## Introduction to Slice Genus Class

A Slice Genus Class is a mathematical term that refers to a generalized constant expression used to fix differential formulas. It has options that are periodic and is essential in differential Class resolving. The Slice Genus function, or "quad," is used in this context.

## The Slice Genus Class Formula

The Slice Genus Class can be represented as Q(x) = -kx2, where Q(x) is the Slice Genus Class and k is a constant. The q part of the Class is the Slice Genus continuous, while the x part is the Slice Genus function. There are four Slice Genus functions with proper solutions: K4, K7, K3, and L4.

## Slice Genus Functions

- K4: The K part of a Slice Genus Class is the Slice Genus function, which can be written in partial fractions as (x2 - y2)/(x + y). To resolve for K4, we increase it by the correct Slice Genus function: k(x) = x2, y2, or x - y.
- K7: The K7 Slice Genus Class has a solution of the kind x4y2 - y4x3 = 0. The Slice Genus function is then increased by x to get x2 + y2 = 0.
- K3: The Slice Genus function Class is K3 + K2 = 0. We then increase by k for K3.
- K3(t): The Slice Genus function equation is K3(t) + K2(t). We multiply by k for K3(t).

## Understanding the Slice Genus Function

The Slice Genus function is also known as "K4" because of the initials of the letters K and 4. K suggests Slice Genus, and the word "quad" is pronounced as "kah-rab." The Slice Genus Class is among the primary techniques of fixing differential formulas.

## Approaches for Solving Slice Genus Equations

The approaches for solving Slice Genus equations include particular worth decay, factorization, ideal algorithm, mathematical service, or the Slice Genus function approximation. The Slice Genus Solution is not the only way to solve Slice Genus formulas; it is essential to learn about trigonometry, which uses the Pythagorean theorem, and then use Slice Genus formulas to resolve issues.

## Online Resources for Slice Genus Equations

There are numerous online resources available that provide free or affordable Slice Genus equations to solve for all the values of x. These resources typically do not require a membership fee or any sort of investment. The answers provided are the result of complex-variable Slice Genus equations that have been resolved.

## Conclusion

In conclusion, the Slice Genus Class is a generalization of a linear Class. It is a term that is an extension of a direct Class, and one benefit of using Slice Genus formulas is that they are more general than the linear formulas. They are easier to solve for all the values of x. The Slice Genus Class has one unknown that can be found with the Slice Genus formula.


# GIS Quest - Part IV

Part I defined a polygon to limit the dataset, and Part II calculated trails completed within that polygon, with a current completion rate of 82%. The next step involves transitioning from an arbitrary polygon to a structure that tiles nicely.

## Squares

Dividing the world into 32 squares and recursively subdividing each into 32 new squares until reaching the required accuracy results in a Geohash system. This system uses a Z-order curve to label each square. To cover the area within the polygon, multiple Geohash codes would be needed, such as `u4ez9t`, `u4ez9v`, `u4ez9k`, `u4ez9s`, `u4ez9u`, `u4ez97`, `u4ez9e`, `u4ez9g`, `u4ez9d`, and `u4ez9f`.

## Hexagons

Hexagons are a suitable alternative, as they tile nicely. Although they do not subdivide as nicely as squares and triangles, geospatial indexing systems like H3 utilize hexagons. Since the polygon defined in Part I was already close to a hexagon and subdivision is not necessary, this approach can be considered a viable option.

### Bestagon 0

The quest continues, exploring the use of hexagons as a potential solution for structuring the dataset.


# Maximal Elements Algorithms

Kosaraju's algorithm and the Floyd-Warshall algorithm can be used to calculate the Schwartz set and the Smith set for an election. When there are N candidates, Kosaraju's algorithm runs in Θ(N^2) time with Θ(N^2) space, while the Floyd-Warshall algorithm runs in Θ(N^3) time with Θ(N^2) space.

## Background

The general problem is identifying the members of the maximal elements of an order generated from a binary relation. The order is the natural partial order generated from the preorder that is the transitive-reflexive closure of the relation. The Schwartz set is associated with the beatpath order, and the Smith set is associated with the beat-or-tie order.

From a graph-theoretic perspective, these algorithms identify which vertices in the maximal strongly connected components of the directed graph of the given relation. The algorithms start with a voting matrix where `Votes[i,j]` is the number of votes for candidate `i` in a pair-wise contest against candidate `j`. The final result is a boolean array, `IsInMaximal`, where `IsInMaximal[i]` is true if candidate `i` is in the maximal cycle equivalence classes.

## Kosaraju's Algorithm

Kosaraju's algorithm performs two depth-first searches (DFS) of the relation. The first search records the search finish times, which are then reversed to be used as a search order for the second search using the transpose of the relation. The second search partitions the candidates into search trees, which are actually cycle equivalence classes, and identifies which candidates are in maximal cycle equivalence classes.

The pseudo-code for Kosaraju's algorithm is as follows:
```python
function KosarajuMaximal(boolean [1..N,1..N] Relation)
  // Initialize variables
  integer [1..N] SearchOrder
  for i from 1 to N
    SearchOrder[i] = i
  StartDFS(Relation, SearchOrder)

  // Second DFS
  integer [1..N] SecondSearchOrder
  for i from 1 to N
    SecondSearchOrder[i] = FinishOrder[N+1 - i]
  boolean [1..N,1..N] RelationTranspose
  for i from 1 to N
    for j from 1 to N
      RelationTranspose[i,j] = Relation[j,i]
  StartDFS(RelationTranspose, SecondSearchOrder)

  // Identify maximal candidates
  bool [1..N] IsInMaximal
  for i from 1 to N
    if TreeConnectsToAnyPriorTree[Tree[i]] == true
      IsInMaximal[i] = false
    else
      IsInMaximal[i] = true
  return IsInMaximal
```

## Floyd-Warshall Algorithm

The Floyd-Warshall algorithm calculates whether there is a path between any two candidates and uses the results to see if each candidate is in a maximal cycle equivalence class.

The pseudo-code for the Floyd-Warshall algorithm is as follows:
```python
function FloydWarshallMaximal(boolean [1..N,1..N] Relation)
  boolean [1..N] IsInMaximal
  for i from 1 to N
    IsInMaximal[i] = true
  boolean [1..N,1..N] HasPath
  for i from 1 to N
    for j from 1 to N
      if i <> j
        if Relation[i,j] == true
          HasPath[i,j] = true
        else
          HasPath[i,j] = false

  // Expand consideration to paths with intermediate nodes
  for k from 1 to N
    for i from 1 to N
      if k <> i
        for j from 1 to N
          if k <> j and i <> j
            if HasPath[i,k] == true and HasPath[k,j] == true
              HasPath[i,j] = true

  // Disqualify non-maximal candidates
  for i from 1 to N
    for j from 1 to N
      if i <> j
        if HasPath[j,i] == true and HasPath[i,j] == false
          IsInMaximal[i] = false
  return IsInMaximal
```

## References

* "[EM] New Voting mailing list: Politicians and Polytopes". Election-Methods mailing list. 2000-04-07.
* "[EM] Beatpath Magnitude Algorithm Revision". Election-Methods mailing list.
* "[EM] Re: bicameral design poll". Election-Methods mailing list.
* Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L. (1990). Introduction to Algorithms, first edition, MIT Press and McGraw-Hill. ISBN 0-262-03141-8. Sections 23.5 and 26.2.


# Kiloamperes to Megaamperes Converter

To convert electric current from kiloamperes to megaamperes, divide the current by 1,000, as 1 megaampere equals 1,000 kiloamperes. The formula is: electric current in megaamperes = electric current in kiloamperes / 1,000.

## How to Convert Kiloamperes to Megaamperes

The conversion ratio is based on the fact that one megaampere is equal to 1,000 kiloamperes. For example, to convert 5,000 kiloamperes to megaamperes: 5,000 kiloamperes / 1,000 = 5 megaamperes.

## What is a Kiloampere?

A kiloampere is a unit of electric current equal to 1,000 amperes, which is the flow of one coulomb per second. It is a multiple of the ampere, the SI base unit for electric current, with "kilo" meaning thousands (10^3). Kiloamperes are abbreviated as kA.

## What is a Megaampere?

A megaampere is a unit of electric current equal to 1,000,000 amperes, or the flow of one coulomb per second. It is also a multiple of the ampere, with "mega" meaning millions (10^6). Megaamperes are abbreviated as MA.

## Kiloampere to Megaampere Conversion Table

| Kiloamperes | Megaamperes |
| --- | --- |
| 1 kA | 0.001 MA |
| 2 kA | 0.002 MA |
| 3 kA | 0.003 MA |
| 4 kA | 0.004 MA |
| 5 kA | 0.005 MA |
| 6 kA | 0.006 MA |
| 7 kA | 0.007 MA |
| 8 kA | 0.008 MA |
| 9 kA | 0.009 MA |
| 10 kA | 0.01 MA |
| 20 kA | 0.02 MA |
| 30 kA | 0.03 MA |
| 40 kA | 0.04 MA |
| 50 kA | 0.05 MA |
| 60 kA | 0.06 MA |
| 70 kA | 0.07 MA |
| 80 kA | 0.08 MA |
| 90 kA | 0.09 MA |
| 100 kA | 0.1 MA |
| 200 kA | 0.2 MA |
| 300 kA | 0.3 MA |
| 400 kA | 0.4 MA |
| 500 kA | 0.5 MA |
| 600 kA | 0.6 MA |
| 700 kA | 0.7 MA |
| 800 kA | 0.8 MA |
| 900 kA | 0.9 MA |
| 1,000 kA | 1 MA |

## More Conversions

- Kiloamperes to amperes
- Kiloamperes to milliamperes
- Kiloamperes to biots
- Kiloamperes to abamperes
- Kiloamperes to statamperes
- Kiloamperes to coulombs per second
- Amperes to megaamperes
- Milliamperes to megaamperes
- Biots to megaamperes
- Abamperes to megaamperes
- Statamperes to megaamperes
- Coulombs per second to megaamperes


Today we'll continue with the Tinkercad Fractals project, refining your first draft designs. If your designs are well-defined fractals, you'll print them in class; otherwise, you'll redo them and print after class.

Discuss the following with your group members:
- Is your object a fractal? Why?
- What level of your fractal is illustrated, and what's the rule for your fractal? If you can't answer these, you might need to redo your design.
- Can you measure the length, area, and/or volume of this level of your fractal? What information would you need to know to do that? If these questions seem unanswerable, you might want to redo your design.
- Will your fractal print successfully? Why or why not? How big will it have to be to print successfully?
- If you've addressed all the above points, you can start printing your fractal now if it will finish before the end of class. Otherwise, come in later with swipe-card access to complete your print.
- Ensure your blog post meets all requirements, including: a featured image, embedded images at full size, section headers, group names, a title that is the name of your fractal, and tagged with the "Tinkercad Fractal" category. Review the Tinkercad Fractal Example post to ensure you've covered everything necessary.

