* Production Management: product mix, production planning, assembly-line balancing, blending problems, and trim loss.
* Financial Management: portfolio selection and profit planning.
* Marketing Management: media selection, traveling salesman problem, and physical distribution.
* Personnel Management: staffing problem, determination of equitable salaries, and job evaluation and selection.

## Advantages of Linear Programming

The advantages of LP include:

* Attaining the optimum use of productive resources.
* Improving the quality of decisions.
* Providing possible and practical solutions.
* Highlighting bottlenecks in production processes.
* Helping in re-evaluation of a basic plan for changing conditions.

## Limitations of Linear Programming

The limitations of LP include:

* The need for a clearly identifiable and measurable objective.
* The requirement for distinct and measurable activities.
* The need for identifiable and measurable resources.
* The assumption of linear relationships between variables.
* The availability of feasible alternative courses of action.
* The potential for non-integer valued solutions.
* The lack of consideration for the effect of time and uncertainty.
* The assumption of constant parameters.

## Situation Analysis

A company, Phang Furniture System Inc. (Fursys), manufactures two models of stools, Potty and Hardy. The company wants to determine the optimal production levels for maximum profit using LP.

## Decision Variables and Objective Function

The decision variables are X1 (number of Potty stools produced daily) and X2 (number of Hardy stools produced daily). The objective function is to maximize the total daily profit: Maximize 10.5X1 + 15X2.

## Constraints

The constraints include:

* Plastic: X1 + 1.5X2 <= 350
* Legs: X1 + X2 <= 300
* Production Time: 15X1 + 24X2 <= 4800
* Non-negativity: X1, X2 >= 0

## Mathematical Model

The mathematical model is: Maximize 10.5X1 + 15X2, subject to the constraints.

## Solution

The solution to the LP problem is X1 = 266.67 and X2 = 33.33, with a maximum profit of $3300 per day.

## Sensitivity Analysis

The sensitivity analysis includes:

* Range of Optimality: The range of values for which the optimal solution remains unchanged.
* Reduced Costs: The amount by which the optimal profit will change per unit increase in a variable.
* Shadow Price: The premium value above the existing unit value for a resource if the need arises to purchase more resources.

## Analysis of Available Solutions

The analysis includes evaluating options such as seeking additional sources of plastic, taking up Yuen Supplies' offer to deliver extra sets of legs, and adding a part-time worker.

## Recommendations

Based on the analysis, the company can consider the options and make informed decisions to maximize profit and optimize resource allocation.


To find the real solutions of the equation log(x+2) x + log(x) (x+2) = 5/2, we start by using the relation log(a) b = 1/log(b) a and log a + log b = log a*b.

Given the equation:
log(x+2) x + log(x) (x+2) = 5/2

We apply the logarithmic relation:
1/log(x) (x+2) + log(x) (x+2) = 5/2

Let log(x) (x + 2) = y, then:
1/y + y = 5/2

Multiplying through by 2y to clear the fraction gives:
2 + 2y^2 = 5y

Rearranging the equation:
2y^2 - 5y + 2 = 0

Factoring the quadratic equation:
2y^2 - 4y - y + 2 = 0
2y(y - 2) - 1(y - 2) = 0
(2y - 1)(y - 2) = 0

This yields two possible values for y:
y = 1/2 and y = 2

Substituting back for y = log(x) (x + 2):
1. For y = 2:
log(x) (x + 2) = 2
x + 2 = x^2
x^2 - x - 2 = 0
Factoring:
(x + 1)(x - 2) = 0
So, x = -1 or x = 2. However, x cannot be -1 because the log of a negative number is undefined.

2. For y = 1/2:
log(x) (x + 2) = 1/2
x + 2 = sqrt(x)
(x + 2)^2 = x
x^2 + 4x + 4 = x
x^2 + 3x + 4 = 0

Applying the quadratic formula:
x = [-3 ± sqrt(9 - 16)] / 2
x = [-3 ± sqrt(-7)] / 2

Since sqrt(-7) is undefined, this equation has no real solutions.

Considering the constraints for the existence of logarithms:
- x > 0
- x + 2 > 0, thus x > -2
- x cannot be 1 (since it would make one of the logarithms undefined)
- x + 2 cannot be 1, thus x cannot be -1

The only valid solution that respects these constraints and solves the equation is x = 2.


# Matrix Decomposition Decoded

This article covers matrix decomposition, eigenvalues, and eigenvectors, and discusses the purpose of matrices and vectors in linear algebra.

## Purpose of Matrices and Vectors

In machine learning, matrices have features as columns. The purpose of a matrix is to "act" on input vectors to give output vectors. This is similar to a function in calculus, where a number is input and a transformed number is output. The matrix transforms the input vector into the output vector.

## Intuition behind Eigenvalues and Eigenvectors

An eigenvector is an input vector that, when transformed by a matrix, results in an output vector that is parallel to the input vector. The eigenvalue represents the magnitude of the transformation. A matrix can be decomposed into a vector and its magnitude, written as a product of both: A.x = λx, where A is a square matrix, x is an eigenvector of A, and λ is the scalar eigenvalue.

## Finding Eigenvectors

To find eigenvectors, we need to solve the equation (A - λI).x = 0, where I is the identity matrix. The determinant of (A - λI) must be zero, so we solve |A - λI| = 0 to find the eigenvalues. We can then substitute each eigenvalue into the equation to find the corresponding eigenvector.

## Properties of Eigenvalues and Eigenvectors

1. Eigenvectors and eigenvalues can only be found for square matrices.
2. A square matrix of shape (n, n) has at most n independent eigenvalues and eigenvectors.
3. The sum of eigenvalues is equal to the trace of the matrix (sum of diagonals).
4. The product of eigenvalues is equal to the determinant of the matrix.
5. For a triangular matrix, the eigenvalues are the diagonal values.
6. Repeated eigenvalues result in one less independent eigenvector.
7. Symmetric matrices produce real eigenvalues, while asymmetric matrices produce complex eigenvalues.

## Applications of Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are useful for calculating powers of matrices. A matrix A can be factorized into three terms: S, Λ, and S^-1, where S is the eigenvector matrix, Λ is the diagonal matrix of eigenvalues, and S^-1 is the inverse of S. This allows us to calculate A^K as S.Λ^K.S^-1.

## Principal Component Analysis (PCA)

PCA is used for feature extraction and dimensionality reduction. It reduces the number of variables in a dataset by projecting them onto a lower-dimensional space. The goal is to preserve the most important information in the data while reducing noise.

## Steps of PCA

1. Normalize the data by subtracting the mean and standardizing.
2. Calculate the covariance matrix of the normalized data.
3. Decompose the covariance matrix into eigenvalues and eigenvectors using np.linalg.eig().
4. Sort the eigenvalues in descending order and select the top k eigenvalues to retain maximum explained variance.
5. The corresponding eigenvectors represent the components of the reduced subspace.

## Importance of Eigenvalues and Eigenvectors in PCA

The eigenvalues and eigenvectors are crucial in PCA as they help to reduce the dimensionality of the data while preserving the most important information. The eigenvectors represent the new axes in the reduced subspace, and the eigenvalues represent the amount of variance explained by each axis. By selecting the top k eigenvalues, we can retain the maximum amount of information in the data.


# Interpolation and Extrapolation

Interpolation is the process of estimating a value between two known values of a function by approximating a line or smooth curve between them. The term "interpolate" originates from the Latin prefix "inter," meaning "between," and "polire," meaning "to adorn or polish," resulting in "to smooth between." Extrapolation, on the other hand, involves estimating values outside the range of known points by extending the interpolation concept.

These techniques are used in mathematical analysis to estimate unknown values of a dependent variable based on known values of an independent variable. This is achieved by finding a function f(x) that closely approximates the known corresponding values of the dependent variable. In interpolation, it is assumed that the relationship holds within the range of known values, whereas in extrapolation, this relationship is assumed to hold outside the known range, which is a less justifiable assumption.

The simplest and most commonly used technique for interpolation and extrapolation is drawing a straight line or curve through a set of points on a graph, assuming it represents a genuine relationship between the variables. Other techniques include using partial difference equations. Interpolation and extrapolation are essential in statistics, particularly when studying patterns over time to predict future events. By applying these techniques, researchers can make informed estimates and predictions based on available data.


# Tensor Calculus Lecture 15: Geodesic Curvature Preview

This video explores the relationship between curves and surfaces in differential geometry, focusing on the curvature normal and its interpretation. The script discusses the formula that links the behavior of a curve embedded in a surface to the ambient Euclidean space, highlighting the concepts of principal curvatures, Gaussian curvature, and geodesic curvature.

