recursal/Fanatic-Fandom huggingface dataset:
Computing Divisors with Python

**By: Tao Steven Zheng (郑涛)**

## **Task**

Write code that asks the user to input a positive integer and computes all the divisors (including 1). The divisors are then put into a list. The code must also determine and report whether the positive integer is prime or composite.

## **Code**

```
n = int(input("Enter a positive integer = "))
divisors = []

        divisors.append(i)
print(divisors)

if len(divisors) == 2: #which will contain only 1 and itself
    print("{} is prime".format(n))
else:
    print("{} is composite".format(n))

```

## **Results**

**For a composite number**

Enter a positive integer = 60

[1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60]

60 is composite

**For a prime number**

Enter a positive integer = 23

[1, 23]

23 is prime
Jiuzhang Suanshu Fangcheng 9

## **Problem**

   今有五羊、四犬、三雞、二兔，直錢一千四百九十六；四羊、二犬、六雞、三兔直錢一千一百七十五；三羊、一犬、七雞、五兔，直錢九百五十八；二羊、三犬、五雞、一兔，直錢八百六十一。問：羊、犬、雞、兔價各幾何？

5 goats, 4 dogs, 3 chicken, and 2 rabbits are worth 1496 coins;

4 goats, 2 dogs, 6 chicken, and 3 rabbits are worth 1175 coins;

3 goats, 1 dog, 7 chicken, and 5 rabbits are worth 958 coins;

2 goats, 3 dogs, 5 chickens, and 1 rabbits are worth 861 coins.

How much is a goat, hound, chicken, and rabbit each?

## **Solution**

Construct the matrix

A

=

[

5

4

3

2

1496

4

2

6

3

1175

3

1

7

5

958

2

3

5

1

861

]

{\displaystyle \mathbf{A} =
 \begin{bmatrix}
 5 & 4 & 3 & 2 & 1496 \\
 4 & 2 & 6 & 3 & 1175 \\
 3 & 1 & 7 & 5 & 958 \\
 2 & 3 & 5 & 1 & 861 \\
 \end{bmatrix}
}

Use the MATLAB/OCTAVE command **rref(A)** to row reduce the matrix. Click here for an online Octave shell. For the full solution all written out, use this site: Linear Algebra Toolkit.

r
r
e
f
(

A

# )

[

1

0

0

0

177

0

1

0

0

121

0

0

1

0

23

0

0

0

1

29

]

{\displaystyle rref(\mathbf{A}) =
 \begin{bmatrix}
 1 & 0 & 0 & 0 & 177 \\
 0 & 1 & 0 & 0 & 121 \\
 0 & 0 & 1 & 0 & 23 \\
 0 & 0 & 0 & 1 & 29 \\
 \end{bmatrix}
}

A goat costs 177 coins, a dog costs 121 coins, a chicken costs 23 coins, a rabbit costs 29 coins.
Jiuzhang Suanshu Junshu 26

## **Problem**

   This classic problem comes from Chapter 6 of the *Jiuzhang Suanshu* (九章算術).

今有池，五渠注之。其一渠開之，少半日一滿；次，一日一滿；次，二日半一滿；次，三日一滿；次，五日一滿。今皆決之，問：幾何日滿池？

Suppose there there is a cistern, and 5 canals fill into it. Open the first canal and the cistern fills in 1/3 day; with the second, it fills in 1 day; with the third, in 2 1/2 days; with the fourth, in 3 days, and with the fifth in 5 days. Now, suppose all the canals are opened. Question: How how many days will cistern be full?

## **Solution**

Let

T

{\displaystyle T}

 be the time (in days) it takes to fill the cistern if all canals are open.

Let

x

{\displaystyle x}

 be the volume of water.

x

1

/

3

x
1

x

5

/

2

x
3

x
5

=

x
T

{\displaystyle
\frac{x}{1/3} + \frac{x}{1} + \frac{x}{5/2} + \frac{x}{3} + \frac{x}{5} = \frac{x}{T}
}

x

(

1

1

/

3

1
1

1

5

/

2

1
3

1
5

)

=

x
T

{\displaystyle
x\left(\frac{1}{1/3} + \frac{1}{1} + \frac{1}{5/2} + \frac{1}{3} + \frac{1}{5}\right) = \frac{x}{T}
}

Divide

x

{\displaystyle x}

 on both sides of the equation.

3
1

2
5

1
3

1
5

=

1
T

{\displaystyle
3 + 1 + \frac{2}{5} + \frac{1}{3} + \frac{1}{5} = \frac{1}{T}
}

74
15

=

1
T

{\displaystyle
\frac{74}{15} = \frac{1}{T}
}

# T

15
74

{\displaystyle
T = \frac{15}{74}
}

If all canals are opened, the cistern will be filled in

15
74

{\displaystyle \frac{15}{74} }

 days.
Integral of Inverse Tangent

This problem is a good exercise of integration by parts and integration by substitution.

## **Problem**

Compute the integral

∫
arctan
⁡
(
x
)

d
x
.

{\displaystyle \int \arctan(x) \,dx. }

# u
arctan
⁡
(
x
)

{\displaystyle u = \arctan(x) }

 and

d
# v
(
1
)

d
x

{\displaystyle dv = (1) \,dx }

. Then

d
# u

1

1

x
2

d
x

{\displaystyle du = \frac{1}{1+x^2} \,dx }

 and

# v
x

{\displaystyle v = x }

.

Subsequently,

∫
arctan
⁡
(
x
)

d
# x
arctan
⁡
(
x
)
⋅
x
−
∫

x

1

x
2

d
x
.

{\displaystyle \int \arctan(x)\,dx = \arctan(x)\cdot x - \int \frac{x}{1+x^2} \,dx. }

Now focus on the integral

∫

x

1

x
2

d
x

{\displaystyle \int \frac{x}{1+x^2} \,dx }

. Use the substitution method for this integral. Let

# u
1

x
2

{\displaystyle u = 1 +x^2 }

, then

d
# u
2
x

d
x

{\displaystyle du = 2x \, dx }

.

Therefore,

∫

x

1

x
2

d
# x

1
2

∫

1
u

d
u

{\displaystyle \int \frac{x}{1+x^2}\, dx = \frac{1}{2}\int \frac{1}{u}\, du }

∫

x

1

x
2

d
# x

1
2

ln
⁡

1

x
2

.

{\displaystyle \int \frac{x}{1+x^2} \,dx = \frac{1}{2} \ln|1+x^2|. }

Consequently,

∫
arctan
⁡
(
x
)

d
# x
x
arctan
⁡
(
x
)
−

1
2

ln
⁡

1

x
2

C
.

{\displaystyle \int \arctan(x)\, dx = x\arctan(x) - \frac{1}{2} \ln|1+x^2| + C. }
Ballistic Pendulum

## **Problem**

A bullet of mass

m

{\displaystyle m}

 is fired into a stationary block of mass

M

{\displaystyle M}

 at some velocity

v
b

{\displaystyle v\_b}

. The block is attached to a wire (assume to be massless and infinitely rigid). The bullet-block system swings to a height

h

{\displaystyle h}

.

**Part 1:** Explain why the conservation of energy cannot be applied before the collision?

**Part 2:** Determine

v
b

{\displaystyle v\_b}

 in terms of

m

{\displaystyle m}

,

M

{\displaystyle M}

, and

h

{\displaystyle h}

.

**Part 3:** Suppose the block is 100 times heavier than the bullet. If the bullet-block system swings to a height of 20 cm, what was the initial velocity of the bullet? Assume this setup is on Earth.

## **Solution**

**Part 1**

Energy is only conserved in perfectly elastic collisions. Since this collision is inelastic, energy is not conserved prior to collision and during the collision. However, kinetic energy of the bullet-block system is conserved post collision, which is converted to gravitational potential energy.

**Part 2**

Conservation of momentum (before and during collision).

m

v
b

M
(
0
# )
(
m
M
)

v
′

{\displaystyle m v\_b + M(0) = (m+M)v' }

Conservation of energy (post collision).

1
2

(
m
M
)
(

v
′

)
2

=
(
m
M
)
g
h

{\displaystyle \frac{1}{2} (m+M) (v')^2 = (m+M)gh }

From the conservation of momentum, we get

v
′

=

2
g
h

.

{\displaystyle v' = \sqrt{2gh}. }

Thus,

v
b

=

m
M

m

2
g
h

.

