recursal/Fanatic-Fandom huggingface dataset:
Latin Quotations

**Compiled by Tao Steven Zheng (郑涛)**

## **Cicero (106 BC – 43 BC)**

Attributed to Julius Caesar by Suetonius when Caesar led his army across the Rubicon.

## **St. Thomas Aquinas (1225 – 1274)**

### Quia parvus error in principio magnus est in fine, secundum philosophum in I caeli et mundi, ens autem et essentia sunt quae primo intellectu concipiuntur, ut dicit Avicenna in principio suae metaphysicae.

A small mistake in the beginning is a big one in the end, according to the Philosopher in the first book of On the Heavens and the Earth. And as Ibn-Sînâ (Avicenna) says in the beginning of his Metaphysics, being and essence are what is first conceived by the intellect.

From one of Thomas Aquinas' thirteen commentaries on Aristotle, *De ente at essentia* (On Being and Essence).

## **Isaac Newton (1643 - 1727)**

### Amicus Plato amicus Aristoteles magis amica veritas.

Plato is my friend, Aristotle is my friend, but my greatest friend is truth.

From Newton's 1661 student notebook entitled *Quaestiones quaedam philosophicae* (*Certain philosophical questions*).
Adiabatic Process

## **Physics Concept**

An **adiabatic process** is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings. Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work.

## **Problem**

Suppose the volume

V

{\displaystyle V}

 and the pressure

P

{\displaystyle P}

 of a gas undergoing adiabatic compression is modelled by

P
V

1.2

=
C

{\displaystyle {PV}^{1.2} = C }

, where

C

{\displaystyle C}

 is constant. When the volume is 64 L, the pressure is 20 kPa.

If at a moment the pressure is decreasing at the rate of 1 kPa per second, find the instantaneous rate of change of the volume.

## **Solution**

What is known? The problems states that the adiabatic process is modelled by the implicit function

P
V

1.2

=
C

{\displaystyle {PV}^{1.2} = C }

 with the conditions

d
P

d
t

=
−
1

k
P
a

/

s

{\displaystyle \frac{dP}{dt} = -1 \; kPa/s }

 at

# P
20

k
P
a

{\displaystyle P = 20 \; kPa }

 and

# V
64

L

{\displaystyle V = 64 \; L }

.

Use implicit differentiation to derive the function with respect to time.

d

d
t

(

P

V

1.2

)

=

d

d
t

(
C
)

{\displaystyle \frac{d}{dt} \left(PV^{1.2} \right) = \frac{d}{dt} (C) }

Since

C

{\displaystyle C}

 is constant,

d
P

d
t

V

1.2

P
(
1.2
)

V

0.2

d
V

d
t

=
0

{\displaystyle \frac{dP}{dt} V^{1.2} + P(1.2) V^{0.2} \frac{dV}{dt} = 0 }

Substitute all the given values and solve for the unknown rate

d
V

d
t

{\displaystyle \frac{dV}{dt} }

:

(
−
1
)

(
64
)

1.2

(
20
)
(
1.2
)

(
64
)

0.2

d
V

d
t

=
0

{\displaystyle (-1){(64)}^{1.2} + (20)(1.2){(64)}^{0.2} \frac{dV}{dt} = 0 }

d
V

d
t

=

64
24

{\displaystyle \frac{dV}{dt} = \frac{64}{24} }

d
V

d
t

=

8
3

{\displaystyle \frac{dV}{dt} = \frac{8}{3} }

Therefore, the volume is increasing at the rate of

8
3

L

/

s

{\displaystyle \frac{8}{3} \; L/s }

.
Atoms in the Sun

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

## **Problem**

Estimate the number of atoms in the sun. Give your answer as an *order of magnitude*.

This problem appeared in the 2014 UBC Physics Olympics Fermi problems event.

## **Solution**

The mass of the Sun is roughly

1.99
×

10

30

k
g

{\displaystyle 1.99 \times {10}^{30} \; kg }

 and the Sun is composed of mainly hydrogen atoms (roughly 90%) whose weight is

1.67
×

10

−
27

k
g

{\displaystyle 1.67 \times {10}^{-27} \; kg }

. Dividing both quantities to give the approximate number of atoms in the Sun.

1.99
×

10

30

k
g

1.67
×

10

−
27

k
g

=
1.2
×

10

57

{\displaystyle \frac{1.99 \times {10}^{30} \; kg}{1.67 \times {10}^{-27} \; kg} = 1.2 \times {10}^{57} }

The order of magnitude is 57.
Interval Set Notation

## **Problem**

Consider two intervals on the real number line:

# A
(
−
1
,
2
]

{\displaystyle A = (-1,2] }

 which means

−
1
<
x
≤
2

{\displaystyle -1 < x \le 2 }

# B
(
1
,
3
)

{\displaystyle B = (1,3) }

 which means

1
≤
x
≤
3

{\displaystyle 1 \leq x \leq 3}

.

**Part 1:** Compute

A
∪
B

{\displaystyle A \cup B}

 and

A
∩
B

{\displaystyle A \cap B}

.

**Part 2:** Verify that

A
C

∩

B
C

=

(
A
∪
B
)

C

{\displaystyle A^C \cap B^C = {(A \cup B)}^{C} }

.

**Part 3:** Verify that

A
C

∪

B
C

=

(
A
∩
B
)

C

{\displaystyle A^C \cup B^C = {(A \cap B)}^{C} }

## **Solution**

**Part 1**

A
∪
B

{\displaystyle A \cup B}

 is the union of the sets. This means all elements of both sets are included.

A
∪
# B
(
−
1
,
3
)

{\displaystyle A \cup B = (-1,3) }

A
∩
B

{\displaystyle A \cap B}

 is the intersection of the sets. This means all elements common to both sets are included.

A
∩
# B
(
1
,
2
)

{\displaystyle A \cap B = (1,2) }

**Part 2**

Compute the complement of each set.

A
C

=
(
−
∞
,
−
1
]
∪
(
2
,
∞
)

{\displaystyle A^C = (-\infty , -1] \cup (2, \infty) }

B
C

=
(
−
∞
,
1
]
∪
[
3
,
∞
)

{\displaystyle B^C = (-\infty , 1] \cup [3, \infty) }

A
C

∩

B
C

=
(
−
∞
,
−
1
]
∪
[
3
,
∞
)

{\displaystyle A^C \cap B^C = (-\infty , -1] \cup [3, \infty) }

We arrive at this result using

(
A
∪
B
)

C

{\displaystyle {(A \cup B)}^{C} }

.

(
A
∪
B
# )
(
−
1
,
3
)

{\displaystyle (A \cup B) = (-1,3) }

(
A
∪
B
)

C

=
(
−
∞
,
−
1
]
∪
[
3
,
∞
)

{\displaystyle {(A \cup B)}^{C} = (-\infty , -1] \cup [3, \infty) }

.
**Part 3**

From part 2,

A
C

=
(
−
∞
,
−
1
]
∪
(
2
,
∞
)

{\displaystyle A^C = (-\infty , -1] \cup (2, \infty) }

B
C

=
(
−
∞
,
1
]
∪
[
3
,
∞
)

{\displaystyle B^C = (-\infty , 1] \cup [3, \infty) }

A
C

∪

B
C

=
(
−
∞
,
1
]
∪
[
2
,
∞
)

{\displaystyle A^C \cup B^C = (-\infty , 1] \cup [2, \infty) }

We arrive at this result using

(
A
∩
B
)

C

{\displaystyle {(A \cap B)}^{C} }

.