## Takeaways

- The video script discusses differential geometry and tensor calculus, focusing on the relationship between curves and surfaces in an ambient Euclidean space.
- The script revisits a complex formula that connects various aspects of curves embedded in surfaces, emphasizing the importance of understanding the behavior of both the curve and the surface.
- The concept of principal curvatures and Gaussian curvature is introduced, using a derived formula to explain these geometric properties.
- The 'V alpha beta' term from the formula represents the curvature tensor for the surface in the Euclidean space, which is key to understanding the surface's curvature.
- The 'geodesic curvature' is highlighted as the main focus of the video, explaining it as the mean curvature for a curve embedded in a surface.
- The script explains how to construct the curvature tensor for a curve embedded in a surface by applying the same logic used for a surface in Euclidean space, but with adjustments for the curved space.
- The 'curvature normal' is described as an invariant vector that represents the rate of change of the unit tangent vector along the curve, providing an intuitive understanding of curvature.
- The script uses an analogy to illustrate the concept of the curvature normal being orthogonal to the axis of a curve, helping to visualize the concept.
- The formula presented in the script decomposes the curvature normal into components related to the surface normal and the tangential normal, providing a deeper understanding of the curve's embedding.
- The concept of 'geodesic' is introduced as the shortest path between two points on a surface, with the script suggesting that the geodesic curvature is zero for the shortest path.
- An intuitive demonstration using an apple as a curved surface helps to visualize the geodesic and its curvature, emphasizing the idea that the geodesic curvature vanishes for the shortest path.

## Q & A

### What is the main topic of discussion in the video script?
The main topic of discussion in the video script is the relationship between the curvatures of a curve embedded in a surface and the surface itself, with a focus on differential geometry and tensor calculus.

### What is the significance of the formula derived in the last video?
The formula derived in the last video is significant as it relates the behavior of a curve with the behavior of the surface in which it is embedded, connecting various aspects of curvatures in the context of differential geometry.

### What are the two main normals mentioned in the script when discussing an embedded curve?
The two main normals mentioned are the unit normal of the surface and the tangential normal of the curve, which lies in the plane tangential to the surface.

### What is the curvature tensor V_alpha_beta mentioned in the script?
V_alpha_beta is the curvature tensor for the surface as embedded in the overall Euclidean space, representing the curvature normal, which is a key concept in the script's discussion.

### What is the geodesic curvature and why is it the focus of the video?
The geodesic curvature is the curvature tensor or mean curvature for the curve as embedded in the surface. It is the focus of the video because it provides an intuitive understanding of how curves behave when embedded in surfaces and is central to the concept of shortest paths or geodesics.

### How is the curvature normal related to the rate of change of the unit tangent vector along a curve?
The curvature normal is directly related to the rate of change of the unit tangent vector along a curve, representing the vector rate of change that points inward and is proportional to the curvature of the curve.

### What is the physical interpretation of the curvature normal?
The physical interpretation of the curvature normal is the acceleration of a material particle moving along the curve with unit speed, where the derivative points in the direction of the inward normal to the curve.

### How does the script relate the curvature normal to the surface normal and the tangential normal?
The script explains that the curvature normal can be decomposed into components along the surface normal and the tangential normal, expressing any vector orthogonal to the curve as a linear combination of these two normals.

### What is the concept of principal curvatures mentioned in the script?
Principal curvatures are the components of the curvature normal in the direction normal to the surface, leading to the concept of in-plane or tangential curvature, which is part of the analysis of the curve embedded in the surface.

### Why is the geodesic curvature considered to be zero for the shortest path between two points on a surface?
The geodesic curvature is considered to be zero for the shortest path, or geodesic, because this path does not meander and goes straight over the surface, lacking an in-plane component to its curvature normal, thus making the total geodesic curvature zero.

### How does the script use an apple to illustrate the concept of geodesics and curvature normals?
The script uses an apple as a curved surface to visually demonstrate the shortest path between two points (geodesic) and how the curvature normal aligns with the surface normal for this path, indicating zero geodesic curvature.

## Introduction to Differential Geometry and Tensor Calculus
The speaker returns to the topic of differential geometry and tensor calculus, focusing on the relationship between curves and surfaces in Euclidean space. They introduce the concept of principal curvatures and Gaussian curvature, reminding viewers of a previously derived formula that connects the behavior of a curve embedded in a surface to the behavior of the surface itself.

## Exploring Curvature Normals and Their Intuitive Interpretation
This section delves into the concept of curvature normals, which represent the rate of change of the unit tangent vector along a curve. The speaker uses an analogy to illustrate how the curvature normal points inward and is proportional to the curve's steepness. They explain that the curvature normal can be decomposed into components related to the surface and the curve's embedding, leading to the idea of principal curvatures.

## Geodesics and the Intuitive Link to Zero Curvature
The speaker uses an apple as a model for a curved surface to explain the concept of geodesics as the shortest paths between two points on a surface. They argue intuitively that the curvature normal for a geodesic should be aligned with the surface's normal, implying that the in-plane component of the curvature normal is zero for geodesics. This results in the geodesic curvature being zero, which is a key point in understanding the relationship between shortest distances on a surface and the curvature properties of that surface.


## What is Energy Density?

Energy density is the amount of energy stored in a given system or region of space per unit volume. It can also refer to energy per unit mass, although the more accurate term for this is specific energy.

## What is Specific Energy of a Fuel?

Specific energy is the energy per unit mass of a fuel, expressed in joule per kilogram (J/kg) or equivalent units. It decreases with increasing oxygen content of the fuel.

## Energy Density of Fuel

Energy density is generally expressed in two ways:
- Volumetric energy density: how much energy a system contains in comparison to its volume, typically expressed in watt-hours per liter (Wh/L) or Megajoules per liter (MJ/L).
- Fuel energy density.

The following table lists the energy density of various fuels:
| Fuel | Energy Density |
| --- | --- |
| Natural gas | 55 MJ/kg |
| Propane | 13.8 Wh/kg, 25.4 MJ/kg |
| Petrol | 13.0 Wh/kg, 34.7 MJ/kg |
| Diesel oil (DERV) | 12.7 Wh/kg, 37.9 MJ/kg |
| Kerosene | 12.8 Wh/kg, 37 MJ/kg |

## Energy Content of Fuels

One liter of diesel fuel has an energy content of approximately 38 MJ, which is equivalent to 10 kWh. However, the efficiency of conversion into kinetic energy is only about 30%. In comparison, petrol has an efficiency of about 25%.

A liter of gasoline contains 31,536,000 joules of energy, which is equivalent to 8.76 kW/hr. The calorific values of different fuels are:
- Diesel: 10.7 kWh/l
- Jet fuel: 10.4 kWh/l
- Petrol: 9.7 kWh/l

The energy content of a gallon of fuel is:
- Diesel Fuel: 139,000 Btu
- Gasoline: 124,000 Btu
- Heating Oil: 139,000 Btu

## Calculating Energy Content

Energy content or calorific value can be calculated from thermodynamical values or measured in a suitable apparatus. A known amount of fuel is burned at constant pressure and under standard conditions, and the heat released is captured in a known mass of water in a calorimeter.

## Energy Density Formula

Energy density is calculated using the formula: Ed = E/V, where Ed is the energy density, E is the energy, and V is the volume.

## Biofuel Energy Content

The energy content of biofuel is a description of the potential energy contained in a given biofuel, measured per unit mass of that fuel as specific energy, or per unit of volume of the fuel as energy density.


# Linear Regression Algorithm Method

Linear regression is a statistical analysis technique used to test for a linear relationship between two factors. It produces a straight line in its output, with the slope indicating the degree of association between the two factors. The y-intercept represents the value of one variable at a value of zero for the other.

Although regression analysis helps identify associations between variables, it cannot prove causation. It is widely used in economics, business, and finance to learn about relationships between variables, such as commodity prices and stocks, and to determine the fair market value of assets.

## Types of Regression

There are several types of regression, including:

* Linear regression algorithm for data analysis
* Stepwise regression with a polynomial model and logistic regression
* Ridge regression
* Lasso regression
* ElasticNet regression

## What is Linear Regression?

Linear regression is a machine learning method that relies on supervised learning. It uses explanatory factors to make predictions and research associations between factors. The technique is useful for making predictions and identifying relationships between variables.

## Algebraic Form of Linear Regression

The algebraic form of linear regression is y = a0 + a1x + ε, where:

* Y is the dependent variable
* X is the independent variable
* a0 is the y-intercept
* a1 is the linear regression constant
* ε is the assumption error

## Methods of Linear Prediction

There are two types of methods used to perform linear regression:

* Single-factor regression: predicts a numerical dependent variable using a single independent variable
* Multiple linear regression: predicts a numerical dependent variable using multiple independent variables

## Single-Factor Regression

Single-factor regression uses a single independent variable to predict a numerical dependent variable. The equation for single-factor regression is Y = β0 + β1*x, where β0 is the intercept and β1 is the slope.

## Multiple Linear Regression

Multiple linear regression uses multiple independent variables to predict a numerical dependent variable. The equation for multiple linear regression is Y = β0 + β1*x1 + β2*x2 + … + βn\*xn, where β0 is the intercept and β1, β2, …, βn are the slopes.

## Purpose of Regression Analysis

The purpose of regression analysis is to identify the line that best fits the data. The best-fit line is obtained by minimizing the sum of the forecast errors.

## Example of Linear Regression

For example, if we have a dataset showing a link between the number of hours studied and marks earned, we can use linear regression to determine which is more crucial. The regression line minimizes errors when analyzing example data.