{\displaystyle v\_b = \frac{m+M}{m} \sqrt{2gh}. }

**Part 3**

Substitute

# M
100
m

{\displaystyle M = 100m }

,

# g
9.8

m

/

s

{\displaystyle g = 9.8 \, m/s }

, and

# h
0.20

m

{\displaystyle h = 0.20 \, m }

 into the formula derived in part 2.

v
b

=

m
100
m

m

2
(
9.8
)
(
0.20
)

=
101

3.92

=
197.99.

{\displaystyle \begin{align}
v\_b &= \frac{m+100m}{m} \sqrt{2(9.8)(0.20)} \\[5 pt]
&= 101\sqrt{3.92} \\[5 pt]
&= 197.99.
\end{align} }

The initial velocity of the bullet roughly

v
b

=
200

m

/

s

{\displaystyle v\_b = 200 \, m/s }

.
Centroid of a Solid

## **Problem**

   Show that the centroid

(

x
¯

,

y
¯

,

z
¯

)

{\displaystyle (\bar{x}, \bar{y}, \bar{z}) }

 of a solid with volume

V

{\displaystyle V}

 is

c
¯

=

1

2
V

∬

∂
V

(

x
2

y
2

z
2

)

n
^

d
S
.

{\displaystyle \bar{c} =\frac{1}{2V}\iint\_{\partial V} (x^2 + y^2 + z^2)\hat{\boldsymbol{n}}\,dS. }

## **Solution**

Let the centroid of the solid be denoted

c
¯

=
(

x
¯

,

y
¯

,

z
¯

)

{\displaystyle \bar{c} = (\bar{x}, \bar{y}, \bar{z}) }

. The centroid of a three-dimensional object can be computed by the vector equation

c
¯

=

1
V

∭
V

r

d
V

{\displaystyle \bar{c} = \frac{1}{V}\iiint\_V \boldsymbol{r} \,dV }

where

# r
x
i
y
j
z
k

{\displaystyle \boldsymbol{r} = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} }

.

It can be shown that

1
2

∇
(
r
⋅
r
# )
r

{\displaystyle \frac{1}{2} \nabla (\boldsymbol{r} \cdot \boldsymbol{r}) = \boldsymbol{r} }

.

r
⋅
# r

x
2

y
2

z
2

{\displaystyle \boldsymbol{r} \cdot \boldsymbol{r} = x^2 + y^2 + z^2 }

∇
(
r
⋅
r
# )
2
x
i
2
y
j
2
z
k

{\displaystyle \nabla(\boldsymbol{r} \cdot \boldsymbol{r}) = 2x\boldsymbol{i} + 2y\boldsymbol{j} + 2z\boldsymbol{k} }

1
2

∇
(
r
⋅
r
# )
x
i
y
j
z
# k
r

{\displaystyle \frac{1}{2} \nabla (\boldsymbol{r} \cdot \boldsymbol{r}) = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} = \boldsymbol{r} }

Subsequently, the centroid is

c
¯

=

1
V

∭
V

r

d
# V

1

2
V

∭
V

∇
(
r
⋅
r
)

d
V
.

{\displaystyle \bar{c} = \frac{1}{V}\iiint\_V \boldsymbol{r} \,dV = \frac{1}{2V}\iiint\_V \nabla (\boldsymbol{r} \cdot \boldsymbol{r}) \,dV. }

Let

# ϕ
r
⋅
r

{\displaystyle \phi = \boldsymbol{r} \cdot \boldsymbol{r} }

 (since the dot product here produces a scalar function). By the property proven in the article Divergence Theorem Reduction

∭
V

∇
ϕ
d
# V

∬

∂
V

ϕ
n
d
S

{\displaystyle \iiint\_V \nabla \phi dV = \iint\_{\partial V} \phi \boldsymbol{n} dS }

it can be shown that

1

2
V

∭
V

∇
(
r
⋅
r
)

d
# V

1

2
V

∬

∂
V

r
⋅
r

n
^

d
S
.

{\displaystyle \frac{1}{2V}\iiint\_V \nabla (\boldsymbol{r} \cdot \boldsymbol{r})\, dV = \frac{1}{2V}\iint\_{\partial V} \boldsymbol{r} \cdot \boldsymbol{r} \,\hat{\boldsymbol{n}} \,dS. }

Since

r
⋅
# r

x
2

y
2

z
2

{\displaystyle \boldsymbol{r} \cdot \boldsymbol{r} = x^2 + y^2 + z^2 }

, we find that

1

2
V

∭
V

∇
(
r
⋅
r
)

d
# V

1

2
V

∬

∂
V

(

x
2

y
2

z
2

)

n
^

d
S
.

{\displaystyle \frac{1}{2V}\iiint\_V \nabla (\boldsymbol{r} \cdot \boldsymbol{r})\, dV = \frac{1}{2V}\iint\_{\partial V} (x^2 + y^2 + z^2)\hat{\boldsymbol{n}} \, dS. }
Rocket League

**Rocket League** – multiplatformowa gra sportowa stworzona i wydana przez studio Psyonix. Jusv do @ uuuuusbh

premiera odbyła się 7 lipca 2015 roku na platformach Windows oraz PlayStation 4, na konsolę Xbox One została wydana dopiero 17 lutego 2016 roku. Rocket League to kontynuacja gry Supersonic Acrobatic Rocket-Powered Battle-Cars. Mimo że rozgrywka w tej grze opiera się głównie na trybie multiplayer, jest w niej też tryb dla jednego gracza.

## Rozgrywka

Rozgrywka w Rocket League jest w większości aspektów taka sama jak u poprzednika. Gracze rozdzieleni są do dwóch przeciwnych drużyn - czerwonej i niebieskiej. Ich celem jest uzyskanie jak największej liczby punktów, które zdobywa się strzelając gole lub trafiając piłką do obręczy w trybie *Hoops*. Tryby wzorowane są na prawdziwych dyscyplinach sportowych, takich jak piłka nożna, czy koszykówka. Jedyną różnicą w rozgrywanych meczach jest to, że gracze kontrolują wyścigówki!

## Tryby

Gra Rocket League ma wiele różnych trybów, przy których gracze mogą się świetnie bawić! Są to:

* Duel
* Doubles
* Standard
* Chaos
* Snow Day
* Rocket Labs
* Hoops
* Rumble

Gra jest stale rozwijana, więc w przyszłości prawdopodobnie pojawią się inne ciekawe tryby!

## Przedmioty

Po rozegranym meczu gracz ma szansę otrzymać jeden z wielu przedmiotów, a dokładniej nowe auta, malowania, opony, rakietowe szlaki, nakrycia lub antenki. Dany przedmiot można założyć od razu po otrzymaniu go lub w zakładce *"Garage"*.

## DLC

* Supersonic Fury DLC Pack
* Revenge of the Battle-Cars DLC Pack
* Chaos Run DLC Pack
* Batman v Superman: Dawn of Justice Car Pack
* NBA Flag Pack

## Galeria

* Zdjęcia
* Trailery

Rocket League® - Announcement TeaserRocket League® - Hoops TrailerRocket League® - Neo Tokyo TrailerRocket League® - Rumble Trailer
Cracking Permutation Locks

**Problem created by Tao Steven Zheng (郑涛).**

## **Problem**

In an escape game, you find a permutation lock with numbers from 1 to 60 with four rotating disks. You have previously found a piece of paper with four clues:

1) Four numbers complete the sequence.

2) No two numbers are the same.

3) The second number is twice the third.

4) The third number is prime.

How many possible permutations exist for the lock?

## **Solution**

There is a specific order to think things through. First consider the last clue. There are 10 prime numbers between 1 and 60 such that its doubled value is also between 1 and 60 (31 is the 11th prime whose double is 62).

The second number would be 1 choice, since it is linked with the third number. Because no two numbers are the same, we do the following product:

58
(
1
)
(
10
)
(
57
# )
33060

{\displaystyle 58(1)(10)(57)=33060}
Arbelos

## **Problem**

In the *Book of Lemmas*[1], Thabit ibn Qurra (c. 826 – 901 AD) describes a shape called the “arbelos”. The “arbelos” in the figure below is the shape bounded by three semicircles with diameters AB, AD, and BD (shaded in blue). Suppose a circle be constructed such that its diameter is CD (shaded in red), determine the ratio of the areas of the arbelos to the circle.