(
A
∩
B
# )
(
1
,
2
)

{\displaystyle (A \cap B) = (1,2) }

(
A
∩
B
)

C

=
(
−
∞
,
1
]
∪
[
2
,
∞
)

{\displaystyle {(A \cap B)}^{C} = (-\infty , 1] \cup [2, \infty) }

.
Arithmetic Series of Higher Order

## **Problem**

Chinese mathematicians of the Song-Yuan period (960 - 1368 AD) investigated finite sums related to the diagonals of Jia Xian's triangle (arithmetic triangle). The following table lists the given names of several finite sums found in Zhu Shijie’s *Suanxue Qimeng* (1299) and *Siyuan Yujian* (1303).

1
2
3
4
.
.
.
# n

1

2
!

n
(
n
1
)

{\displaystyle 1+2+3+4+...+n = \frac{1}{2!} n(n+1)}
 落一形垛

1
3
6
10
.
.
.

1

2
!

n
(
n
1
# )

1

3
!

n
(
n
1
)
(
n
2
)

{\displaystyle 1+3+6+10+...+\frac{1}{2!}n(n+1) = \frac{1}{3!}n(n+1)(n+2) }
 撒星形垛

1
4
10
20
.
.
.

1

3
!

n
(
n
1
)
(
n
2
# )

1

4
!

n
(
n
1
)
(
n
2
)
(
n
3
)

{\displaystyle 1+4+10+20+...+\frac{1}{3!}n(n+1)(n+2) = \frac{1}{4!}n(n+1)(n+2)(n+3)}
 撒星更落一形垛

1
5
15
35
.
.
.

1

4
!

n
(
n
1
)
(
n
2
)
(
n
3
# )

1

5
!

n
(
n
1
)
(
n
2
)
(
n
3
)
(
n
4
)

{\displaystyle 1+5+15+35+...+ \frac{1}{4!}n(n+1)(n+2)(n+3)=\frac{1}{5!}n(n+1)(n+2)(n+3)(n+4)}

These finite sums are called **arithmetic series of higher order**, which follow the general pattern

∑

# i
1

n

1

p
!

i
(
i
1
)
(
i
2
)
.
.
.
(
i
p
−
1
# )

1

(
p
1
)
!

n
(
n
1
)
(
n
2
)
.
.
.
(
n
p
)

{\displaystyle \sum\_{i=1}^{n}\frac{1}{p!}i(i+1)(i+2)...(i+p-1) = \frac{1}{(p+1)!}n(n+1)(n+2)...(n+p) }

Alternatively this can be expressed as

∑

# i
1

n

(

i
p
−
1

p

)

=

(

n
p

p
1

)

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \binom{n+p}{p+1}}

Prove this identity.

## **Solution**

Write out the terms of the sum on the left hand side:

∑

# i
1

n

(

i
p
−
1

p

)

=

(

p
p

)

(

p
1

p

)

(

p
2

p

)

.
.
.

(

n
p
−
1

p

)

.

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \binom{p}{p} + \binom{p+1}{p} + \binom{p+2}{p} + ... + \binom{n+p-1}{p}. }

Observe that

(

p
p

)

=

(

p
1

p
1

)

{\displaystyle \binom{p}{p} = \binom{p+1}{p+1} }

, hence,

∑

# i
1

n

(

i
p
−
1

p

)

=

[

(

p
1

p
1

)

(

p
1

p

)

]

(

p
2

p

)

.
.
.

(

n
p
−
1

p

)

.

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \left[\binom{p+1}{p+1} + \binom{p+1}{p} \right] + \binom{p+2}{p} + ... + \binom{n+p-1}{p}.
}

By the binomial identity

(

n
−
1

k

)

(

n
−
1

k
−
1

)

=

(

n
k

)

{\displaystyle \binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k} }

, we get

∑

# i
1

n

(

i
p
−
1

p

)

=

(

p
2

p
1

)

(

p
2

p

)

.
.
.

(

n
p
−
1

p

)

.

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \binom{p+2}{p+1} + \binom{p+2}{p} + ... + \binom{n+p-1}{p}. }

Following this step-pattern gives

∑

# i
1

n

(

i
p
−
1

p

)

=

[

(

p
2

p
1

)

(

p
2

p

)

]

(

p
3

p

)

.
.
.

(

n
p
−
1

p

)

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \left[\binom{p+2}{p+1} + \binom{p+2}{p} \right] + \binom{p+3}{p} + ... + \binom{n+p-1}{p}
}

until we reach the final term

∑

# i
1

n

(

i
p
−
1

p

)

=

[

(

n
p
−
1

p
1

)

(

n
p
−
1

p

)

]

.

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \left[\binom{n+p-1}{p+1} + \binom{n+p-1}{p} \right]. }

Therefore,

∑

# i
1

n

(

i
p
−
1

p

)

=

(

n
p

p
1

)

.

{\displaystyle \sum\_{i=1}^{n} \binom{i+p-1}{p} = \binom{n+p}{p+1}.}
Brahmagupta's Identity

## **Problem**

   The Indian mathematician and astronomer Brahmagupta (597 - 668 AD) used the identity

(

a
2

N

b
2

)
(

c
2

N

d
2

# )
(
a
c
∓
N
b
d

)
2

N
(
a
d
±
b
c

)
2

{\displaystyle (a^2+Nb^2)(c^2+Nd^2) = (ac\mp Nbd)^2 + N(ad \pm bc)^2 }

for solving quadratic indeterminate equations[1] of the form

A

x
2

# 1

y
2

{\displaystyle Ax^2 + 1 = y^2}

, where

A
,
x
,
y

{\displaystyle A,x,y}

 are integers.

**Question:** Verify that this identity is valid.

[1] In Europe these equations are called **Pell's equations**, named after English mathematician John Pell (1611 – 1685 AD). John Pell was not the first European to study these equations! In fact, it was first proposed by Pierre de Fermat (1607 - 1665 AD) in 1657. William Brouncker (1620 – 1684 AD) was the first European to solve Pell's equation. The name of Pell's equation arose from Leonhard Euler (1707 – 1783 AD) mistakenly attributing Brouncker's solution of the equation to John Pell.

## **Solution**

Expand the left hand side.

(

a
2

N

b
2

)
(

c
2

N

d
2

)

=

a
2

c
2

N

a
2

d
2

N

b
2

c
2

N
2

b
2

d
2

=

a
2

c
2

N
2

b
2

d
2

N
(

a
2

d
2

b
2

c
2

)

{\displaystyle \begin{align} (a^2+Nb^2)(c^2+Nd^2) &= a^2c^2 + Na^2d^2 + Nb^2c^2 + N^2b^2d^2 \\[5 pt]
&= a^2c^2 + N^2b^2d^2 + N(a^2d^2 + b^2c^2)
\end{align} }

Expand the right hand side.