## Multiple Linear Regression

Multiple linear regression explains a dependent variable using multiple independent variables. It is used to identify the dependent variable from a collection of independent variables and to quantify the level of intercorrelation between the independent variables.

## Differences Between Linear and Multiple Regression

Linear regression is used to identify the relationship between two factors, while multiple regression is used to identify the relationships between multiple factors. For example, an analyst can use linear regression to find a relationship between stock price changes and trading activity, and multiple regression to determine which factors, such as P/E ratio, dividends, and inflation, affect the stock price and to what degree.


Addition of literals follows the same properties as addition of numbers. The sum of two literals, such as y and 7, is denoted by y + 7, which can be read as "y plus 7" or "7 more than y." Similarly, the sum of two literals x and y is written as x + y, which can also be read as the sum of x and y.

The expression (x + y) + z means that the sum of literals x and y is added to the literal z, whereas x + (y + z) means that the literal x is added to the sum of literals y and z. For example, the sum of m and 25 is written as m + 25, and the sum of m and n is written as m + n. The sum of m and m is written as m + m = 2m, and the sum of m, n, and 5 is written as m + n + 5.

Properties of Addition of Literals:
1. Commutativity: For any two literals a and b, a + b = b + a. For example, 2 + x = x + 2.
2. Associativity: For any three literals a, b, and c, (a + b) + c = a + (b + c). For example, (3 + x) + y = 3 + (x + y).
3. Identity: For any literal a, a + 0 = a = 0 + a. For example, 5 + 0 = 5 = 0 + 5, where 0 is the additive identity.

Examples of writing phrases using numbers, literals, and the basic operation of addition:
1. The sum of x and 3: x + 3
2. The sum of 10 and z: 10 + z
3. The sum of x and y: x + y
4. 3 more than a number x: x + 3
5. 100 more than a number p: p + 100
6. x added to 9: x + 9
7. m added to 50: m + 50
8. Increase x by 4: x + 4
9. Increase z by 10: z + 10
10. The sum of x and 5 added to y: (x + 5) + y
11. The sum of m and 25 added to n: (m + 25) + n
12. The sum of x and y added to z: (x + y) + z
13. The sum of a and b added to 5: (a + b) + 5
14. x added to the sum of y and 4: x + (y + 4)
15. 5 added to the sum of a and b: 5 + (a + b)
16. a added to the sum of y and z: a + (y + z)


To calculate joules of energy, enter the total kilowatts and total time into the calculator.

## KW to Joules Formula

The formula to calculate joules from kilowatts is: J = KW * T * 1000, where J is joules, KW is kilowatts, and T is time in seconds. This formula is used because kilowatts measure power, which is energy per unit time, and joules measure energy. To convert between the two, power must be multiplied by time.

## Converting KW to Joules

KW is a measure of power, and joules are a measure of energy. The conversion involves multiplying the power by the time, as energy equals power multiplied by time.

## Calculating Joules from KW

An example problem illustrates the steps to calculate joules from kW. First, determine the total power, which in this example is 10 kW. Next, determine the total time, which is 500 seconds. Then, apply the formula: J = KW * T * 1000. Substituting the given values: J = 10 * 500 * 1000 = 5,000,000 joules. This calculation demonstrates how to find the total amount of joules from a given amount of kilowatts and time.


## Proportional Valve Control Calculation. Part 2.

**Part 2. Proportional Control in the Regeneration Mode.**

The goal of this calculation is to determine the cylinder velocity at the current operating voltage at the valve's solenoid. To perform the calculation, we need to know the following parameters:

- Cylinder and valve parameters, which can be obtained from the manufacturer's catalog
- Values of forces at the cylinder rod (load)
- Pressure drop at the return line, which can be assumed as zero if unknown

**Summary:**

1. **Regeneration Mode Schematic**
2. **Regeneration Mode: Cylinder Extension Calculation**
   2.1. **Balance of Forces at the Cylinder**: This step involves analyzing the forces acting on the cylinder.
   2.2. **Spool Lands Ratio and Pressure Drops at the Spool Lands**: Understanding the spool lands ratio and the pressure drops is crucial.
   2.3. **Hydraulic Cylinder Extension. System Pressure Calculation**: Calculate the system pressure during cylinder extension.
   2.4. **Flow through Spool Lands: Max. and at Current Voltage**: Determine the flow rates through the spool lands.
3. **Hydraulic Cylinder Retraction Calculation**
4. **Final Words**


The radiance of frequency $f$ from a blackbody of temperature $T$ is given by $R(f,T) = \gamma Tf^2$ for $f \leq T$ and $R(f,T) = 0$ for $f > T$. The total radiative transfer $R_{True}$ to a blackbody 1 of temperature $T_1$ from a blackbody 2 of temperature $T_2 > T_1$ is given by integration over frequencies as follows: $R_{True} = \int_{T_1}^{T_2} \gamma T_2 f^2 df + \int_0^{T_1} \gamma (T_2 - T_1) f^2 df \equiv I_1 + I_2$.

The first integral $I_1$ represents the heating effect from blackbody 2 above the cut-off of blackbody 1, and the second integral represents the net heating from blackbody 2 below the cut-off. $R_{True}$ is the sum of two integrals with positive integrands, indicating that it is the sum of many small positive contributions, all with transfer of heat from blackbody 2 to blackbody 1.

False-SB arises by rewriting $I_1$ as follows: $I_1 = \int_0^{T_2} \gamma T_2 f^2 df - \int_0^{T_1} \gamma T_2 f^2 df = \sigma T_2^4 - \sigma T_2 T_1^3$, where $\sigma = \frac{\gamma}{3}$. This gives False-SB on the form $R_{False} = \sigma T_2^4 - \sigma T_1^4$, expressing the transfer of energy from blackbody 2 to blackbody 1 as the difference of two gross flows with different signs.

The use of a simplified high-frequency cut-off is not supported by observational data, as real blackbody spectra do not exhibit a cut-off. Planck's law does not have a cut-off, but rather an exponential decline. The concept of a cut-off is a simplification of this decline, but it is not a physical reality.

The second law of thermodynamics states that heat transfer is from hot to cold, and net radiative energy flow can be in the opposite direction. However, this does not imply a violation of the second law, as the net radiative energy flow is not the same as heat transfer. The second law is often formulated in terms of entropy, which is a statistical concept. However, the use of statistics in physics is not universally accepted, and some argue that physics should be based on deterministic laws rather than statistical mechanics.

The concept of analog computation is proposed as an alternative to statistical mechanics, where atoms are seen as performing finite precision computations rather than throwing dice. This idea is based on the concept of finite precision computation, where atoms chop decimals rather than having infinite precision. However, this idea is not widely accepted and is seen as a speculative approach to physics.

The discussion of the second law of thermodynamics and the concept of entropy is complex and often ambiguous. The use of statistics in physics is not universally accepted, and some argue that it is not a fundamental aspect of physics. The concept of analog computation is proposed as an alternative, but it is not widely accepted and requires further development and clarification.


About 15 years ago, a model was developed to predict the NBA's MVP Award winner in a given season. The goal was to model the behavior of the voters, not to determine the "most deserving" player. The model uses data from the 1980-81 season as its starting point, when the voting shifted from the players to the media. Every MVP Award winner since then has met the following criteria: played for one team, played in at least 80% of their team's games, and averaged at least 30 minutes per game.

These standards establish a player pool, with 62 eligible players in the last season, averaging over two players per team. A logistic regression model was built with an indicator for MVP Award winner as the dependent variable. The independent variables considered were basic statistics such as points per game, rebounds per game, and assists per game, while advanced metrics like Player Efficiency Rating and Win Shares were not.

The model identified four statistically significant factors: points per game, rebounds per game, assists per game, and team winning percentage. The logistic regression model has the following parameters: Intercept = –37.7, Points per game = 0.40, Rebounds per game = 0.46, Assists per game = 0.49, and Team winning percentage = 26.6.

Using Nikola Jokic's 2023-24 season as an example, with 26.4 PPG, 12.4 RPG, and 9.0 APG for a team with a winning percentage of .695, the model yields a logit of 1.39, which is converted into a probability of 0.801. After adjusting the probabilities to sum to one, Jokic's adjusted probability is 0.316.

The top players with an adjusted probability greater than 0.05 last season were: Nikola Jokic (0.316), Luka Doncic (0.295), Jayson Tatum (0.191), Giannis Antetokounmpo (0.094), and Shai Gilgeous-Alexander (0.066). The model correctly predicts the MVP Award winner in 32 out of 44 seasons (72.7%), including 15 out of the last 18 (83.3%).

The model's results are summarized in Q&A format:

#### How often does the model identify the winner?
The model correctly predicts the MVP Award winner in 32 out of 44 seasons (72.7%), including 15 out of the last 18 (83.3%). The only misses over the last 18 seasons are Derrick Rose in 2010-11, Russell Westbrook in 2016-17, and Nikola Jokic in 2021-22.

#### What's the biggest miss by the model?
In 2005-06, MVP Award recipient Steve Nash had the ninth-highest P value (0.013), with the top P value belonging to Dirk Nowitzki. The second-biggest miss was Nash in 2004-05, with the seventh-highest P value (0.047), while the model pegged Amar'e Stoudemire as the probable winner.