To introduce the concept of "dimension", we'll watch the first part of the 3Blue1Brown video (up to 7:45) in class, pausing occasionally to work on math problems on the board.

For the next class, you'll:
- Fix any issues in your blog post
- Print and document a copy of your Tinkercad Fractal first draft
- Watch a video about the perimeter and area of the Koch snowflake (see the Assignments page for details)


**Misc Math / Re: Bonus Web Problem**

Solution is attached up to converting PDE to ODE.

A question was raised about plugging λ=t^(-1/2) to reduce u to a function of one variable.

Another user provided an explanation for the method of continuation and thanked the professor.

Regarding Question 3 part a, it was noted that p should be ((y-x+2kαt) / (√ 2kt)) instead of ((y-x+2kαt) / (√ 4kt)) in Peishan's solution. Additionally, there is a missing g(y) in the final solution for both conditions.

A user inquired about finding the IVP solution to the heat equation by separation of variables and then applying IBVP, or using the IVP solution from the lecture notes and applying IBVP. The response was that the IVP solution from the lecture notes can be used and applied to IBVP.

A user, Aida, thanked the professor for remembering her and requested consideration for her posted solutions, which she spent time resizing. She plans to use a scanner next time.

Solutions to various parts of questions, including question 1 and question 6, are attached. Full solutions to other questions are also attached.


In the Taylor series expansion of e^{x} about x = 2, the coefficient of (x- 2)^{4} is determined by the fourth derivative of e^{x} evaluated at x = 2, which is e^{2}.

Given the differential equation $\ddot x+3x=0,$ with initial conditions $x\left(0\right)=1,\dot x\left(0\right)=0$, we can solve for x(1) using the general solution of the differential equation, which is $x(t) = A\cos(\sqrt{3}t) + B\sin(\sqrt{3}t)$. Applying the initial conditions yields $x(t) = \cos(\sqrt{3}t)$, so $x(1) = \cos(\sqrt{3})$.

The limit $\underset{x\to 8}{\mathrm{lim}}\frac{{x}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}-2}{\left(x-8\right)}$ can be evaluated by recognizing it as the derivative of $x^{1/3}$ at x = 8, which is $\frac{1}{3}x^{-2/3}$ evaluated at x = 8, resulting in $\frac{1}{3}(8)^{-2/3} = \frac{1}{12}$.

A coin is tossed 4 times, and the probability of getting heads exactly 3 times can be calculated using the binomial probability formula $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$, where n = 4, k = 3, and p = 0.5. This yields $P(X = 3) = \binom{4}{3}(0.5)^3(0.5)^1 = 4 \times \frac{1}{8} \times \frac{1}{2} = \frac{1}{4}$.

The matrix $\left[\begin{array}{ccc}1& 2& 4\\ 3& 0& 6\\ 1& 1& p\end{array}\right]$ has one eigenvalue equal to 3. The sum of the other two eigenvalues is equal to the trace of the matrix minus the known eigenvalue. The trace is $1 + 0 + p$, and since one eigenvalue is 3, the sum of the other two eigenvalues is $1 + 0 + p - 3 = p - 2$.

The divergence of the vector field $\left(x-y\right)\hat{i}+\left(y-x\right)\hat{j}+\left(x+y+z\right)\hat{k}$ is calculated as $\nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x-y) + \frac{\partial}{\partial y}(y-x) + \frac{\partial}{\partial z}(x+y+z) = 1 - 1 + 1 = 1$.

The transverse shear stress acting in a beam of rectangular cross-section, subjected to a transverse shear load, is given by the formula $\tau = \frac{VQ}{It}$, where V is the shear force, Q is the first moment of area, I is the moment of inertia, and t is the thickness of the beam.

To calculate the resulting change in diameter of a rod of Length L and diameter D subjected to a tensile load P, we need to know the Poisson's ratio of the material, as the change in diameter is related to the change in length and the original dimensions through the formula $\Delta D = -\nu \frac{\Delta L}{L}D$, where $\nu$ is Poisson's ratio.

A straight rod of length L(t), hinged at one end and freely extensible at the other end, rotates through an angle $\mathit{\theta}\left(t\right)$ about the hinge. The magnitude of the velocity at the other end of the rod can be found using the formula $v = \sqrt{(\dot{L})^2 + (L\dot{\theta})^2}$, which yields $v = \sqrt{(1)^2 + (1 \times 1)^2} = \sqrt{2}$ m/s.

For the cantilever type gate hinged at Q with the mass of the cantilever part being 75 kg, to achieve static balance, the mass of the counterweight must be such that the moments about the hinge are balanced. This requires the counterweight to have a mass that can provide a moment equal to the moment due to the cantilever's weight.

A planar mechanism has 8 links and 10 rotary joints. Using Gruebler’s criterion, the number of degrees of freedom of the mechanism is given by $F = 3(n-1) - 2j$, where n is the number of links and j is the number of joints. Substituting the given values yields $F = 3(8-1) - 2 \times 10 = 21 - 20 = 1$.

The presence of an axial residual compressive stress on the outer surface of a rotating shaft subjected to bending can increase the fatigue life of the shaft. The residual compressive stress helps to reduce the tensile stresses that occur during loading, thus improving the fatigue performance.

When 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen, the change in entropy due to mixing, per mole of oxygen, can be calculated using the formula $\Delta S = -R \ln \frac{P_f}{P_i}$, where $P_f$ and $P_i$ are the final and initial pressures. However, since the final total pressure and temperature of the mixture are the same as those of the individual constituents at their initial states, and assuming ideal gas behavior, the change in entropy per mole due to mixing can be simplified to $\Delta S = R \ln 2$.

For flow of fluid over a heated plate, given the hydrodynamic boundary layer thickness is 1 mm, the thermal boundary layer thickness at the same location can be estimated using the relation $\frac{\delta_t}{\delta} = \frac{1}{\sqrt{Pr}}$, where $\delta_t$ is the thermal boundary layer thickness, $\delta$ is the hydrodynamic boundary layer thickness, and $Pr$ is the Prandtl number. For most fluids, $Pr$ is around 0.7, so $\delta_t = \frac{1}{\sqrt{0.7}} \times 1 \approx 1.2$ mm.

The continuity equation $\overrightarrow\nabla\bullet\overrightarrow V=0$ implies that the fluid is incompressible, which is a necessary condition for this equation to be valid.

In an air-standard Otto cycle, necessary assumptions include that the working fluid is air, which behaves as an ideal gas, and that the process is adiabatic. The assumption that is NOT necessary for the air-standard Otto cycle is that the specific heats of air are constant.

In an M/M/1 queuing system, the probability density function f(t) of the inter-arrival time is given by $f(t) = \lambda e^{-\lambda t}$, where $\lambda$ is the arrival rate.

A set of 5 jobs is to be processed on a single machine to minimize the total inventory cost. The optimal schedule can be determined by sequencing the jobs based on their processing times and holding costs, typically using the Shortest Processing Time (SPT) rule or other more complex algorithms that consider the specific costs and constraints of the problem.

For generating a Coon’s surface, we require a set of boundary curves that define the shape of the surface.

Internal gear cutting operation can be performed by a boring machine or a gear hobbing machine equipped with a special internal gear cutting tool.


The international unit of length is the meter, also spelled metre. Multiples and submultiples of the meter include kilometer, millimeter, micrometer, and others. In the metric system, units of length are millimeters, centimeters, meters, and kilometers. In contrast, old-fashioned units for length include inches, feet, yards, and miles.

Length is a scalar quantity, meaning it only measures magnitude, not direction. It is a fundamental quantity, like mass and time, and is used to derive other quantities, such as volume. The longest side of any perimeter is considered the length.

Strain is a dimensionless quantity because it is the ratio of the same quantities, such as change in length to original length, or change in volume to original volume. For example, tensile strain is calculated as the change in length divided by the original length, resulting in a unitless value.

In physics, length is distinct from displacement, which is a vector quantity that includes both magnitude and direction. Distance, on the other hand, describes the length of an object's path, regardless of the path taken. Units of length include femtometers, which are used to measure very small distances.

Key concepts related to length include:

* Scalar quantity: a quantity with only magnitude, not direction
* Vector quantity: a quantity with both magnitude and direction
* Fundamental quantity: a quantity that cannot be derived from other quantities, such as length, mass, and time
* Derived quantity: a quantity that can be derived from other quantities, such as volume, which is derived from length

Q: What is the quantity of length?
A: Length is a scalar quantity that measures magnitude, but not direction.

Q: Is length a vector quantity?
A: No, length is a scalar quantity, but displacement, which includes direction, is a vector quantity.


To compare circular arcs, it is essential to position them with the same center and a common endpoint, lying on the same side of this endpoint. This allows for the comparison of arcs from the same circle or equal circles, similar to comparing line segments.

An arc is considered greater than another if, starting from the common endpoint and moving along the arc, the next point is encountered before the final point. For instance, if arcs AB and AC are in the same position, arc AB is greater than arc AC if point C is encountered before point B when moving along arc AB from point A. Conversely, arc AB is smaller than arc AC if the order of points is ABC.

The sum of two arcs, AB and BC, on the same circle or equal circles can be determined by moving them end-to-end. A circular arc can be divided into two or more equal parts, similar to a line segment. The direction of motion is crucial when comparing arcs, as it determines the order in which points are encountered. This is because points A and B divide the circle into two arcs, and the direction of motion affects the order of encountering points B and C.


The table of the first 280 prime numbers shows each prime number with its remainder when divided by 6. Analyzing the remainders reveals a pattern.

A notable prime number is 1741. Its factors are 1 and 1741, and it has exactly two factors. The prime factorization of 1741 is simply 1741, as it is a prime number. The square root of 1741 cannot be simplified because it has no exponents greater than 1 in its prime factorization.

To verify that 1741 is prime, we check divisibility by prime numbers less than or equal to its square root. Since 1741 is not divisible by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, or 41, it is indeed a prime number.

Another way to confirm the primality of 1741 is to use the fact that its last two digits divided by 4 leave a remainder of 1, and it can be expressed as the sum of two squares (30² + 29² = 1741) with no common prime factors. Since 1741 is not divisible by any prime number Pythagorean triple hypotenuse less than or equal to its square root, it is prime.

The number 1741 is also the hypotenuse of a Pythagorean triple: 59-1740-1741, calculated from 30² – 29², 2(30)(29), 30² + 29². Additionally, 1741 is the difference of two consecutive squares: 871² – 870² = 1741.

In contrast, 1625 is a composite number with prime factorization 1625 = 5³ × 13. It has exactly eight factors and can be expressed as the sum of two squares in four different ways. The number 1625 is also the hypotenuse of ten Pythagorean triples.

