## 13.33 Derived Colimits

In a triangulated category, there is a notion of derived colimit.

### Definition 13.33.1
Let $\mathcal{D}$ be a triangulated category and $(K_n, f_n)$ a system of objects of $\mathcal{D}$. An object $K$ is a *derived colimit* or *homotopy colimit* of the system $(K_n)$ if the direct sum $\bigoplus K_n$ exists and there is a distinguished triangle:

\[ \bigoplus K_n \to \bigoplus K_n \to K \to \bigoplus K_n[1] \]

where the map $\bigoplus K_n \to \bigoplus K_n$ is given by $1 - f_n$ in degree $n$. If this is the case, we sometimes indicate this by the notation $K = \text{hocolim} K_n$.

By TR3, a derived colimit, if it exists, is unique up to (non-unique) isomorphism. Moreover, by TR1, a derived colimit of $K_n$ exists as soon as $\bigoplus K_n$ exists. The derived category $D(\textit{Ab})$ of the category of abelian groups is an example of a triangulated category where all homotopy colimits exist.

### Lemma 13.33.4
Let $\mathcal{D}$ be a triangulated category and $(K_n, f_n)$ a system of objects of $\mathcal{D}$. Let $n_1 < n_2 < n_3 < \ldots$ be a sequence of integers. Assume $\bigoplus K_n$ and $\bigoplus K_{n_i}$ exist. Then there exists an isomorphism $\text{hocolim} K_{n_i} \to \text{hocolim} K_n$ such that:

\[ \xymatrix{ K_{n_i} \ar[r] \ar[d]_{\text{id}} & \text{hocolim} K_{n_i} \ar[d] \\ K_{n_i} \ar[r] & \text{hocolim} K_n } \]

commutes for all $i$.

### Proof
Let $g_i : K_{n_i} \to K_{n_{i+1}}$ be the composition $f_{n_{i+1} - 1} \circ \ldots \circ f_{n_i}$. We construct commutative diagrams:

\[ \vcenter { \xymatrix{ \bigoplus \nolimits _i K_{n_i} \ar[r]_{1 - g_i} \ar[d]_b & \bigoplus \nolimits _i K_{n_i} \ar[d]^a \\ \bigoplus \nolimits _n K_n \ar[r]^{1 - f_n} & \bigoplus \nolimits _n K_n } } \quad \text{and} \quad \vcenter { \xymatrix{ \bigoplus \nolimits _n K_n \ar[r]_{1 - f_n} \ar[d]_d & \bigoplus \nolimits _n K_n \ar[d]^c \\ \bigoplus \nolimits _i K_{n_i} \ar[r]^{1 - g_i} & \bigoplus \nolimits _i K_{n_i} } } \]

as follows. Let $a_i = a|_{K_{n_i}}$ be the inclusion of $K_{n_i}$ into the direct sum. In other words, $a$ is the natural inclusion. Let $b_i = b|_{K_{n_i}}$ be the map:

\[ K_{n_i} \xrightarrow {1, f_{n_i}, f_{n_i + 1} \circ f_{n_i}, \ldots, f_{n_{i+1} - 2} \circ \ldots \circ f_{n_i}} K_{n_i} \oplus K_{n_i + 1} \oplus \ldots \oplus K_{n_{i+1} - 1} \]

If $n_{i-1} < j \leq n_i$, then we let $c_j = c|_{K_j}$ be the map:

\[ K_j \xrightarrow {f_{n_i - 1} \circ \ldots \circ f_j} K_{n_i} \]

We let $d_j = d|_{K_j}$ be zero if $j \not= n_i$ for any $i$ and we let $d_{n_i}$ be the natural inclusion of $K_{n_i}$ into the direct sum. In other words, $d$ is the natural projection. By TR3, these diagrams define morphisms:

\[ \varphi : \text{hocolim} K_{n_i} \to \text{hocolim} K_n \quad \text{and} \quad \psi : \text{hocolim} K_n \to \text{hocolim} K_{n_i} \]

Since $c \circ a$ and $d \circ b$ are the identity maps, we see that $\varphi \circ \psi$ is an isomorphism by Lemma 13.4.3. The other way around, we get the morphisms $a \circ c$ and $b \circ d$. Consider the morphism $h = (h_j) : \bigoplus K_n \to \bigoplus K_n$ given by the rule: for $n_{i-1} < j < n_i$ we set:

\[ h_j : K_j \xrightarrow {1, f_j, f_{j+1} \circ f_j, \ldots, f_{n_i - 1} \circ \ldots \circ f_j} K_j \oplus \ldots \oplus K_{n_i} \]

Then the reader verifies that $(1 - f) \circ h = \text{id} - a \circ c$ and $h \circ (1 - f) = \text{id} - b \circ d$. This means that $\text{id} - \psi \circ \varphi$ has square zero by Lemma 13.4.5. In other words, $\psi \circ \varphi$ differs from the identity by a nilpotent endomorphism, hence is an isomorphism. Thus $\varphi$ and $\psi$ are isomorphisms as desired.

### Lemma 13.33.5
Let $\mathcal{A}$ be an abelian category. If $\mathcal{A}$ has exact countable direct sums, then $D(\mathcal{A})$ has countable direct sums. In fact, given a collection of complexes $K_i^\bullet$ indexed by a countable index set $I$, the termwise direct sum $\bigoplus K_i^\bullet$ is the direct sum of $K_i^\bullet$ in $D(\mathcal{A})$.

### Proof
Let $L^\bullet$ be a complex. Suppose given maps $\alpha_i : K_i^\bullet \to L^\bullet$ in $D(\mathcal{A})$. This means there exist quasi-isomorphisms $s_i : M_i^\bullet \to K_i^\bullet$ of complexes and maps of complexes $f_i : M_i^\bullet \to L^\bullet$ such that $\alpha_i = f_i s_i^{-1}$. By assumption, the map of complexes:

\[ s : \bigoplus M_i^\bullet \longrightarrow \bigoplus K_i^\bullet \]

is a quasi-isomorphism. Hence, setting $f = \bigoplus f_i$, we see that $\alpha = f s^{-1}$ is a map in $D(\mathcal{A})$ whose composition with the coprojection $K_i^\bullet \to \bigoplus K_i^\bullet$ is $\alpha_i$. We omit the verification that $\alpha$ is unique.

### Lemma 13.33.6
Let $\mathcal{A}$ be an abelian category. Assume colimits over $\mathbf{N}$ exist and are exact. Then countable direct sums exist and are exact. Moreover, if $(A_n, f_n)$ is a system over $\mathbf{N}$, then there is a short exact sequence:

\[ 0 \to \bigoplus A_n \to \bigoplus A_n \to \mathop{\mathrm{colim}}\nolimits A_n \to 0 \]

where the first map in degree $n$ is given by $1 - f_n$.

### Proof
The first statement follows from $\bigoplus A_n = \mathop{\mathrm{colim}}\nolimits (A_1 \oplus \ldots \oplus A_n)$. For the second, note that for each $n$ we have the short exact sequence:

\[ 0 \to A_1 \oplus \ldots \oplus A_{n-1} \to A_1 \oplus \ldots \oplus A_n \to A_n \to 0 \]

where the first map is given by the maps $1 - f_i$ and the second map is the sum of the transition maps. Take the colimit to get the sequence of the lemma.

### Lemma 13.33.7
Let $\mathcal{A}$ be an abelian category. Let $L_n^\bullet$ be a system of complexes of $\mathcal{A}$. Assume colimits over $\mathbf{N}$ exist and are exact in $\mathcal{A}$. Then the termwise colimit $L^\bullet = \mathop{\mathrm{colim}}\nolimits L_n^\bullet$ is a homotopy colimit of the system in $D(\mathcal{A})$.

### Proof
We have an exact sequence of complexes:

\[ 0 \to \bigoplus L_n^\bullet \to \bigoplus L_n^\bullet \to L^\bullet \to 0 \]

by Lemma 13.33.6. The direct sums are direct sums in $D(\mathcal{A})$ by Lemma 13.33.5. Thus the result follows from the definition of derived colimits in Definition 13.33.1 and the fact that a short exact sequence of complexes gives a distinguished triangle (Lemma 13.12.1).

### Lemma 13.33.8
Let $\mathcal{D}$ be a triangulated category having countable direct sums. Let $\mathcal{A}$ be an abelian category with exact colimits over $\mathbf{N}$. Let $H : \mathcal{D} \to \mathcal{A}$ be a homological functor commuting with countable direct sums. Then $H(\text{hocolim} K_n) = \mathop{\mathrm{colim}}\nolimits H(K_n)$ for any system of objects of $\mathcal{D}$.

### Proof
Write $K = \text{hocolim} K_n$. Apply $H$ to the defining distinguished triangle to get:

\[ \bigoplus H(K_n) \to \bigoplus H(K_n) \to H(K) \to \bigoplus H(K_n[1]) \to \bigoplus H(K_n[1]) \]

where the first map is given by $1 - H(f_n)$ and the last map is given by $1 - H(f_n[1])$. Apply Lemma 13.33.6 to see that this proves the lemma.

### Lemma 13.33.9
Let $\mathcal{D}$ be a triangulated category with countable direct sums. Let $K \in \mathcal{D}$ be an object such that for every countable set of objects $E_n \in \mathcal{D}$ the canonical map:

\[ \bigoplus \mathop{\mathrm{Hom}}\nolimits_\mathcal{D}(K, E_n) \longrightarrow \mathop{\mathrm{Hom}}\nolimits_\mathcal{D}(K, \bigoplus E_n) \]

is a bijection. Then, given any system $L_n$ of $\mathcal{D}$ over $\mathbf{N}$ whose derived colimit $L = \text{hocolim} L_n$ exists, we have that:

\[ \mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{Hom}}\nolimits_\mathcal{D}(K, L_n) \longrightarrow \mathop{\mathrm{Hom}}\nolimits_\mathcal{D}(K, L) \]

is a bijection.