(
a
c
∓
N
b
d

)
2

N
(
a
d
±
b
c

)
2

=

a
2

c
2

∓
2
N
a
b
c
d

N
2

b
2

d
2

N
(

a
2

d
2

±
2
a
b
c
d

b
2

c
2

)

=

a
2

c
2

∓
2
N
a
b
c
d

N
2

b
2

d
2

N

a
2

d
2

±
2
N
a
b
c
d
N

b
2

c
2

=

a
2

c
2

N
2

b
2

d
2

N
(

a
2

d
2

b
2

c
2

)

{\displaystyle \begin{align} (ac\mp Nbd)^2 + N(ad \pm bc)^2 &= a^2c^2 \mp 2Nabcd + N^2b^2d^2 + N(a^2d^2 \pm 2abcd + b^2c^2)\\[5 pt]
&= a^2c^2 \mp 2Nabcd + N^2b^2d^2 + Na^2d^2 \pm 2Nabcd + Nb^2c^2\\[5 pt]
&= a^2c^2 + N^2b^2d^2 + N(a^2d^2 + b^2c^2)
\end{align}}

The expanded results on the left hand side matches the right hand side indeed.
Generating Streamplots with Python

**By: Tao Steven Zheng (郑涛)**

## **Code**

```
import numpy as np
import matplotlib.pyplot as plt

# Generating a coordinate grid using meshgrid
L = 5
nx, ny = 100, 100
x = np.linspace(-L, L, nx)
y = np.linspace(-L, L, ny)
X, Y = np.meshgrid(x, y)

# Generating a streamplot for a vector field
fig = plt.figure(figsize = (8,8))
plt.streamplot(x, y, np.sin(X), np.sin(Y), linewidth=1, density=2, arrowstyle='->', arrowsize=1.5)

```
Bond Order

## **Problem**

Explain how the bond order of the following diatomic gases (oxygen, nitrogen, fluorine) is calculated using electron configurations. Also determine whether each diatomic gas is diamagnetic or paramagnetic.

## **Solution**

The covalent bonding of atoms can be explained by the overlapping of their orbitals. In these examples, the overlapping of the

2
s

{\displaystyle 2s}

 orbitals produces sigma-bonds

(
σ
)

{\displaystyle (\sigma) }

 , while the overlapping of

2
p

{\displaystyle 2p}

 orbitals produces pi-bonds

(
π
)

{\displaystyle (\pi) }

. Note that for each bond, there is a corresponding anti-bond that is situated at a higher energy. These anti-bonds(denoted with an asterisk) weakens the covalent bond caused by its counterpart. Thus, if a certain bond, for example a sigma bond is filled, and the counterpart sigma anti-bond is also filled, the bonding is negated.
To calculate the bond order, halve the difference of the number of the boding electrons to the number of anti-boding electrons.

B
# O

B
−
A

2

{\displaystyle BO = \frac{B - A}{2}}

**Diatomic nitrogen**

For diatomic nitrogen,

N

2

{\displaystyle \ce{N2}}

, there are five valence electrons per nitrogen atom, thus providing ten valence electrons in total. This is enough to completely fill in the following bonds:

σ

2
s

{\displaystyle {\sigma}\_{2s} }

,

σ

∗

2
s

{\displaystyle {{\sigma}^{\*}}\_{2s} }

,

σ

2
p

{\displaystyle {\sigma}\_{2p} }

, and two

π

2
p

{\displaystyle {\pi}\_{2p} }

.

B
# O

8
−
2

2

=
3

{\displaystyle BO = \frac{8 - 2}{2} = 3 }

A bond order of 3 indicates a triple bond.
Since there are no unpaired electron spins, diatomic nitrogen is **diamagnetic**.

**Diatomic oxygen**

For diatomic oxygen,

O

2

{\displaystyle \ce{O2 }}

, there are six valence electrons per oxygen atom, thus providing twelve valence electrons in total. This is enough to completely fill in the following bonds:

σ

2
s

{\displaystyle {\sigma}\_{2s} }

,

σ

∗

2
s

{\displaystyle {{\sigma}^{\*}}\_{2s} }

,

σ

2
p

{\displaystyle {\sigma}\_{2p} }

, and two

π

2
p

{\displaystyle {\pi}\_{2p} }

. The

π

∗

2
p

{\displaystyle {{\pi}^{\*}}\_{2p} }

 anti-bond is only partially filled.

B
# O

8
−
4

2

=
2

{\displaystyle BO = \frac{8 - 4}{2} = 2 }

A bond order of 2 indicates a double bond.
Since there are two unpaired electron spins, diatomic oxygen is **paramagnetic**.

**Diatomic fluorine**

For diatomic fluorine,

F

2

{\displaystyle \ce{F2 }}

, there are seven valence electrons per oxygen atom, thus providing fourteen valence electrons in total. This is enough to completely fill in the following bonds:

σ

2
s

{\displaystyle {\sigma}\_{2s} }

,

σ

∗

2
s

{\displaystyle {{\sigma}^{\*}}\_{2s} }

,

σ

2
p

{\displaystyle {\sigma}\_{2p} }

, two

π

2
p

{\displaystyle {\pi}\_{2p} }

, and

π

∗

2
p

{\displaystyle {{\pi}^{\*}}\_{2p} }

.

B
# O

8
−
6

2

=
1

{\displaystyle BO = \frac{8 - 6}{2} = 1 }

A bond order of 1 indicates a single bond.

Since there are no unpaired electron spins, diatomic fluorine is **diamagnetic**.
Difference of Consecutive Squares

## **Problem**

Show that the sum of two consecutive integers is equal to the difference of their perfect squares.

**Example:**

3
# 4

4
2

−

3
2

{\displaystyle 3+4=4^2-3^2 }

## **Solution**

Let

n

{\displaystyle n}

 and

n
1

{\displaystyle n+1}

 be the two consecutive integers.

n
(
n
1
# )
(
n
1

)
2

−

n
2

{\displaystyle n+(n+1)=(n+1)^2-n^2 }

2
n
# 1

n
2

2
n
1
−

n
2

{\displaystyle 2n+1=n^2+2n+1-n^2 }

2
n
# 1
2
n
1

{\displaystyle 2n+1=2n+1 }

Alternatively we may use the strategy

a
2

−

b
2

=
(
a
b
)
(
a
−
b
)

{\displaystyle a^2-b^2=(a+b)(a-b)}

n
(
n
1
# )
(
n
1

)
2

−

n
2

{\displaystyle n+(n+1)=(n+1)^2-n^2 }

2
n
# 1
(
n
1
n
)
(
n
1
−
n
)

{\displaystyle 2n+1=(n+1+n)(n+1-n) }

2
n
# 1
2
n
1

{\displaystyle 2n+1=2n+1 }
Gravity Powered Collider

## **Problem**

A block of mass

m
1

{\displaystyle m\_1 }

, initially at rest, slides down a frictionless track of height

h

{\displaystyle h}

. The mass collides elastically with another block of mass

m
2

{\displaystyle m\_2 }

 at the bottom of the track (also initially a rest). Calculate the velocity of each block immediately after the collision in terms of

m
1

,

m
2

,
h

{\displaystyle m\_1, m\_2, h}

.