#### Who had the highest P value?
LeBron James had a P value of 0.909 in 2009-10, with Kobe Bryant a distant second at 0.033. The gap of 0.876 is the largest between the top two candidates.

#### What's the highest P value by a non-winner?
Karl Malone had a P value of 0.699 in 1997-98, finishing second in the voting to Michael Jordan. Malone's Utah Jazz tied Jordan's Chicago Bulls for the NBA lead in wins, with Malone averaging more rebounds and assists than Jordan.

#### What's the lowest P value by a winner?
Steve Nash in 2005-06 (0.013) and 2004-05 (0.047) had the lowest P values by a winner, followed by Allen Iverson in 2000-01 (0.063).

#### Who was "snubbed" the most times?
Larry Bird is the only player to finish with the highest P value multiple times in seasons in which he did not win the award, with the highest P values in 1980-81, 1981-82, and 1987-88.

#### Who has the most seasons with the highest P value?
Larry Bird finished with the highest P value six times, followed by LeBron James with five, and Michael Jordan with four.

#### What's the smallest gap between the top two?
David Robinson held a slim 0.007 edge over Hakeem Olajuwon in 1993-94, with Olajuwon picking up his first and only MVP Award. Robinson finished second in the voting but won the award the following season.

#### What was the deepest race?
Four players had a P value of 0.200 or higher in 2021-22: Giannis Antetokounmpo (0.294), Nikola Jokic (0.277), Joel Embiid (0.203), and Devin Booker (0.202). Those players finished in the top four in the actual voting.

#### What's the highest P value by a player who received no MVP votes?
Klay Thompson had a P value of 0.046 in 2015-16 but did not receive any MVP votes, while his backcourt mate Stephen Curry had the highest P-value that season (0.531) and became the first unanimous MVP Award winner.


A matrix consisting entirely of zeros is called a zero matrix and is denoted by O. The nxn matrix consisting of 1's on its main diagonal and 0's elsewhere is called the identity matrix of order n. A matrix is said to be a zero matrix, if all its entries are zero and is denoted by O.


# What Percent is 25 of 30? Calculate a Percentage Using Decimals and Fractions

To calculate what percent 25 is of 30, we need to understand how to calculate percentages using decimals and fractions. This involves multiplying the percentage value by the quantity to find its value, then multiplying the result by 100.

## Calculating 25 of 30

The percentage formula is an algebraic equation involving three values: P (percentage), V1 (first value), and V2 (result of the percentage operating on V1). To calculate 25 of 30, we can use the formula: (25/30) * 100.

## Calculating Fractions

A fraction is a part of something greater than or equal to another number, with a numerator representing the parts and a denominator representing the total parts. For example, 25/30 can be simplified to 5/6 by finding the greatest common factor (GCF) between 25 and 30, which is 5.

To simplify fractions, we can use equivalent fractions by multiplying the numerator and denominator by the same number. We can also use a calculator to find equivalent fractions. Converting fractions to decimals involves dividing the numerator by the denominator, which can be done by hand or using a calculator.

## Calculating Decimals

To find the decimal equivalent of 25/30, we can use long division or a calculator. The result is approximately 0.83. To convert this decimal to a percentage, we multiply by 100, which gives us 83%.

Using a calculator is a great way to convert percentages to decimals. We can simply multiply the amount by the percentage, then add the percent symbol after the result. For example, 25/30 = 0.83, which is approximately 83% when multiplied by 100.

In summary, to calculate what percent 25 is of 30, we can use the formula: (25/30) * 100, which gives us approximately 83%. This can be done using decimals, fractions, or a combination of both.


### Abstract

The distance-two constrained labeling of graphs is a problem that arises in the context of frequency assignment in mobile and wireless networks. This problem involves assigning frequencies to network stations to minimize interference between nearby stations while exploiting frequency reusability. It was first formulated as a graph coloring problem by Hale, who introduced the concept of T-coloring. In 1988, Roberts proposed a variation of the channel assignment problem, which led to the study of L(2, 1)-labeling of graphs. This concept has attracted significant interest due to its practical and theoretical applications.

The L(2, 1)-labeling of a graph involves assigning labels to vertices such that vertices at distance two have labels that differ by at least two. This problem has been studied extensively, and many aspects of it remain to be explored. In this thesis, an upper bound for the L(2, 1)-labeling number, denoted as λ, is established for a special class of graphs called n-cubes (Qn), where n = 2k + 1. Additionally, the L(1, 1)-labeling number of Qn is studied, and a lower bound for λ1(Qn) is provided.

A related problem, L(2, 1)-choosability, is also investigated. This problem involves determining the minimum number of labels required such that each vertex can be assigned a label from a given list, subject to the L(2, 1)-labeling constraint. The L(2, 1)-choosability of paths, cycles, Cartesian product graphs, and generalized Petersen graphs is studied, and upper bounds for the L(2, 1)-choice numbers are established for these graphs.

Key concepts in this study include distance-two labeling, λ-number, L(2, 1)-labeling, L(d, 1)-labeling, list-L(2, 1)-labeling, choosability, and L(2, 1)-choice number. The graphs studied include paths, cycles, n-cubes, spiders, Cartesian product graphs, and generalized Petersen graphs.

### Details

* Date of Award: 2013
* Original language: English
* Supervisor: Wai Chee SHIU

### Keywords

* Graph theory
* Graph labelings
* Distance-two labeling
* λ-number
* L(2, 1)-labeling
* L(d, 1)-labeling
* List-L(2, 1)-labeling
* Choosability
* L(2, 1)-choice number
* Path
* Cycle
* n-cube
* Spider
* Cartesian product graph
* Generalized Petersen graph


**Growing Tax Burden on a per Capita Basis**

Owen Sound's population declined in 2011 and 2016, according to Statistics Canada. The 2021 census showed a slight rebound to 21,612, still less than the 2011 population of 21,688.

To calculate the cost of running the city per person, total expenses from Figure 2 were divided by the population data for census years. Figure 4 shows the cost per person increased from $1,851 in 2011 to $2,522 in 2021, a $671 increase (36%) over ten years. This exceeds the Consumer Price Index (CPI) increase rate. Despite a declining population, the cost per person rose. Statistics Canada reported 141 fewer people in Owen Sound in 2021 than in 2011, yet the city spent $13.2 million (32.7%) more to deliver the same services.

The question arises: "Why does it cost more to deliver city services to fewer people?" The trend line for this data can be calculated using the equation y=$67x-$133,058. Estimating the cost per person for 2026 and 2031 yields $2,684 and $3,019, respectively. This raises another question: if the population declined from 2011 to 2020, why did expenses increase by $12.3 million (30.7%)?

The "Cost per Person" graph illustrates that increasing expenses were passed directly to a declining population during the study period. Owen Sound's census data is available for further reference.


**SPIRALS**

The session began with a series of questions about spirals, including what a spiral is, whether it can be constructed with a compass, and whether it can have straight lines. The students also explored how to make a Baravelle spiral.

**EUCLID REVISITED AGAIN**

The class then moved on to Euclid cards, which are based on the definitions from Euclid's Elements Book I. The students drew cards and attempted to define the terms, construct the items with a compass and straightedge, or ask for help. The cards drawn included *#4 Straight Line*, *#11 Obtuse Angle*, *#2 Line*, and *#14 Figure*. The students discussed the differences between a line and a straight line, and whether all lines are straight. They also explored the concept of a figure, which must be enclosed by a boundary, and whether a spiral is a figure.

The students refined their definition of a spiral to "almost a circle, not a figure, the radius gets bigger." They also discussed whether a ghost can be considered a figure, and whether Euclid had fans in his time.

**THE BEAR PROBLEM**

The class then tackled the infamous "Bear Problem," which states: "A woman goes for a walk. She walks 3 miles due south, then 3 miles due west, then 3 miles due north. She notices that she is back where she started, and she sees a bear. What color is the bear?" The students were initially perplexed but eventually realized that the problem could be solved using math. One student announced that the bear was white, and the class discussed how this answer was related to the earth being round.

The students used dramatization to understand the problem, with one student acting as the globe and an eraser representing the woman's path. Eventually, one student explained why the bear was white using mathematics. The class then discussed whether the path the woman walked was a triangle and why Euclid did not define a triangle in general in his definition list.

The session ended with the students showing their compass art and discussing potential future math classes. The instructor reflected on the session, noting that one student had become upset when another student announced the answer to the Bear Problem, and wondering how this incident could have been handled differently.

Multiple choice questions were not provided in the original text, but some questions that can be derived from the text are:
- What is the definition of a spiral?
A) A circle with a constant radius
B) A shape with a constantly increasing radius
C) Almost a circle, not a figure, the radius gets bigger
D) A straight line

Answer: C) Almost a circle, not a figure, the radius gets bigger

- What is a figure according to Euclid?
A) A diagram
B) A shape enclosed by a boundary
C) A straight line
D) A curve

Answer: B) A shape enclosed by a boundary

- What color is the bear in the Bear Problem?
A) Black
B) White
C) Brown
D) Gray

Answer: B) White


Dan Krejsa is attempting to define an isomorphism from `c2` to itself using cubical Agda. The `c2` type represents a circle broken into two halves, with `b0` at noon and `b1` at 6 o'clock. The circle has two paths: `p0` moving clockwise from `b0` to `b1` and `p1` moving clockwise from `b1` to `b0`.