### Proof
Consider the defining distinguished triangle:

\[ \bigoplus L_n \to \bigoplus L_n \to L \to \bigoplus L_n[1] \]

Apply the cohomological functor $\mathop{\mathrm{Hom}}\nolimits_\mathcal{D}(K, -)$ (see Lemma 13.4.2). By elementary considerations concerning colimits of abelian groups, we get the result.


My brother and I were learning the basics of mathematics and had some confusion about the behavior of zero. After researching, we found that its behavior depends on the context in which it is used.

We wondered why 1/0 is considered positive infinity rather than negative infinity when approaching zero from the negative axis.

The answer is that 1/0 is actually undefined. Similarly, the limit of 1/x as x approaches 0 is also undefined. However, when taking the limit of 1/x as x approaches zero from the left, the result is negative infinity, and when approaching from the right, the result is positive infinity.

Source: Link, Question by Rishav Mehta, Answer by S.C.B.


**What is Euclidean Geometry?**

Geometry is a branch of mathematics concerned with questions of shape, size, and relative position of figures. The term "Geometry" is derived from the Greek words "Geo" (Earth) and "Metron" (to measure). Euclidean Geometry is the study of geometry based on definitions, terms, and assumptions of the mathematician Euclid (330 B.C.). It is the geometry of flat surfaces, also known as plane geometry, which is the study of lines and shapes on a flat surface.

Euclid's work, "The Elements," is a comprehensive discussion of geometry, consisting of 13 books. It is credited with developing the first comprehensive deductive system of geometry. Euclidean Geometry can be illustrated on flat paper or a chalkboard, where certain principles hold true, including:

- The shortest distance between two points is a straight line.
- The angles in any triangle add up to 180 degrees.
- A perpendicular line intersects another line at 90-degree angles.

However, Euclidean Geometry has its limitations, as it cannot describe all physical space, such as curved space. Non-Euclidean Geometry, on the other hand, describes curved space.

**Euclid’s 5 Postulates (Axioms)**

The foundation of Euclidean Geometry is based on five postulates, which are:

1. It is possible to draw a straight line from any point to any point.
2. If you have a straight line, it is possible to extend it in any direction to infinity.
3. It is possible to draw a circle given any center and a radius.
4. All right angles are equal (congruent).
5. If you have two straight lines and a third line crossing them, and the sum of the interior angle measure of the two lines is less than two 90 degrees, then if you extend the lines, they will eventually cross on that side.

These postulates are essential to understanding Euclidean Geometry, as they provide the basis for points, lines, and planes.


Christina is trying to prove the correctness of a matrix transpose specification using TLAPS. She has written a PlusCal algorithm and a TLA+ specification, but is unsure how to use TLAPS to prove the theorem `diagseq == diag = tdiag`.

Stephan Merz responded with the following advice:

1. Generate the TLA+ specification from the PlusCal algorithm by using the "Translate PlusCal Algorithm" option in the Toolbox.
2. State the theorem as a consequence of the specification in the form `THEOREM diagseq == Spec => (diag = tdiag)`.
3. The proof will look like "BY DEF Spec, Init", but note that the assertion only talks about the initial state.

Christina's original specification is as follows:
```
---- MODULE transp ----
EXTENDS Naturals, TLC
CONSTANT n, m
(* --algorithm transpose
variables i = 1, j = 1, array = <<<<>>, <<>>>>, transp = <<<<>>, <<>>>>, diag = <<>>, tdiag = <<>>,
begin
  while j <= m do
    i := 1;
    while i <= n do
      transp[i][j] := array[j][i];
      if i = j then
        diag[j] := array[j][i];
        tdiag[j] := transp[i][j];
      end if;
      i := i + 1;
    end while;
    j := j + 1;
  end while;
  print array;
  print transp;
end algorithm *)
THEOREM diagseq == diag = tdiag
PROOF OBVIOUS
```
However, Christina is getting an error that `tdiag` is unknown. Stephan Merz explained that the reason for this error is that the assertion only talks about the initial state, and the proof is just comparing two empty lists.

To fix this, Christina needs to modify her theorem and proof to take into account the entire specification, not just the initial state. The correct theorem should be stated as `THEOREM diagseq == Spec => (diag = tdiag)`, and the proof will involve showing that the specification implies the desired property.

The key concepts involved in this problem are:

* PlusCal algorithms and their translation to TLA+ specifications
* TLAPS and its use in proving theorems about TLA+ specifications
* The importance of considering the entire specification, not just the initial state, when proving theorems.

By following Stephan Merz's advice and modifying her theorem and proof, Christina should be able to successfully prove the correctness of her matrix transpose specification using TLAPS.


## Exam-Style Question on Bearings

A fishing boat is somewhere along the line from A to C.

(a) By measuring an angle, write down the three-figure bearing of the boat from A.

(b) The coastguard at B sees the ship on a bearing of 047°. Draw accurately the line showing a bearing of 047° from B.

(c) Mark the position of the fishing boat, F, on the diagram.

(d) Measure the length of the line AB on the diagram in centimeters.

(e) The distance from A to B is 12 kilometers. Calculate the scale of the map in the form 1 : n.

(f) A lighthouse at A has a range of 9.6 kilometers. Using the scale, draw the locus of points 9.6 kilometers from A.

(g) The boat sails straight for a lobster pot at D. Draw the line FD on the diagram. How far is the boat from the lobster pot when the light from the lighthouse is first seen? Give the answer in kilometers, correct to the nearest 100 meters.

(h) If the boat does not alter course, it will arrive at D after 30 minutes. Calculate the speed of the boat in kilometers per hour.

(i) Convert the speed from part (h) to knots, given that 1 knot is 1 nautical mile per hour and 1 nautical mile is 1.85 kilometers. Give the answer to one decimal place.

The solutions to these exam-style questions are available to Transum subscribers, who can access step-by-step solutions, screen shots of calculator procedures, and ad-free browsing.

This question is similar to one that appeared on a GCSE Higher paper in 2005. The use of a calculator is allowed.

To find more exam-style questions and worked solutions, visit Transum.org/Maths/Exam/.


Here are a few of the best Halloween math problems:

1. An adult skeleton has 206 bones, with 27 bones in a hand and 26 bones in a foot. To find the number of bones not in the hands or feet, subtract the total bones in the hands and feet from the total number of bones. Since there are two hands and two feet, the calculation is: 206 - (2 * 27) - (2 * 26) = 206 - 54 - 52 = 100.

2. Count Juan Toothree's party had the following guest activity: 10 guests arrived, 2 left, 12 arrived, 3 left, 6 arrived, 8 left, 10 arrived, 1 left, 2 arrived, and 8 left. To find the total number of guests at midnight, calculate the net change in guests: 10 - 2 + 12 - 3 + 6 - 8 + 10 - 1 + 2 - 8 = 18.

3. The World Championship Punkin Chunkin Association aims to hurl a pumpkin a full mile, which is 5,280 feet. The current record is 4,484 feet, so they need to cover an additional 5,280 - 4,484 = 796 feet. To calculate the time it would take for the pumpkin to travel a mile at 1000 mph, use the formula time = distance / speed. Convert the speed to feet per second: 1000 mph * 5280 feet/mile / 3600 seconds/hour = 1466.67 feet/second. Then, calculate the time: 5280 feet / 1466.67 feet/second = 3.6 seconds.

4. Anoka, MN, the Halloween Capital of the World, created the first Halloween parade in 1920. To find the number of years until their 100th anniversary, subtract the starting year from the target year: 2020 - 1920 = 100 years, which has already passed. However, if we consider the problem as asking for the time until the next anniversary, we can calculate the years until 2120: 2120 - 2020 = 100 years. To find the number of months, multiply the years by 12: 100 years * 12 months/year = 1200 months. For the number of days, multiply the years by 365: 100 years * 365 days/year = 36,500 days.

5. A class of 18 students had 162 pumpkin seeds to divide evenly. To find the number of seeds each student received, divide the total seeds by the number of students: 162 seeds / 18 students = 9 seeds per student.


To start a project on Probability, let's relate it to a common object like a deck of cards. Probability is the likelihood that a particular event will occur, calculated using the formula:

Probability = Number of possible outcomes / Total number of items

In a deck of cards, there are 52 cards, which can be calculated by multiplying the 13 cards in each suit by the 4 suits (4 x 13 = 52).

The first problem is finding the probability of getting a red card from the deck. Since there are 2 red suits (hearts and diamonds) with 13 cards each, the total number of red cards is 26 (13 x 2 = 26). The probability is:

Number of red cards (26) / Total number of cards (52) = 26/52 = 1/2

Another example is finding the probability of getting a face card. There are 3 face cards (Jack, Queen, King) in each of the 4 suits, making a total of 12 face cards (3 x 4 = 12). The probability is:

Number of face cards (12) / Total number of cards (52) = 12/52 = 3/13

These concepts can be applied to various card games, and understanding probability can be helpful. Additionally, probability can be illustrated using tools like Venn diagrams to show the outcomes. By remembering these tips, you can better figure out the probability of getting the cards you need in a game.


The simplest form of 2032240 is 12715. To simplify the fraction 2032/240, find the Greatest Common Divisor (GCD) of the numerator and denominator. The GCD of 2032 and 240 is 16. Divide both the numerator and denominator by the GCD:
2032 ÷ 16 = 127
240 ÷ 16 = 15
The reduced fraction is 127/15. Therefore, 2032/240 simplified to lowest terms is 127/15.
Equivalent fractions include 4064/480, 1016/120, and 6096/720. Other related fractions are 4064/240, 2032/480, 6096/240, 2032/720, 2033/240, 2032/241, 2031/240, and 2032/239.