## **Solution**

The potential energy of the first block is converted to kinetic energy at the bottom of the track.

m
1

g
# h

1
2

m
1

v
2

{\displaystyle m\_1 gh = \frac{1}{2} m\_1 v^2 }

# v

2
g
h

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

Since this collision is elastic (which we assume to be perfectly elastic here), both linear momentum and kinetic energy are conserved.

**Conservation of momentum**

m
1

# v

m
1

v
1

m
2

v
2

{\displaystyle m\_1 v = m\_1 v\_1 + m\_2 v\_2 }

**Conservation of energy**

1
2

m
1

v

2

=

1
2

m
1

v
1

2

1
2

m
2

v
2

2

{\displaystyle \frac{1}{2} m\_1 {v}^{2} = \frac{1}{2} m\_1 {v\_1}^{2} + \frac{1}{2} m\_2 {v\_2}^{2} }

Thus **equation 1** is

v
1

=

2
g
h

−

(

m
2

m
1

)

v
2

{\displaystyle v\_1 = \sqrt{2gh} - \left(\frac{m\_2}{m\_1}\right) v\_2 }

and **equation 2** is

2
g
# h

v
1

2

m
2

v
2

2

.

{\displaystyle 2gh = {v\_1}^{2} + m\_2 {v\_2}^{2}. }

Substitute **equation 1** into **equation 2** and solve for

v
2

{\displaystyle v\_2}

.

2
g
# h

[

2
g
h

−

(

m
2

m
1

)

v
2

]

2

m
2

v
2

2

{\displaystyle 2gh = {\left[\sqrt{2gh} - \left(\frac{m\_2}{m\_1}\right) v\_2 \right]}^{2} + m\_2 {v\_2}^{2} }

2
g
# h
2
g
h
−
2

(

m
2

m
1

)

2
g
h

v
2

(

m
2

m
1

)

2

v
2

2

m
2

v
2

2

{\displaystyle 2gh = 2gh - 2\left(\frac{m\_2}{m\_1}\right)\sqrt{2gh} v\_2 + {\left(\frac{m\_2}{m\_1}\right)}^{2}{v\_2}^{2} + m\_2 {v\_2}^{2} }

[

(

m
2

m
1

)

2

m
2

]

v
2

2

−
2

(

m
2

m
1

)

2
g
h

v
2

=
0

{\displaystyle \left[{\left(\frac{m\_2}{m\_1}\right)}^{2}+ m\_2 \right] {v\_2}^{2} -2\left(\frac{m\_2}{m\_1}\right)\sqrt{2gh} v\_2 = 0 }

v
2

(

[

(

m
2

m
1

)

2

m
2

]

v
2

−
2

(

m
2

m
1

)

2
g
h

)

=
0

{\displaystyle v\_2\left(\left[{\left(\frac{m\_2}{m\_1}\right)}^{2}+ m\_2 \right] v\_2 -2\left(\frac{m\_2}{m\_1}\right)\sqrt{2gh} \right) = 0 }

Reject the zero-velocity solution because the first block should recoil. Solve for the non-zero velocity

v
2

{\displaystyle v\_2}

:

v
2

=

2

m
2

2
g
h

m
1

⋅

m
1

2

m
2

(

m
1

m
2

)

=

2
g
h

(

2

m
1

m
1

m
2

)

.

{\displaystyle \begin{align}
v\_2 &= \frac{2m\_2 \sqrt{2gh}}{m\_1} \cdot \frac{{m\_1}^{2}}{{m\_2} \left(m\_1 + m\_2 \right)}\\[5 pt]
&= \sqrt{2gh} \left(\frac{2 m\_1}{m\_1 + m\_2} \right).
\end{align} }

Substitute

v
2

{\displaystyle v\_2}

 into **equation 1** to solve for

v
1

{\displaystyle v\_1}

:

v
1

=

2
g
h

−

(

m
2

m
1

)

⋅

2

2
g
h

m
1

m
1

m
2

=

2
g
h

(

1
−

2

m
2

m
1

m
2

)

.

{\displaystyle \begin{align}
v\_1 &= \sqrt{2gh} - \left(\frac{m\_2}{m\_1}\right) \cdot \frac{2\sqrt{2gh} m\_1}{m\_1 + m\_2}\\[5 pt]
&= \sqrt{2gh}\left(1 - \frac{2 m\_2}{m\_1 + m\_2} \right).
\end{align} }
Hailstone Numbers with Python

**By: Tao Steven Zheng (郑涛)**

## **Task**

A Hailstone sequence is generated from a starting positive integer

n

{\displaystyle n}

 by the following program:

* If

n

{\displaystyle n}

 is **1** then the sequence ends.
* If

n

{\displaystyle n}

 is **even** then the next number of the sequence

n

/

2

{\displaystyle n/2}
* If

n

{\displaystyle n}

 is **odd** then the next number of the sequence

(
3
∗
n
)
1

{\displaystyle (3 \* n) + 1 }

The unproven Collatz conjecture states that the hailstone sequence for any starting positive integer always terminates (by reaching 1). This sequence was named by Lothar Collatz (1910 - 1990) in 1937.

Create a function to generate the hailstone sequence for any integer number (make sure to exclude non-positive integers).

## **Code**

```
def hailstone(n):
    sequence = [n]
    if n < 1:
        print("Cannot generate sequence. Not a positive integer!")
    if n == 1:
        print([1])
                n = 3 \* n + 1
                sequence.append(n)
            if n % 2 == 0:
                n = n // 2
                sequence.append(n)
        return sequence

```

## **Experiments and Results**

```
'''Experiments'''
hailstone(-10)
hailstone(1)
print(hailstone(10))
print(len(hailstone(10)))
print(hailstone(9))
print(len(hailstone(9)))

'''Results'''
Cannot generate sequence. Not a positive integer!
[1]
[10, 5, 16, 8, 4, 2, 1]
7
[9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1]
20

```
Infinite Series of Nicole Oresme

## **Problem**

**Nicole Oresme** (c. 1320 – 1382) was an influential French philosopher, theologian, and mathematician during the late Middle Ages. In mathematics, Oresme made important contributions to the study of proportion and infinite series. Among his most famous mathematical works are *Tractatus de configurationibus qualitatum et motuum* and *Algorismus proportionum*.

Determine the following infinite series he created:

∑

# k
1

∞

k

2
k

=

1
2

⋅
1

1
4

⋅
2

1
8

⋅
3
⋯

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = \frac{1}{2} \cdot 1 + \frac{1}{4} \cdot 2 + \frac{1}{8} \cdot 3 + \cdots }

## **Solution**

Nicole Oresme gave a geometric argument for solving this infinite series. The geometric argument up to the first three terms is shown in the figure below. We will demonstrate Oresme's method arithmetically using modern mathematical notation.