The goal is to define two functions, `f` and `g`, where `f` is the identity function and `g` maps each point on the circle to the opposite point 180 degrees away. Dan wants to show that `g` is a section of `f` and a retract of `f` by defining `fgid` and `gfid`.

The `fgid` function should be a path that goes clockwise 180 degrees from `f (g y)` to `y`, while the `gfid` function should be a path that goes counterclockwise 180 degrees from `g (f x)` to `x`. However, Dan is having trouble defining `fgid` and `gfid` due to issues with definitional equality.

The `f` and `g` functions are defined as follows:
```agda
f : c2 → c2
f x = x

g : c2 → c2
g b0 = b1
g b1 = b0
g (p0 i) = p1 i
g (p1 i) = p0 i
```
The `fgid` function is partially defined, but the definitions for `b0` and `b1` are missing:
```agda
fgid : (y : c2) → f (g y) ≡ y
fgid b0 = {!!}
fgid b1 = {!!}
fgid (p0 i) = (λ j → p1 (i ∨ j)) ∙ (λ j → p0 (i ∧ j))
fgid (p1 i) = (λ j → p0 (i ∨ j)) ∙ (λ j → p1 (i ∧ j))
```
The `gfid` function is also undefined:
```agda
gfid : (x : c2) → g (f x) ≡ x
gfid x = {!!}
```
The main challenge is to define `fgid` and `gfid` in a way that satisfies the definitional equality constraints.

Possible approaches to solve this problem include:
1. Using a different definition for `fgid` and `gfid` that avoids the definitional equality issues.
2. Using a lemma or theorem to prove that `g` is a section of `f` and a retract of `f`.
3. Using a different approach to define the isomorphism between `c2` and itself.

Note: The original email discussion is available at http://lists.chalmers.se/pipermail/agda/attachments/20220411/56304aa9/attachment.html.


The constant proportionality in the equation y = 5x is 5.

To understand this, we need to know about proportionality. There are two types of proportion: direct and indirect proportion. In direct proportion, an increase or decrease in one quantity causes an increase or decrease in the other quantity. This can be represented mathematically as y ∝ x.

When we remove the proportionality symbol, we introduce a proportionality constant, resulting in the equation y = kx, where k is the proportionality constant. Comparing this general form to the given equation y = 5x, we see that k = 5.

Therefore, 5 is the proportionality constant in the given equation. Key concepts related to this include proportion and constant.


Math students celebrate Pi Day on the 14th day of the 3rd month. The symbol π (pronounced paɪ in English) is the sixteenth letter of the Greek alphabet, representing a real number of special significance. When π is written in decimal notation, it begins 3.14, suggesting the date 3/14. In fact, the decimal expansion of π begins 3.1415.

π is defined as the ratio between the circumference of a circle and the diameter of that circle. This ratio is the same for any size circle, making it intrinsically attached to the idea of circularity. People have been studying this fundamental ratio since at least the ancient Babylonians.

You can see that π is greater than 3 by looking at a hexagon inscribed within a circle, and less than 4 by looking at the square that circumscribes a circle. Experimentation with a measuring tape and a dinner plate suggests that π might be 22/7, but this is only an approximation. The fraction 355/113 is much closer to π but still not exactly equal to it.

In 1761, Swiss mathematician Johann Lambert proved that no fraction exactly equals π, implying its decimal expansion is never-ending with no repeated pattern. The German mathematician Ferdinand Lindemann proved in 1882 that π is transcendental, meaning it does not solve any polynomial equation with integer coefficients.

π appears in many different mathematical contexts and across all mathematical areas. It is the ratio of the area of a circle to the area of the square built on the radius of the circle, and the ratio of the surface area of a sphere to the area of the square built on the diameter of the square. The area under the bell-shaped curve y=1/(1+x²) is π, and the probability that a pin crosses one of the lines on a sheet of lined paper is 2/π.

There are thousands of formulas for π, including one discovered by Srinivasan Ramanujan in 1914. π is a universal constant that is ubiquitous across mathematics, existing prior to the Big Bang and remaining permanent and unchanging. The celebration of Pi Day seems silly, as the connections between π and human-made inventions like the Gregorian calendar and pies are arbitrary.

No one knows whether each of the ten digits – 0 through 9 – appears with equal frequency in the decimal expansion of π. This question is out of reach of modern mathematics, and it's unclear how to attack it effectively. π is equal to … π, and its exact nature remains a mystery. Despite being able to compute trillions of decimal digits of π, machines do not help us understand what π is exactly.


**3-Digit Subtraction**

3-digit subtraction is the subtraction of numbers where the minuend consists of 3 digits and the subtrahend can be of either 1, 2, or 3 digits. Subtraction is an operation in math that gives the difference between two numbers. The number from which we subtract is called the minuend, and the number that is subtracted from the minuend is called the subtrahend.

**What is 3-Digit Subtraction?**

In 3-digit subtraction, we need to subtract the given numbers after placing them correctly according to their place values. We need to ensure that the bigger number is placed in the upper row, while the smaller number is placed below it. After aligning them into columns of ones, tens, and hundreds, we start the process of subtraction.

**Steps to Perform 3-Digit Subtraction:**

1. Write the given numbers one below the other in such a way that the bigger number is placed up and the smaller number is placed below it. They should be correctly placed under the columns of ones, tens, and hundreds.
2. Start subtracting the numbers from the ones column, followed by the tens column, and then the hundreds column.
3. After all the columns are subtracted, we get the difference of the given numbers.

**3-Digit Subtraction Without Regrouping**

3-digit subtraction without regrouping means subtracting a set of numbers without borrowing any number from the preceding digit. This type of subtraction can be easily done when all the digits in the minuend are bigger than the digits in the subtrahend.

**Example:**

Subtract 342 from 754.

1. Subtract the numbers under the ones column: 4 - 2 = 2. Write the difference (2) in ones column.
2. Subtract the numbers under the tens column: 5 - 4 = 1. Write the difference (1) in the tens column.
3. Subtract the numbers under the hundreds column: 7 - 3 = 4. Write the difference (4) in the hundreds column.
The difference between the given numbers is 412.

**3-Digit Subtraction With Regrouping**

3-digit subtraction with regrouping means we need to borrow a number from the preceding digit. This type of subtraction is necessary when a digit in the upper row is smaller than the digit in the lower row.

**Example:**

Subtract 167 from 283.

1. Write the given numbers according to their place values, one below the other.
2. Start subtracting the numbers from the ones column. Since 3 is smaller than 7, borrow 1 from the tens column, making it 13. Now, 13 - 7 = 6. Write the difference (6) under the ones column.
3. After giving 1 to the ones column, the '8' in the tens column changes to 7. Now, subtract the digits at the tens place: 7 - 6 = 1. Write the difference (1) under the tens column.
4. In the hundreds column, subtract 1 from 2: 2 - 1 = 1. Write the difference (1) in the hundreds column.
The difference between the given numbers is 116.

**3-Digit Addition and Subtraction**

3-digit addition and subtraction are similar to 2-digit addition and subtraction. When adding 3-digit numbers, we write the given numbers one below the other, ensuring they are correctly placed under the columns of ones, tens, and hundreds. Then, we add each column starting from the right-hand side. When subtracting 3-digit numbers, we write the bigger number on top and the smaller number below it, then subtract each column one by one, starting from the ones column.

**Examples:**

1. Subtract 769 - 245.
Solution: 524
2. Find the difference between 463 and 174.
Solution: 289
3. Write true or false for the following statements:
a) 3-digit subtraction with regrouping means we can borrow a number from the preceding column. (True)
b) 3-digit subtraction means both the minuend and the subtrahend should be of 3 digits. (False)

**FAQs on 3-Digit Subtraction:**

1. What is 3-digit subtraction in math?
3-digit subtraction is the subtraction of numbers where the minuend consists of 3 digits and the subtrahend can be of either 1, 2, or 3 digits.
2. What is 3-digit subtraction without regrouping?
3-digit subtraction without regrouping means subtracting a set of numbers without borrowing any number from the preceding digit.
3. How to do 3-digit subtraction?
In 3-digit subtraction, we need to subtract the given numbers after placing them correctly according to their place values.
4. What is 3-digit subtraction with borrowing?
3-digit subtraction with borrowing means we need to regroup the numbers in the minuend because they are smaller than the respective subtrahend.
5. Is 3-digit subtraction with regrouping the same as 3-digit subtraction with borrowing?
Yes, 3-digit subtraction with regrouping means 3-digit subtraction with borrowing.
6. How to do 3-digit subtraction with zeros?
In 3-digit subtraction, when there are zeros in the subtrahend, it can be easily subtracted. However, when there are zeros in the minuend, we need to regroup and borrow 1 from the preceding column and then subtract the given numbers.