[1] It is uncertain whether the Book of Lemmas was authored by Thabit ibn Qurrah or Archimedes because the earliest edition is written in Arabic and refers to Archimedes in the third person.

## **Solution**

**Area of the circle**

A
D

C
D

=

C
D

B
D

{\displaystyle \frac{AD}{CD} = \frac{CD}{BD} }

C

D
2

=
A
D
⋅
B
D

{\displaystyle CD^2 = AD \cdot BD }

A

c
i
r
c
l
e

=

π
4

A
D
⋅
B
D

{\displaystyle {A}\_{circle} = \frac{\pi}{4} AD \cdot BD }

**Area of the arbelos**

A

a
r
b
e
l
o
s

=

π
2

(

A
B

2

)

2

−

π
2

(

A
D

2

)

2

−

π
2

(

B
D

2

)

2

{\displaystyle {A}\_{arbelos} = \frac{\pi}{2} \left(\frac{AB}{2} \right)^2 - \frac{\pi}{2} \left(\frac{AD}{2} \right)^2 - \frac{\pi}{2} \left(\frac{BD}{2} \right)^2}

Since

A
# B
A
D
B
D

{\displaystyle AB = AD + BD }

A

B
2

=
A

D
2

2
A
D
⋅
B
D
B

D
2

{\displaystyle AB^2 = AD^2 + 2AD\cdot BD + BD^2 }

A

a
r
b
e
l
o
s

=

π
8

(

A

D
2

2
A
D
⋅
B
D
B

D
2

)

−

π
8

A

D
2

−

π
8

B

D
2

{\displaystyle {A}\_{arbelos} = \frac{\pi}{8} \left(AD^2 + 2AD\cdot BD + BD^2 \right) - \frac{\pi}{8} AD^2 - \frac{\pi}{8} BD^2 }

A

a
r
b
e
l
o
s

=

π
4

A
D
⋅
B
D

{\displaystyle {A}\_{arbelos} = \frac{\pi}{4} AD \cdot BD }

Therefore,

A

a
r
b
e
l
o
s

=

A

c
i
r
c
l
e

{\displaystyle {A}\_{arbelos} = {A}\_{circle} }

and

A

a
r
b
e
l
o
s

:

A

c
i
r
c
l
e

=
1
:
1

{\displaystyle {A}\_{arbelos}:{A}\_{circle} = 1:1 }
Basic Chemical Reactions

## **Problem**

Consider the following chemical reactions:

(1)

KClO

3

(
s
)

⟶

KCl

(
s
)

O

2

(
g
)

{\displaystyle \ce{{KClO3}\_{(s)} -> {KCl}\_{(s)} + {O2}\_{(g)}}}

(2)

Sn

(
s
)

Cu

(

NO

3

)

2

(
aq
)

⟶

Sn

(

NO

3

)

2

(
aq
)

Cu

(
s
)

{\displaystyle \ce{{Sn}\_{(s)} + {Cu(NO3)2}\_{(aq)} -> {Sn(NO3)2}\_{(aq)} + {Cu}\_{(s)}}}

(3)

Pb

(

NO

3

)

2

(
aq
)

NaI

(
aq
)

⟶

PbI

2

(
s
)

NaNO

3

(
aq
)

{\displaystyle \ce{{Pb(NO3)2}\_{(aq)} + {NaI}\_{(aq)} -> {PbI2}\_{(s)} + {NaNO3}\_{(aq)}}}

(4)

K

(
s
)

Cl

2

(
g
)

⟶

KCl

(
s
)

{\displaystyle \ce{{K}\_{(s)} + {Cl2}\_{(g)} -> {KCl}\_{(s)}}}

Balance each chemical reaction and name the type of reaction (synthesis, decomposition, single replacement, double replacement).

Note that the states of matter of each substance are included: (s) solid, (l) liquid, (g) gas, and (aq) aqueous.

## **Solution**

(1)

2

KClO

3

(
s
)

⟶
2

KCl

(
s
)

3

O

2

(
g
)

{\displaystyle \ce{2{KClO3}\_{(s)} -> 2{KCl}\_{(s)} + 3{O2}\_{(g)}}}

**Reaction type:** Decomposition

(2)

Sn

(
s
)

Cu

(

NO

3

)

2

(
aq
)

⟶

Sn

(

NO

3

)

2

(
aq
)

Cu

(
s
)

{\displaystyle \ce{{Sn}\_{(s)} + {Cu(NO3)2}\_{(aq)} -> {Sn(NO3)2}\_{(aq)} + {Cu}\_{(s)}}}

This reaction is already balanced!

**Reaction type:** Single replacement

(3)

Pb

(

NO

3

)

2

(
aq
)

2

NaI

(
aq
)

⟶

PbI

2

(
s
)

2

NaNO

3

(
aq
)

{\displaystyle \ce{{Pb(NO3)2}\_{(aq)} + 2{NaI}\_{(aq)} -> {PbI2}\_{(s)} + 2{NaNO3}\_{(aq)}}}

**Reaction type:** Double replacement

(4)

2

K

(
s
)

Cl

2

(
g
)

⟶
2

KCl

(
s
)

{\displaystyle \ce{2{K}\_{(s)} + {Cl2}\_{(g)} -> 2{KCl}\_{(s)}}}

**Reaction type:** Synthesis
Emperor Kangxi on Mathematics and Astronomy

Emperor Kangxi (1654 - 1722, reigned 1661 - 1722) was an avid learner.  He learned mathematics and astronomy from the Jesuit missionaries who visited Peking (Beijing). The following excerpt is from a diary of Emperor Kangxi called the *Qing Shengzu Shilu* (《清聖祖實錄》, *Records of Emperor Kangxi*).

**《清聖祖實錄》**
尔等惟知朕[1]算术之精，却不知我学算之故。朕幼时，钦天监[2]汉官与西洋人不睦，互相参劾，几至大辟。杨光先、汤若望\*于午门外九卿[4]前当面赌测日影，奈九卿中无一知其法者。朕思己不知，焉能断人之是非，因自愤而学焉。
**【今译】**
你们只知道我数学很好，但不知道我学数学的原因。我小时候，钦天监的汉族官员和西洋人不和，互相指责得很厉害，几乎闹出人命！后来，杨光先和汤若望\*在午门外当着九卿的面前预测太阳的影子，而他们九卿没有一个人懂计算的方法。我想如果自己都不知道，又怎么能判断谁对谁错呢？因此，我开始发奋学习西方数学了。

***Records of Emperor Kangxi***
You all only know that I am good at mathematics but you do not know why I study mathematics. When I was young, the Han officials at the Imperial Bureau of Astronomy did not get along with the Westerners. They accused each other to the point that someone could have been executed! Later, Yang Guangxian and Tang Ruowang (Johann Adam Schall von Bell)\* predicted the sun's shadow in front of the Nine Ministers outside the Meridian Gate, and not one of them understood the calculation methods. I then thought, if I don't know, how can I judge who is right and who is wrong? This is why I put great effort in learning mathematics.

\*应该是南怀仁。Emperor Kangxi mistook the Jesuit missionary Ferdinand Verbiest (南怀仁; Nan Huairen) for Johann Adam Schall von Bell (汤若望; Tang Ruowang).

**【注释】**

[1] 朕 zhèn：皇帝的人称代词。Personal pronoun of the Chinese emperor.

[2] 钦天监 qīn tiān jiàn：明清两朝观察天文、天象、历法的机构。Imperial Bureau of Astronomy.

[3] 午门 wǔ mén：紫禁城的正门。Meridian Gate (main gate of the Forbidden City).

[4] 九卿 jiǔ qīng：中国古代朝廷的九个高级官职。Nine high-ranking officials of Imperial China.
Magnetic Compass 指南针

**《梦溪笔谈·卷24杂志（一）18》**
方家以磁石磨针锋，则能指南，然常微偏东，不全南也。水浮多荡摇，指爪及碗唇上皆可为之，运转尤速，但坚滑易坠，不若缕悬为最善。其法取新纩（kuàng）中独茧（jiǎn）缕，以芥子许蜡缀于针腰，无风处悬之，则针常指南。其中有磨而指北者。予家指南、 北者皆有之。磁石之指南犹柏之指西，莫可原其理。

***Mengxi Bitan* Chapter 24 Zazhi (Part I) 18**
Diviners can make a needle point to the south by rubbing it with magnetite. However the needle often inclines to the southeast direction, not pointing to due south. If the needle is floating on water, it often sways. The needle may be placed on a nail or the edge of a bowl, which may make it rotate a bit more flexibly, but the hard and smooth surface may make it easily fall off. The best way is to hang the needle with a silk thread. If we take a newly-reeled raw silk and paste it to the middle of the needle with wax similar to a seed of leaf mustard, and hang up the needle in a windless place, the needle will often point to the south. Some needles may point to the north after being rubbed with magnetite, for instance, in my house some needles point to the south while others point to the north. When the needle rubbed with magnetite points to the south, just as the cypress tree leans to the west. Nobody knows the reason why it is so.