∑

# k
1

∞

k

2
k

=

1
2

⋅
1

1
4

⋅
2

1
8

⋅
3
⋯

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = \frac{1}{2} \cdot 1 + \frac{1}{4} \cdot 2 + \frac{1}{8} \cdot 3 + \cdots }

∑

# k
1

∞

k

2
k

=

1
2

(

1
4

1
4

)

(

1
8

1
8

1
8

)

⋯

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4} \right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \cdots }

∑

# k
1

∞

k

2
k

=

(

1
2

1
4

1
8

⋯

)

(

1
4

1
8

1
16

⋯

)

(

1
8

1
16

1
32

⋯

)

⋯

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \right) + \left(\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \right) + \left(\frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots \right) + \cdots }

∑

# k
1

∞

k

2
k

=

∑

# k
1

∞

1

2

k

∑

# k
1

∞

1

2

k
1

∑

# k
1

∞

1

2

k
2

⋯

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = \sum\_{k=1}^{\infty} \frac{1}{{2}^{k}} + \sum\_{k=1}^{\infty} \frac{1}{{2}^{k+1}} + \sum\_{k=1}^{\infty} \frac{1}{{2}^{k+2}} + \cdots }

∑

# k
1

∞

k

2
k

=

∑

# i
0

∞

(

∑

# k
1

∞

1

2

k
i

)

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = \sum\_{i=0}^{\infty} \left(\sum\_{k=1}^{\infty} \frac{1}{{2}^{k+i}} \right) }

∑

# k
1

∞

k

2
k

=
1

1
2

1
4

⋯

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = 1+ \frac{1}{2} + \frac{1}{4} + \cdots }

∑

# k
1

∞

k

2
k

=
2

{\displaystyle \sum\_{k=1}^{\infty} \frac{k}{2^k} = 2 }
Ball and Wall

## **Problem**

A 0.5 kg ball is thrown at a wall horizontally at a speed of 5 m/s. The ball recoils at the same height

h

{\displaystyle h}

, and at a speed of 5 m/s.

**Part 1:** What is the work done by the wall?

**Part 2:** What is the impulse exerted by the wall on the rubber ball?

**Part 3:** What is the magnitude of force exerted on the ball by the wall if the collision time lasts 2 milliseconds?

## **Solution**

**Part 1**

Since the speed of the ball is unchanged (the velocity is changed), the kinetic energy is also unchanged. Thus, the work done by the wall is also zero. Mathematically speaking,

K

E
i

P

E
i

# W
K

E
f

P

E
f

{\displaystyle KE\_i + PE\_i + W = KE\_f + PE\_f }

1
2

m

v
i

2

m
g

h
i

# W

1
2

m

v
f

2

m
g

h
f

{\displaystyle \frac{1}{2} m{v\_i}^{2} + mgh\_i + W = \frac{1}{2}m{v\_f}^{2} + mgh\_f }

For this problem, substitute the values

h
i

=

h
f

=
h

{\displaystyle h\_i = h\_f = h }

,

v
i

=
5

m

/

s

{\displaystyle v\_i = 5 \; m/s}

, and

v
f

=
−
5

m

/

s

{\displaystyle v\_f = -5 \; m/s}

:

1
2

(
0.5
)

(
5
)

2

m
g
h
# W

1
2

(
0.5
)

(
−
5
)

2

m
g
h

{\displaystyle \frac{1}{2} (0.5){(5)}^{2} + mgh + W = \frac{1}{2} (0.5){(-5)}^{2} + mgh}

6.25
# W
6.25

{\displaystyle 6.25 + W = 6.25 }

# W
0

J

{\displaystyle W = 0 \;J }

**Part 2**

Although the magnitude of the momentum is unchanged, the direction is. Because momentum and impulse are a vector quantities, the impulse is non-zero. Here,

p

{\displaystyle p}

 denotes the momentum and

J

{\displaystyle J}

 is the impulse (change in momentum).

J

=

p
f

−

p
i

[
5
p
t
]

=
m

v
f

−
m

v
i

[
5
p
t
]

=
m
(

v
f

−

v
i

)

[
5
p
t
]

{\displaystyle \begin{align}
J &= p\_f - p\_i \\ [5 pt]
&= mv\_f - mv\_i \\ [5 pt]
&= m (v\_f - v\_i) \\ [5 pt]
\end{align} }

Substitute the given mass and velocities yields:

J

=
0.5
(
−
5
−
5
)

=
−
5

k
g
⋅
m

s

{\displaystyle \begin{align}
J &= 0.5 (-5 - 5) \\[5 pt]
&= -5 \; \frac{kg\cdot m}{s}
\end{align} }

**Part 3**

Impulse can be calculated in two ways:

# J
m
Δ
v

{\displaystyle J = m \Delta v }

, or

# J
F
Δ
t

{\displaystyle J = F \Delta t }

. Solve for the force using the second equation:

# J
F
Δ
t

{\displaystyle J = F \Delta t }

−
# 5
F
(
2
×

10

−
3

)

{\displaystyle -5 = F (2 \times {10}^{-3}) }

# F
−
2500

N

{\displaystyle F = -2500 \; N }

The magnitude of force is

2500

N

{\displaystyle 2500 \; N }

.
Force on the Oceans

## **Problem**

Calculate the force with which the atmosphere of the Earth pushes down on all the oceans. Note that the Earth's oceans occupy roughly 70% of the global surface area.

**Important numbers**

 Radius of the Earth

## **Solution**

Force is equal to pressure multiplied by area. For the all the oceans' surface,

# F
P

(

0.7
⋅
4
π

r
2

)

{\displaystyle F = P \left(0.7 \cdot 4 \pi r^2 \right) }

.

# F

(

101300

P
a

)

(

0.7
⋅
4
π

(

6.37
×

10

6

m

)

2

)

{\displaystyle F = \left(101300 \; Pa \right)\left(0.7 \cdot 4 \pi {\left(6.37 \times {10}^{6} \; m \right)}^{2} \right) }

# F
3.62
×

10

19

N

{\displaystyle F = 3.62 \times {10}^{19} \; N }
Current Carrying Wires

## **Problem**

Consider two current-carrying wires, one with current

I
1

{\displaystyle I\_1}

, and one with current

I
2

{\displaystyle I\_2}

. Using the right-hand rule, one can determine the direction of the magnetic fields across each wire. The right-hand rule is also used to determine the direction of the magnetic forces exerted on each current-carrying wire. Try it!

**Part 1:** If the currents through each wire are travelling in the same direction, will the wires attract or repel?

**Part 2:** Determine the magnitude of the magnetic force density (magnetic force per unit length) exerted on each wire.

## **Solution**

**Part 1**

Refer to the diagram on the left. Using the right-hand rule, if the two currents are travelling in the same direction, the magnetic fields will be curling in the same direction (counter-clockwise in the diagram because both currents point up). Thus the generated forces are pointing towards each other, which means the wires will attract.

**Part 2**

First determine the magnitude of the magnetic field for each wire. One can use Ampere's Law:

∮
C

B

⋅
d
# l

μ
0

I

{\displaystyle \oint\_C \mathbf{B} \cdot d\boldsymbol{l} = \mu\_0 I }

B
1

2
π
# r

μ
0

I
1

{\displaystyle |\mathbf{B\_1}| 2\pi r = \mu\_0 I\_1 }

B
1

=

μ
0

I
1

2
π
r

.

{\displaystyle |\mathbf{B\_1}| = \frac{\mu\_0 I\_1}{2\pi r}. }

Similarly,

B
2

=

μ
0

I
2

2
π
r

.