#### Answer

D (0,0)
E (2a, 0)
F (a, b)

#### Work Step by Step

D is the origin, with coordinates (0, 0). E lies on the x-axis, resulting in coordinates (2a, 0). F is located on the perpendicular bisector of DE. Consequently, its x-coordinate is 'a'. Since the height of the triangle is unknown, we introduce another variable 'b' for the y-coordinate: F (a, b).


**1. A Thin, Uniform, Metal Bar**

A 2.5 m long, 90 N metal bar hangs vertically from the ceiling by a frictionless pivot. It is struck 1.3 m below the ceiling by a 4-kg ball traveling horizontally at 14 m/s. The ball rebounds at 7 m/s.

**Part A. Find the Angular Speed of the Bar**

To solve, use conservation of angular momentum:
m1v_0d = -m1vd + (1/3)m2L^2(omega)
where m1 = 4 kg, v_0 = 14 m/s, d = 1.3 m, v = 7 m/s, m2 = 90 N / 9.82 m/s^2, and L = 2.5 m.

**Calculation:**
(4 kg)(14 m/s)(1.3 m) = -(4 kg)(7 m/s)(1.3 m) + (1/3)(90 N / 9.82 m/s^2)(2.5 m)^2(omega)
72.8 = -36.4 + (1/3)(9.18)(6.25)(omega)
109.2 = 20.42(omega)
omega = 5.33 rad/s

**2. A Ball Rolling Uphill**

A uniform, solid sphere (bowling ball) rolls without slipping up a ramp sloping upward at an angle beta to the horizontal.

**Part A. Acceleration of the Center of Mass**

The force equation is:
- (Wsin(beta) + muWcos(beta)) = ma_cm
where W = mg, and a_cm = acceleration of the center of mass.

**Solution:**
a_cm = -g(sin(beta) + mucos(beta))

**Part B. Minimum Coefficient of Static Friction**

The minimum coefficient of static friction to prevent slipping is given by:
mu = tan(beta)


### Evaluating dπ¦ by dπ₯ at π₯ equals one

To evaluate dπ¦ by dπ₯ at π₯ equals one, we first need to find dπ¦ by dπ₯ by differentiating the function π¦ = (6π₯² - 2π₯ - 3^(-3))(π₯² - 2^5). Since π¦ is a product of two functions, we use the product rule, which states that if π¦ = π'π'', then dπ¦ by dπ₯ = π' dπ'' dπ₯ + π'' dπ' dπ₯.

Let π' = 6π₯² - 2π₯ - 3^(-3) and π'' = π₯² - 2^5. To apply the product rule, we need to find dπ'' by dπ₯ and dπ' by dπ₯.

To find dπ' by dπ₯, we use the chain rule, which states that if π' = π of π and π = π of π₯, then dπ' by dπ₯ = dπ' by dπ multiplied by dπ by dπ₯. Let π = 6π₯² - 2π₯ - 3. Then dπ by dπ₯ = 12π₯ - 2. Since π' = π^(-3), dπ' by dπ = -3π^(-4). Applying the chain rule, we get dπ' by dπ₯ = -3π^(-4) multiplied by 12π₯ - 2.

Substituting π = 6π₯² - 2π₯ - 3 into the expression for dπ' by dπ₯, we get dπ' by dπ₯ = -3(6π₯² - 2π₯ - 3)^(-4) multiplied by 12π₯ - 2.

To find dπ'' by dπ₯, we again use the chain rule. Let π = π₯² - 2. Then dπ by dπ₯ = 2π₯. Since π'' = π^5, dπ'' by dπ = 5π^4. Applying the chain rule, we get dπ'' by dπ₯ = 5π^4 multiplied by 2π₯. Substituting π = π₯² - 2 into the expression for dπ'' by dπ₯, we get dπ'' by dπ₯ = 5(π₯² - 2)^4 multiplied by 2π₯, which simplifies to 10π₯(π₯² - 2)^4.

Now that we have dπ' by dπ₯ and dπ'' by dπ₯, we can apply the product rule: dπ¦ by dπ₯ = π' dπ'' dπ₯ + π'' dπ' dπ₯. Substituting the expressions for π', π'', dπ' by dπ₯, and dπ'' by dπ₯, we get dπ¦ by dπ₯ = (6π₯² - 2π₯ - 3^(-3))(10π₯(π₯² - 2)^4) + (π₯² - 2^5)(-3(6π₯² - 2π₯ - 3)^(-4) multiplied by 12π₯ - 2).

We are asked to evaluate dπ¦ by dπ₯ at π₯ equals one. Substituting π₯ = 1 into the expression for dπ¦ by dπ₯, we get dπ¦ by dπ₯ = (6(1)² - 2(1) - 3^(-3))(10(1)(1² - 2)^4) + (1² - 2^5)(-3(6(1)² - 2(1) - 3)^(-4) multiplied by 12(1) - 2).

Simplifying, we have dπ¦ by dπ₯ = (6 - 2 - 1/27)(10(1)(-1)^4) + (-31)(-3(6 - 2 - 3)^(-4) multiplied by 10). This further simplifies to dπ¦ by dπ₯ = (4 - 1/27)(10) + (-31)(-3(-1)^(-4) multiplied by 10), which becomes dπ¦ by dπ₯ = (10 - 10/27) + 930.

Finally, evaluating the expression gives us dπ¦ by dπ₯ = (270/27 - 10/27) + 930 = 260/27 + 930 = 40.

Therefore, the value of dπ¦ by dπ₯ at π₯ equals one is 40.


I've recently revisited the 3D printed models designed by my students in my Math 341 Introduction to Topology class in Fall 2014. With my increased knowledge, I decided to review, edit, and reprint the models, and publish them on Thingiverse.

The first two models I examined were Candace Bethea's ('15) \(6_2\) knot and Hayley Archer-McClelland's ('15) three interlocking trefoils. We utilized the Seifert View program, which normally doesn't allow exporting computed curves and surfaces, but Professor Aaron Abrams arranged for this feature to be available. This software has been invaluable for 3D printing knots and Seifert surfaces.

The \(6_2\) knot was adjusted in SeifertView, exported, and then opened in Cinema4D. The surface required fixing due to overlapping polygons, which was resolved by deleting the overlaps and filling the gaps using the *Stitch and Sew* tool in *Edge mode*. The knot was initially printed in orange on the Projet-260 3D Systems printer and later in clear resin on the FormLabs 1+ printer.

The three interlocking trefoils were designed entirely in Cinema4D using parametric equations: \(x(t)= (2+\cos(3t))\cos(2t), y=(2+\cos(3t))\sin(2t), z=-\sin(3t)\) for \(t\in[-\pi, \pi]\). The curve was created with the *Formula* tool for \( t\in[-3.14, 3.14]\), and a *SweepNurb* with a 0.2 cm radius circle was added around the curve. The gap between the tube ends was sealed using the *Stitch and Sew* tool. The trefoils were initially printed in pink, pale green, and blue on the Projet-260 3D Systems printer and later in clear resin on the FormLabs 1+ printer.

Mithra Muthukrishnan ('16) used Seifert View to design a figure-8 knot and its Seifert surface. After tweaking, the surface was exported to Cinema4D, where it underwent a complex design process to color the knot and two sides of the surface differently. The surface was extruded, and three copies were made, each with different parts deleted to allow for separate coloring. The knot was colored red/pink, and the Seifert surface sides were colored white and yellow. The model was initially printed in color on the Projet-260 3D Systems printer and later in bright white on the MakerBot 2x printer.

The octopus model, designed by DanJoesph Quijada ('15) in Cinema4D, used parametric equations like \(x(t) = a\sqrt{2}t, y(t)=0, z(t) = b^2 cos^2(ct)e^{-dt^2}\) for the legs, which were thickened using a *SweepNurbs*. The head was created by extruding a box and applying the *Subdivision Surface* tool. The model was initially printed in pink on the Projet-260 3D Systems printer and later in bright white on the MakerBot 2x printer, although the legs posed some challenges. All models can be found on Thingiverse.


**Calculating the Circumference of a Circle**

A circle is a shape with all points equidistant from the center. The distance around a circle is called the circumference, which can also be referred to as the perimeter. The distance across a circle through the center is called the diameter (d). Half of the diameter is called the radius (r).

The circumference of a circle is the distance around its outer edge. To find it, multiply pi (π = 3.14) by the diameter. For example, a circle with a diameter of 4 has a circumference of (3.14 x 4) ≈ 12.56. If the radius is known, the diameter is twice as large (d = 2r).

Circumference, diameter, and radius are measured in linear units such as inches and centimeters. A circle has multiple radii and diameters, all passing through the center. A real-life example of a radius is a bicycle wheel spoke.


Public Key cryptography solves the problem of securely sharing encryption/decryption keys. It uses one key to encrypt and a different key to decrypt, allowing the public key to be shared with the world while keeping the private key secure. This enables encryption, signing, non-repudiation, and more.

The math behind Public Key cryptography is based on a simplified version of RSA encryption. To illustrate this, let's use small numbers:

Choose two random prime numbers, p and q:
**p** = 13
**q** = 7

Multiply the numbers together to get the modulus, N:
**N** = pq = 13*7 = 91

This is known as a trapdoor function, making it easy to work out N if you know pq, but very difficult to discover p and q if you only know N.