**【今译】**

看风水的人用磁石去磨针尖，针尖就能指出南方，不过常常略为偏东，并不完全指向正南。把 磁针浮在水面上，常晃荡不停。也可以把磁针放在指甲上和碗口上，磁针转动更灵快些，但这上面又 硬又滑，磁针容易掉落，不如用丝线悬挂为最好。其方法是从新丝棉中抽取一根蚕丝，用芥菜籽大小的一点蜡粘连在磁针的腰部，悬挂在没有风的地方，磁针就常常指向南方。磁针也有经磨后指向北的。我家的磁针，指南指北的都有。磁石的指南，如同柏树的指西，不能推究出其中的道理。
Electric Field of a Charged Disk

## **Problem**

Find the electric ﬁeld a distance

x

{\displaystyle x}

 along the axis from a disc of radius

R

{\displaystyle R}

 and uniform charge density

σ

{\displaystyle \sigma}

.

## **Hint**

(1) Think of this disk as a bunch of concentric rings. Begin by ﬁnding the electric field a distance

x

{\displaystyle x}

 along the axis up from a thin ring of charge

d
Q

{\displaystyle dQ }

 and radius

a

{\displaystyle a}

.

(2) You may use the integral

∫

x
d
x

(

x
2

a
2

)
3

=
−

1

x
2

a
2

C
.

{\displaystyle \int \frac{xdx}{(\sqrt{x^2 + a^2})^3} = -\frac{1}{\sqrt{x^2 + a^2}} + C. }

## **Solution**

Consider the ring problem first. Due to the symmetry of this geometry, there is a cancellation effect in the y-direction. Therefore, all contributions to the electric field in the x-direction. The distance of a point on the x-axis from the ring is

# r

x
2

a
2

.

{\displaystyle r = \sqrt{x^2 + a^2}. }

Hence, the differential element of electric field (along the x-axis) is

d

E
x

=
d
E
cos
⁡

(
θ
)

{\displaystyle dE\_x = dE \cos{(\theta)}}

d

E
x

=

1

4
π

ϵ

0

d
Q

x
2

a
2

x

x
2

a
2

.

{\displaystyle dE\_x = \frac{1}{4\pi {\epsilon}\_{0}} \frac{dQ}{x^2 + a^2} \frac{x}{\sqrt{x^2 + a^2}}. }

Integrating the differential yields:

E
x

=

1

4
π

ϵ

0

x

(

x
2

a
2

)
3

∫

0

Q

d
Q
:

{\displaystyle E\_x = \frac{1}{4\pi {\epsilon}\_{0}} \frac{x}{(\sqrt{x^2 + a^2})^3} \int\_{0}^{Q} dQ
:}

# E

1

4
π

ϵ

0

x
Q

(

x
2

a
2

)
3

i
.

{\displaystyle E = \frac{1}{4\pi {\epsilon}\_{0}} \frac{xQ}{(\sqrt{x^2 + a^2})^3} \boldsymbol{i}. }

For the disk problem, replace the differential charge with

d
# Q
σ
2
π
r
d
r

{\displaystyle dQ = \sigma 2\pi r dr }

 because the areal charge density is defined as

# σ

d
Q

d
A

{\displaystyle \sigma = \frac{dQ}{dA} }

 and

# A
π

r
2

{\displaystyle A = \pi r^2 }

 for a disk.

Since we are integrating with respect to the variable

r

{\displaystyle r}

 now, the variable radii

a

{\displaystyle a}

 is replaced with

r

{\displaystyle r}

. Therefore, the electric field is modified to

d

E
x

=

1

4
π

ϵ

0

σ
2
π
r
d
r

x
2

r
2

x

x
2

r
2

.

{\displaystyle dE\_x = \frac{1}{4\pi {\epsilon}\_{0}} \frac{\sigma 2\pi rdr}{x^2 + r^2} \frac{x}{\sqrt{x^2 + r^2}}. }

Using the given integral yields,

E
x

=

1

4
π

ϵ

0

∫

0

R

(
σ
2
π
r
d
r
)
x

(

x
2

r
2

)
3

{\displaystyle E\_x = \frac{1}{4\pi {\epsilon}\_{0}} \int\_{0}^{R} \frac{(\sigma 2\pi rdr) x}{(\sqrt{x^2 + r^2})^3} }

E
x

=

σ
x

2

ϵ

0

(

−

1

x
2

R
2

1
x

)

.

{\displaystyle E\_x = \frac{\sigma x}{2 {\epsilon}\_{0}} \left( -\frac{1}{\sqrt{x^2 + R^2}} + \frac{1}{x} \right). }

Therefore, the electric ﬁeld a distance

x

{\displaystyle x}

 along the axis from a disc of radius

R

{\displaystyle R}

 and uniform charge density

σ

{\displaystyle \sigma}

is

# E

σ

2

ϵ

0

(

1
−

x

x
2

R
2

)

i
.

{\displaystyle E = \frac{\sigma}{2 {\epsilon}\_{0}} \left(1 -\frac{x}{\sqrt{x^2 + R^2}} \right) \boldsymbol{i}. }
Geometric Series

## **Problem**

**Part 1:** Prove that the arithmetic series can be computed with the formula

∑

# k
0

n

a

r
k

=

a
(
1
−

r

n
1

)

1
−
r

.

{\displaystyle \sum\_{k=0}^{n} ar^k = \frac{a(1-r^{n+1})}{1-r}. }

**Part 2:** Show that if

−
1
<
r
<
1

{\displaystyle -1<r<1}

, then

∑

# k
0

∞

a

r
k

=

a

1
−
r

.

{\displaystyle \sum\_{k=0}^{\infty} ar^k = \frac{a}{1-r}. }

## **Solution**

**Part 1**

Let

S
n

=
a
a
r
a

r
2

⋯
a

r
n

{\displaystyle S\_n = a + ar + ar^2 + \cdots + ar^n }

then

r

S
n

=
a
r
a

r
2

a

r
3

⋯
a

r

n
1

.

{\displaystyle rS\_n = ar + ar^2 + ar^3 + \cdots + ar^{n+1}. }

Subtracting the two equations above yields

S
n

−
r

S
n

=
a
−
a

r

n
1

.

{\displaystyle S\_n - rS\_n = a - ar^{n+1}. }

Therefore,

S
n

(
1
−
r
# )
a
(
1
−

r

n
1

)

{\displaystyle S\_n (1-r) = a(1 - r^{n+1}) }

and

S
n

=

a
(
1
−

r

n
1

)

1
−
r

.

{\displaystyle S\_n = \frac{a(1 - r^{n+1})}{1-r}. }

**Part 2**

Let

S

∞

=

∑

# k
0

∞

a

r
k

{\displaystyle S\_{\infty} = \sum\_{k=0}^{\infty} ar^k }

 represent the infinite geometric series. If

−
1
<
r
<
1

{\displaystyle -1<r<1}

, then

lim

n
→
∞

r

n
1

=
0

{\displaystyle \lim\_{n \to \infty} r^{n+1} = 0 }

. Consequently,

S

∞

=

a

1
−
r

.

{\displaystyle S\_{\infty} = \frac{a}{1-r}.
 }
טארוט: המג

I The Mage

Meaning: Will

Communication, Inherent Ability, Memory, Clarity of thought and feeling, Organization, Invention, Originality

Association: Virtual Adepts

"What is will but the projection of energy into the world of matter?"

With an act of conscious thought, Dante grasps the wand of Prime, closing the link between the energy that gives him the power of magic and the tools that lie before him on the table of the world.

The building blocks of his desires - the Primordial, Questing, Dynamic and Pattern Essences - contain within them the remaining eight Spheres of magic; the key to knowledge unlocks them all.

The infinite hovers, crown-like, above him.