In the United States, the most common way of measuring length is in inches, with 12 inches in a foot and 36 inches in a yard. One inch is equal to 2.54 centimeters, and 1/8 inch is equal to 3.175 millimeters.

To convert inches to millimeters, first convert inches to centimeters by multiplying by 2.54, then convert centimeters to millimeters by multiplying by 10. For example, 12 inches is equal to 12 x 2.54 = 30.48 centimeters, which is equal to 30.48 x 10 = 304.8 millimeters.

Converting 1 3/8 inches to millimeters involves first converting 1 3/8 inches to a decimal, which is 1.375 inches. Then, multiply 1.375 inches by 2.54 to get 3.4935 centimeters, and finally multiply 3.4935 centimeters by 10 to get 34.935 millimeters.

The equivalent of 1/8 inch in millimeters is 3.175 mm. To confirm, 1/8 inch is not equal to 5mm, as 5mm is a metric measurement and 1/8 inch is an imperial measurement. There are exactly 3.175 mm in an 1/8 inch.

To measure 1/4 inch in mm, multiply 1/4 by 25.4, which gives 6.35 mm. Measuring to the nearest 1/8 inch is commonly done in the United States, where fractions are used to refer to inches.

In conclusion, converting fractions of inches to millimeters is a useful skill, especially when working with metric measurements. The conversion of 1/8 inch to millimeters is 3.175 mm, which can be remembered or calculated using a ruler or conversion method.

Multiple choice questions:
1. What is the equivalent of 1/8 inch in millimeters?
A) 5mm
B) 3.175mm
C) 6.35mm
D) 25.4mm

Answer: B) 3.175mm

2. Is 5mm the same as 1/8 inch?
A) Yes
B) No
C) Maybe
D) Depends on the context

Answer: B) No

3. How many mm is an 1/8 inch?
A) 25.4 mm
B) 5 mm
C) 3.175 mm
D) 6.35 mm

Answer: C) 3.175 mm

4. How do you measure 1/4 inch in mm?
A) Multiply 1/4 by 2.54
B) Multiply 1/4 by 25.4
C) Divide 1/4 by 2.54
D) Divide 1/4 by 25.4

Answer: B) Multiply 1/4 by 25.4


To determine the internal forces and bending moment caused by the applied loads at the specified sections, we first find the reactions at the supports from the whole beam's equilibrium. We then use the method of sections to find the internal forces and moments at the specified locations.

Given the dimensions of and loading on the beam, our objective is to find the axial force P, the shear V, and the bending moment M. We assume equilibrium and Saint-Venant's principle.

Our strategy involves replacing the distributed load with its equivalent concentrated load for the purpose of finding reaction forces. The magnitude of this concentrated load is the area under the distributed load: W = 1/2 * (8 k/ft) * (6 ft) = 24 kips. It acts at the centroid of the triangular area, 2 ft from the left end of the beam.

Using the free-body diagram, we calculate the reactions:
∑M_B = 0 = -R_A * (8 ft) + (24 k) * (6 ft) → R_A = 18 kips
∑M_A = 0 = -(24 k) * (2 ft) + R_B * (8 ft) → R_B = 6 kips

We verify the result using ∑F_z = 0: R_A + R_B = 24 kips.

Next, we consider sections a-a and b-b. For section a-a, we choose the simpler side to calculate, the portion of the beam to the right of a-a. Applying the equilibrium equations:
∑F_x = 0 = N_a
∑F_z = 0 = -V_a + R_B - 6 k → V_a = 0 kips
∑M_about a = 0 = M_a - (6 k) * (1 ft) + R_B * (5 ft) → M_a = -24 ft-kips

For section b-b, we use the portion of the beam to the right of b-b:
∑F_x = 0 = P_b
∑F_z = 0 = -V_b + R_B → V_b = 6 kips
∑M_about b = 0 = R_B * (2 ft) + M_b → M_b = -12 ft-kips

Note that the negative signs on the moments indicate they are opposite to the direction drawn in the FBDs. We assume positive shear and bending moment when constructing FBDs, with negative signs representing negative values.

Questions:
7.9: Although we have learned about bars subject to axial loading, how do internal forces and bending moments differ in beams?
7.8: The first of the equilibrium equations represents which type of force balance?
7.7: We begin with a general statement of the fact that internal forces and moments are related to the external loads; how do we apply this relationship to beams?
7.6: First, we need to consult our FBD and find the reaction forces; what is the next step in determining internal forces and moments?
7.5: We obtained a formula for shear stress at a given point; how does this relate to the overall shear force and bending moment in the beam?
7.4: The gage at A, in the upper portion of the cross section, measures what type of stress?
7.3: The symmetry of the cross section suggests what about the distribution of internal forces and moments?
7.2: In each case, we will first find the external reactions; what is the purpose of these reactions in determining internal forces and moments?


# What Is a Factor of a Number?

A factor of a number in math is a number that divides the given number without any remainder.

## Answer: A factor is a number that divides the given number without any remainder.

A factor is a divisor of the given number, dividing it evenly with no remainder. Key properties of factors include:
- The number of factors of a number is finite.
- A factor of a number is always less than or equal to the given number.
- Every number except 0 and 1 has at least two factors, 1 and itself.
- Division and multiplication are used to find factors.

For example, the factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Factors can be calculated using methods such as prime factorization and the factor tree method.


Abel's theorem states that if a power series converges for some nonzero value x₀, then it converges absolutely for any value of x, for which |x| < |x₀|. This concept is named after the Norwegian mathematician Niels Henrik Abel.

Absolute measurement is a measurement in which the comparison is directly with quantities whose units are basic units of the system. For example, measuring speed by measuring distance and time is an absolute measurement. Note that absolute measurement implies nothing about precision or accuracy.

The Balmer decrement refers to the intensity ratio among adjacent Balmer lines, which have well-known theoretical values. These ratios are used to determine interstellar extinction.

Bayes' theorem is a concept in probability theory concerned with determining the conditional probability of an event when another event has occurred. The theorem allows revision of the original probability with new information, and its simplest form is: P(A|B) = P(B|A) P(A)/P(B).

Bernoulli's theorem states that the quantity (P/ρ) + gz + (V²/2) is constant along any streamline, where P is the fluid pressure, V is the fluid velocity, ρ is the mass density of the fluid, g is the acceleration due to gravity, and z is the vertical height. This equation affirms that if the internal velocity of the flow increases, the internal pressure must drop.

The binomial theorem is a rule for writing an equivalent expansion of an expression such as (a + b)ⁿ without having to perform all multiplications involved. The general expression is (a + b)ⁿ = Σ (n,k)aⁱbⁿ⁻ⁱ, where the summation is from k = 0 to n.

Birkhoff's theorem states that for a four-dimensional space-time, the Schwarzschild metric is the only solution of Einstein's field equations that describes the gravitational field created by a spherically symmetrical distribution of mass.

Bremsstrahlung refers to the electromagnetic radiation emitted by a fast-moving charged particle when it passes within the strong electric field of an atomic nucleus and is decelerated.

Cauchy's theorem states that if f(x) and φ(x) are two continuous functions on the interval [a,b] and differentiable within it, and φ'(x) does not vanish anywhere inside the interval, there will be found, in [a,b], some point x = c, such that [f(b) - f(a)] / [φ(b) - φ(a)] = f'(c) / φ'(c).

The central limit theorem states that the distribution of the means of random samples from a given statistical population is equal to the mean of the population from which the samples were drawn. Moreover, if the original population has a normal distribution, the sampling distribution of means will also be normal.

The convolution theorem states that the Fourier transform of the convolution of f(x) and g(x) is equal to the product of the Fourier transform of f(x) and g(x): F{f*g} = F{f}.F{g}.

Coriolis theorem states that the absolute acceleration of a point P, which is moving with respect to a local reference frame that is also in motion, is equal to the vector sum of the Coriolis effect.

Decrement refers to the amount lost in the process of decreasing.

The divergence theorem, also known as Gauss's theorem, states that the flux of a vector field through a closed surface is equal to the divergence of the vector field integrated over the volume enclosed by the surface.

An existence theorem asserts the existence of at least one object, such as the solution to a problem or equation.

Extreme refers to something that is farthest from the center or middle, outermost, or exceeding the bounds of moderation. Examples include extreme adaptive optics, extreme HB star, and extreme ultraviolet.

Extreme adaptive optics is an adaptive optics system with high-contrast imaging and spectroscopic capabilities, enabling the detection of faint objects close to bright sources.

An extreme HB star, also known as an extreme horizontal branch star, is the hottest variety of stars on the horizontal branch, with temperatures ranging from 20,000 to 40,000 K. These stars have undergone extreme mass loss during their first ascent up the giant branch, resulting in a very thin hydrogen envelope surrounding the helium-burning core.

Extreme infrared refers to a portion of the far infrared radiation, including wavelengths between 100 and 1,000 microns.


# How To Bell Shaped Bar Graph: 4 Strategies That Work

A single peak over the center is called bell-shaped. A graph with two peaks is called bimodal. Understanding bar graphs involves describing a graph using the information learned.

To visually represent data using a bar graph, follow these steps:
1. Decide the title of the bar graph.
2. Draw the horizontal axis and vertical axis.
3. Label the horizontal axis.

The bell-shaped curve is symmetric around its center. If we disregard the two extreme outliers, the histogram of the NBS-10 data is roughly bell-shaped.