Choose a public key, e:
**e** = 5 (generally chosen from {3, 5, 17, 257, 65537}, which are also prime numbers)

To compute the associated private key, d, first compute φ (phi):
**φ** = (p-1)(q-1) = (13-1)(7-1) = 12*6 = 72

Then compute the private key, d:
**d** = (1/e) mod φ or ed = 1 mod φ

In English, this means finding a whole number, d, which, when multiplied by 'e' and then divided by 'φ', leaves a remainder of 1.

Substituting the known values, we get:
5d = 1 mod 72, so d = 29 (because 5*29/72 = 2 remainder 1)

We now have all the required parts to encrypt and decrypt a message:
The public key is (N, e) = (91, 5)
The private key is (N, d) = (91, 29)
The key pair is written ((N,e), d) – in our case ((91, 5), 29)

To encrypt a message, we need to convert it from letters to numbers using the standard Unicode Transformation Format 8-bit (UTF-8) encoding:

A = 65 | G = 71 | M = 77 | S = 83 | Y = 89 |
B = 66 | H = 72 | N = 78 | T = 84 | Z = 90 |
C = 67 | I = 73 | O = 79 | U = 85 |
D = 68 | J = 74 | P = 80 | V = 86 |
E = 69 | K = 75 | Q = 81 | W = 87 |
F = 70 | L = 76 | R = 82 | X = 88 |
– a space would be represented by 32

The message "ATTACK" would be encoded as 65, 84, 84, 65, 67, 75.

To encrypt the plaintext message, m, into ciphertext, c:
c = m^e mod N

A would be 65^5 mod 91 = 1,160,290,625 mod 91 = **39**
T would be 84^5 mod 91 = 4,182,119,424 mod 91 = **28**
C would be 67^5 mod 91 = 1,350,125,107 mod 91 = **58**
K would be 75^5 mod 91 = 2,373,046,875 mod 91 = **17**

Our encrypted message is now 39, 28, 28, 39, 58, 17.

To decrypt the ciphertext, c, back to the plaintext, m:
m = c^d mod N

39 would be 39^29 mod 91 = **65**
28 would be 28^29 mod 91 = **84**
58 would be 58^29 mod 91 = **67**
17 would be 17^29 mod 91 = **75**

Our decrypted message is 65, 84, 84, 65, 67, 75, which decodes to "ATTACK" using the UTF-8 table.

Note: When working with large numbers, we can keep them smaller by dividing by 91 and keeping just the remainder as we go along. For example, 39^29 mod 91 can be calculated step by step:
1: 39*39 = 1,521 – divide by 91 to get 16 remainder 65
2: 65*39 = 2,535 – divide by 91 to get 27 remainder 78
3: 78*39 = 3,042 – divide by 91 to get 33 remainder 39
...
27: 65*39 mod 91 = 78
28: 78*39 mod 91 = 39
29: 39*39 mod 91 = 65

This results in the same answer as 39^29 mod 91.


# Dao's Archimedean Twins

## Introduction

Circles with a radius of $\displaystyle r=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$ are known as Archimedean. This concept is based on one of Archimedes' discoveries.

## Problem Statement

Given semicircles $(O_1)$ and $(O_2)$ drawn on diameters $AC$ and $BC$ respectively, with $C$ being a point on segment $AB$, and $D$ and $E$ being the midpoints of the arcs $AC$ and $BC$. Lines $AE$ and $CD$ intersect at $F$, while $BD$ and $CE$ intersect at $G$. The projections of $F$ and $G$ on $AB$ are $F'$ and $G'$. Additionally, $I$ is the second intersection of $AE$ with $(O_2)$, and $H$ is the second intersection of $BD$ with $(O_1)$.

## Key Findings

It is given that $FF'=GG'=CF'=CG'=r$. Circles $F(r)$ and $G(r)$ centered at $F$ and $G$ with radius $r$ are Archimedean and tangent at $P$. Furthermore, circles $H(P)$ and $I(P)$ centered at $H$ and $I$ through $P$ are also Archimedean.

## First Pair of Circles

Triangles $AFF'$ and $AEO_2$ are similar, which leads to the equation $\displaystyle\frac{FF'}{AF'}=\frac{EO_2}{AO_{2}}$. By substituting $CF'=x$, this equation becomes $\displaystyle\frac{x}{2R_{1}-x}=\frac{R_2}{2R_{1}+R_{2}}$, resulting in $x=\displaystyle\frac{R_{1}R_{2}}{R_{1}+R_{2}}=r$. Similarly, $GG'=r$.

## Second Pair of Circles

Using the Power of a Point theorem, it can be proven that points $C, F, H, I, G$ lie on circle $P(r)$ centered at $P$ with radius $r$. The similarity of triangles $AFF'$ and $AEO_{2}$ gives $\displaystyle\frac{AF}{AE}=\frac{FF'}{EO_{2}}=\frac{r}{R_2}$. Multiplying this by $AI\cdot AE=AC\cdot AB=4R_{1}(R_{1}+R_{2})$ yields $\displaystyle AI\cdot AF=4R_{1}^{2}$, indicating that $I$ and $F$ belong to a circle with $AC$ as a tangent. The same applies to $G$ and $H$. As $F$ and $G$ also belong to circle $P(r)$, all five points lie on this circle. Consequently, $IP=HP=r$.

## Conclusion

The above circles were discovered by Dao Thanh Oai. The first pair of Archimedean twins is not new, but Dao's construction is shorter and highlights new properties. The second pair, known as Dao's twins, appears to be new.


**Introduction**

The dilemma between dynamic and static scope is a frequent topic of discussion in the Lisp community. Many Lisp dialects have tried different approaches to address this issue. Dynamic scope fits better with the main Lisp design goals, as it allows symbols and expressions passed as arguments to retain their semantics.

**The Problem**

Dynamic scope can lead to unintended interactions between functions, making it more difficult to understand their behavior. This can result in called functions showing behavior not intended by the programmer. Newlisp contexts provide lexical scope, allowing programmers to avoid these problems. However, if a programmer wants to use dynamic scope, they need to be aware of the potential issues.

**Example**

Consider a function `sum` that evaluates `f(1)+...+f(n)` for a given function `f` and integer `n`. If we define `sum` using dynamic scope, it can lead to unexpected behavior when used with other functions that also use dynamic scope.

**(set 'sum (lambda(f n)(set 'result 0)(for (i 1 n)(inc result (f i)))result))**

**The Issue**

When using `sum` with another function that also uses dynamic scope, the variables `i` and `result` can interact in unintended ways, leading to incorrect results.

**The Solution - The Weak One**

One simple technique to avoid this issue is to use a naming convention for variables. For example, we can define `sum` using a prefix for its variables:

**(set 'sum (lambda(sum.f sum.n)(set 'sum.result 0)(for (sum.i 1 sum.n)(inc sum.result (sum.f sum.i)))sum.result))**

By following this convention, programmers can avoid using variable names that might conflict with those used in other functions.

**The Problem Persists**

However, this solution does not completely eliminate the problem. Even with a naming convention, programmers can still unintentionally use the same variable names in different functions, leading to conflicts.

**The Solution - The Stronger One**

A more robust solution is to use a function that generates a safe version of the original function. This safe version uses unique variable names for each call, avoiding conflicts.

**(set 'safet (lambda(safe-name unsafe-function vars k)(list 'let(map (lambda(v)(list v(list 'quote(sym (append (string safe-name)"."(string v)"."(string k))))))vars)(append (list 'expand)(list unsafe-function)(map (lambda(v)(list 'quote v))vars))))**

This function takes the original function, a list of variable names, and a counter as input. It generates a new function with unique variable names and returns it.

**Automating the Solution**

To automate the process of generating safe functions, we can define a function `protect` that takes the name of the safe function, the original function, and a list of variable names as input. It generates a safe version of the original function and defines it under the given name.

**(set 'protect (lambda(safe-name unsafe-name vars)(let ((counter-name (sym (append (string safe-name)".counter"))))(set counter-name 0)(eval (expand '(set 'safe-name(lambda-macro()(inc counter-name)(eval (cons (eval (safet 'safe-name unsafe-name 'vars counter-name))(args))))'safe-name 'unsafe-name 'vars 'counter-name)))))**

This function allows programmers to easily generate safe versions of their functions, avoiding the potential issues with dynamic scope.

**Conclusion**

Dynamic scope can be a powerful tool in Lisp programming, but it requires careful attention to variable naming and scope. By using a naming convention and generating safe versions of functions, programmers can avoid unintended interactions and ensure that their code behaves as expected.


**Right Circular Cone**

A right circular cone is a type of cone whose axis falls perpendicular to the plane of the base. It has a flat circular surface and a curved surface that meet at a point toward the top, called the apex or vertex. The base of a right circular cone is in the shape of a circle.

**Parts of a Right Circular Cone**

The three elements of a right circular cone are its radius, height, and slant height. The radius is the distance between the center of the circular base to any point on the circumference of the base. The height is the distance from the apex of the cone to the center of the circular base. The slant height is the distance between the apex of the cone to any point on the circumference of the cone.