{\displaystyle |\mathbf{B\_2}| = \frac{\mu\_0 I\_2}{2\pi r}. }

The magnetic force of a current-carrying wire is

F

=
I
l
×

B

.

{\displaystyle \mathbf{F} = I\boldsymbol{l} \times \mathbf{B}. }

The magnitude of the magnetic force is thus

F

=

I
l

B

sin
⁡
θ
.

Therefore, the magnetic force is

F

=

I
1

l

μ
0

I
2

2
π
r

sin
⁡

90

∘

and the magnetic force density (magnetic force per unit length) is

F

l

=

μ
0

I
1

I
2

2
π
r

.

{\displaystyle \frac{|\mathbf{F}|}{l} = \frac{\mu\_0 I\_1 I\_2}{2\pi r}. }
Empirical Formula

## **Problem**

A blue powdered substance is found to be made of 25.3% copper, 12.9% sulfur, 25.7% oxygen, and 36.1% water.

Determine the empirical formula and name of this compound.

## **Hints**

(1) Begin by considering only 100 grams of this compound.

(2) Construct a table that includes the mass, molar mass, and the number moles for each component of the compound.

(3) This compound is not dissolved in water.

## **Solution**

Consider only 100 grams of this mystery compound. Its contents would be 25.3 grams of copper , 12.9 grams of sulfur, 25.7 grams of oxygen, and 36.1 grams of water. With this information, construct a table including the molar masses of each component, and the number of moles (to be calculated).

 Molar mass (g/mol)
 Moles (mol)

To calculate the moles, simply divide the masses of each component by its respective molar mass.

 Molar mass (g/mol)
 Moles (mol)

There is approximately 0.400 moles of copper, 0.400 moles of sulfur, 1.6 moles of oxygen, and 2 moles of water. Look at the lowest number of moles, in this case copper and sulfur, and determine the multiplier.

To calculate the multiplier, simply divide 1 by the lowest number of moles.

1
0.4

=
2.5

{\displaystyle \frac{1}{0.4} = 2.5 }

Use this multiplier to determine the ratio of each component of this mystery compound.

Cu
:
S
:
O
:

H

2

O

=

1

:
1

:
4

:
5

{\displaystyle \ce{Cu:S:O:H2O = 1:1:4:5}}

Since this compound is not dissolved in water, it should be a hydrate; hence, the empirical formula is

CuSO

4

⋅
5

H

2

O

{\displaystyle \ce{CuSO4 \* 5H2O}}

or **copper (II) sulfate pentahydrate**.
Magnifying Glass

**By: Tao Steven Zheng (鄭濤)**

## **Problem**

A magnifying glass uses a convex lens to magnify objects. Suppose you find a magnifying glass that focuses sunlight about 2.5 cm from the lens. How far should you hold the lens in front of a book if you want to double the size of the type?

## **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, use the magnification formula to calculate the ratio of the distance of the image to the distance of the object.

# M
−

d
i

d
o

=

h
i

h
o

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

2

=
−

d
i

d
o

d
i

=
−
2

d
o

.

{\displaystyle \begin{align}
2 &= -\frac{d\_i}{d\_o} \\[5 pt]
d\_i &= -2d\_o.
\end{align} }

Next, use the thin lens formula to solve for the distance of the object

1

d
o

1

d
i

=

1
f

.

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

1

d
o

1

−
2

d
o

=

1

2.5

c
m

1

2

d
o

=

0.4

c
m

d
o

=
1.25

c
m

{\displaystyle \begin{align}
\frac{1}{d\_o} + \frac{1}{-2d\_o} &= \frac{1}{2.5 \; cm} \\[5 pt]
\frac{1}{2d\_o} &= \frac{0.4}{\; cm} \\[5 pt]
d\_o &= 1.25 \; cm
\end{align} }
Attractor Numbers

## **Problem**

This problem originated from my investigations on the Collatz Conjecture (December 23, 2010 - April 5, 2012). I defined two concepts called the **Attractor Numbers** and **Continuous Rays**: an *attractor number* is a positive integer greater than 4 in the form

6
n
4

{\displaystyle 6n + 4}

; a *continuous ray* is the set of positive integers that are powers of two.

Determine all attractor numbers that are also powers of two. In other words, solve the congruence

2
i

≡
4

(
mod

6
)

{\displaystyle 2^i \equiv 4 \pmod 6}

where are positive integers

i
>
2

{\displaystyle i > 2}

.

## **Solution**

Convert the congruence

2
i

≡
4

(
mod

6
)

{\displaystyle 2^i \equiv 4 \pmod 6}

into the equation

6
n
# 4

2
i

{\displaystyle 6n + 4 = 2^i}

. Hence,

n

=

2
i

−
4

6

=

4

i

/

2

−
4

6

=

2
3

(

4

i

/

2
−
1

−
1
)
.

{\displaystyle \begin{align}
n &= \frac{2^i - 4}{6} \\[5 pt]
&= \frac{4^{i/2} - 4}{6} \\[5 pt]
&= \frac{2}{3} (4^{i/2 - 1} - 1).
\end{align}}

Observe that this expression is a geometric series.

2
3

(

4

i

/

2
−
1

−
1
# )
2
(
1
4
16
⋯

4

i

/

2
−
2

)

{\displaystyle \frac{2}{3} (4^{i/2-1} - 1)= 2(1 + 4 + 16 + \cdots + 4^{i/2-2}) }

Make the substitution

# t
i

/

2
−
1

{\displaystyle t = i/2 -1}

; thus,

2
3

(

4
t

−
1
# )
2
(
1
4
16
⋯

4

t
−
1

)

{\displaystyle \frac{2}{3} (4^t - 1)= 2(1 + 4 + 16 + \cdots + 4^{t-1}) }

2
3

(

2

2
t

−
1
# )
2
(
1
4
16
⋯

2

2
t
−
2

)

{\displaystyle \frac{2}{3} (2^{2t} - 1)= 2(1 + 4 + 16 + \cdots + 2^{2t-2}) }

2
3

(

2

2
t

−
1
# )
2
8
32
⋯

2

2
t
−
1

.

{\displaystyle \frac{2}{3} (2^{2t} - 1)= 2 + 8 + 32 + \cdots + 2^{2t-1}. }

Therefore,

# k
0

t
−
1

2

2
k
1

.

{\displaystyle n = \sum\_{k=0}^{t-1} 2^{2k+1}.
}

Substitute this result into the first equation.

6

(

∑

# k
0

t
−
1

2

2
k
1

)

# 4

2
i

{\displaystyle 6\left(\sum\_{k=0}^{t-1} 2^{2k+1} \right) + 4 = 2^i}

6

(

2
3

(

2

2
t

−
1
)

)

4

=

2
i

4
(

2

2
t

−
1
)
4

=

2
i

2

2
t
2

=

2
i

{\displaystyle \begin{align}
6\left(\frac{2}{3} (2^{2t} - 1) \right) + 4 &= 2^i \\[5 pt]
4 (2^{2t} - 1) + 4 &= 2^i \\[5 pt]
2^{2t+2} &= 2^i \\[5 pt]

\end{align}}

We find that

# i
2
t
2

{\displaystyle i = 2t + 2}

, where

t
≥
1
,

t
∈

Z

{\displaystyle {t \ge 1, \, t \in \mathbb{Z}}}

.