The number 1405 is a composite number with prime factorization 1405 = 5 × 281. It has exactly four factors and can be expressed as the sum of two squares in two ways. The number 1405 is also the hypotenuse of four Pythagorean triples.

A chart of numbers from 1101 to 1200 shows their prime factorizations and the number of factors for each. The numbers with simplifiable square roots have exponents in their prime factorizations.

The number 1201 is a prime number with exactly two factors. Its prime factorization is simply 1201, and the square root of 1201 cannot be simplified. To verify its primality, we check divisibility by prime numbers less than or equal to its square root. Since 1201 is not divisible by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, or 31, it is indeed a prime number.

Another way to confirm the primality of 1201 is to use the fact that its last two digits divided by 4 leave a remainder of 1, and it can be expressed as the sum of two squares (25² + 24² = 1201) with no common prime factors. Since 1201 is not divisible by any prime number Pythagorean triple hypotenuse less than or equal to its square root, it is prime.

The number 1201 is also the hypotenuse of a primitive Pythagorean triple: 49-1200-1201, calculated from 25² – 24², 2(25)(24), 25² + 24².

The number 1105 is a composite number with prime factorization 1105 = 5 × 13 × 17. It has exactly eight factors and can be expressed as the sum of two squares in four different ways. The number 1105 is also the hypotenuse of thirteen Pythagorean triples, with four of them being primitive.

The number 1105 is the smallest number that is the product of three different Pythagorean hypotenuses. It gets one triple for each of its three individual factors, two triples for each of the three ways the factors can pair up with each other, and four primitive triples for the one way they can all three be together.

The square number 841 has several unique properties. It is the sum of the 29th and the 28th triangular numbers and the sum of the first 29 odd numbers. The number 841 is also the sum of two consecutive square numbers: 20² + 21² = 841 = 29².

The number 841 is a composite number with prime factorization 841 = 29². It has exactly three factors and is the hypotenuse of two Pythagorean triples. The number 841 can also be expressed as the sum of the nine prime numbers from 73 to 109 and the sum of three consecutive primes: 277 + 281 + 283 = 841.

The number 761 is a prime number with exactly two factors. Its prime factorization is simply 761, and the square root of 761 cannot be simplified. To verify its primality, we check divisibility by prime numbers less than or equal to its square root. Since 761 is not divisible by 2, 3, 5, 7, 11, 13, 17, 19, or 23, it is indeed a prime number.

Another way to confirm the primality of 761 is to use the fact that its last two digits divided by 4 leave a remainder of 1, and it can be expressed as the sum of two squares (20² + 19² = 761) with no common prime factors. Since 761 is not divisible by any prime number Pythagorean triple hypotenuse less than or equal to its square root, it is prime.

The number 761 is also the hypotenuse of a primitive Pythagorean triple: 39-760-761, calculated from 20² – 19², 2(20)(19), 20² + 19². Additionally, 761 can be expressed as the sum of three squares in ten different ways.

The number 685 is a composite number with prime factorization 685 = 5 × 137. It has exactly four factors and can be expressed as the sum of two squares in two ways. The number 685 is also the hypotenuse of four Pythagorean triples, with two of them being primitive.

The number 685 is the 19th centered square number because 18 and 19 are consecutive numbers and 19² + 18² = 685. There are 685 small squares of various colors in the corresponding graphic.


This is the third post in the Eedi Summer School series, where insights are shared from a cohort of Year 6 students participating in a seven-week program. The focus is on identifying potential areas of struggle for Year 7 students.

The subject of Week 3's quiz is Negative numbers. The group of 10 and 11-year-olds scored an average of 76% across 10 questions. The proportion of students getting each question correct, in reverse order, shows that Question 9 caused the most problems.

Question 9 is:
-4 - (-5) = ?

Students' results show that a significant proportion struggled with this question, with answers A and C being the most popular incorrect choices.

For answer A, -9, students' explanations included:
- Because if you did the inverse (-4 add 5) the answer is -9
- Because -9 - 5 is -4

These approaches, although incorrect, demonstrate attempts to apply inverse operations or solve the problem as a missing numbers style question.

For answer C, 9, students' reasons included:
- I think it is C because if you do 5+4 you would get 9 and if you do 9-5 you would get -4
- Two minuses make a positive
- The difference between 5 and -4 is 9

These explanations reveal misapplied rules and attempts to think about the question in terms of differences, which is problematic in this context.

To address these misconceptions, it's essential to ask:
- Why are students going wrong?
- What can be done about it?

Negative numbers, particularly operations involving negative numbers, is a challenging area of mathematics. Teachers often have strong views on how to approach it, using methods such as rules, number lines, temperature, or analogies. However, these approaches may not yield satisfactory results.

A recommended approach is to expose students to a carefully selected sequence of examples, holding key aspects constant to draw attention to variations and their impact on outcomes. This sequence is available for download.

Another approach is to challenge students to modify the original question to make each incorrect answer correct. For example, how could the question be changed to make option A, -9, the correct answer? This helps students focus on the impact of changes and move away from misremembered or misapplied rules.

The second and third worst-answered questions from the quiz, along with the proportion of students opting for each answer, are also available for analysis.

The whole quiz and Insights page with percentages and student explanations can be accessed online.

Next week's focus will be on Decimals, with more insights based on students' responses.


Euclidean Geometry is a mathematical study of geometry involving undefined terms, such as points, planes, and lines. Despite some investigation results about Euclidean Geometry having been carried out by Greek Mathematicians, Euclid is highly honored for producing a comprehensive deductive system. Euclid's mathematical approach in geometry mainly involves furnishing theorems from a finite number of postulates or axioms.

Euclidean Geometry is essentially a study of plane surfaces, where most geometrical concepts are illustrated by drawings on a piece of paper or on a chalkboard. Key concepts include the shortest distance between two points, the theory of the perpendicular to a line, and the concept of angle sum of a triangle, which always adds up to 180 degrees. Euclid's fifth axiom, also known as the parallel axiom, states that if a straight line traversing any two straight lines forms interior angles on one side less than two right angles, the two straight lines will meet on that same side.

In modern mathematics, the parallel axiom is stated as: through a point outside a line, there is only one line parallel to that exact line. Euclid's geometrical concepts remained unchallenged until the early 19th century when other principles in geometry began to emerge. These new geometrical concepts are known as non-Euclidean geometries and are used as alternatives to Euclid's geometry.

Non-Euclidean geometry is a type of geometry that contains an axiom equivalent to that of Euclidean parallel postulate. There are several non-Euclidean geometry studies, including:

## Riemannian Geometry

Riemannian geometry, also known as spherical or elliptical geometry, was named after the German Mathematician Bernhard Riemann. Riemann identified some shortcomings of Euclidean Geometry in 1889 and built upon the work of Girolamo Sacceri, an Italian mathematician. Riemann geometry states that when there is a line l and a point p outside the line l, then there will be no parallel lines to l passing through point p. This type of geometry deals with the study of curved surfaces and is an advancement of Euclidean theory.

Riemann geometry introduces several concepts on curved surfaces, including the angle sum of any triangle, which is greater than 180 degrees, and the fact that there are no lines on a spherical surface. The shortest distance between any two points on a curved surface, known as a geodesic, is not unique. For example, there are several geodesics between the south and north poles on the earth's surface that are not parallel.

## Hyperbolic Geometry

Hyperbolic geometry, also known as saddle geometry or Lobachevsky, states that when there is a line l and a point p outside the line l, then there are at least two parallel lines to line p. This geometry is named after the Russian Mathematician Nicholas Lobachevsky, who advanced non-Euclidean geometrical concepts. Hyperbolic geometry has several applications in science, including orbit prediction, astronomy, and space travel.

Hyperbolic geometry introduces several concepts, including:
i. The absence of similar triangles on a hyperbolic surface,
ii. The angle sum of a triangle, which is less than 180 degrees,
iii. The surface areas of any set of triangles having the same angle are equal, and
iv. The possibility of drawing parallel lines on an hyperbolic space.

### Conclusion

Due to advanced studies in the field of mathematics, it is necessary to replace Euclidean geometrical concepts with non-Euclidean geometries. Euclidean geometry is limited in that it is only useful when analyzing a point, line, or flat surface. Non-Euclidean geometries can be used to analyze any type of surface, making them a crucial part of modern mathematics.


The blog-post "Capacity Planning on a Cocktail Napkin" is a misleading explanation of capacity planning. The author applies the equation E = λ without explaining the symbols. Eventually, it is revealed that E stands for Erlangs, a teletraffic measure of capacity. However, the author fails to clarify that the service time (S) does not include buffering or wait time (W).

The connection between S and W metrics is the residence time (R), where R = S + W. This is important because the author uses only S, which is typically shorter than R, and may underestimate capacity. The equation E = λ is a version of Little's Law, which can be rewritten as ρ = λS, where ρ is the utilization of the server or switch.

Little's Law also includes buffering time (W) and is expressed as N = λR, where N is the level of concurrency or requests in the entire system. Applying Little's Law to load average data, the average number of resident processes in the system during a 20-minute window is N = 3.845. This result is obtained by adding up the blocks belonging to the 20 brown columns (1-minute load average data) between 21:30 and 21:50.

The author's explanation jumps between Pareto, Poisson, Exponential, and Normal distributions, causing confusion. If arrivals are Pareto distributed, the Poisson/Exponential assumption cannot apply. However, the author uses the Exponential distribution anyway. The Exponential distribution has properties that do not apply to the Normal distribution, such as the mean being equal to the standard deviation.

The author claims that the Normal distribution approximates the Exponential distribution, which is incorrect. The Normal distribution is symmetric about its mean, while the Exponential distribution is highly asymmetric. Using a Normal approximation can lead to a large underestimation error. The 95th percentile of the Exponential distribution is approximately 3 times the mean, while the Normal distribution would yield a value near 2.

The Exponential distribution is important for capacity planning because it offers a higher degree of variation, promoting the detection of unanticipated failures. It also provides a convenient opportunity to compare with other performance tools. The correct load profile for modeling a Help Center or teletraffic system is the Exponential distribution.

Bursts of Exponential arrivals should not be confused with Pareto-type correlations. The Exponential distribution can exhibit clusters of arrivals, but these are balanced out by gaps, resulting in statistically independent arrivals. The author's use of a Normal approximation can lead to incorrect conclusions about capacity planning.

In response to a comment, it is possible to emulate an Exponential distribution by combining multiple runs of a tool in parallel using a different distribution. However, this approach may be thwarted by the central limit theorem, which states that adding collection metrics will tend to emulate a Normal distribution. A more effective approach is to use the inverse transform method to generate Exponential variates.