**Right Circular Cone vs Oblique Cone**

A cone can be classified into two types based on the alignment of the apex in comparison to the base: the right circular cone and the oblique cone. A right circular cone's axis is perpendicular to its base, whereas an oblique cone's axis is not perpendicular to the base.

**Properties of a Right Circular Cone**

The properties of a right circular cone include:

* A circular base
* The axis is a line that joins the vertex to the center of the base
* The slant height of the cone is measured from the vertex to the edge of the circular base
* The altitude or height of a right cone coincides with the axis of the cone
* If a right triangle is rotated with the perpendicular side as the axis of rotation, a right circular cone is constructed
* Any horizontal section of the right circular cone parallel to the base produces the cross-section of a circle

**Surface Areas of a Right Circular Cone**

The surface area of a right circular cone is defined as the total region covered by the surface of the 3D dimensional shape. It can be of two types: curved surface area or total surface area.

**Curved Surface Area of a Right Circular Cone**

The curved surface area of a right circular cone is defined as the region occupied by the curved surface of the cone. It can be calculated using the formula: πrs = πr√(r² + h²), where r is the radius, h is the height, and s is the slant height.

**Total Surface Area of a Right Circular Cone**

The total surface area of a right circular cone is defined as the region or area of the complete surface of the cone, including the base area. It can be calculated using the formula: πr² + πrs = πr² + πr√(r² + h²), where r is the radius, h is the height, and s is the slant height.

**Volume of a Right Circular Cone**

The volume of a right circular cone is defined as the total space occupied by the object in a 3-dimensional plane. It can be calculated using the formula: (1/3) × πr² × h, where r is the radius and h is the height.

**Example 1: Finding the Volume of a Right Circular Cone**

Given that the height of a right circular cone is three times the radius, find the approximate volume of the cone in terms of its radius.

Solution: Volume of a Cone(V) = (1/3)πr²h = (1/3)πr²(3r) = πr³

**Example 2: Finding the Slant Height of a Right Circular Cone**

Given the radius and height of a right circular cone, find the slant height.

Solution: (Slant height)² = (radius)² + (height)² = 3² + 4² = 9 + 16 = 25, so slant height = √25 = 5

**Example 3: Finding the Volume and Curved Surface Area of a Right Circular Cone**

Given the radius, height, and slant height of a right circular cone, find the volume and curved surface area.

Solution: Volume of a cone = (1/3) × π × (4)² × 6 = 32π in³, Curved surface area of a cone = π × 4 × 8 = 32π in²

**FAQs on Right Circular Cone**

* What is meant by a right circular cone? A right circular cone is one whose axis is perpendicular to the plane of the base.
* What is the volume of a right circular cone? Volume of a right circular cone = (1/3) × πr²h, where r is the radius and h is the height.
* What is the surface area or total surface area of a right circular cone? Total surface area of a cone = πr² + πrs, where r is the radius, s is the slant height, and h is the height.
* What is the curved surface area of a right circular cone? Curved surface area of a cone = πrs = πr√(r² + h²), where r is the radius, s is the slant height, and h is the height.
* How can you find the radius of a right circular cone using volume? The radius of a cone can be found using its volume and height: Volume of a cone = (1/3) × πr²h, where r is the radius and h is the height.
* How many vertices are there in a right circular cone? There is only one vertex in a right circular cone.
* How do you find the radius of a right circular cone? The radius of a right circular cone can be found using the formula: s = √(r² + h²), where r is the radius, s is the slant height, and h is the height.
* How do you find the side length/slant height of a right cone? The height of the cone, the slant height, and the radius of the base form a right triangle, so the Pythagorean theorem can be used to find the slant height.
* Is it possible to find the right circular cone with the same height and slant height? A right circular cone cannot have the same height as its slant height, as the length of the hypotenuse is greater than the lengths of the remaining two sides of the triangle.


# kfoldPredict

Predict responses for observations in cross-validated regression model

## Syntax
yFit = kfoldPredict(CVMdl)
[yFit,ySD,yInt] = kfoldPredict(CVMdl)
___ = kfoldPredict(CVMdl,'Name,Value')

## Description
kfoldPredict predicts responses for observations in a cross-validated regression model.

## Examples
### Compute Cross-Validation Loss Manually
Load the carsmall data set.
X = [Cylinders Displacement Horsepower Weight];
Y = MPG;
Train a cross-validated regression tree model.
CVMdl = fitrtree(X,Y,'CrossVal','on');
Compute the 10-fold cross-validation MSE.
L = kfoldLoss(CVMdl)
Predict the responses by using the cross-validated regression model.
yfit = kfoldPredict(CVMdl);
Compute the mean squared error between yfit and the true responses CVMdl.Y.

## Input Arguments
CVMdl — Cross-validated partitioned regression model
RegressionPartitionedModel object | RegressionPartitionedEnsemble object | RegressionPartitionedGAM object | RegressionPartitionedGP object | RegressionPartitionedNeuralNetwork object | RegressionPartitionedSVM object
Name-Value Arguments
Alpha — Significance level
0.05 (default) | numeric scalar in [0,1]
IncludeInteractions — Flag to include interaction terms
true | false
PredictionForMissingValue — Predicted response value to use for observations with missing predictor values
"median" | "mean" | numeric scalar

## Output Arguments
yFit — Predicted responses
numeric vector
ySD — Standard deviations of response variable
column vector
yInt — Prediction intervals of response variable
two-column matrix

## Extended Capabilities
### GPU Arrays
Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox.

## Version History
Introduced in R2011a
R2023b: Specify predicted response value to use for observations with missing predictor values
R2023a: GPU support for RegressionPartitionedSVM models


Conversion of Measurement Units: Megalitre to Milliliter

To convert megalitre to milliliter, note that 1 megalitre is equal to 1,000,000,000 milliliters. The conversion factor can be used to convert between these units.

Key Conversion Factors:
- 1 cubic meter is equal to 0.001 megalitre or 1,000,000 milliliters.
- 1 megalitre is equal to 1,000,000 liters.
- 1 milliliter is equal to one thousandth of a liter or 1 cubic centimeter (cm³).

Quick Conversion Chart:
- 1 megalitre = 1,000,000,000 milliliters
- 2 megalitre = 2,000,000,000 milliliters
- 3 megalitre = 3,000,000,000 milliliters
- 4 megalitre = 4,000,000,000 milliliters
- 5 megalitre = 5,000,000,000 milliliters
- 6 megalitre = 6,000,000,000 milliliters
- 7 megalitre = 7,000,000,000 milliliters
- 8 megalitre = 8,000,000,000 milliliters
- 9 megalitre = 9,000,000,000 milliliters
- 10 megalitre = 10,000,000,000 milliliters

Definition of Units:
- A megalitre (Ml or ML) is one million liters in the metric system, equivalent to 1000 m³.
- A milliliter (ml or mL) is a metric unit of volume equal to one thousandth of a liter, equivalent to 1 cubic centimeter (cm³).

For reverse conversion or to convert between other units, use an online conversion calculator. Note that rounding errors may occur, so always check the results.


A function is a relation that associates each element of a set with an element of another set. It appears in various contexts, such as a social security number identifying a person, income tax rate depending on income, and a final letter grade determined by test scores and homework.

Formally, a function is defined as a special type of binary relation. A function f from set A to set B is a relation that satisfies two conditions:
1. For each element a in A, there is an element b in B such that <a, b> is in the relation.
2. If <a, b> and <a, c> are in the relation, then b = c.

The set A is called the domain of the function, and B is its codomain. A function covers the domain and is single-valued. The relation between a and b is denoted as f(a) = b, where b is the image of a under f.

The set of images of elements of a set S under a function f is called the image of S under f, denoted by f(S) = {f(a) | a ∈ S}. The image of the domain under f is called the range of f.

Example: Let f be a function from natural numbers N to N that maps each natural number x to x^2. The domain and codomain are N, the image of 3 is 9, and the range is the set of squares {0, 1, 4, 9, 16, ...}.

The sum and product of functions f and g from set A to real numbers R are defined as:
(f + g)(x) = f(x) + g(x) and (f*g)(x) = f(x)*g(x).

Example: Let f(x) = 3x + 1 and g(x) = x^2. Then (f + g)(x) = x^2 + 3x + 1, and (f*g)(x) = 3x^3 + x^2.

A function f is one-to-one (injective) if f(x) = f(y) implies x = y. Example: f(x) = x^2 from N to N is one-to-one.

A function f from set A to set B is onto (surjective) if for every element y in B, there is an element x in A such that f(x) = y. Example: f(x) = 2x from N to nonnegative even numbers E is onto.

A function is a bijection if it is both onto and one-to-one. Every bijection has an inverse function. Example: f(x) = 2x from N to E is a bijection, and its inverse is f^(-1)(x) = 1/2x from E to N.

A composite function fg is defined as fg(x) = f(g(x)) for all x in A, where g is a function from A to B, and f is a function from B to C. Example: Let f(x) = x^2 and g(x) = x + 1. Then f(g(x)) = (x + 1)^2.

Test Your Understanding of Function:
Indicate which of the following statements are correct and which are not.
Click True or False, then Submit. There are two sets of questions.