Reality, virtual or otherwise, belongs to him.
Hot Air Balloon

## **Problem**

The density of air is

1.29

k
g

/

m
3

{\displaystyle 1.29 \; kg/m^3 }

 and the density of helium is

0.18

k
g

/

m
3

{\displaystyle 0.18 \; kg/m^3 }

. The mass of the balloon is

20

k
g

{\displaystyle 20 \; kg }

.

What volume of helium would a hot air balloon needed to lift a

180

k
g

{\displaystyle 180 \; kg }

 crate?

## **Solution**

The balloon must be able to lift the crate and its own mass (balloon mass and mass of helium gas).

F
B

=
(

m

H
e

180
20
)
g

{\displaystyle F\_B = ({m}\_{He} + 180 + 20)g }

ρ

a
i
r

V
# g

(

ρ

H
e

V
200

)

g

{\displaystyle {\rho}\_{air} Vg = \left({\rho}\_{He} V + 200 \right) g }

# V

200

ρ

a
i
r

−

ρ

H
e

{\displaystyle V = \frac{200}{{\rho}\_{air} - {\rho}\_{He}} }

# V

200

1.29
−
0.18

{\displaystyle V = \frac{200}{1.29 - 0.18} }

# V
180

m
3

{\displaystyle V = 180 \; m^3 }
Laplacian Condition

## **Problem**

Suppose

f
″

−
α
# f
0

{\displaystyle \; f'' - \alpha f = 0 }

,

g
″

−
β
# g
0

{\displaystyle \; g'' - \beta g = 0 }

, and

h
″

−
γ
# h
0

{\displaystyle \; h'' - \gamma h = 0 }

. Determine the condition in the form of an equation that relates

α
,
β
,
γ

{\displaystyle \alpha, \beta, \gamma}

 such that the function

F
(
x
,
y
,
z
# )
f
(
x
)

g
(
y
)

h
(
z
)

{\displaystyle F(x,y,z) = f(x)\,g(y)\,h(z) }

satisfies the three-dimensional Laplacian equation (in Cartesian coordinates)

F

x
x

F

y
y

F

z
z

=

{\displaystyle {F}\_{xx} + {F}\_{yy} + {F}\_{zz} = 0. }

## **Solution**

Take the second derivatives of each variable

F

x
x

=

f
″

(
x
)

g
(
y
)

h
(
z
)

F

y
y

=
f
(
x
)

g
″

(
y
)

h
(
z
)

F

z
z

=
f
(
x
)

g
(
y
)

h
″

(
z
)
.

{\displaystyle \begin{align}
{F}\_{xx} &= f''(x)\,g(y)\,h(z) \\[5 pt]
{F}\_{yy} &= f(x)\,g''(y)\,h(z) \\[5 pt]
{F}\_{zz} &= f(x)\,g(y)\,h''(z).
\end{align} }

Subsequently,

F

x
x

F

y
y

F

z
z

=
0

{\displaystyle {F}\_{xx} + {F}\_{yy} + {F}\_{zz} = 0 }

becomes

f
″

(
x
)
g
(
y
)
h
(
z
)
f
(
x
)

g
″

(
y
)
h
(
z
)
f
(
x
)
g
(
y
)

h
″

(
z
# )

{\displaystyle f''(x)g(y)h(z) + f(x)g''(y)h(z) + f(x)g(y)h''(z) = 0. }

Since

f
″

−
α
# f
0

{\displaystyle f'' - \alpha f = 0 }

,

g
″

−
β
# g
0

{\displaystyle g'' - \beta g = 0 }

, and

h
″

−
γ
# h
0

{\displaystyle h'' - \gamma h = 0 }

, it follows that

f
″

=
α
f

g
″

=
β
g

h
″

=
γ
h
.

{\displaystyle \begin{align}
f'' &= \alpha f \\[5 pt]
g'' &= \beta g \\[5 pt]
h'' &= \gamma h.
\end{align} }

Substituting the above relations into the modified Laplacian equation yields

α
f
(
x
)

g
(
y
)

h
(
z
)
f
(
x
)

β
g
(
y
)

h
(
z
)
f
(
x
)

g
(
y
)

γ
h
(
z
# )
0

{\displaystyle \alpha f(x)\,g(y)\,h(z) + f(x)\,\beta g(y)\,h(z) + f(x)\,g(y)\,\gamma h(z) = 0 }

or

(
α
β
γ
)
f
(
x
)

g
(
y
)

h
(
z
# )

{\displaystyle (\alpha + \beta + \gamma) f(x)\,g(y)\,h(z) = 0. }

Since the product of functions

f
(
x
)

g
(
y
)

h
(
z
)

{\displaystyle f(x)\,g(y)\,h(z) }

 is not necessarily equal to zero for all coordinates

(
x
,
y
,
z
)

{\displaystyle (x, y, z)}

, it must be necessary true that

(
α
β
γ
# )
0

{\displaystyle (\alpha + \beta + \gamma) = 0 }

.
Math Learning Resources

Math is Fun - An easy-to-use resource for learning elementary and high school level math.

Paul's Online Math Notes - A great site dedicated to teaching basic university level mathematics.

Math Insight - An index of university-level math concepts explained with interactive applets.

UBC Math Exam Resources - A repository of past exams for math exams of select courses at the University of British Columbia.

Introduction to Probability, Statistics and Random Processes - An excellent online resource for learning about probability statistics, and random processes.
Jiuzhang Suanshu Shanggong 8

## **Problem**

答曰：三千八百四十尺。

術曰：方自乘，以高乘之，即積尺。
Answer: 3840 [cubic] *chi*.

Method: Multiply the edge of the square by itself, and multiply by the height, the result is the volume in *chi*.

1 Literally translates to “square watchtower”.

## **Derivation**

The *fang bao dao* (方堡壔, "square watchtower" ) is a rectangular prism with a square base (Figure 1). The volume of a rectangular prism whose length, width, and height are

a
,
b
,
c

{\displaystyle a,b,c}

 is

# V
a
b
c
.

{\displaystyle V = abc. }

Since the *fang bao dao* is a rectangular prism with a square base, the length and width are equal

# a
b

{\displaystyle a = b}

. Let

h

{\displaystyle h}

 denote the height. The volume is therefore

# V

a
2

h
.

{\displaystyle V = a^2h. }

## **Calculation**

Convert the base length and the height of the rectangular prism (方堡壔 *fang bao dao*) to *chi.*

**Base length:** 1 *zhang* 6 *chi* = 16 *chi*

**Height:** 1 *zhang* 5 *chi* = 15 *chi*

Substitute the base length, and height of the fang bao dao into the formula.

V

=

a
2

h

=

16
2

×
15

=
3940

{\displaystyle \begin{align}
V &= a^2h \\[5 pt]
& = 16^2 \times 15 \\[5 pt]
&= 3940
\end{align}}

The volume of the rectangular prism (方堡壔 *fang bao dao*) is 3840 cubic *chi*.
Epitaph of Diophantus

## **Problem**

This is the modern rendition of a problem by Metrodorus (c.550 AD).

Diophantus's youth lasted

1
6

{\displaystyle \frac{1}{6}}

 of his life. He grew a beard after

1
12

{\displaystyle \frac{1}{12}}

 of his life (after his youth). Diophantus married after

1
7

{\displaystyle \frac{1}{7}}

 more of his life. Five years later, he had a son. His son lived half as long as his father. Diophantus died four years after his son's death. How long did Diophantus live?

## **Solution**

Let

x

{\displaystyle x}

 represent the age of Diophantus. Then

# x

x
6

x
12

x
7

5

x
2

4

{\displaystyle x = \frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 }

.
Simplifying this equation yields

3
28

# x
9

{\displaystyle \frac{3}{28}x = 9 }

and the answer is

# x
84

{\displaystyle x = 84 }

.

According to this solution, his youth ended (according to Roman-Byzantine tradition) at age 14, he grew a beard when he was 21, he married when he was 33, he had a son when he was 38, his son died when he was 80 years old at the age of 42.

## **Historical Note**

Diophantus was a Roman mathematician who flourished during the 3rd century AD. There is no consensus on his ethnicity, but he was probably of Greek descent. His most influential work is the *Arithmetica (Arithmetic)*. Despite this title, the book is a collection of problems on algebra and number theory.