# WIPP-Salt Model: Hydrostatic Compression Test

The hydrostatic compression test validates the WIPP-Salt model in FLAC3D. The test involves one zone at an initially elastic state of equilibrium, allowed to creep for 23 days while monitoring fractional density and comparing it to analytical calculations.

The confining pressure is held constant, with the rate of change of density given by the WIPP-Salt model. The fractional density is defined as \(F_d = \rho / \rho_f\), where \(\rho_f\) is the density of intact salt. Integration of the equation yields the fractional density over time.

Numerical results for fractional density versus creep time are compared to analytical calculations in Figure 1 for a stress level of 8.97 MPa. The initial density is 1350 kg/m^{3}, and the final density is 2300 kg/m^{3}.

The test is set up with the following parameters:
- Zone size: 1x1x1 with an edge of 5
- Constitutive model: WIPP-Salt
- Properties:
  - Bulk: 0.1186e9
  - Shear: 0.0714e9
  - Activation-energy: 12000
  - Constant-a: 4.56
  - Constant-b: 127
  - Constant-d: 5.79e-36
  - Creep-rate-critical: 5.39e-8
  - Exponent: 4.9
  - Temperature: 300
- Boundary conditions:
  - Velocity-normal: 0.0 at 'Bottom', 'West', and 'South'
  - Stress-normal: [valsig] at 'Top', 'East', and 'North'
- Initial conditions:
  - Density: [valrho]
  - Stress: [valsig] in all directions
- Histories:
  - Creep time-total
  - Fish histories for fractional density, error, and other parameters
- Test:
  - Large-strain on
  - Creep active on
  - Creep timestep starting at 5.e-5, minimum 5.e-5, maximum 500
  - Solve time-total: 2e6

The data file for this test is provided, detailing the setup and parameters used in the hydrostatic compression test for the WIPP-Salt model.


The movie "The Man Who Knew Infinity" explores the life of Indian mathematician Srinivasa Ramanujan and his profound understanding of math. A particular scene in the movie caught my attention, where Ramanujan's mentor, G.H. Hardy, comments on taxi number 1729, calling it "a rather dull one." Ramanujan replies, "No, Hardy, it's a very interesting number! It's the smallest number expressible as the sum of two cubes in two different ways."

This problem, known as Ramanujan's Taxi, can be expressed as a³ + b³ = c³ + d³. I wondered how to implement this equation in code and started with a brute-force implementation in Scala. The code had a performance of O(n⁴) due to the four loops used to calculate all values of a³, b³, c³, and d³.

To improve performance, I asked myself if all values of a³, b³, c³, and d³ need to be calculated. By solving for d³, I got d³ = a³ + b³ - c³, which allowed me to replace the fourth loop with a direct calculation. This improved the performance to O(n³).

Further optimization was possible by storing every combination of the sum of a³ + b³ and the corresponding pair (a³, b³) in a Hashmap. If a sum has two pairs, it means we have found a³ + b³ = c³ + d³. This implementation has a performance of O(n²) since only two loops are needed to calculate the combinations for a³ and b³.

I suspect there is a fourth optimization where we only need to calculate values of a³ and derive b³ from a³, achieving O(n) performance. Another challenge is to re-write the implementations as a functional programming pattern.

Ramanujan's Taxi, almost a century old, is still making new discoveries. Mathematicians at Emory University have found that the number 1729 relates to elliptic curves and K3 surfaces, objects important in string theory and quantum physics. The beauty of Ramanujan's genius is still being explored, and I expect we have only scratched the surface of his amazing work.

The equation a³ + b³ = c³ + d³ can be solved using different approaches:

1. Brute-force implementation: O(n⁴) performance
2. Improved implementation: O(n³) performance by replacing the fourth loop with a direct calculation
3. Optimized implementation: O(n²) performance by storing combinations in a Hashmap

Note: The original text included a brief introduction to the author, Geoffrey Bourne, but it was not directly related to the main topic of Ramanujan's Taxi and has been removed to maintain focus on the subject.


Lesson 8: Solving Word Problems with the Design Recipe

Overview
Students will practice the Design Recipe with word problems, designing functions to solve them and writing contracts for complex scenarios.

Agenda
- Getting Started
- Activity
- Anchor Standard: Common Core Math Standards, specifically F.BF.1, which involves writing a function that describes a relationship between two quantities.

Objectives
- Design functions to solve word problems.
- Practice writing contracts with more complex scenarios.

Teaching Guide
- Introduction: Students will fill in missing pieces of contracts, ensuring examples are complete to avoid error messages.
- Activity: Use the Design Recipe to create functions that solve word problems in Code Studio, stage 11.

Standards Alignment
- Common Core Math Standards:
  - F.BF.1: Write a function describing a relationship between two quantities.
  - F.BF.2: Write arithmetic and geometric sequences, model situations, and translate between forms.
  - A.CED.1-4: Create equations and inequalities, represent constraints, and rearrange formulas.
  - 6.EE.9 and 7.EE.4: Use variables to represent quantities, construct equations, and solve problems.
  - F.IF.1-9: Understand functions, use function notation, interpret key features of graphs and tables, and compare properties of functions.
  - F.LE.1-2: Distinguish between linear and exponential functions, construct functions given graphs or descriptions.
- Math Practices (MP.1-8): Students will explain problems, make conjectures, use tools, communicate precisely, and look for patterns.

Key Concepts and Skills
- Designing functions for word problems
- Writing contracts for complex scenarios
- Using the Design Recipe
- Aligning with Common Core Math Standards and Math Practices

By focusing on these key areas, students will develop a deeper understanding of solving word problems with the Design Recipe and enhance their mathematical proficiency.


A couple of years ago, a blog post was written about finding the closest point on a cubic Bezier to a given point, which involved reducing the problem to solving a quintic equation. The post provided an implementation in C# using Laguerre's Method to find the roots of the quintic. Additionally, it mentioned a fancier algorithm for solving quintics described in Doyle and McMullen's "Solving the Quintic by Iteration" and speculated that it might be the fastest way to find the closest point to a Bezier.

Recently, the Doyle and McMullen algorithm was implemented in C++ as a numeric algorithm. The algorithm is interesting because, although there is no general formula for the roots of a 5th degree equation, it can represent numbers with radical expressions that are arbitrarily close to these roots. Each root can be defined exactly in terms of the limit of a recursive function converging on as the depth of recursion approaches infinity.

The implementation is a numeric algorithm, using ordinary floating-point numbers, as opposed to a symbolic math package. This is an important point, as the algorithm defined in the original paper is over real numbers with arbitrary precision. The implementation was tested on randomly generated general quintics with coefficients uniformly distributed between -1000 and 1000, and the results were compared to the implementation of Laguerre's Method from "Data Structures and Algorithms in C++". The Doyle and McMullen algorithm seemed to have merit as a numeric algorithm at double precision, returning results that were about as good or better than Laguerre's Method, but about 500 times faster.

The implementation works as follows:

1. Convert a general quintic to principal form using the resultant and solving for c₁ and c₂.
2. Convert the principal quintic to Brioschi form.
3. Find two solutions to the Brioschi form quintic via iteration as described in the original paper.
4. Convert the two roots back to the general quintic.
5. Test both roots, and if necessary, perform iterations of Halley's Method to refine the roots.
6. If two good roots are found, perform synthetic division and solve the resulting cubic via radicals. If only one good root is found, divide the quintic by (z-v) and solve the resulting quartic.

The implementation is a header-only C++17 library, parametrized on the specific floating-point type used. Double precision works, but single precision is not sufficient. The library uses the cubic solving procedure described in "Numerical Recipes" and the quartic formula as described in the original paper.

Key facts and numbers:

* The algorithm solves quintic equations, which are 5th degree equations.
* The implementation uses double precision floating-point numbers.
* The algorithm is about 500 times faster than Laguerre's Method for randomly generated general quintics with coefficients between -1000 and 1000.
* The algorithm is more accurate and about 100 times faster than Laguerre's Method when the coefficients are between -10 and 10.
* The implementation uses a recursive function to converge on the roots of the quintic equation.
* The library is parametrized on the specific floating-point type used, allowing for flexibility in the choice of precision.


Definite Integral

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# Quadratic Response

This section provides directives for calculating quadratic response functions.

## General Control Statements

### Print Level
The print level is specified using the `.PRINT` directive, with a default value of 0.

## Definition of Quadratic Response Function

### Dipole Operators
The `.DIPLEN` directive specifies dipole operators for A, B, and C. The `.A OPERATOR`, `.B OPERATOR`, and `.C OPERATOR` directives specify the A, B, and C operators, respectively.

### Frequencies
The `.B FREQ` directive specifies frequencies of operator B, with a default value of 1 frequency at 0.0. The `.C FREQ` directive specifies frequencies of operator C.

### Evaluating Response Functions
The `.ALLCMB` directive evaluates all nonzero quadratic response functions, disregarding overall permutational symmetry. By default, only unique, nonzero response functions are evaluated.

## Excited State Properties

First-order properties of excited states can be computed from the quadratic response function using the `.EXCPRP` directive, which specifies the number of "left" and "right" states in each boson symmetry.

## Control Variational Parameters

### Excluding Rotations
The `.SKIPEE` directive excludes rotations between occupied positive-energy and virtual positive-energy orbitals, while the `.SKIPEP` directive excludes rotations between occupied positive-energy and virtual negative-energy orbitals.

## Control Reduced Equations

### Maximum Iterations
The `.MAXITR` directive sets the maximum number of iterations, with a default value of 30. The `.MAXRED` directive sets the maximum dimension of the matrix in the reduced system, with a default value of 100. The `.THRESH` directive sets the threshold for convergence of the reduced system, with a default value of 1.0D-5.

## Control Integral Contributions

### Integral Flags
The `.INTFLG` directive specifies which two-electron integrals to include. The `.CNVINT` directive sets the threshold for convergence before adding SL and SS integrals to SCF iterations, with default values of very large numbers. The `.ITRINT` directive sets the number of iterations before adding SL and SS integrals to SCF iterations, with default values of 1.

## Control Trial Vectors

### Normalization
The `.XQRNRM` directive normalizes trial vectors, with a default value of using un-normalized vectors.

## Advanced/Debug Flags

### Preconditioning
The `.NOPREC` directive disables preconditioning of initial trial vectors, with a default value of preconditioning trial vectors. The `.RESFAC` directive sets the residual factor, with a default value of 1000.0.


## Introduction to the Deflated Sharpe Ratio

The Deflated Sharpe Ratio (DSR) is a statistical measure designed to help differentiate a good investment strategy from statistical flukes. It was introduced by Marcos López de Prado and David H. Bailey in their research paper. The DSR is essential in addressing the Multiple Testing Problem and Selection Bias that arise when evaluating numerous investment strategies.