Use SPSS to do the following with the NBS-10 data:
- Delete the outliers.
- Plot a histogram with a superimposed normal curve.

The empirical rule calculator is a tool for finding the ranges that are 1 standard deviation, 2 standard deviations, and 3 standard deviations from the mean.

With symmetric data, the mean and median are close together. A histogram can take on many different shapes, including bell-shaped, which resembles a “bell” curve and has one single peak in the middle of the distribution.

A symmetrical bell-shaped curve represents the distribution of a set of values, also called a normal distribution. A graph that represents data in groups is similar to a bar graph, but unlike in a bar graph, there is no spacing between the columns that represent the data.

A bar chart plots numeric values for levels of a categorical feature as bars. Each categorical value claims one bar, and the length of each bar corresponds to the bar’s value. Bars are plotted on a common baseline to allow for easy comparison.

Notice the bell-shaped distribution of the data. The area under a histogram is 1. Accurate histograms can display either cut-points or midpoints of the interval.

In a survey, we asked 1000 respondents which car they would purchase if they had a choice between an Audi, a Mazda, a Toyota, or a Ford. A frequency distribution for a categorical variable groups the data into categories and records the number of observations that fall into each category.

A histogram graph is a bar graph representation of data. It is a representation of a range of outcomes into columns formation along the x-axis. The number count or multiple occurrences in the data for each column is represented by the y-axis.

The t-distribution is a type of continuous probability distribution that takes random values on the whole real line. The main properties of the t-distribution are:
- It is continuous.
- It is "bell-shaped", in the same way the normal curves are bell-shaped.

A bell curve (also known as a normal distribution curve) is a way to plot and analyze data that looks like a bell curve. In the bell curve, the highest point is the one that has the highest probability of occurring, and the probability of occurrences goes down on either side of the curve.

The shape of a distribution refers to the shape of a frequency or relative frequency histogram for quantitative data. A shape may be described by its symmetry, skewness, and/or modality.

A symmetric histogram is also called a bell-shaped histogram. When you draw the vertical line down the center of the histogram, and the two sides are identical in size and shape, the histogram is said to be symmetric.

To create a histogram, you need to use the Data Analysis Tool. To use the Data Analysis Tool, you need to have a Bin range. We set a bin range by studying our dataset’s lowest and highest values.

A normal distribution has some interesting properties: it has a bell shape, the mean and median are equal, and 68% of the data falls within 1 standard deviation.

Let us find the normal distribution using Excel’s mean and standard deviation.
1. Add Data.
2. Calculate the Mean.
3. Calculate the Standard Deviation.

A histogram is symmetric if you cut it down the middle and the left-hand and right-hand sides resemble mirror images of each other. The above graph shows a symmetric data set; it represents the amount of time each of 50 survey participants took to fill out a certain survey.

The shape of a distribution is described by its number of peaks and by its possession of symmetry, its tendency to skew, or its uniformity. Graphs often display peaks, or local maximums. It can be seen from the graph that the data count is highest at the center and decreases as we move away from the center.


Geometry is a branch of mathematics concerned with the properties of and relationships between points, lines, planes, and figures, as well as generalizations of these concepts. There are several types of geometry, including:

1. **Euclidean geometry**: the elementary geometry of two and three dimensions, based on the Elements of Euclid (c. 300 BC). It was later developed into **analytic geometry** by Rene Descartes (1637), who used numbers to describe points and express geometric relations in algebraic form.
2. **Descriptive geometry**: developed by Gaspard Monge in the late 18th century to represent three-dimensional objects on a two-dimensional surface.
3. **Differential geometry**: founded by Monge and C. F. Gauss in the late 18th and early 19th centuries, which applies calculus concepts to curves, surfaces, and other geometric objects.
4. **Non-Euclidean geometry**: developed by N. I. Lobachevsky (1826), Janos Bolyai (1832), and Bernhard Riemann (1854), which replaces Euclid's postulate regarding parallel lines and changes theorems accordingly.
5. **Projective geometry**: a more general type of geometry that includes metric geometries as a special case and makes no reference to lengths or angle measurements.
6. **Absolute geometry**: a general metric geometry that consists of all Euclidean geometry except the part dependent on the parallel postulate, and is valid for both Euclidean and non-Euclidean geometry.
7. **Affine geometry**: includes Euclid's parallel postulate but disregards two other postulates concerning circles and angle measurement, and is valid in the four-dimensional geometry of spacetime used in the theory of relativity.
8. **Ordered geometry**: consists of all propositions common to both absolute geometry and affine geometry, and includes the notion of intermediacy ("betweenness") but not measurement.
9. **Topology**: a type of geometry concerned with properties invariant under continuous transformations, which carry neighborhoods of points into neighborhoods of their images.

The different geometries are classified and related to one another in various ways. The Erlangen program, proposed by Felix Klein in 1872, classifies geometries with respect to the geometric properties that are left unchanged under a given group of transformations. For example:

* Euclidean geometry is the study of properties unchanged by similarity transformations.
* Affine geometry is concerned with properties invariant under linear transformations that preserve parallelism.
* Projective geometry studies invariants under projective transformations.

The axiomatic approach to geometry, developed by Euclid, involves defining basic terms, stating axioms and postulates, and deriving theorems through deductive logic. This method has been adopted throughout mathematics and other fields. However, the close examination of Euclidean geometry's axioms and postulates in the 19th century revealed that the logical basis of geometry was not as firm as previously thought, leading to the development of new axiom and postulate systems by mathematicians such as David Hilbert (1899).


**Introduction to Rational Expressions**

A rational expression is a fraction that has a polynomial in the numerator, denominator, or both. For example, $\frac{x}{x-2}$ is a rational expression. To work with rational expressions, it's essential to understand the concepts of factors, terms, and expressions.

**Key Concepts**

* **Factors**: Factors are the building blocks of multiplication. In rational expressions, factors are algebraic expressions that completely divide the rational expression.
* **Terms**: Terms are single numbers or variables and numbers connected by multiplication. Examples of terms include 6, -5, and $x^3$.
* **Expressions**: Expressions are groups of terms connected by addition and subtraction. For example, $4x^2 - 7$ is an expression.

**Simplifying Rational Expressions**

To simplify a rational expression, follow these steps:

1. Factorize the numerator and denominator.
2. Cancel out common factors in the numerator and denominator.
3. Rewrite the remaining factors in the numerator and denominator.

**Example**: Simplify the rational expression $\frac{3x^2- 3x}{3x^3- 6x^2+3x}$.

**Solution**: Factor out the common factor $3x$ from the numerator and denominator to get $\frac{3x(x-1)}{3x(x^2-2x+1)}$. Then, cancel out the common factor $3x$ to get $\frac{x-1}{x^2-2x+1}$. Finally, factor the denominator to get $\frac{x-1}{(x-1)(x-1)}$, which simplifies to $\frac{1}{x-1}$.

**Adding and Subtracting Rational Expressions**

To add or subtract rational expressions, follow these steps:

1. Find a common denominator.
2. Combine the fractions as a single fraction with the common denominator.
3. Simplify the expression in the numerator.
4. Reduce the fraction to lowest terms.

**Example 1**: Solve $\frac{x}{x-1} + \frac{2-x}{x-1}$.

**Solution**: Since the denominators are the same, add the numerators to get $\frac{x+2-x}{x-1} = \frac{2}{x-1}$.

**Example 2**: Solve $\frac{x+7}{2x+12} + \frac{6}{x^2- 36}$.

**Solution**: Find the least common multiple (LCM) of the denominators, which is $2(x+6)(x-6)$. Multiply each fraction to get equivalent fractions with the LCM as the denominator. Then, combine the fractions and simplify to get $\frac{x-5}{2(x+6)}$.

**General Rules and Common Errors**

* Rational expressions can only be added or subtracted if they have a common denominator.
* The numerator and denominator of a fraction may be multiplied by the same quantity to get an equivalent fraction.
* When subtracting rational expressions, subtract the entire numerator of the second fraction, and change the sign of each term in the numerator.

By following these rules and concepts, you can work with rational expressions and simplify them to their lowest terms.


**Exercise 6.1: Squares and Square Roots**

**Q.1. Unit Digit of Squares**
Find the unit digit of the squares of the following numbers:
(i) 81
(ii) 272
(iii) 799
(iv) 3853
(v) 1234
(vi) 26387
(vii) 52698
(viii) 99880
(ix) 12796
(x) 55555

**Q.2. Non-Perfect Squares**
State why the following numbers are not perfect squares:
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050

**Q.3. Odd Squares**
Which of the following numbers have odd squares?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004

**Q.4. Pattern: Squares of 1**
Observe the pattern and find the missing digits:
11^2 = 121
101^2 = 10201
1001^2 = 1002001
10001^2 = 1 …… 2 …… 1
1000001^2 = ………………………

**Q.5. Pattern: Squares of 101**
Observe the pattern and supply the missing numbers:
11^2 = 1 2 1
101^2 = 1 0 2 0 1
10101^2 = 1020301
1010101^2 = ……………………………
^2 = 10203040504030201

**Q.6. Pattern: Sum of Squares**
Using the given pattern, find the missing numbers:
1^2 + 2^2 + 2^2 = 3^2
2^2 + 3^2 + 6^2 = 7^2
3^2 + 4^2 + 12^2 = 13^2
4^2 + 5^2 + _^2 = 21^2
5^2 + _^2 + 30^2 = 31^2
6^2 + 7^2 + _^2 = __^2

**Q.7. Sum of Odd Numbers**
Find the sum without adding:
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

**Q.8. Express as Sum of Odd Numbers**
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

**Q.9. Numbers between Squares**
Find how many numbers lie between the squares of the following numbers:
(i) 12 and 13
(ii) 25 and 26
(iii) 99 and 100


# Equatorial Bulge

The Earth has an equatorial bulge of 42.72 km due to its rotation, resulting in a diameter of 12,756.28 km across the equatorial plane, which is 42.72 km more than the diameter between the poles (12,713.56 km).