## **Examples**

(
n
,
i
)

{\displaystyle (n,i)}

2
i

=
6
n
4

{\displaystyle 2^i = 6n + 4}

# k
0

t
−
1

2

2
k
1

{\displaystyle n = \sum\_{k=0}^{t-1} 2^{2k+1}}

(
2
,
4
)

{\displaystyle (2,4)}

2
4

=
6
⋅
2
4

{\displaystyle 2^4 = 6\cdot 2 + 4}

# 2

2
1

{\displaystyle 2 = 2^1}

(
10
,
6
)

{\displaystyle (10,6)}

2
6

=
6
⋅
10
4

{\displaystyle 2^6 = 6\cdot 10 + 4}

# 10

2
1

2
3

{\displaystyle 10 = 2^1 + 2^3}

(
42
,
8
)

{\displaystyle (42,8)}

2
8

=
6
⋅
42
4

{\displaystyle 2^8 = 6\cdot 42 + 4}

# 42

2
1

2
3

2
5

{\displaystyle 42 = 2^1 + 2^3 + 2^5}

(
170
,
10
)

{\displaystyle (170,10)}

2

10

=
6
⋅
170
4

{\displaystyle 2^{10} = 6\cdot 170 + 4}

# 170

2
1

2
3

2
5

2
7

{\displaystyle 170 = 2^1 + 2^3 + 2^5 + 2^7}
Balancing with Geometry

## **Problem**

An isosceles triangle is to be cut from a square plate of uniform density

ρ

{\displaystyle \rho}

, and width

a

{\displaystyle a}

 (as shown below). What is the height of the removed isosceles triangle such that the resulting shape balances at the tip of the triangle?

## **Solution**

Partition the shape into two zones: the rectangle, and the triangles. The pivot point is the tip of the removed isosceles triangle. The two triangles forms an isosceles triangle equal in area to the removed segment. Due to its symmetry, this problem is equivalent to inverting the triangle regions. Therefore, to balance the torques on both sides, we calculate:

τ

r
e
c
t
a
n
g
l
e

=

τ

t
r
i
a
n
g
l
e

.

{\displaystyle {\tau}\_{rectangle} = {\tau}\_{triangle}.}

To do so, we must know where the center-of-mass of the rectangle, and the center-of-mass of the isosceles triangle are located. Since the material is of uniform density, the torques are proportional to the areas of each shape. Let

a

{\displaystyle a}

 be the width of the square,

h

{\displaystyle h}

 be the height of the triangle, and

a
−
h

{\displaystyle a - h }

 be the height of the rectangle. The center-of-mass of an isosceles triangle is located at

1
3

h

{\displaystyle \frac{1}{3}h }

 from its base. Relative to the pivot, the center-of-mass for the isosceles triangle is

2
3

h

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

. It is visually obvious (perhaps intuitive is a better word) that the center-of-mass of the rectangle is located at

1
2

(
a
−
h
)

{\displaystyle \frac{1}{2}(a - h) }

. Consequently,

τ

r
e
c
t
a
n
g
l
e

=

τ

t
r
i
a
n
g
l
e

{\displaystyle {\tau}\_{rectangle} = {\tau}\_{triangle}}

ρ

(

1
2

a
h

)

(

2
3

h

)

=
ρ

(

a
(
a
−
h
)

)

(

1
2

(
a
−
h
)

)

.

{\displaystyle \rho \left(\frac{1}{2}ah\right) \left(\frac{2}{3}h\right) = \rho \left(a(a-h)\right) \left(\frac{1}{2}(a - h)\right).}

This torque balancing equation reduces to

1
3

h
2

=

1
2

(
a
−
h

)
2

{\displaystyle \frac{1}{3}h^2 = \frac{1}{2}(a - h)^2}

# 0
3

a
2

−
6
a
h
2

h
2

.

{\displaystyle 0 = 3a^2 - 6ah + 2h^2. }

The solutions for the height of the triangle is

# h
a

(

3
±

3

2

)

{\displaystyle h = a\left(\frac{3 \pm \sqrt{3}}{2}\right) }

. However, the plus sign solution makes no sense because it would imply that the pivot is located outside the plate. Therefore, the height of the isosceles triangle is

# h
a

(

3
−

3

2

)

.

{\displaystyle h = a\left(\frac{3 - \sqrt{3}}{2}\right). }
Euler's Gem

## **Problem**

First prove the formula

e

i
x

=
cos
⁡
(
x
)
i
sin
⁡
(
x
)

{\displaystyle {e}^{ix} = \cos(x) + i\sin(x) }

. Then discover the beautiful identity

e

i
π

# 1
0

{\displaystyle {e}^{i\pi} + 1 = 0 }

.

Hint: Use the Maclaurin series of exponential and sinusoidal functions.

## **Solution**

Begin with the Maclaurin series of

e
x

,
sin
⁡
(
x
)
,
cos
⁡
(
x
)

{\displaystyle e^x, \sin(x), \cos(x) }

.

e
x

=
1
x

x
2

2
!

x
3

3
!

.
.

# n
0

∞

x
n

n
!

{\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... = \sum\limits\_{n=0}^\infty \frac{x^n}{n!} }

sin
⁡
(
x
# )
x
−

x
3

3
!

x
5

5
!

−
.
.

# n
0

∞

(
−
1

)
n

x

2
n
1

(
2
n
1
)
!

{\displaystyle \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... = \sum\limits\_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!} }

cos
⁡
(
x
# )
1
−

x
2

2
!

x
4

4
!

−
.
.

# n
0

∞

(
−
1

)
n

x

2
n

(
2
n
)
!

{\displaystyle \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - ... = \sum\limits\_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!} }

Replace

x

{\displaystyle x}

 with

i
x

{\displaystyle ix}

, where

i

{\displaystyle i}

 is the imaginary unit.

e

i
x

=
1
i
x

(
i
x

)
2

2
!

(
i
x

)
3

3
!

(
i
x

)
4

4
!

(
i
x

)
5

5
!

.
.

# n
0

∞

(
i
x

)
n

n
!

{\displaystyle e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} +... = \sum\limits\_{n=0}^\infty \frac{(ix)^n}{n!} }

The powers of

i

{\displaystyle i}

 follow a pattern:

i

4
n

=
1

{\displaystyle {i}^{4n} = 1 }

, for

# n
0
,
1
,
2
,
.
.
.

{\displaystyle n = 0, 1, 2,...}

i

4
n
1

=
i

{\displaystyle {i}^{4n+1} = i }

, for

# n
0
,
1
,
2
,
.
.
.

{\displaystyle n = 0, 1, 2,...}

i

4
n
2

=
−
1

{\displaystyle {i}^{4n+2} = -1 }

, for

# n
0
,
1
,
2
,
.
.
.

{\displaystyle n = 0, 1, 2,...}

i

4
n
3

=
−
i

{\displaystyle {i}^{4n+3} = -i }

, for

# n
0
,
1
,
2
,
.
.
.

{\displaystyle n = 0, 1, 2,...}

Thus,

e

i
x

=
1
i
x
−

x
2

2
!