## Context: Probabilistic Sharpe Ratio

The Probabilistic Sharpe Ratio (PSR) is a measure that indicates the probability a strategy's true Sharpe Ratio (SR) will be greater than a given benchmark. The PSR calculation depends on the estimated SR, its variance, and the desired benchmark SR. The variance of the estimated SR is influenced by the skewness and kurtosis of the returns and the length of the track record.

## Multiple Testing Problem

The Multiple Testing Problem occurs when the probability of choosing a false positive increases as more strategies are tested. Even with a high estimated SR and a high PSR, a strategy might not perform well in the future due to the influence of chance rather than an actual edge. This problem can lead to overestimation of a strategy's potential.

## Selection Bias

Selection Bias arises when a strategy is selected based on its performance in a specific period, without considering how many other strategies were tested and failed. The Hold-out method, which involves dividing data into in-sample and out-of-sample subsets, can partially address this issue but does not account for the number of trials attempted before selecting a model.

## Expected Maximum Sharpe Ratio

The expected maximum SR after numerous independent trials can be approximated using a formula that takes into account the variance of the estimated SRs, the number of trials, and the Euler-Mascheroni constant. This formula shows that even with a true SR of zero, the expected maximum SR can be significantly high after a large number of trials.

## Solution: Deflated Sharpe Ratio

The Deflated Sharpe Ratio adjusts the PSR to reflect the multiplicity of trials, providing a more accurate assessment of a strategy's significance. It computes the probability that an estimated SR is statistically significant, controlling for the effects of multiple trials, non-normal returns, and shorter sample lengths.

## Calculating Independent Trials

To calculate the number of independent trials (N), one approach is to estimate the average correlation of all trials and then adjust the total number of trials (M) based on this correlation. The formula for N considers both the correlated and uncorrelated aspects of the trials.

## Practical Example

A practical example involves launching thousands of simulations with random weights assigned to a set of ETFs. Even with completely random trading decisions, some simulations will yield high SRs and PSRs. However, calculating the Deflated Sharpe Ratio can reveal whether these results are likely due to chance or an actual strategy edge.

## Conclusions

The Deflated Sharpe Ratio is a critical tool for evaluating investment strategies, especially when multiple strategies are tested. It helps in distinguishing between true strategy performance and statistical flukes. By considering the multiplicity of trials and adjusting the rejection threshold, the DSR provides a more reliable measure of a strategy's potential for future success.

## Bonus: When to Stop Testing

The theory of Optimal Stopping offers guidance on when to stop testing strategies to avoid unnecessarily increasing the probability of finding a false positive. This involves sampling a portion of the strategies, measuring their results, and then testing the remaining strategies until one is found that outperforms all previously tested ones.

## Key Takeaways

- The Deflated Sharpe Ratio is essential for determining the probability that a discovered strategy is not a false positive.
- Multiple testing exercises should be planned carefully to avoid running an unnecessarily large number of trials.
- The DSR helps in addressing the Multiple Testing Problem and Selection Bias by accounting for the number of independent trials and adjusting the significance threshold accordingly.


The probability of winning a game with multiple trials is 1 - p^n, where p is the probability of winning one trial, and n is the number of trials. To illustrate this, consider a slot jackpot with odds of 1 in 216,000. If you play 216,000 spins, your odds of hitting the jackpot are 1 - (215,000 ÷ 216,000)^216,000 = 0.63, or 63%.

Applying this to the Texas Lotto, which uses balls numbered 1-54, the chances of matching one number with one pick are 1/54. With six balls drawn and three tickets, each containing six numbers, there are 18 trials to try to match the first ball drawn. The chances of not matching the first number in 18 tries are (53/54)^18 = 0.7143.

Since there are six balls drawn, the probability of not matching any of the six numbers is 0.7143^6 = 0.1328, or approximately 1 in 7.53. Alternatively, the probability of not matching the first number is 36/54, or 2/3, since there are 36 numbers that don't match any on the ticket. For the subsequent numbers, the probability changes as there are fewer numbers remaining that can be drawn.

The correct calculation involves combinations, where the probability of not matching any of the six numbers is C(36,6) / C(54,6), or about 1 in 13.26. This method takes into account that any number on the ticket could match any ball, whereas considering each ball drawn as a separate game may lead to incorrect results.


## An Overview of Simplifying Fractions

Simplifying fractions involves finding the simplest form of a fraction by dividing both the numerator and denominator by their greatest common factor (GCF). This process is essential in making fractions easier to work with, especially when performing operations like addition, subtraction, multiplication, and division.

To simplify a fraction, one must first understand the concept of equivalent fractions. Equivalent fractions are fractions that have the same value but different numerators and denominators. For example, 4/8 and 1/2 are equivalent fractions because they both represent the same value.

## Dividing by the Greatest Common Factor (GCF)

The key to simplifying fractions is to find the GCF of the numerator and denominator and divide both by this common factor. For instance, to simplify the fraction 36/54, we first find the GCF of 36 and 54, which is 18. Then, we divide both the numerator and denominator by 18, resulting in the simplified fraction 2/3.

## Factoring Completely as a Product of Prime Numbers

Another approach to simplifying fractions is to factor both the numerator and denominator completely into their prime factors. Then, any common prime factors can be canceled out. For example, to simplify the fraction 36/54, we factor 36 into 2*2*3*3 and 54 into 2*3*3*3. We can then cancel out the common factors of 2 and 3*3, leaving us with 2/3.

## Dividing by One Factor at a Time

It's not always necessary to find the GCF or factor completely into prime factors. Sometimes, it's easier to simplify a fraction by dividing by one common factor at a time. For example, to simplify 36/54, we can first divide both numbers by 2, getting 18/27. Then, we can divide both numbers by 3, resulting in 6/9. Finally, we can divide both numbers by 3 again, giving us the simplified fraction 2/3.

## Understanding Why Simplification Works

Simplifying fractions is essentially a process of canceling out common factors between the numerator and denominator. This can be thought of as "un-multiplying" the fraction. When we multiply fractions, we multiply the numerators together to get the new numerator and the denominators together to get the new denominator. Simplifying is the reverse process, where we find common factors that can be divided out, leaving us with a simpler fraction.

## Recognizing When a Fraction Cannot Be Reduced

Not all fractions can be simplified. To determine if a fraction is already in its simplest form, we need to check if there are any common factors between the numerator and denominator. If there are no common factors other than 1, the fraction is already simplified. For example, the fraction 18/35 cannot be simplified because there are no common factors between 18 and 35.

## Finding the Greatest Common Factor (GCF)

The GCF is the largest number that divides both the numerator and denominator without leaving a remainder. To find the GCF, we can list all the factors of both numbers and find the largest common factor. Alternatively, we can use the Euclidean algorithm, which involves dividing the larger number by the smaller number and finding the remainder. We repeat this process until the remainder is 0, and the last non-zero remainder is the GCF.

## Simplifying Fractions with Large Numbers

When dealing with large numbers, simplifying fractions can be more challenging. However, we can use the same strategies as before, such as finding common factors or using the Euclidean algorithm. It's also helpful to simplify the fraction step by step, dividing by common factors as we go, rather than trying to find the GCF all at once.

## Using the Euclidean Algorithm

The Euclidean algorithm is a method for finding the GCF of two numbers. It involves dividing the larger number by the smaller number and finding the remainder. We repeat this process until the remainder is 0, and the last non-zero remainder is the GCF. This algorithm can be particularly useful when dealing with very large numbers, as it allows us to find the GCF without having to list all the factors of both numbers.

## Practice Problems

1. Simplify the fraction 24/30.
2. Simplify the fraction 18/24.
3. Simplify the fraction 36/48.
4. Simplify the fraction 20/25.
5. Simplify the fraction 32/40.

## Answers

1. 4/5
2. 3/4
3. 3/4
4. 4/5
5. 4/5

By following these steps and practicing with different fractions, you can become proficient in simplifying fractions and make working with them easier and more efficient.


This technology blog is dedicated to exploring the underlying philosophy and mathematical backbone of technological advancements, specifically in the context of the net and IT revolution. The focus will be on the intersection of technology and higher mathematics, rather than gadgets or general computer tips.

As technology evolves, its dependence on mathematics will grow. This is not yet the era of technology, but rather a period of development, similar to the early stages of physics. The marriage of physics and mathematics was cemented by Newton and Einstein, while technology is still in its "tinkering" stage, with occasional interactions with mathematics.

Examples of the power of mathematics in technology include Google's use of mathematical algorithms to harness the power of the net in its search engine. The paper by Larry Page and Sergey Brin, co-founders of Google, provides further insight into this relationship.

The beauty of mathematics is exemplified by Euclid's proof of the infinitude of primes:

Theorem: There are infinitely many primes.

Proof: Suppose that p1 = 2 < p2 = 3 < … < pr are all of the primes. Let P = p1*p2*…*pr + 1 and let pr be a prime dividing P; then pr cannot be any of p1, p2, …, pr, otherwise pr would divide the difference P - p1*p2*…*pr = 1, which is impossible. Hence either P itself is a prime number or any prime other than p1, p2, …, pr. Now if P is a prime then we get a contradiction as we assumed that pr is the largest prime. If there is any other prime number which divides P, say Q, then Q is other than p1, p2, …, pr and hence greater than pr, because all the primes less than pr are already excluded. Hence Q is a prime greater than pr - again a contradiction. Hence there are infinitely many primes.

This proof showcases the force and beauty of mathematics, and its delicate balance. The relationship between mathematics and technology will continue to evolve, and this blog will explore the underlying philosophy and mathematical concepts that drive technological advancements.


Lesson 11: Percentage Contexts

**Problem 1: Calculating Retail Price**
A car dealership buys a car for $8,350 and marks up the price by 17.4%. What is the retail price of the car?

**Problem 2: Finding Original Price**
A store has a 20% off sale on pants, resulting in a price of $15.20 before tax. What was the original price of the pants?
A) $3.04
B) $12.16
C) $18.24
D) $19.00

**Problem 3: Calculating Commission and Store Earnings**
A salesperson earns a 5.5% commission on the selling price of a couch. If the couch sells for $495, how much commission will the salesperson earn, and how much will the store get from the sale?

**Problem 4: Calculating Loan Balance**
A college student takes out a $7,500 loan with annual interest rates of 3.8% and 5.3%. What will the loan balance be after one year without any payments?

**Problem 5: Matching Situations with Equations**
Match the following situations with the corresponding equations:
1. Mai slept for x hours, and Kiran slept for 1/10 less.
2. Kiran practiced piano for x hours, and Mai practiced for 2/5 less.
3. Mai drank x oz of juice, and Kiran drank 4/3 more.
4. Kiran spent x dollars, and Mai spent 1/4 less.
5. Mai ate x grams of almonds, and Kiran ate 1.5 times more.
6. Kiran collected x pounds of recycling, and Mai collected 3/10 less.
7. Mai walked x kilometers, and Kiran walked 3/8 more.
8. Kiran completed x puzzles, and Mai completed 3/5 more.
Equations:
A) y = 2.33x
B) y = 1.375x
C) y = 0.6x
D) y = 0.9x
E) y = 0.75x
F) y = 1.6x
G) y = 0.7x
H) y = 2.5x


Students are required to figure out which operation to apply given the problem context. Complete the test and get an award. There are 8 green apples. Explore all of our math word problem worksheets, from kindergarten through grade 5.