## Kinetic Energy and Potential Energy

A device with a springy metal band illustrates the physics of equatorial bulge. When spun, the band's kinetic energy increases, and some of this energy is converted to elastic potential energy. Once the device reaches a constant rotation rate, the energy is divided equally between kinetic energy and elastic potential energy. This division occurs because the energy transforms between kinetic energy and elastic potential energy as the band bulges out or contracts.

## Actual Planet

For a planet, the lowest possible state of gravitational potential energy is a perfect sphere. The equatorial bulge of a spinning celestial body is a balance between kinetic energy and gravitational potential energy.

## Discussion

The Earth's equatorial bulge is often misunderstood as a force equilibrium between centripetal and centrifugal forces. However, it is more accurately described as an energy equilibrium between kinetic energy and gravitational potential energy.

## Differences in Gravitational Acceleration

The gravitational acceleration is less at the equator than at the poles due to the Earth's rotation. At the equator, the effective gravitational acceleration is 9.7805 m/s², while at the poles it is 9.8322 m/s². This difference is primarily due to two factors: objects at the equator are about 21 kilometers further from the Earth's center of mass (contributing 70% to the difference), and the non-spherical shape of the Earth (contributing 30%).

## Centre of Gravitation

For a satellite in an equatorial orbit, the center of gravitational attraction is about 10 kilometers away from the Earth's geometrical center. This means that the effective center of gravitational attraction circumnavigates the geometrical center at a distance of about 10 kilometers.

## Satellite Orbits

The Earth's slightly non-spherical gravitational field affects satellite orbits, causing them to deviate from pure ellipses. This is particularly important for GPS satellites.

## Other Celestial Bodies

The planet with the largest known equatorial bulge is Saturn, with a bulge of 11,808 km. Many rotating astronomical bodies exhibit equatorial bulges due to their rotation.


Absolute values are used in mathematics, physics, and engineering to determine the size or magnitude of an object, such as a number or a vector, regardless of its direction. The absolute value of a number is its distance from zero on the number line, without considering whether it is positive or negative.

## Defining the Absolute Value

The absolute value of a number x can be defined as a piecewise function:

| x | = x if x ≥ 0
| x | = -x if x < 0

This function returns the value of x without its sign. For example, |12| = 12 and |-12| = 12.

## Implementing the Absolute Value Function in Python

The absolute value function can be implemented in Python from scratch using a piecewise function:

```python
def absolute_value(x):
    if x >= 0:
        return x
    else:
        return -x
```

Alternatively, it can be implemented using a conditional expression:

```python
def absolute_value(x):
    return x if x >= 0 else -x
```

Or using the built-in `max()` function:

```python
def absolute_value(x):
    return max(x, -x)
```

The algebraic definition of an absolute value can also be implemented in Python:

```python
from math import sqrt
def absolute_value(x):
    return sqrt(pow(x, 2))
```

Or using exponentiation:

```python
def absolute_value(x):
    return (x**2) ** 0.5
```

## Using the Built-in `abs()` Function

Python has a built-in `abs()` function that can be used to calculate the absolute value of a number. This function is more efficient and versatile than the custom implementations:

```python
>>> abs(-12)
12
>>> abs(-12.0)
12.0
```

The `abs()` function can also be used with complex numbers:

```python
>>> z = complex(3, 2)
>>> abs(z)
3.605551275463989
```

And with fractions and decimals:

```python
>>> from fractions import Fraction
>>> abs(Fraction("1/3"))
Fraction(1, 3)
>>> abs(Fraction("-3/4"))
Fraction(3, 4)
>>> from decimal import Decimal
>>> abs(Decimal("0.3333333333333333"))
Decimal('0.3333333333333333')
>>> abs(Decimal("-0.75"))
Decimal('0.75')
```

## Calling `abs()` on Other Python Objects

The `abs()` function can be used with NumPy arrays and pandas series:

```python
>>> import numpy as np
>>> temperature_readings = np.array([1, -5, 1, -4, -1, -8, 0, -7, 3, -5, 2])
>>> abs(temperature_readings)
array([1, 5, 1, 4, 1, 8, 0, 7, 3, 5, 2])
>>> import pandas as pd
>>> df = pd.DataFrame({
...     "A": [1, -5, 1, -4, -1, -8, 0, -7, 3, -5, 2],
...     "B": [2, -3, 2, -2, 2, -1, 2, -4, 2, -3, 2]
... })
>>> abs(df)
   A  B
0   1  2
1   5  3
2   1  2
3   4  2
4   1  2
5   8  1
6   0  2
7   7  4
8   3  2
9   5  3
10  2  2
```

## Customizing the Behavior of `abs()`

The behavior of `abs()` can be customized for custom classes by implementing the special `.__abs__()` method:

```python
>>> import math
>>> class Vector:
...     def __init__(self, *coordinates):
...         self.coordinates = coordinates
...
...     def __abs__(self):
...         return math.hypot(*self.coordinates)
>>> snowflake_velocity = Vector(0.42, 1.5, 0.87)
>>> abs(snowflake_velocity)
1.7841804841439108
```

This allows the `abs()` function to work with instances of the custom class.


The football team F, basketball team B, and track team T decide to form a varsity club V. Given that $n\left(F\right)\,=\,25,\,n\left(B\right)\,=\,12,\,n\left(T\right)\,=\,30$ and no person belongs to two teams, the total number of members in V is $n\left(F\right)+n\left(B\right)+n\left(T\right)\,=\,67$.

If $n\left(F\cap T\right)\,=\,6$ but there are no members of B who are in F or T, then $n\left(V\right)\,=\,67 - 6\,=\,61$.

If $n\left(F\cap T\right)\,=\,6$, $n\left(T\cap B\right)\,=\,4$, and $n\left(F\cap B\right)\,=\,0$, then $n\left(V\right)\,=\,67 - 10\,=\,57$.

If $n\left(F\cap T\right)\,=\,6$, $n\left(T\cap B\right)\,=\,4$, $n\left(F\cap B\right)\,=\,3$, and $n\left(F\cap B\cap T\right)\,=\,2$, then the number of persons belonging to two teams is $6 + 4 + 3 - 2\cdot2\,=\,9$. The number of persons belonging to all three teams is $2$. Therefore, $n\left(V\right)\,=\,67 - 9 - 2 + 2\,=\,58$.

The correct calculations for the last part are:
- Total members in two teams without counting the intersection twice: $6 + 4 + 3 = 13$
- However, since 2 members are in all three teams, they were counted three times (in each pair), so we need to subtract them twice (not thrice, because we still count them once as members) to correct the count of members in exactly two teams: $13 - 2\cdot2 = 9$
- Then subtract this corrected count and add back the ones in all three teams once (since they're still members) from the total: $67 - 9 - 2 + 2 = 58$

Thus, the answers are:
a. $n\left(V\right)\,=\,61$
b. $n\left(V\right)\,=\,57$
c. $n\left(V\right)\,=\,58$


This chapter explores applications of the Discrete Fourier Transform (DFT). It begins with a series of spectrum analysis examples using an efficient DFT implementation in Matlab or Octave. The Fourier theorems serve as a foundation for understanding spectral analysis elements. The chapter then covers linear systems theory, utilizing the convolution theorem and complex number properties. Additionally, it introduces DFT applications in statistical signal processing, including cross-correlation, matched filtering, system identification, power spectrum estimation, and coherence function measurement. A secondary focus is on practical Matlab usage for signal processing, such as displaying signals and spectra.


What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?

A code is based on two different prime numbers less than 10. Multiply them together and shift the alphabet forwards by the result to decipher the code.

A clockmaker's wife cut up his birthday cake to look like a clock face. Determine who received each piece.

Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.

Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside is written on each envelope. Determine the possible numbers inside the envelopes.

Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.

Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. Find the total and determine how this can be done.

Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. Determine the possible amounts Lolla could have paid for the balloon.

Put eight beanbags in the hoops so that there are four in the blue hoop, five in the red, and six in the yellow. Find all the ways of doing this.

Cassandra, David, and Lachlan are brothers and sisters, ranging in age between 1 year and 14 years. Figure out their exact ages from the clues.

Ram divided 15 pennies among four small bags. He could then pay any sum of money from 1p to 15p without opening any bag. Determine the number of pennies Ram put in each bag.

There are 4 jugs which hold 9 litres, 7 litres, 4 litres, and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.

A magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell, and go out a total of 15.

There are 78 prisoners in a square cell block of twelve cells. The prison warder arranged them so there were 25 along each wall of the prison block. Determine how he did it.

Exactly 195 digits have been used to number the pages in a book. Find the number of pages the book has.

Place six toy ladybirds into the box so that there are two ladybirds in every column and every row.

Three children are going to buy some plants for their birthdays. They will plant them within circular paths. Determine how they could do this.

Place the "worm" on the 100 square and find the total of the four squares it covers. Keeping its head in the same place, find the other totals that can be made.

Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it.

Put the numbers 1, 2, 3, 4, 5, 6 into the squares so that the numbers on each circle add up to the same amount. Find the rule for giving another set of six numbers.

Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Explain what happens.

You have 5 darts and your target score is 44. Find the different ways you could score 44.

Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them.

Fill in the numbers to make the sum of each row, column, and diagonal equal to 34.

There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.

Can you put plus signs in so this is true: 1 2 3 4 5 6 7 8 9 = 99? Find the number of ways you can do it.

Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees, and jelly belly beans. Determine the number of different ways she could make a jolly jelly feast with 32 legs.

Investigate some interesting dates yourself, such as 03.06.09 or 08.01.09.

Find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column.

Using the statements, work out how many of each type of rabbit there are in these pens.

Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.

Can you use the information to find out which cards were used?

This task follows on from Build it Up and takes the ideas into three dimensions.

Find all the ways to get 15 at the top of this triangle of numbers.

A dice train has been made using specific rules. Determine the number of different trains that can be made.

Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference. Explain what happens.

A group of children are using measuring cylinders but they lose the labels. Help relabel them.

Tim had nine cards each with a different number from 1 to 9 on it. Determine how he could have put them into three piles so that the total in each pile was 15.

Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.

Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.

Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether.

There were chews for 2p, mini eggs for 3p, Chocko bars for 5p, and lollypops for 7p in the sweet shop. Determine what each of the children could buy with their money.

I was looking at the number plate of a car parked outside. Using my special code, S208VBJ adds to 65. Crack my code and use it to find out what both of these number plates add up to.


A population refers to a group of people, objects, events, or measurements, representing the standard deviation of the mean within a dataset. The standard error is a measure of the accuracy of a statistical sample population, functioning as a way to determine the accuracy of the sample or multiple samples by analyzing deviation within the means. It is the standard deviation of the sampling distribution for a statistic.

The standard error is inversely proportional to the sample size; the larger the sample size, the smaller the standard error, as the statistic will approach the actual value. However, if the sample size is very large (e.g., greater than 1,000), virtually any statistical result calculated on that sample will be statistically significant. The standard deviation measures the variability of the sample, and the standard error accounts for the differences between the means across datasets.

The standard error is not the only measure of dispersion and accuracy of the sample statistic. Other measures, such as effect size, p-value, and sample size, are also useful tools for understanding the accuracy of statistics calculated on random samples. A t-test is used to determine the difference between two sample means from two normally distributed populations with unknown variance.

Standard error statistics are a class of statistics provided as output in many inferential statistics, functioning as descriptive statistics. They are used to determine the accuracy of the sample or multiple samples. The smaller the spread among the variables, the more accurate the dataset. The standard error is a useful tool for researchers to understand the accuracy of statistics calculated on random samples, especially when used in conjunction with other measures such as effect size and sample size.


I want to examine the issue from a different perspective. Suppose the minus and plus inputs to the op-amp are not equal, where V_minus = V_plus + delta_v. How will the op-amp make them equal again?


**Introduction to Numerical Inaccuracies**

Computers typically use floating point numbers, which can cause substantial rounding errors in complex calculations. These small inaccuracies can accumulate and lead to significant errors. Arbitrary precision arithmetic offers a solution to this problem.

## Numerical Inaccuracies in Practice

Using the Windows Calculator in scientific mode, we can observe the following examples:

* √4 - 2 = **-8.1648465955514287168521180122928e-39** (instead of 0)
* 10^{-37} + 1 = **1** (instead of a more precise result)

In programming languages, similar effects occur when using floating point numbers. For example, in Java:

```java
float intermediateResult = (3.0f / 5000.0f) * 5000.0f;
System.out.println("Intermediate result: " + intermediateResult);
float wrongResult = 1.0f / (intermediateResult - 3.0f);
System.out.println("Final result: " + wrongResult);
```

This code returns an intermediate result of **3.0000002** and a final result of **4194304**. Using the `double` datatype (64-bit double-precision) instead of `float` (32-bit single-precision) does not improve the result significantly.

## Examples of Numerical Inaccuracies

* Solving linear equation systems can be unstable numerically, as seen in the example of Hilbert-Matrices on the JLinAlg website.
* Algorithms to approximate mathematical constants like π may not converge to the correct result when using floating point numbers.
* The failure of the Patriot missile defense system in Dhahran, Saudi Arabia, in 1991 was caused by a calculation error due to floating point numbers.

## Floating Point Numbers

Floating point numbers are designed to perform numerical calculations as exactly as possible with a limited number of bits. They use scientific notation to represent numbers in a compact form, which can lead to rounding errors. For example:

* Planck time: 5.391·10^{-44} seconds (as a fixed point number: 0.000,000,000,000,000,000,000,000,000,000,000,000,000,000,053,91 seconds)
* Mass of the Earth: 5.974 · 10^{24} kg (as a fixed point number: 5,974,000,000,000,000,000,000,000 kg)

## Arbitrary Precision Arithmetic

Arbitrary precision arithmetic offers a way to avoid rounding errors. JLinAlg is an open-source project that provides datatypes with arbitrary precision. Using rational numbers, we can achieve more precise results:

```java
Rational one = Rational.FACTORY.get(1.0);
Rational three = Rational.FACTORY.get(3.0);
Rational fiveThousand = Rational.FACTORY.get(5000.0);
Rational intermediateResult = (three.divide(fiveThousand)).multiply(fiveThousand);
System.out.println("Intermediate result: " + intermediateResult);
one.divide(intermediateResult.subtract(three));
```

This code returns an intermediate result of **3** and an error message when trying to divide by zero.

## Comparison of Precision

The result of 116 / 406 can be calculated using different methods:

* Single precision floating point numbers (Float): **0.2857143**
* Double precision floating point numbers (Double): **0.2857142857142857**
* Arbitrary precision (Rational): **2 / 7**

The last result is the most precise and easiest to handle.

## Conclusion

Rounding errors can occur when using floating point numbers, and it is essential to be aware of them. In some cases, sacrificing computational performance for arbitrary precision arithmetic is worth it to achieve more accurate results.


Connecting a three-phase motor to a single-phase sector can be achieved through several solutions, including the use of a capacitor assembly. This method, however, results in a drop in power output, with an average loss of 30% of useful power and 50% at startup. It is suitable for machines that do not start under load, such as a drum or a polisher, but not recommended for engines that start charging or require high power, like a back stand or pestle.

To implement this solution, a capacitor with a voltage greater than 230V and designed for AC (non-polarized) and continuous motor operation is required. The capacitor value is typically expressed in microfarads (µf) and can be calculated by multiplying the motor's power in horsepower (HP) by 50. For example, a 250W (0.25kW) motor first needs to be converted into horsepower: 250W / 736W/HP = 0.34 HP. Then, the capacitor value is 0.34 HP * 50 = 17 µf.

If the capacitor value is less than calculated, the motor may not operate efficiently or at all. Conversely, a higher value than necessary does not negatively impact performance. The formula to remember is C = 50 * P, where C is the capacitor value in µf and P is the engine horsepower.

For conversions, to go from kW to W, multiply by 1000 (e.g., 0.25 kW * 1000 = 250 W), and to convert W to HP, divide by 736 (e.g., 250 W / 736 = 0.34 HP).

When connecting the motor, a triangle coupling is used for a 380V three-phase motor to a 220V single-phase sector. The phase is connected to "u", the neutral to "v", and a capacitor is added between "v" and "w". To reverse the motor's direction, cross the two wires of the sector.

Capacitors can be sourced from new appliances or salvaged from old engines or washing machines, with a preference for plastic-wrapped (self-healing) capacitors over those packed in aluminum. If the exact capacitor value is not available, multiple capacitors can be connected in parallel or series to achieve the required value. In parallel, the values are added (e.g., 10 µf + 15 µf = 25 µf), and in series, the inverse of the values is added and then inverted (e.g., 1/(1/10 + 1/15) = 6 µf for 10 µf and 15 µf capacitors).


# Increasing or Decreasing? Variation 1

## Task

Consider the expression $$\frac{1}{\displaystyle\frac1{R_1} + \frac1{R_2}}$$ where $R_1$ and $R_2$ are positive. Suppose we increase the value of $R_1$ while keeping $R_2$ constant. Does the value of the expression above increase, decrease, or stay the same? Explain in terms of the structure of the expression.

## Solution

An increase in $R_1$ causes $\displaystyle \frac{1}{R_1}$ to decrease. Since the denominator of the expression is the sum of $\displaystyle \frac{1}{R_1}$ and a constant, $\displaystyle \frac{1}{R_2}$, the denominator also decreases. A decrease in the denominator results in an increase in the value of the expression.

To understand this, consider a numerical example: if $R_1$ and $R_2$ are both equal to 1, then $\displaystyle \frac{1}{R_1} + \frac{1}{R_2} = 1 + 1 = 2$, so $$\frac{1}{\displaystyle\frac1{R_1} + \frac1{R_2}} = \frac12.$$ If $R_1$ is increased to $2$, while $R_2$ is kept constant at 1, then $\displaystyle \frac{1}{R_1} + \frac{1}{R_2} = \frac12 + 1 = \frac32$, so $$\frac{1}{\displaystyle\frac1{R_1} + \frac1{R_2}} = \frac23,$$ which is greater than $1/2$.

This task illustrates Standard for Mathematical Practice 7, Look for and make sense of structure. Students should be able to see meaning in both pieces of an expression and the expression as a whole, and reason about how changes in the variables affect the overall value of the expression.

What happens to the value of the expression when $R_1$ is increased?
A) It increases
B) It decreases
C) It stays the same

Answer: A) It increases


## Uncrossed Knight’s Tours

The 2018 International Collegiate Programming Contest (ICPC) featured a challenging problem proposed by Derek Kisman. The problem involves finding the longest possible tour of a knight on a rectangular m × n chessboard without crossing its own path. The knight must visit as many squares as possible and return to its starting square.

The problem is an extension of the classic knight's tour puzzle on an 8 × 8 chessboard. However, the added constraint of not crossing its own path makes it impossible to visit every square. The goal is to maximize the number of visited squares.