Metrodorus was a Greek grammarian from the 6th century AD who compiled many mathematical problems that are now part of the *Greek Anthology*.
Artificial Neural Networks

## **Introduction**

Artificial neural networks (ANN) is a computational model that aims to mimic the architecture of the brain. The architecture and biological mechanisms of neurons (brain cells) and their connections are thus mathematically described using nodes with associated numerical values, and connections with assigned weightings. The nodes represent each individual neuron and the connections represent synaptic strengths.
Covariance and Expectation

## **Problem**

**Covariance** provides a measure of the strength of correlation between two or more random variables. For two random variables, the covariance is a measure of the relationship between the two random variables and to what extent they change together.

The covariance of two random variables

X
,
Y

{\displaystyle X, Y}

 is defined using expectation

c
o
v
(
X
,
Y
# )
E
[
(
X
−
E
[
X
]
)
(
Y
−
E
[
Y
]
)
]

{\displaystyle cov(X,Y) = E[(X - E[X])(Y - E[Y])] }

.
**Part 1:** Show that

c
o
v
(
X
,
Y
# )
E
[
X
Y
]
−
E
[
X
]
E
[
Y
]

{\displaystyle cov(X,Y) = E[XY] - E[X]E[Y] }

.

**Part 2:** Show that

c
o
v
(
X
,
X
# )
V
a
r
(
X
)

{\displaystyle cov(X,X) = Var(X) }

.

## **Solution**

**Part 1**

c
o
v
(
X
,
Y
# )
E
[
(
X
−
E
[
X
]
)
(
Y
−
E
[
Y
]
)
]

{\displaystyle cov(X,Y) = E[(X - E[X])(Y - E[Y])] }

c
o
v
(
X
,
Y
# )
E
[
X
Y
−
X
(
E
[
Y
]
)
−
Y
(
E
[
X
]
)
E
[
X
]
E
[
Y
]
]

{\displaystyle cov(X,Y) = E[XY - X(E[Y]) - Y(E[X]) + E[X]E[Y]] }

c
o
v
(
X
,
Y
# )
E
[
X
Y
]
−
E
[
X
(
E
[
Y
]
)
]
−
E
[
Y
(
E
[
X
]
)
]
E
[
E
[
X
]
E
[
Y
]
]

{\displaystyle cov(X,Y) = E[XY] - E[X(E[Y])] - E[Y(E[X])] + E[E[X]E[Y]] }

Since

E
[
X
]

{\displaystyle E[X] }

 and

E
[
Y
]

{\displaystyle E[Y] }

 are constants,

c
o
v
(
X
,
Y
# )
E
[
X
Y
]
−
E
[
X
]
E
[
Y
]
−
E
[
X
]
E
[
Y
]
E
[
X
]
E
[
Y
]
E
[
1
]

{\displaystyle cov(X,Y) = E[XY] - E[X]E[Y] - E[X]E[Y] + E[X]E[Y]E[1] }

c
o
v
(
X
,
Y
# )
E
[
X
Y
]
−
E
[
X
]
E
[
Y
]
−
E
[
X
]
E
[
Y
]
E
[
X
]
E
[
Y
]

{\displaystyle cov(X,Y) = E[XY] - E[X]E[Y] - E[X]E[Y] + E[X]E[Y] }

c
o
v
(
X
,
Y
# )
E
[
X
Y
]
−
E
[
X
]
E
[
Y
]

{\displaystyle cov(X,Y) = E[XY] - E[X]E[Y] }

.
**Part 2**

Knowing that

V
a
r
(
X
# )
E
[

X
2

]
−

(
E
[
X
]
)

2

{\displaystyle Var(X) = E[X^2] - {(E[X])}^2 }

 (proof here), it is obvious from the result in **part 1** that

c
o
v
(
X
,
Y
# )
E
[
X
X
]
−
E
[
X
]
E
[
X
]

{\displaystyle cov(X,Y) = E[XX] - E[X]E[X] }

c
o
v
(
X
,
Y
# )
E
[

X
2

]
−

(
E
[
X
]
)

2

{\displaystyle cov(X,Y) = E[X^2] - {(E[X])}^2 }

c
o
v
(
X
,
X
# )
V
a
r
(
X
)

{\displaystyle cov(X,X) = Var(X) }

.
Jacobi Identity

## **Problem**

Prove Jacobi’s identity

[
A
,
[
B
,
C
]
]
[
B
,
[
C
,
A
]
]
[
C
,
[
A
,
B
]
# ]
0

{\displaystyle [A, [B, C]] + [B, [C, A]] + [C, [A,B]] = 0 }

.

Remember that

[
A
,
B
# ]
A
B
−
B
A

{\displaystyle [A, B] = AB - BA }

 and the operators cannot be assumed to be commutative.

## **Solution**

[
A
,
[
B
,
C
]
# ]
[
A
,
B
C
−
C
B
]

{\displaystyle [A,[B,C]]=[A,BC-CB] }

[
A
,
[
B
,
C
]
# ]
A
(
B
C
−
C
B
)
−
(
B
C
−
C
B
)
A

{\displaystyle [A,[B,C]]=A(BC-CB)-(BC-CB)A }

[
A
,
[
B
,
C
]
# ]
A
B
C
−
A
C
B
−
B
C
A
C
B
A

{\displaystyle [A,[B,C]]=ABC-ACB-BCA+CBA }

Repeating this process for the other two brackets will lead to:

[
B
,
[
C
,
A
]
# ]
B
C
A
−
B
A
C
−
C
A
B
A
C
B

{\displaystyle [B,[C,A]]=BCA-BAC-CAB+ACB }

[
C
,
[
A
,
B
]
# ]
C
A
B
−
C
B
A
−
A
B
C
B
A
C

{\displaystyle [C,[A,B]]=CAB-CBA-ABC+BAC }

.
Note that the order of the operators matter. Nevertheless, every product has a negative pair, so

[
A
,
[
B
,
C
]
]
[
B
,
[
C
,
A
]
]
[
C
,
[
A
,
B
]
# ]
0

{\displaystyle [A, [B, C]] + [B, [C, A]] + [C, [A,B]] = 0 }

.
Falling Ball

## **Problem**

A ball is dropped off the top of the Leaning Tower of Pisa. The tower is 56 m tall . If the ball is initially at rest. Calculate the time it takes to reach the ground. Calculate the impact speed. Neglect air resistance.

## **Solution**

The acceleration due to gravity near the surface of the Earth is

# a
−
9.8
m

/

s
2

{\displaystyle a = -9.8 m/s^2 }

. The initial velocity is

v
i

=
0
m

/

s

{\displaystyle v\_i = 0 m/s }

.

To solve for the time, use the formula

x
f

=

x
i

v
i

t

1
2

a

t
2

{\displaystyle x\_f = x\_i + v\_i t + \frac{1}{2}at^2 }

.
For this problem

x
f

=
56
−

1
2

9.8

t
2

{\displaystyle x\_f = 56 -\frac{1}{2}9.8t^2 }

.
Since the height of the tower is 56 m tall, the initial position is 56 m and the final height is 0 m.

# 0
56
−

1
2

9.8

t
2

{\displaystyle 0 = 56 -\frac{1}{2}9.8t^2 }

.
Therefore,

# t

2
(
56
)

9.8

{\displaystyle t = \sqrt{\frac{2(56)}{9.8}} }

# t
3.38
s

{\displaystyle t = 3.38 s }

.
The impact velocity can be calculated in two ways.

**Method 1**

v
f

=

v
i

a
t

{\displaystyle v\_f = v\_i + at }

v
f

=
0
(
−
9.8
)
(
3.38
)

{\displaystyle v\_f = 0 + (-9.8)(3.38) }

v
f

=
−
33.12
m

/

s

{\displaystyle v\_f = -33.12 m/s }

Thus, the impact speed is 33.12 m/s.

**Method 2**

v
f

2

=

v
i

2

2
a
(

x
f

−

x
i

)

{\displaystyle {v\_f}^{2} = {v\_i}^{2} + 2a(x\_f - x\_i) }

v
f

2

=

0

2

2
(
−
9.8
)
(
0
−
56
)

{\displaystyle {v\_f}^{2} = {0}^{2} + 2(-9.8)(0 - 56) }

v
f

=
33.12
m

/

s

{\displaystyle v\_f = 33.12 m/s }
Derivation of the Quadratic Formula

# y
A

x
2

B
x
C

{\displaystyle y = Ax^2 + Bx + C }

. This quadratic equation is in standard form.

**Part 1:** Use the method of completing the square to determine the coordinate of the vertex in terms of

A
,
B
,
C

{\displaystyle A, B, C}

.