The following worksheets contain a mix of grade 3 addition, subtraction, multiplication, and division word problems. These word problems are relatively short, requiring minimal reading. They do not have any extra or superfluous data, so the principal challenge for the student is to choose between addition, subtraction, multiplication, or division in order to arrive at the answer.

These third-grade subtraction word problem worksheets cover simple subtraction and column form subtraction. The problems are relatively simple, but emphasize the use of variables and the writing of equations. Students should be reasonably proficient at multiplication in columns before attempting the more difficult problems.

Long division questions are relatively simple (numbers 1-100) with answers expressed with remainders. Time and elapsed time word problems; intervals to the nearest minute. Length and height word problems in inches, feet, centimeters, and meters. Mass and weights in ounces, pounds, grams, and kilograms. Volume and capacity in fluid ounces, cups, gallons, milliliters, and liters.

Word problems where students use reasoning and critical thinking skills to solve each problem. These word problems help deepen a student's understanding of mathematical concepts by relating mathematics to everyday life. Mixing math word problems is the ultimate test of understanding mathematical concepts, as it forces students to analyze the situation rather than mechanically apply a solution.

John saved 1,500 dollars in February. A washing machine cost $135 less than the oven. The cost of the washing machine was $725. Find the change received by Ben if he paid the cashier $1000 for the washing machine. How much more money did John save in February than in January?

Janine owns a catering service company. She was hired to cater for the mayor’s 50th birthday. For the appetizers, she needs to make 750 mini meat pies. She divided her crew into 3 teams.

These 3rd-grade fraction word problems cover identifying fractions and adding and subtracting fractions. Fraction word problems nº 1, nº 2, and nº 3. These worksheets are appropriate for 3rd Grade, 4th Grade, 5th Grade, 6th Grade, and 7th Grade.

These addition word problem worksheets cover both simple addition and addition in columns, as well as some mixed addition and subtraction word problems. Simple real-life scenarios form the basis of these addition word problem worksheets. Have 'total' fun by adding up a wide range of addends displayed in these worksheets!

Add and subtract numbers up to 50, up to 100, whole tens, whole hundreds, 2-digit numbers, 3-digit numbers, 4-digit numbers. More challenging computations, typically 2 or 3 digit numbers. No conversions of units. No conversion of units.

These third-grade division word problem worksheets cover both simple division and long division. The problems are relatively simple, but emphasize the use of variables and the writing of equations. Students should be reasonably proficient at multiplication in columns before attempting the more difficult problems.

These percentage word problems worksheets are appropriate for 3rd Grade, 4th Grade, 5th Grade, 6th Grade, and 7th Grade. Word problems with variables (variable is chosen for the student), Writing variables to solve word problems (student chooses the variable).

Math word problems help deepen a student's understanding of mathematical concepts by relating mathematics to everyday life. Mixing math word problems is the ultimate test of understanding mathematical concepts, as it forces students to analyze the situation rather than mechanically apply a solution.

These worksheets are best attempted after a student has studied the underlying skill; for example, our 'addition in columns" word problem worksheets should not be attempted until students are comfortable with addition in columns. In many of our word problems, we intentionally include superfluous data or information, so that students need to read and think about the questions carefully, rather than simply applying a computation pattern to solve the problems.

We help your children build good study habits and excel in school. K5 Learning offers free worksheets and inexpensive workbooks for kids in kindergarten to grade 5. WorksheetWorks.com is an online resource used every day by thousands of teachers, students, and parents.


# Improper Riemann Integrals

Improper Riemann integrals are a special type of Riemann integral.

Definition: Let $f$ be a function defined on an unbounded interval $I = [a, \infty)$ where $a \in \mathbb{R}$, and assume $f$ is Riemann integrable on $[a, b]$ for all $b > a$. The Improper Riemann Integral of $f$ on $I$ is denoted $\displaystyle{\int_a^{\infty} f(x) \: dx}$, and $f$ is said to be Improper Riemann Integrable on $I$ if $\displaystyle{\lim_{b \to \infty} \int_a^b f(x) \: dx}$ exists. In this case, $\displaystyle{\int_a^{\infty} f(x) \: dx = \lim_{b \to \infty} \int_a^b f(x) \: dx}$.

For $I = (-\infty, a]$ where $a \in \mathbb{R}$, the improper Riemann integral is defined as $\displaystyle{\int_{-\infty}^a f(x) \: dx = \lim_{b \to -\infty} \int_{b}^a f(x) \: dx}$. If $f$ is improper Riemann integrable on $(-\infty, a]$ and $[a, \infty)$, then $f$ is improper Riemann integrable on all of $\mathbb{R}$.

Consider the function $f(x) = \frac{1}{x^2}$ on $[1, \infty)$. Since $f$ is nonnegative, the improper integral represents the area under $f$ and above the $x$-axis on $[1, \infty)$. We have:

$\displaystyle{\int_1^{\infty} \frac{1}{x^2} \: dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2} \: dx}$.

Evaluating the integral gives:

$\displaystyle{\int_1^b \frac{1}{x^2} \: dx = \left[ -\frac{1}{x} \right]_1^b = -\frac{1}{b} + 1}$.

Taking the limit as $b \to \infty$ yields:

$\displaystyle{\lim_{b \to \infty} \left( -\frac{1}{b} + 1 \right) = 1}$.

Thus, $f(x) = \frac{1}{x^2}$ is improper Riemann integrable on $[1, \infty)$, and $\displaystyle{\int_1^{\infty} \frac{1}{x^2} \: dx = 1}$.


# Calculus 1

## Limits and Continuity
Limits intro, estimating limits from graphs and tables, formal definition of limits (epsilon-delta), properties of limits, limits by direct substitution, and limits using algebraic manipulation are fundamental concepts. The squeeze theorem, types of discontinuities, continuity at a point, continuity over an interval, removing discontinuities, infinite limits, limits at infinity, and the intermediate value theorem are also crucial.

## Derivatives: Definition and Basic Rules
The average vs. instantaneous rate of change, secant lines, derivative definition, estimating derivatives, differentiability, power rule, and derivative rules (constant, sum, difference, and constant multiple) are key topics. Combining the power rule with other derivative rules, derivatives of cos(x), sin(x), 𝑒ˣ, and ln(x), product rule, quotient rule, and derivatives of tan(x), cot(x), sec(x), and csc(x) are also important.

## Derivatives: Chain Rule and Other Advanced Topics
The chain rule, implicit differentiation, implicit differentiation (advanced examples), differentiating inverse functions, derivatives of inverse trigonometric functions, strategy in differentiating functions, differentiation using multiple rules, second derivatives, disguised derivatives, and logarithmic differentiation are advanced concepts.

## Analyzing Functions
The mean value theorem, extreme value theorem and critical points, intervals on which a function is increasing or decreasing, relative (local) extrema, absolute (global) extrema, concavity and inflection points intro, analyzing concavity and inflection points, second derivative test, sketching curves, connecting f, f', and f'', solving optimization problems, and analyzing implicit relations are vital for understanding functions.

## Integrals
Accumulations of change introduction, approximation with Riemann sums, summation notation review, Riemann sums in summation notation, defining integrals with Riemann sums, fundamental theorem of calculus and accumulation functions, interpreting the behavior of accumulation functions, properties of definite integrals, and fundamental theorem of calculus and definite integrals are integral concepts. Reverse power rule, indefinite integrals of common functions, definite integrals of common functions, integrating with u-substitution, integrating using long division and completing the square, and integrating using trigonometric identities are also essential.

## Applications of Integrals
Average value of a function, straight-line motion, non-motion applications of integrals, area (vertical area between curves, horizontal area between curves, curves that intersect at more than two points), volume (squares and rectangles cross sections, triangles and semicircles cross sections, disc method, washer method) are practical applications of integrals.


The problem involves a smaller circle rolling around a larger circle. To understand how many revolutions the smaller circle makes, we can start by imagining it rolling on a line the same length as the circumference of the larger circle. In this case, the smaller circle clearly makes rotations equal to the length of the line divided by its own circumference.

Next, we consider the difference between sliding along a straight line and doing the same along the circumference of a circle. If you slide a certain distance along a straight line, you arrive at your destination without changing orientation. However, if you do the same along the circumference of a circle, you will have made a full rotation by the time you return to your starting point.

When the small circle rolls along the circumference of the larger circle, two kinds of movement occur simultaneously: revolution and rotation. The four movements initially considered are the four revolutions, but rotation is more difficult to grasp. The key is to understand that the small circle makes four rolling rotations plus one sliding revolution, for a total of five.

This can also be understood by considering the center of the small circle, which needs to travel around a circle of radius equal to the sum of the radii of the two circles. The center of the small circle travels a distance equal to the circumference of this larger circle, which results in the small circle making five rotations.

An alternative way to think about this problem is to imagine the small circle rolling on the exterior of a regular polygon, such as a hexagon. As the small circle rolls along each side of the hexagon, it does not get to roll over the complete length of each side due to the convexity of the interior of the hexagon. This results in some rotation being "lost" because the small circle completes a rotation along one side when it is tangent to two sides of the hexagon.

The mathematical formula for the number of rotations is Xπr / Yπr = X/Y, where X is the circumference of the larger circle and Y is the circumference of the smaller circle. However, this formula only accounts for the rolling rotations and does not include the additional sliding revolution.

In reality, no decomposition of the movement into rotation and revolution is taking place. Just as a human's heart and lungs work simultaneously, rotation and revolution occur together. Separating revolution from rotation is helpful for understanding, but it does not provide a fundamental solution.

The problem has been discussed in various forms, including in the book "Aha! Gotcha: Paradoxes to Puzzle and Delight" by Martin Gardner and in Scientific American in 1868. Alternative proofs and explanations for this problem are welcome.

Some people have suggested that the question is vague and that there are several correct answers depending on the definition of rotation and revolution. However, the standard definition of a rotation is a rotation around the center of the circle, and a revolution is a circular movement around an external point.

Others have pointed out that the problem can be understood by considering the limit of regular polygons as the number of sides tends to infinity. This can help to provide a better intuitive sense of why an additional revolution takes place.

In conclusion, the smaller circle makes five rotations when rolling around the larger circle. This can be understood by considering the movement of the center of the small circle, the rolling and sliding rotations, and the mathematical formula for the number of rotations. The problem is a classic example of a mathematical paradox that requires careful consideration of the definitions and assumptions involved.