The input consists of two integers, m and n, representing the dimensions of the chessboard, where 1 ≤ m ≤ 8 and 1 ≤ n ≤ 10^15. The problem requires dynamic programming and the realization that tours have repeating patterns.

Derek Kisman's solution involves rephrasing the problem as a graph problem, where the goal is to find the highest sum of weights for a path of length N from the start to the end node. The problem becomes periodic, and the highest-ratio cycle dominates all paths.

To compute results, Derek's search program observes when the cycle starts, as all states of the dynamic programming have values that are just a constant added to all states of the dynamic programming X turns ago. The cycle occurs much faster than the theoretical limit.

The problem has been solved for m up to 9 and n up to infinity, with the discovery of cycles for the functions f_m(n) and g_m(n), which represent the largest size of an uncrossed knight's closed and open tours, respectively.

The solution has implications for mathematical theory, as it provides a new approach to solving problems with periodic behavior. The discovery of the cycles for the functions f_m(n) and g_m(n) allows for the calculation of these functions for any n, even if n is extremely large.

The problem has also been discussed in the context of a 2014 Math Festival in Moscow, where a similar problem was presented to 7th graders. The problem involved finding the longest possible path inside a 5 × 8 rectangle using diagonals of 1 × 2 rectangles. The solution to this problem is equivalent to finding the largest uncrossed knight's closed tour on a 6 × 9 grid.

Overall, the problem of uncrossed knight's tours is a challenging and interesting problem that requires dynamic programming and mathematical theory. The solution provides a new approach to solving problems with periodic behavior and has implications for mathematical theory and computer science.


## 1. Introduction to Independent Component Analysis

Independent Component Analysis (ICA) is a statistical technique used in signal processing and machine learning to filter signals. This tutorial explains the ICA concept, provides a simple example of the problem it solves, and explores its applications.

## 2. The Cocktail Party Problem

The "cocktail party problem" illustrates the ICA technique. Imagine two people conversing at a party, recorded by two microphones at different volumes, along with background noise. The challenge is to separate the two voices from the noisy recordings. ICA can solve this problem by separating a multivariate signal into additive subcomponents.

## 3. Definition of Independent Component Analysis

ICA is a method for converting a set of vectors into a maximally independent set. It can be defined using the following equations:

- Measured signals as a linear combination of independent components: **x = As**
- Independent components as a linear combination of measured signals: **s = Wx**

where **A** is a weight matrix, **W** is an unmixing matrix, **x** are measured signals, and **s** are independent components.

The goal is to find a matrix **W** such that the source signals **s** are maximally independent. This can be achieved by:

- Minimizing mutual information between independent components
- Maximizing non-Gaussianity between independent components

## 3.1 Assumptions for Independent Component Analysis

To apply ICA, three assumptions must be made:

1. Each measured signal is a linear combination of the sources.
2. The source signals are statistically independent of each other.
3. The values in each source signal have a non-Gaussian distribution.

Statistical independence between two signals **x** and **y** means their joint distribution is equal to the product of their individual probability distributions: **p(x, y) = p(x) * p(y)**.

## 4. Independent Component Analysis Algorithms

To estimate one of the source signals, a linear combination of signals can be used: **y = w^T x**, where **w** is a weight vector. Maximizing the non-Gaussianity of **y** can help recover one of the independent components.

Kurtosis is a measure of non-Gaussianity, which can be used to maximize the absolute value of kurtosis: **|Kurt(y)|**. The FastICA algorithm is an iterative method that uses non-linear optimization to find the independent components.

Other ICA algorithms include:

- Infomax: maximizes mutual information between mixed signals and estimated independent components
- Joint Approximated Diagonalization of Eigenmatrices (JADE): separates mixed signals into source signals using fourth-order moments
- Particle Swarm Optimization (PSO): a heuristic optimization algorithm that searches for the mixing matrix

## 5. Applications of Independent Component Analysis

ICA has various applications in:

- Signal processing for speech, audio, or image separation
- Neuroscience: separating neural signals into independent components
- Finance: identifying hidden features in financial time series for forecasting
- Data mining: finding patterns and correlations in large datasets

## 6. Conclusion

ICA is a powerful technique for signal analysis, capable of uncovering hidden patterns. This tutorial provided an introduction to ICA, its definition, assumptions, algorithms, and applications, demonstrating its potential in various fields.


The terms "average," "mean," and "mode" are often used in statistics to describe a set of numbers. The "mean" is a general term that is most commonly used as an abbreviation for "arithmetical mean." To calculate the arithmetical mean, also known as the average, you add all the elements in a set of numbers and then divide by the number of elements. For example, given the set {5, 6, 1, 1, 1, 7, 4, 3}, the arithmetical mean is (5+6+1+1+1+7+4+3)/8 = 28/8 = 3.5.

In addition to the arithmetical mean, there are other types of means, including the geometric mean and the harmonic mean. The geometric mean is calculated by multiplying all the elements in a set and then taking the nth root, where n is the number of elements. Using the same set {5, 6, 1, 1, 1, 7, 4, 3}, the geometric mean is (5*6*1*1*1*7*4*3)^(1/8) = 2.6618. The harmonic mean is calculated by taking the reciprocal of the arithmetical mean of the reciprocals of the elements. For the set {5, 6, 1, 1, 1, 7, 4, 3}, the harmonic mean is 8/(1/5+1/6+1/1+1/1+1/1+1/7+1/4+1/3) = 1.9546.

The mode is another important concept in statistics, which refers to the number that appears most frequently in a set. In the set {5, 6, 1, 1, 1, 7, 4, 3}, the mode is 1, since it appears three times, more than any other number. The mean, whether arithmetical, geometrical, or harmonic, is often used as a "representative" value for a set of data, but it's not always the best choice. Sometimes, the mode is a better representative value, as it can provide more insight into the characteristics of the data.

In summary, the key concepts are:
- Arithmetical mean (average): the sum of all elements divided by the number of elements.
- Geometric mean: the nth root of the product of all elements.
- Harmonic mean: the reciprocal of the arithmetical mean of the reciprocals of the elements.
- Mode: the number that appears most frequently in a set.
Understanding these concepts can help in choosing the best representative value for a set of data.


**Lesson 3: Understanding Addends and Sum**

* **Addends** are the numbers being added in an equation. For example, in 4+1=5, 4 and 1 are the addends.
* The **Sum** is the answer to an addition problem.

When adding three numbers, follow these steps:
1. Add the first two addends.
2. Add their sum to the third addend.

Example: 6 + 3 + 5 =
1. 6 + 3 = 9
2. 9 + 5 = 14

**Finding Missing Addends**

To find a missing addend:
1. Add the two known numbers.
2. Count on to reach the sum.

Example: 5 + 2 + __ = 15
1. 5 + 2 = 7
2. Count on from 7 to 15: 7 + 8 = 15
So, 5 + 2 + 8 = 15

**Lesson 4: Order of Addends**

Changing the order of addends does not change the sum. Numbers can be added in any order.

Examples:
* 8 + 4 = 12 and 4 + 8 = 12
* 7 + 1 + 8 = 16 and 8 + 1 + 7 = 16

**Lesson 5: Related Facts**

**Related facts** are equations with the same numbers.
Example: Related facts for 6, 7, and 13 are:
* 6 + 7 = 13 and 13 - 6 = 7
* 7 + 6 = 13 and 13 - 7 = 6

To create related facts:
* For subtraction, the largest number goes first.
* For addition, the largest number goes last.


Irregular Sinusoidal Curves

A function can be constructed to graph a sinusoidal curve with constant periodicity but non-equidistant maximum and minimum values. For example, a curve with a period of 24 hours, where the time span from the maximum value to the minimum value is 16 hours and the span from minimum value to maximum value is 8 hours, can be achieved by varying parameters in the basic sinusoidal function.

The basic sinusoidal function is of the form y = A*sin[2*pi*t/T + phi(t)], where T is the period. To make the maximum and minimum occur at specific times, the phase phi can be made time-dependent. If phi is a constant, the graph of y is shifted horizontally, resulting in a sine wave with equal half-wavelengths between maximum and minimum and between minimum and maximum.

To force the maximum to occur at t = tmax = +8 hr and the minimum to occur at t = tmin = -8 hr, the argument of the sine function can be set to pi/2 and -pi/2, respectively. This results in two equations that specify the values of phi at the maximum and minimum: phi(T/3) = -pi/6 and phi(-T/3) = +pi/6.

A simple function phi that passes through these two points and is T-periodic can be chosen:
- phi(t) = +pi*(t/T + 1/2) for -T/2 <= t < -T/3
- phi(t) = -pi*t/(2*T) for -T/3 <= t <= T/3
- phi(t) = -(pi/6) + pi*(t/T - 1/3) for T/3 < t <= T/2

Using this expression for phi(t) in the equation for y forces the maximum to occur at t = tmax = 8 hr and the minimum to occur at t = tmin = -8 hr, resulting in imbalanced durations between min-max and max-min. To reverse the intervals, the function can be inverted by multiplying it by –1: y = -sin[2*pi*t/T + phi(t)].

This method is an example of “phase modulation” (PM), a technique used in communication systems. Other parameters in the sinusoidal wave y = A(t) sin[2*pi*f(t)*t + phi(t)] can also be varied. If A varies with t, it is called “amplitude modulation”, and if f varies with t, it is called “frequency modulation”. These terms are commonly used in the context of AM and FM radio.


**Baccarat Money Formula with a Fixed Ratio and Consistent Profits**

Most Baccarat money formulas focus on extreme profits with extreme risks. However, this approach can be risky and may not be suitable for all players. A more conservative approach is to use a fixed ratio formula that allows for consistent profits with lower risks.