**Part 2:** Use the vertex form to derive the quadratic formula for this quadratic equation.

## **Solution**

**Part 1**

Factor out the leading coefficient from the first two terms:

# y
A

(

x

2

B
A

x

)

C

{\displaystyle y=A\left(x^{2}+{\frac {B}{A}}x\right)+C}

.
Divide away the leading coefficient and complete the square:

# y
A

(

x

2

B
x

2
A

B

2

4

A

2

)

C
−
A

(

B

2

4

A

2

)

{\displaystyle y=A\left(x^{2}+{\frac {Bx}{2A}}+{\frac {B^{2}}{4A^{2}}}\right)+C-A\left({\frac {B^{2}}{4A^{2}}}\right)}

# y
A

(

x

B

2
A

)

2

C
−

B
2

4
A

{\displaystyle y = A{\left(x+\frac{B}{2A}\right)}^{2} + C - \frac{B^2}{4A} }

.
The vertex form of a parabola is

# y
A
(
x
−
P

)

2

Q

{\displaystyle y=A(x-P)^{2}+Q}

, where

(
P
,
Q
)

{\displaystyle (P,Q)}

 is the coordinate of the vertex. Therefore,

# y
A

(

x
−

(

−
B

2
A

)

)

2

C
−

B

2

4
A

{\displaystyle y=A{\left(x-\left({\frac {-B}{2A}}\right)\right)}^{2}+C-{\frac {B^{2}}{4A}}}

and the coordinate of the vertex must be

(
P
,
Q
# )

(

−
B

2
A

,
C
−

B

2

4
A

)

{\displaystyle (P,Q)=\left({\frac {-B}{2A}},C-{\frac {B^{2}}{4A}}\right)}

.
**Part 2**

Set

# y
0

{\displaystyle y=0}

 and isolate

(

x

B

2
A

)

2

{\displaystyle {\left(x+{\frac {B}{2A}}\right)}^{2}}

:

# y
A

(

x

B

2
A

)

2

C
−

B
2

4
A

{\displaystyle y = A{\left(x+\frac{B}{2A}\right)}^{2} + C - \frac{B^2}{4A} }

# 0
A

(

x

B

2
A

)

2

4
A
C
−

B

2

4
A

{\displaystyle 0=A{\left(x+{\frac {B}{2A}}\right)}^{2}+{\frac {4AC-B^{2}}{4A}}}

(

x

B

2
A

)

2

=

B

2

−
4
A
C

4

A

2

{\displaystyle {\left(x+{\frac {B}{2A}}\right)}^{2}={\frac {B^{2}-4AC}{4A^{2}}}}

.
Take the square root on both sides of the equation:

x

B

2
A

=
±

B

2

−
4
A
C

4

A

2

{\displaystyle x+{\frac {B}{2A}}=\pm {\sqrt {\frac {B^{2}-4AC}{4A^{2}}}}}

.
Isolate for

x

{\displaystyle x}

 and simplify:

# x
−

B

2
A

±

B

2

−
4
A
C

2
A

{\displaystyle x=-{\frac {B}{2A}}\pm {\frac {\sqrt {B^{2}-4AC}}{2A}}}

# x

−
B
±

B

2

−
4
A
C

2
A

{\displaystyle x={\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}}

.
Doppler Beats

## **Problem**

A man standing at a bus stop when two cyclists travelling in the same direction toward him are each blasting a radio with sound of a single frequency. Each cyclist is travelling at 5 m/s. As the first cyclist passes him but before the second cyclist does, the man hears a beat frequency of 5 Hz.

What is the frequency of the sound played by the cyclists?

Use the speed of sound of 343 m/s in air.

## **Solution**

This problem involves beat frequencies and Doppler shift. The formula for beat frequency is :

f

b
e
a
t

=

f

1

−

f

2

{\displaystyle {f}\_{beat}=|f\_{1}-f\_{2}|}

.

Let the frequency from the second cyclist (as heard by the man) be

f
1

{\displaystyle f\_1}

, and the frequency from the first cyclist (as heard by the man) be

f
2

{\displaystyle f\_2}

.

Since the man is not moving, the beat frequency must be

f

b
e
a
t

=

f

0

(

v

v
−

v

s

)

−

f

0

(

v

v

v

s

)

{\displaystyle {f}\_{beat}=f\_{0}\left({\frac {v}{v-v\_{s}}}\right)-f\_{0}\left({\frac {v}{v+v\_{s}}}\right)}

.
Solving for the original frequency from the radio,

f

0

=

f

b
e
a
t

(

v

2

−

v

s

2

2
v

v

s

)

{\displaystyle f\_{0}={f}\_{beat}\left({\frac {v^{2}-{v\_{s}}^{2}}{2v{v}\_{s}}}\right)}

.
Thus

f

0

=
5

(

343

2

−

5

2

2
(
343
)
(
5
)

)

{\displaystyle f\_{0}=5\left({\frac {343^{2}-5^{2}}{2(343)(5)}}\right)}

f

0

=
171.46

H
z

{\displaystyle f\_{0}=171.46\;Hz}

.
Integration by Geometry

## **Problem**

Evaluate the following integral without using the integration by trigonometric substitution technique:

∫

−
2

2

(
x
3
)

4
−

x
2

d
x
.

{\displaystyle \int\_{-2}^{2} (x+3)\sqrt{4-x^2} \, dx. }

## **Solution**

Split the integral in two:

∫

−
2

2

x

4
−

x
2

d
x

∫

−
2

2

3

4
−

x
2

d
x
.

{\displaystyle \int\_{-2}^{2} x\sqrt{4-x^2} \,dx + \int\_{-2}^{2} 3\sqrt{4-x^2} \,dx.}

The function

f
(
x
# )
x

4
−

x
2

{\displaystyle f(x) = x\sqrt{4-x^2} }

 is an odd function by simple verification.

f
(
−
x
# )
(
−
x
)

4
−
(
−
x

)
2

=
−
x

4
−

x
2

{\displaystyle f(-x) = (-x)\sqrt{4 - (-x)^2} = -x\sqrt{4-x^2} }

f
(
−
x
# )
−
f
(
x
)
.

{\displaystyle f(-x) = -f(x). }

For an odd function, the integral

∫

−
a

a

f
(
x
)

d
# x
0

{\displaystyle \int\_{-a}^{a} f(x) \,dx = 0 }

; therefore,

∫

−
2

2

x

4
−

x
2

d
# x

{\displaystyle \int\_{-2}^{2} x\sqrt{4-x^2} \,dx = 0. }

The function

f
(
x
# )

4
−

x
2

{\displaystyle f(x) = \sqrt{4-x^2} }

 is the top half of a semicircle with radius 2.

This shows that

∫

−
2

2

4
−

x
2

d
# x

π
(
2

)
2

2

=
2
π

{\displaystyle \int\_{-2}^{2} \sqrt{4-x^2} \, dx = \frac{\pi (2)^2}{2} = 2\pi}

 and

∫

−
2

2

3

4
−

x
2

d
# x
6
π
.

{\displaystyle \int\_{-2}^{2} 3\sqrt{4-x^2} dx = 6\pi. }

Therefore,

∫

−
2

2

(
x
3
)

4
−

x
2

d
# x
6
π
.

{\displaystyle \int\_{-2}^{2} (x+3)\sqrt{4-x^2}\, dx = 6\pi. }
Deriving the Big Three

## **Problem**

   **Part 1:** The average acceleration is defined as

# a

Δ
v

Δ
t

.

{\displaystyle a= \frac {\Delta v}{\Delta t}.}

Rearrange this equation (with defined quantities) to show that

v
f

=

v
i

a
t

{\displaystyle v\_f = v\_i + at }

.

**Part 2:** The average velocity is define as

# v

Δ
x

Δ
t

.

{\displaystyle v={\frac {\Delta x}{\Delta t}}.}

When the average acceleration is constant, the average velocity is

# v

v

i

v

f

2

.

{\displaystyle v={\frac {v\_{i}+v\_{f}}{2}}.}

Use the result in part 1 and the above formulas for average velocity to show that

x

f

=

x

i

v

i

t

1
2

a

t

2

.

{\displaystyle x\_{f}=x\_{i}+v\_{i}t+{\frac {1}{2}}at^{2}.}

**Part 3:** Use the results in part 1 and part 2 to show that

v

f

2

=

v

i

2

2
a
(

x

f

−

x

i

)
.