I am looking for tools within the Coq logic that can perform induction smartly, similar to ATPs (Automated Theorem Provers). Currently, I am using sauto and coqhammer, but they have limitations. For instance, I want to use a "hammer" like coq_simp_atp to solve simple proofs, such as:

have: forall n: my_nat, O [+] n = n [+] O. by coq_simp_atp.

However, sauto and hammer cannot solve this proof without induction:

have: forall n: my_nat, O [+] n = n [+] O. induction n. by sauto. by sauto.

I previously discussed this issue in Coq hammer's GitHub issues, specifically Issue #125, regarding extending hammer/sauto to use induction.

The context of my question involves the following Coq code:

From Coq Require Import ssreflect ssrfun ssrbool.
From Hammer Require Import Tactics Hammer.

I defined a custom natural number type my_nat and a fixpoint function add_left for addition:

Inductive my_nat : Type :=
  O : my_nat
  | S : my_nat -> my_nat.

Fixpoint add_left (n m : my_nat) : my_nat :=
  match n with
  | O => m
  | S n' => S (add_left n' m)
  end.

Notation "x [+] y" := (add_left x y) (at level 50, left associativity).

My goal is to prove the commutativity of addition:

Theorem my_add_comm:
  forall n m : my_nat, n [+] m = m [+] n.
Proof.
  intros. induction n as [ n' IH].

  have: forall n: my_nat, O [+] n = n [+] O. induction n. by sauto. by sauto.
  by sauto.

Are there any tools or techniques that can simplify this process, especially for simple cases like the one above?


### Problem 7-113

Write an equation for the given problem, where *x* represents the number of brownies and *y* represents the number of cookies. The equations are:

*x* − 3 = *y*
3*x* + 2.5*y* = 218

To solve for *x*, substitute *y* in the second equation:

3*x* + 2.5(*x* − 3) = 218
3*x* + 2.5*x* − 7.5 = 218
5.5*x* = 225.5
*x* = 41

This solution finds the value of *x*, the number of brownies.


## Exercise 2: Non-Homogenous Differential Equation (Sinusoid)

This exercise involves solving a non-homogeneous second-order differential equation using the Laplace transform, where the right-hand side function is a sinusoid. The goal is to apply the Laplace transform method to find the solution to this specific type of differential equation.

The Laplace transform is a powerful tool for solving differential equations, especially those that are non-homogeneous, meaning they have a non-zero right-hand side. In the case of a sinusoidal right-hand side function, the solution requires careful application of the Laplace transform properties and the ability to handle sinusoidal functions within the transform domain.

To solve such an equation, one would typically start by applying the Laplace transform to both sides of the differential equation, then use the properties of the Laplace transform to simplify and solve for the transformed solution. Finally, the inverse Laplace transform would be applied to obtain the solution in the original domain.

This method is particularly useful for equations where traditional methods of solving differential equations, such as undetermined coefficients or variation of parameters, become cumbersome or impractical. The use of the Laplace transform provides a systematic approach to solving these equations, making it a valuable technique in the study of differential equations.


**The Ideal Gas Equation**

The ideal gas equation combines the three gas laws to form a single master equation for the behavior of gases: PV = nRT, where R is the gas constant. This equation describes the relationship among the four variables P, V, T, and n.

**Important Notes of Ideal Gas**

1. An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation.
2. The molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container.
3. Although there is no such thing in nature as an ideal gas, the ideal gas approximation works rather well for most reasonable temperature and pressure ranges.
4. We can safely use the ideal gas equation to solve many gas problems.

**The Gas Constant**

To apply the ideal gas equation to a real system, we must evaluate the gas constant R. At 0°C (273.15 K) and 1 atm pressure, many real gases behave like an ideal gas. Experiments show that under these conditions, 1 mole of an ideal gas occupies 22.414 L. The conditions 0°C and 1 atm are called standard temperature and pressure (STP). The value of R is 0.0821 L·atm/K·mol.

**Example 1: Calculating Pressure**

Calculate the pressure exerted by 1.82 moles of sulfur hexafluoride (SF6) in a steel vessel of volume 5.43 L at 69.5°C. Using the ideal gas equation, we can rearrange it to solve for pressure: P = nRT/V.

**Example 2: Calculating Volume at STP**

Calculate the volume occupied by 7.40 g of NH3 at STP. Recognizing that 1 mole of an ideal gas occupies 22.41 L at STP, we can use the molar mass of NH3 (17.03 g) to find the volume.

**Combined Gas Law Equation**

The combined gas law equation is useful for problems that involve changes in P, V, T, and n for a gas sample. The equation is: P1V1/T1 = P2V2/T2. If n1 = n2, the equation becomes P1V1/T1 = P2V2/T2.

**Example 3: Calculating Final Volume**

An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? Using the combined gas law equation, we can solve for the final volume.

**Example 4: Calculating Final Pressure**

Argon is an inert gas used in light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure. Using the combined gas law equation, we can solve for the final pressure.

**Example 5: Calculating Final Volume**

A small bubble rises from the bottom of a lake, where the temperature and pressure are 8°C and 6.4 atm, to the water's surface, where the temperature is 25°C and the pressure is 1.0 atm. Calculate the final volume of the bubble if its initial volume was 2.1 mL. Using the combined gas law equation, we can solve for the final volume.

**Density of Gas**

The density of a gas can be calculated using the ideal gas equation: d = MP/RT, where d is the density, M is the molar mass, P is the pressure, R is the gas constant, and T is the temperature.

**Example 6: Calculating Density**

Calculate the density of carbon dioxide (CO2) in grams per liter (g/L) at 0.990 atm and 55°C. Using the ideal gas equation, we can solve for the density. The molar mass of CO2 is 44.01 g. The density of CO2 is 1.62 g/L.


Tabulation – Orderly arrangement of data in rows and columns. The different forms of presentation of data are textual and descriptive presentation, tabular presentation, and diagrammatic presentation. Tabulation means presenting data in tabular form. It is a table that shows more than one characteristic of the data.

The purpose of collection of data is to collect evidence for reaching a sound and clear solution to a problem. Data collection is an important aspect of research. Data has to be collected from appropriate sources. The term data refers to any kind of information researchers obtain on the subjects, respondents or participants of the study.

Data can be collected in two different ways. Primary data, quantitative data, and secondary data are a different forms of data collection. Primary data is original data collected by a researcher for a specific purpose. Quantitative data is expressed in numbers. Secondary data is data collected by someone else, such as government agencies or other researchers.

Surveys, questionnaires, and observation are methods of data collection. Surveys are used to collect data from a sample of people. Questionnaires are used to collect data from a large number of people. Observation is used to collect data by observing people or things.

The advantages of tabular presentation of data are it helps in comparing data, tabulated data can be easily presented in the form of diagrams and graphs, and tabulated data can be easily understand and interpreted. The disadvantages of tabular presentation of data are it is difficult to present a large amount of data in a table, and it is not easy to understand and interpret the data.

A histogram is a graphical representation of data that shows the distribution of data. It is used to present time series and frequency distributions. A histogram is a type of bar chart that shows the frequency of data. The horizontal or x-axis represents the classes or categories of data, and the vertical or y-axis represents the frequency of data.

A frequency distribution is a table that shows the frequency of data. It is used to present the distribution of data. A frequency distribution is a type of table that shows the frequency of data. The table has two columns, one for the classes or categories of data and one for the frequency of data.

A cumulative frequency distribution is a table that shows the cumulative frequency of data. It is used to present the distribution of data. A cumulative frequency distribution is a type of table that shows the cumulative frequency of data. The table has two columns, one for the classes or categories of data and one for the cumulative frequency of data.

A less than type cumulative frequency curve is a graphical representation of data that shows the cumulative frequency of data. It is used to present the distribution of data. A less than type cumulative frequency curve is a type of curve that shows the cumulative frequency of data. The horizontal or x-axis represents the classes or categories of data, and the vertical or y-axis represents the cumulative frequency of data.

The lower point of the curve is joined to the X-axis at the lower limit of the first class interval. The points are joined by a free-hand smooth curve. The curve is called the cumulative frequency curve or ogive.

The advantages of diagrammatic presentation of data are it helps in understanding the data easily, it is easy to compare the data, and it is easy to understand and interpret the data. The disadvantages of diagrammatic presentation of data are it is difficult to present a large amount of data in a diagram, and it is not easy to understand and interpret the data.

A bar chart is a graphical representation of data that shows the frequency of data. It is used to present the distribution of data. A bar chart is a type of chart that shows the frequency of data. The horizontal or x-axis represents the classes or categories of data, and the vertical or y-axis represents the frequency of data.

A pie chart is a graphical representation of data that shows the proportion of data. It is used to present the distribution of data. A pie chart is a type of chart that shows the proportion of data. The chart is divided into sectors, each representing a proportion of the data.

The advantages of graphical presentation of data are it helps in understanding the data easily, it is easy to compare the data, and it is easy to understand and interpret the data. The disadvantages of graphical presentation of data are it is difficult to present a large amount of data in a graph, and it is not easy to understand and interpret the data.

In conclusion, data collection is an important aspect of research. Data can be collected in two different ways, primary and secondary. The different forms of presentation of data are textual and descriptive presentation, tabular presentation, and diagrammatic presentation. Each form of presentation has its advantages and disadvantages. The choice of presentation depends on the type of data and the purpose of the research.


# Data Structures

In this lesson, we will study how data can be collected and stored in memory, focusing on vectors, matrices, n-dimensional arrays, tuples, and dictionaries.

## Arrays

### Vectors

A vector is a list of ordered data that shares a common type. It is a one-dimensional array, and often "vector" and "array" are used as synonyms. In Julia, a vector is created using the syntax:
```julia
a = [1, 2, 3, 4, 5]
b = [1.2, 3, 4, 5]
c = ["Hello", "it's me", "TechyTok"]
```
Vector elements can be accessed using the indexing syntax `array_name[element_number]`. Note that in Julia, vectors start at 1. To append an element to an array, use the `append!` function.

### Matrices

Matrices are two-dimensional arrays. They can be defined using the syntax:
```julia
mat1 = [1 2 3; 4 5 6]
```
Rows are separated by semicolons, and columns are separated by spaces. Matrix elements can be accessed using the notation `mat1[row_index, column_index]`.

### N-Dimensional Arrays

N-dimensional arrays can be created using the `zeros` function. For example, to create a 2x3x4 table:
```julia
table = zeros(2, 3, 4)
```
The array can be filled manually or using a loop.

### Slices

A convenient notation called slicing can be used to access a subset of elements from an array. For example:
```julia
a = [1, 2, 3, 4, 5, 6]
b = a[2:5]
```
Slicing can also be used to access a subset of a matrix.