**What is the Fixed Ratio Baccarat Formula?**

The fixed ratio formula divides the available capital into units, each worth a percentage of the total capital. For example, if you have 1,000 baht and want to divide it into units worth 5% each, the value of each unit would be 50 baht (1,000 x 0.05).

The formula not only covers investment units but also profits. The goal is to adjust the investment units when a 20% profit is achieved. For instance, if you start with 1,000 baht and reach 1,200 baht, you would adjust the investment units to 60 baht (1,200 x 0.05). The profit target for the next round would be 240 baht (20% of 1,200 baht).

If the game results in a loss, the same principle applies. For example, if you lose 20% of your capital, you would stop and adjust the investment unit again. If you start with 1,000 baht and lose 20%, you would be left with 800 baht. You would then adjust the investment unit to 40 baht (800 x 0.05).

**Key Principles:**

* Determine the capital to be used for playing each day.
* Bet 5% of the available funds per round.
* When a 20% profit is achieved, take profits and combine them with the capital to create a new capital.
* Bet 5% of the new capital per round.
* If a 20% loss is incurred, adjust the new bet to 5% of the remaining capital.

By following these principles, players can use the fixed ratio Baccarat formula to achieve consistent profits with lower risks. The formula allows for adjustments to be made based on profits and losses, ensuring that players can manage their capital effectively and minimize their risks.


# Section 1.3 - Methods and Techniques

The concept of "related rates" refers to a method for finding relationships between the rates of change of variables in a situation that is changing over time. For example, consider a cone filling up with water, where both the height of the liquid and its volume are changing with respect to time.

In traditional problem-solving, formulas are often viewed as static, where inputting numbers yields a specific output. However, when dealing with related rates, it's essential to consider each variable as dynamic, changing over time. This perspective is crucial in understanding how different variables, such as the radius and area of a ripple spreading through a pond, change at distinct rates.

By utilizing implicit differentiation, we can establish relationships between these rates of change. This concept has numerous applications in real-world situations, allowing us to analyze and understand complex, dynamic systems.


For exercises 1 and 2, translate each phrase into an algebraic expression.
1. The difference of a number and sixteen
A. n + 16
B. n – 16
C. 16n
D. n/16
2. Five times k
A. 5k
B. k + 5
C. k ÷ 5
D. 5 – k

For exercises 3 and 4, evaluate each expression when s = 9 and t = 12.
3. 3(3t + s)
A. 9
B. 24
C. 45
D. 135
4. 2s – t
A. –6
B. 6
C. 15
D. 30

To translate the phrase "the difference of a number and sixteen" into an algebraic expression, we need to identify the variable representing the number, 'n'. The word "difference" usually means subtraction, so the algebraic expression is B. n – 16.

To translate the phrase "five times k" into an algebraic expression, we represent the variable 'k' multiplied by five. The word "times" implies multiplication, so the algebraic expression is A. 5k.

To evaluate the expression 3(3t + s) when s = 9 and t = 12, we substitute the given values into the algebraic expression and simplify: 3(3*12 + 9) = 3(36 + 9) = 3(45) = 135. Therefore, the answer is D. 135.

To evaluate the expression 2s – t when s = 9 and t = 12, we substitute the values into the algebraic expression and simplify: 2*9 – 12 = 18 – 12 = 6. Therefore, the answer is B. 6.

The correct answers are:
1. B
2. A
3. D
4. B


To solve for x in the equation (x+yi) + (2(x+yi) + 3i) = 9, apply algebraic steps:

1. Apply the distributive property: (x + yi) + 2(x + yi) + 3i = 9
2. Simplify by adding terms: x + yi + 2x + 2yi + 3i = 9
3. Combine like terms: 3x + 3yi + 3i = 9
4. Combine the imaginary terms: 3x + (3y + 3)i = 9
5. Since 9 is a real number, the imaginary part must equal 0: 3y + 3 = 0
6. Solve for y: 3y = -3, y = -1
7. Substitute y back into the equation and solve for x: 3x = 9, x = 3

The solution involves recognizing that the real and imaginary parts on both sides of the equation must be equal. Thus, for the real parts: 3x = 9, and for the imaginary parts: 3y + 3 = 0. Solving these gives x = 3 and y = -1.

Given the original task was to solve for x, x = 3.


I previously used the Mann-Whitney test to determine the differences between two sample distributions: pictures that included me and those that didn’t. The samples had different variances, necessitating the use of this test. However, I didn't fully understand the test, so I decided to learn more about it.

The Mann-Whitney U test is used to reject or not reject the null hypothesis that two samples come from the same population. The test assumes that the probability of one sample being greater than the other is the same as the probability of the other sample being greater. In modern computing, simulations can be used instead of statistical tests.

My two samples didn't have the same number of observations, making it difficult to calculate the difference between them using the Mann-Whitney test. Instead, I simulated the data to get an actual probability as the p-value, without making any assumptions. I randomly sampled from the overall population, assigning each post to either sample X or Y.

The data consists of Instagram posts with variables such as image number, number of people, whether I am included, and the number of likes. To simulate the data, I first determined the sizes of the two groups: posts with pictures of me (nx) and posts without me (ny).

```r
nx <- nrow(instaData[includes_me<1 & num_people>0,])
ny <- nrow(instaData[num_people>0]) - nx
```

I created a function to perform the simulation:

```r
picSimulation <- function(data, n, nx, depVar, idVar) {
  meanList <- list()
  for(i in 1:n){
    sampleX <- data[sample(1:nrow(data), nx, replace = F)]
    data[['sampleX']] <- ifelse(data[[idVar]] %in% sampleX[[idVar]], 1, 0)
    meanX = mean(data[sampleX=='1'][[depVar]])
    meanY = mean(data[sampleX=='0'][[depVar]])
    diffMeans = meanY - meanX
    meanList[i] = diffMeans
  }
  return(meanList)
}
```

Using this function, I simulated the data 1000 times to get a distribution of mean differences. The true difference in means between the two samples is 8.795. I calculated the probability of getting a mean difference of this size or greater, given that the two samples come from the same population.

The probability is calculated as:

```r
nrow(df[unlisted_mean_list > diff])/nrow(df)
```

The result is 0.003, indicating that it's highly unlikely (less than 1%) that the two samples come from the same population. This suggests that the two samples are statistically significantly different from each other.

The simulation approach allows for a more intuitive understanding of the data, without relying on statistical tests. The results provide strong evidence that the presence of me in the pictures has a significant impact on the number of likes.


In information theory, selfinformation is a measure of the information content associated with the outcome of a random variable, expressed in units such as bits, nats, or hartleys. The amount of selfinformation contained in a probabilistic event depends on its probability, with smaller probabilities resulting in larger selfinformation.

The measure of selfinformation has the property that if an event C is composed of two mutually independent events A and B, the amount of information associated with event C is the sum of the amounts of information associated with events A and B. The selfinformation I(ω_{n}) associated with outcome ω_{n} is defined using the binary logarithm function as:

I(ω_{n}) = -log2(P(ω_{n}))

This definition complies with the conditions that the unit of selfinformation is in bits when using the logarithm of base 2, in nats when using the logarithm of base e, and in hartleys when using the logarithm of base 10. Selfinformation is also known as surprisal, representing the "surprise" of seeing an outcome, with highly probable outcomes being less surprising.

The information entropy of a random event is the expected value of its selfinformation, and selfinformation is an example of a proper scoring rule. The concept of selfinformation was elaborated by Claude E. Shannon in 1948, and the term "surprisal" was coined by Myron Tribus in his 1961 book Thermostatics and Thermodynamics.

Key properties of selfinformation include:
- Dependence on probability: the smaller the probability of an event, the larger the selfinformation associated with it.
- Additivity: the selfinformation associated with a composite event is the sum of the selfinformation associated with its independent components.
- Unit of measurement: bits, nats, or hartleys, depending on the base of the logarithm used.

Understanding selfinformation is crucial in information theory, as it provides a quantitative measure of the information content associated with random events, allowing for the calculation of information entropy and the evaluation of proper scoring rules.


You have worked with quadratic functions in three forms:
1. Vertex Form: y = a(x – h)² + k
2. Standard Form: y = ax² + bx + c
3. Factored Form: y = a(x – p)(x – q)

Given the following functions in vertex form, rewrite each in standard form and then, if possible, in factored form:
- y = (x – 3)² – 1
- y = (x – 2)² + 4

Sketch a rough graph of each equation and decide whether the graph has x-intercepts:
- y = (x – 5)² – 3
- y = 2(x + 4)² – 7
- y = –3(x + 1)² + 2
- y = 0.5(x – 2)²
- y = (x – 1)² + 1

Use a graphing calculator to check your decisions.

To determine the number of x-intercepts for a quadratic function in vertex form y = a(x – h)² + k, consider the following rules:
- If k < 0 and a > 0, the graph has two x-intercepts.
- If k > 0 and a > 0, the graph has no x-intercepts.
- If k = 0, the graph has one x-intercept.

For each equation that has x-intercepts, rewrite it in factored form and explain how factored form tells you the x-intercepts:
- y = (x – 3)² – 4
- y = (x + 2)² – 25
- y = (x – 3)² – 9
- y = (x – 5)² + 1
- y = (x – 3)² – 25

Use graphing calculators to check your work.

Consider the graph of y = (x – 7)(x + 3) and use the x-intercepts to help locate the vertex. Experiment with other equations in factored form to find a way to locate the vertex of a parabola when you know the x-intercepts. Apply this method to:
- y = (x – 8)(x – 10)
- y = 5(x + 2)(x + 4)

Describe a method for figuring out the vertex for the graph of equations in the form y = (x – p)(x – q).