{\displaystyle v\_{f}^{2}=v\_{i}^{2}+2a(x\_{f}-x\_{i}).}

## **Solution**

**Part 1**

Let

Δ
# t
t

{\displaystyle \Delta t=t}

 be a time interval and

Δ
# v

v
f

−

v
i

{\displaystyle \Delta v=v\_f - v\_i}

:

# a

v

f

−

v

i

t

.

{\displaystyle a={\frac {v\_{f}-v\_{i}}{t}}.}

Isolating for

v
f

{\displaystyle v\_f}

 yields

v

f

=

v

i

a
t
.

{\displaystyle v\_{f}=v\_{i}+at.}

**Part 2**

Let

Δ
# t
t

{\displaystyle \Delta t=t}

 be a time interval and

Δ
# x

x

f

−

x

i

{\displaystyle \Delta x=x\_{f}-x\_{i}}

:

# v

x

f

−

x

i

t

.

{\displaystyle v={\frac {x\_{f}-x\_{i}}{t}}.}

When the average acceleration is constant

# v

v

i

v

f

2

.

{\displaystyle v={\frac {v\_{i}+v\_{f}}{2}}.}

Thus

x

f

=

x

i

(

v

i

v

f

2

)

t
.

{\displaystyle x\_{f}=x\_{i}+\left({\frac {v\_{i}+v\_{f}}{2}}\right)t.}

Since

v
f

=

v
i

a
t

{\displaystyle v\_f = v\_i + at }

 for constant average acceleration:

x

f

=

x

i

(

v

i

v

i

a
t

2

)

t

{\displaystyle x\_{f}=x\_{i}+\left({\frac {v\_{i}+v\_{i}+at}{2}}\right)t}

x

f

=

x

i

v

i

t

1
2

a

t

2

.

{\displaystyle x\_{f}=x\_{i}+v\_{i}t+{\frac {1}{2}}at^{2}.}

**Part 3**

Isolate for time from the result of part 1

# t

v

f

−

v

i

a

.

{\displaystyle t={\frac {v\_{f}-v\_{i}}{a}}.}

From part 2

x

f

−

x

i

=

v

i

(

v

f

−

v

i

a

)

1
2

a

(

v

f

−

v

i

a

)

2

{\displaystyle x\_{f}-x\_{i}=v\_{i}\left({\frac {v\_{f}-v\_{i}}{a}}\right)+{\frac {1}{2}}a\left({\frac {v\_{f}-v\_{i}}{a}}\right)^{2}}

x

f

−

x

i

=

v

i

v

f

−

v

i

2

a

1
2

v

f

2

−
2

v

i

v

f

v

i

2

a

{\displaystyle x\_{f}-x\_{i}={\frac {v\_{i}v\_{f}-v\_{i}^{2}}{a}}+{\frac {1}{2}}{\frac {v\_{f}^{2}-2v\_{i}v\_{f}+v\_{i}^{2}}{a}}}

2
a
(

x

f

−

x

i

# )
2

v

i

v

f

−
2

v

i

2

v

f

2

−
2

v

i

v

f

v

i

2

{\displaystyle 2a(x\_{f}-x\_{i})=2v\_{i}v\_{f}-2v\_{i}^{2}+v\_{f}^{2}-2v\_{i}v\_{f}+v\_{i}^{2}}

2
a
(

x

f

−

x

i

# )
−

v

i

2

v

f

2

.

{\displaystyle 2a(x\_{f}-x\_{i})=-v\_{i}^{2}+v\_{f}^{2}.}

Thus

v

f

2

=

v

i

2

2
a
(

x

f

−

x

i

)
.

{\displaystyle v\_{f}^{2}=v\_{i}^{2}+2a(x\_{f}-x\_{i}).}
Concave Lens

## **Problem**

The image formed by a concave lens for an object 40 cm away is reduced to one-fifth its size. Determine is the focal length of the lens.

## **Hint**

(1) Thin lens formula:

1

d
o

1

d
i

=

1
f

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f} }

.

(2) Magnification formula:

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

.

## **Solution**

Since we know the magnification of the object and the distance of the object to the lens, we can use the magnification formula to calculate the distance of the image to the lens.

# M
−

d
i

d
o

=

h
i

h
o

{\displaystyle M = -\frac{d\_i}{d\_o} = \frac{h\_i}{h\_o} }

1
5

=
−

d
i

40

c
m

d
i

=
−
8

c
m
.

{\displaystyle \begin{align}
\frac{1}{5} &= -\frac{d\_i}{40 \; cm} \\[5 pt]
d\_i &= -8 \; cm.
\end{align} }

Next, use the thin lens formula to solve for the focal length

1

d
o

1

d
i

=

1
f

.

{\displaystyle \frac{1}{d\_o} + \frac{1}{d\_i} = \frac{1}{f}. }

1

40

c
m

1

−
8

c
m

=

1
f

1
f

=

−
1

10

c
m

f

=
−
10

c
m

{\displaystyle \begin{align}
\frac{1}{40 \; cm} + \frac{1}{-8 \; cm} &= \frac{1}{f} \\[5 pt]
\frac{1}{f} &= \frac{-1}{10 \; cm} \\[5 pt]
f &= -10 \; cm
\end{align} }
Linearizing Power Functions

## **Problem**

**Part 1:** Use logarithms to transform the power function

# y
k

x
m

{\displaystyle y = kx^m }

into the linear function

# Y
m
X
b
.

{\displaystyle Y = mX + b. }

**Part 2:** Transform the linear function

# Y
−
2
X
3

{\displaystyle Y = -2X + 3 }

 into a power function. Assume base 10 log.

## **Solution**

**Part 1**

Take the logarithm of both sides of the equation

# y
k

x
m

.

{\displaystyle y = kx^m. }

log
⁡
(
y
)

=
log
⁡
(
k

x
m

)

log
⁡
(
y
)

=
log
⁡
(
k
)
log
⁡
(

x
m

)

log
⁡
(
y
)

=
log
⁡
(
k
)
m
log
⁡
(
x
)

log
⁡
(
y
)

=
m
log
⁡
(
x
)
log
⁡
(
k
)

{\displaystyle
\begin{align} \log(y) &= \log(kx^m) \\[5 pt]
\log(y) &= \log(k) + \log(x^m) \\[5 pt]
\log(y) &= \log(k) + m\log(x) \\[5 pt]
\log(y) &= m\log(x) + \log(k)
\end{align} }

Letting

# Y
log
⁡
(
y
)

{\displaystyle Y = \log(y) }

,

# X
log
⁡
(
x
)

{\displaystyle X = \log(x) }

, and

# b
log
⁡
(
k
)

{\displaystyle b = \log(k) }

, we get the result

# Y
m
X
b
.

{\displaystyle Y = mX + b. }

**Part 2**
For the equation

# Y
−
2
X
3

{\displaystyle Y = -2X + 3 }

, we have

# m
−
2

{\displaystyle m = -2 }

 and

# 3
log
⁡
(
k
)

{\displaystyle 3 = \log(k) }

. Since the logarithm is base 10 in this problem

# k

10

3

=
1000

{\displaystyle k = {10}^{3} = 1000 }

. Therefore,

# y
1000

x

−
2

{\displaystyle y = 1000{x}^{-2} }

or

# y

1000

x
2

.

{\displaystyle y = \frac{1000}{x^2}. }
D'Alembert's Probability Paradox

## **Problem**

In his article *Croix ou Pile* (English: Cross or Tail) in Denis Diderot's the Encyclopédie, **Jean-Baptiste le Rond d'Alembert** (1717 – 1783 AD) considered the following problem:

*In two tosses of a fair coin, what is the probability that heads will appear at least once?*
Determine this probability.

## **Solution**

D'Alembert gave the answer

2
3

{\displaystyle \frac{2}{3}}

. He reasoned that in real life no one would continue the experiment after heads showed up on the first toss; hence, he considered only three cases: heads on the first or second throw, or not at all.

H
,

H
T
,

T
T

{\displaystyle H, \,HT, \,TT}

In actuality there are four cases: first and second throw heads; first throw heads, second throw tails; first throw tails, second throw heads; first and second throw tails.

H
H
,

H
T
,

T
H
,

T
T

{\displaystyle HH,\,HT,\,TH,\,TT}

This gives a probability of

3
4

{\displaystyle \frac{3}{4}}

.