### Views

Arrays are pointers to locations in memory. When assigning one array to another, the elements of the original array can be modified by accessing the new array. To make a copy of an array, use the `copy` function.

## Tuples

A tuple is a fixed-size group of variables that may share a common type but don't need to. Tuples are created using the syntax:
```julia
a = (1, 2, 3)
b = 1, 2, 3
```
Tuples can be unpacked over multiple values and can be used to emulate multiple return values from functions.

### Splatting

Tuples can be unpacked and passed as arguments to a function using the syntax:
```julia
function splat_me(a, b, c)
    return a * b * c
end
tuple1 = (1, 2, 3)
splat_me(tuple1...)
```
### Named Tuples

Named tuples are like tuples but with a name identifier for a single value. They can be created using the syntax:
```julia
namedTuple1 = (a = 1, b = "hello")
```
Named tuples can be accessed using the syntax `namedTuple1[:a]`.

## Dictionaries

A dictionary is a collection of keys and values. They are unordered and can be used to organize datasets. Dictionaries can be created using the syntax:
```julia
person1 = Dict("Name" => "Aurelio", "Phone" => 123456789, "Shoe-size" => 40)
```
Dictionaries can contain other dictionaries and can be modified after creation.

## Conclusion

In this lesson, we learned how to operate on arrays, tuples, and dictionaries. These structures are the basics of in-memory storage for any data. Julia has a built-in garbage collector, so memory management is not a concern. Arrays are lightweight and useful solutions for passing blocks of data.


**Constructing Angle Bisectors**

Constructing angle bisectors divides a given angle into two equal halves. The term "bisect" refers to dividing into two equal parts. For example, when an angle bisector is constructed for an angle of 70°, it divides the angle into two equal angles of 35° each. Angle bisectors can be constructed for acute, obtuse, or right angles.

## Constructing an Angle Bisector with a Compass

To construct an angle bisector, we need a ruler, pencil, compass, and protractor (if the angle measure is given). Let's consider angle AOB. We don't need a protractor if the angle measure is not given.

**Step 1:** Draw two arcs with the compass, centered at O, intersecting rays OA and OB at points C and D. Note that OC = OD, as they are radii of the same circle.

**Step 2:** Without changing the compass width, draw two arcs centered at C and D, intersecting at point E. Note that CE = DE, as the arcs have the same radius.

**Step 3:** Join ray OE, which is the required angle bisector of angle AOB.

## Proving an Angle Bisector

The constructed angle bisector creates two similar triangles, ΔOCE and ΔODE. We can prove that these triangles are congruent using the SSS criterion:

1. OC = OD (radii of the same circle)
2. CE = DE (arcs of equal radii)
3. OE = OE (common)

By the SSS criterion, ΔOCE ≅ ΔODE, which means ∠COE = ∠DOE. Thus, ray OE is the angle bisector of ∠AOB.

## FAQs on Constructing Angle Bisectors

### How Do You Construct an Angle Bisector?

To construct an angle bisector, follow these steps:

1. Draw an angle of any measure.
2. With one end of the horizontal ray as the center, draw an arc intersecting the two rays of the angle at two points.
3. With the intersection points as centers, draw two arcs intersecting each other.
4. Join the point of intersection with the vertex of the angle.

### Can We Construct Angle Bisectors for Angles of Any Measure?

Yes, an angle bisector can be constructed for angles of any measure, including acute, obtuse, and right angles.

### When an Angle Bisector is Constructed for a Right Angle, What is the Measure of the Two Angles Formed?

The two angles formed are congruent, each measuring 45°.

### How to Construct an Angle Bisector with a Protractor and a Compass?

If the angle measure is given, use a protractor to draw the angle and then follow the same steps as before.

### Are There Any Triangles Formed by Constructing an Angle Bisector?

Yes, two congruent triangles are formed, similar to the SSS congruence of triangles. The common side shared by these triangles is the angle bisector line.

### When Constructing an Angle Bisector, Why Must the Arcs Intersect?

The arcs must intersect to form two congruent triangles, which proves that the angle bisector line bisects the given angle.


**Option #1**

**Question 1: Future and Present Values**

1. If you invest $100 at an interest rate of 15%, calculate the future value after 8 years.
2. To save $10,000 in 5 years at an interest rate of 5%, determine the present value.
3. Calculate the present value of $374 paid in year 9, given a cost of capital of 9%.
4. If the present value of a 3-year annuity of $10 is $24.65, find the 3-year annuity factor.

**Question 2: Compounding**

1. Calculate the value of a $10 million investment after 4 years with interest compounded annually, monthly, and continuously at 6%.
2. Compare investments with interest rates of 12% compounded annually, 11.7% compounded semiannually, and 11.5% compounded continuously.
3. Determine the preferred investment and calculate its value after 1, 5, and 20 years.

**Question 3: IRR Rule**

Mr. Clotz must choose between two projects:

| Project | C0 | C1 | C2 | IRR (%) |
| --- | --- | --- | --- | --- |
| A | -400 | 250 | 300 | 23 |
| B | -200 | 140 | 179 | 36 |

1. Explain why choosing the project with the higher IRR is incorrect.
2. Adapt the IRR rule to select the best project.
3. Show that the chosen project also has the higher NPV.

**Question 4: Capital Rationing**

Given the following projects and $20,000 available for investment:

| Project | C0 | C1 | C2 | NPV |
| --- | --- | --- | --- | --- |
| W | -10,000 | -10,000 | 0 | 6,700 |
| X | 0 | -20,000 | 5,000 | 9,000 |
| Y | -10,000 | 5,000 | 5,000 | 0 |
| Z | -15,000 | 5,000 | 4,000 | -1,500 |

1. Set up the problem as a linear program and solve it.
2. Calculate and interpret the shadow prices on the capital constraints.

**Option #2**

**Question 1: Present Values**

Calculate the present value of an oil field with a cash flow of $2 million in the first year and an opportunity cost of capital of 12% for the following cases:

1. Cash flows continue forever with no growth or decline.
2. Cash flows continue for 20 years with no growth or decline.
3. Cash flows continue forever with 3% annual growth.
4. Cash flows continue for 20 years with 3% annual growth.

**Question 2: Continuous Compounding**

1. Calculate the future value of a $100 investment after 20 years with annual compounding at 15%.
2. Calculate the future value of a $100 investment after 20 years with continuous compounding at 15%.

**Question 3: IRR Rule**

Given the cash flows for two projects:

| Project | C0 | C1 | C2 | C3 |
| --- | --- | --- | --- | --- |
| A | -100 | 60 | 60 | 0 |
| B | -10,000 | 140 | 0 | 0 |

1. Calculate the NPV of each project for discount rates of 0%, 10%, and 20%.
2. Plot the NPV against the discount rate and approximate the IRR for each project.
3. Determine the circumstances under which project A should be accepted.

**Question 4: Capital Rationing**

Given the following investment opportunities and $90,000 available for investment:

| Project | NPV | Investment |
| --- | --- | --- |
| 1 | 15,000 | 10,000 |
| 2 | 25,000 | 5,000 |
| 3 | 10,000 | 90,000 |
| 4 | 15,000 | 60,000 |
| 5 | 15,000 | 75,000 |
| 6 | 3,000 | 15,000 |

1. Determine which projects to invest in.
2. Solve the capital rationing problem and interpret the results.


Horizontal Projectile Motion Calculator
This calculator solves a particular case of projectile motion where an object is launched horizontally from an elevated plane. By inputting any two values, the calculator will compute the rest and display the trajectory.

Horizontal Projectile Motion Equations
Since the launch angle is 0° (parallel to the ground), we only have one component of initial velocity: $V_x = V$, while $V_y = 0$. The starting point is at the origin. The equations of motion are:

* Distance:
  - Horizontal distance: $x = V t$
  - Vertical distance: $y = – \frac{1}{2}g t^2$, where $g$ is gravity acceleration and $h$ is elevation.
* Velocity:
  - Horizontal velocity: $V$
  - Vertical velocity: $–g t$
* Acceleration:
  - Horizontal acceleration: 0
  - Vertical acceleration: $g$ (due to gravity)

The horizontal projectile motion equations can be combined to eliminate $t$. The trajectory equation is:
$x = V \sqrt{\frac{2h}{g}}$ for $y = -\frac{1}{2}gt^2$ and $x = Vt$.

Time of Flight
To find the time of flight, we calculate when the projectile hits the ground ($y = h$): $\frac{1}{2}g t^2 = h$. Solving for $t$, we get: $t = \sqrt{\frac{2h}{g}}$.

Range of the Projectile
The range is the total horizontal distance traveled during the flight time: $R = V \times t = V \times \sqrt{\frac{2h}{g}}$.

Example of Horizontal Projectile Motion Calculations
A ball is thrown from the Eiffel Tower (276 m high) with a horizontal speed of 7 m/s. Using the calculator, we find that the time of flight is 7.5 seconds and the horizontal displacement is approximately 52.52 m.

FAQ
1. How do I calculate the horizontal distance in projectile motion?
   - Method 1: $x = Vt$
   - Method 2: $x = V \sqrt{\frac{2h}{g}}$
2. How do I calculate the time of flight in horizontal projectile motion?
   - $t = \sqrt{\frac{2h}{g}}$
3. Is there horizontal acceleration in projectile motion?
   - No, horizontal acceleration is 0.
4. What is the vertical acceleration of a projectile when it is projected horizontally?
   - $g = 9.8 m/s²$ (downward direction)


**Solution to Codility Lesson 2-Exercise 3: Missing Integer Problem**

**Problem Definition:**
Define a function `solution` with input `A`, such that `solution` returns the minimal positive integer (greater than 0) that does not belong to `A`. The expected worst-case time complexity is `O(N)`, and the expected worst-case space complexity is `O(N)`, beyond input storage.

**Analysis:**
To find the minimal positive integer missing from `A`, we first count every element of `A` that belongs to the sequence `{1, ..., N}`. This is achieved by iterating through `A` and incrementing the corresponding count in the `count` array. The time complexity for counting all elements of an array is `O(N)`, as we must traverse all elements of the array.

Next, we iterate through the `count` array to find the first element with a count of 0, which represents the minimal missing integer. If all elements in the `count` array have a count greater than 0, it means `A` contains all integers from 1 to `N`, and the minimal missing integer is `N + 1`.

**Complete Solution:**
```python
def solution(A):
    N = len(A)
    count = [0] * (N + 1)

    # Count all elements of A that belong to sequence {1, ..., N}
    for k in range(N):
        if N >= A[k] > 0:
            count[A[k]] += 1

    # Search for the lesser integer that does not belong to A
    for k in range(1, N + 1):
        if count[k] == 0:
            return k

    # If A has all elements from 1 to N, N + 1 is the minimal integer
    return N + 1
```
**Time Complexity:** `O(N)`
**Space Complexity:** `O(N)`