−

i

x
3

3
!

x
4

4
!

i

x
5

5
!

.
.

# n
0

∞

(
i
x

)
n

n
!

{\displaystyle e^{ix} = 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} +... = \sum\limits\_{n=0}^\infty \frac{(ix)^n}{n!} }

e

i
x

=

(

1
−

x
2

2
!

x
4

4
!

−
.
.
.

)

(

i
x
−

i

x
3

3
!

i

x
5

5
!

.
.
.

)

{\displaystyle e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -...\right) + \left(ix - \frac{ix^3}{3!} + \frac{ix^5}{5!} + ...\right) }

e

i
x

=

(

1
−

x
2

2
!

x
4

4
!

−
.
.
.

)

i

(

x
−

x
3

3
!

x
5

5
!

.
.
.

)

{\displaystyle e^{ix} = \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} -...\right) + i \left(x - \frac{x^3}{3!} + \frac{x^5}{5!} + ...\right) }

This shows that

e

i
x

=
cos
⁡
(
x
)
i
sin
⁡
(
x
)

{\displaystyle {e}^{ix} = \cos(x) + i\sin(x) }

.

Setting

# x
π

{\displaystyle x = \pi}

 yields

e

i
π

=
cos
⁡
(
π
)
i
sin
⁡
(
π
)

{\displaystyle {e}^{i\pi} = \cos(\pi) + i\sin(\pi) }

e

i
π

=
−
1

{\displaystyle {e}^{i\pi} = -1 }

.
Therefore,

e

i
π

# 1
0

{\displaystyle {e}^{i\pi} + 1 = 0 }

.
박상현

## 1. 기본 정보

15학번 경제학부. 2016년도 회장.

2016년 셰르파의 부흥에 정말 큰 역할을 담당한 위대한 학회장이다. 그 눈부신 후광으로 인해 도로교통법이 수정되었을 수 있으며 국제재판소에 이 문제가 회부되었을 수도 있다. ~~셰르파의 진흥왕~~

발제를 담당할 때면 늘 자연스러운 진행을 자랑한다. 물론 발제 담당이 아닐 때도 어느새 자연스러운 진행을 맡고 있는 경우가 있다. ~~그럼...거수를 한 번 해 볼까요?~~

'개돼지'로서의 민중 역할을 충실히 하며 사회 비판을 하기 위해모범적으로 놀고 먹는 학회장이다. 하지만 세미나와 티에스 등의 학회 행사들을 적시에 기획하고 일을 분배하는 철두철미한 ~~두려운~~ 면모 또한 가지고 있다. ~~군대를 다녀와서도 셰르파에 몸 담기 위해 이렇게 굴린다카더라~~ ~~일해라 기획단~~

셰르파에 이상한 문화를 전파시키고 있다. ~~전통을 만드는 중입니다만~~ ~~정말 어느새 전통으로 자리잡았다! 모 학우는 한 교양 수업에서 들은 '만들어진 전통'이 떠오른다고...~~

방수가 안 되는 가죽 자켓을 가지고 있다.

사투리 이야기가 나오면 갑작스레 표준어를 구사하기도 하니 너무 놀라지 말 것.~~정말 안 놀란다~~ ~~나 사투리 안 쓰지 않→아↗?~~

## 2. 발제

1학기 오픈 세미나: 헤겔&마르크스 (2016)

푸코&하버마스 (2016)

몽테스키외&토크빌 (2016)

여름 방학 TS 기획 ~~용사의 용사~~ (2016)

2학기 오픈 세미나: 앙가주망 (2016)
Jiuzhang Suanshu Shanggong 14

## **Problem**

答曰：四萬六千五百尺。

術曰：廣袤相乘，以高乘之，二而一。

Answer: 46500 [cubic] *chi*.

Method: Multiply the width by the length, then multiply by the height, divide by 2.

1 Literally translates to “moat rampart”.

## **Derivation**

The *qian du* (塹堵, “moat rampart”) is a triangular prism (Figure 1). It is formed by cutting rectangular prism in half along a diagonal plane; hence, its two parallel faces are right triangles (Figure 2). The volume of the rectangular prism is

# V
a
b
c

{\displaystyle V=abc }

, where

a
,
b
,
c

{\displaystyle a,b,c}

 denote the width, length, and height respectively. Thus, it follows that the volume of the right triangular prism is

V

q
i
a
n
d
u

=

1
2

a
b
c
.

{\displaystyle V\_{qiandu} = \frac{1}{2} abc. }

## **Calculation**

Convert the width, length, and height of the right triangular prism (塹堵 *qian du*) to *chi.*

**Width:** 2 *zhang* = 20 *chi*

**Length:** 18 *zhang* 6 *chi* = 186 *chi*

**Height:** 2 *zhang* 5 *chi* = 25 *chi*

Substitute the base length, and height of the rectangular prism into the formula.

V

=

1
2

a
b
c

=

1
2

×
20
×
186
×
25

=
46500

{\displaystyle \begin{align}
V &= \frac{1}{2}abc \\[5 pt]
& = \frac{1}{2} \times 20 \times 186 \times 25 \\[5 pt]
&= 46500
\end{align}}

The volume of the right triangular prism (塹堵 *qian du*) is 46500 cubic *chi*.
Jiuzhang Suanshu Shanggong 16

## **Problem**

答曰：二十三尺、少半尺。

術曰：廣袤相乘，以高乘之，六而一。

Answer:

23

1
3

{\displaystyle 23 \frac{1}{3}}

 [cubic] *chi*.

Method: Multiply the width by the length, then multiply by the height, divide by 6.

## **Derivation**

Two solids are produced by making an oblique cut into the *qian du* (塹堵), a triangular prism (Figure 1). The first solid is called the *yang ma* (陽馬), a rectangular pyramid. The second solid is called the *bie nao* (鱉臑), a tetrahedron. Given a *qian du* (triangular prism) of length, width, and height

a
,
b
,
c

{\displaystyle a,b,c}

, its volume is

V

q
i
a
n
d
u

=

1
2

a
b
c

{\displaystyle V\_{qiandu} = \frac{1}{2} abc}

. The volume of the *yang ma* (rectangular pyramid) cut from the *qian du* is

V

y
a
n
g
m
a

=

1
3

a
b
c

{\displaystyle V\_{yangma} = \frac{1}{3} abc}

. To obtain the volume of the *bie nao* (tetrahedron), subtract the volume of the *yang ma* from the *qian du*.

V

b
i
e
n
a
o

=

V

q
i
a
n
d
u

−

V

y
a
n
g
m
a

=

1
2

a
b
c
−

1
3

a
b
c

=

1
6

a
b
c

{\displaystyle \begin{align}
V\_{bienao} &= V\_{qiandu} - V\_{yangma} \\[5 pt]
&= \frac{1}{2} abc - \frac{1}{3} abc \\[5 pt]
&= \frac{1}{6} abc
\end{align} }

## **Calculation**

Substitute the base length, and height of the rectangular prism into the formula.

V

=

1
6

a
b
c

=

1
6

×
5
×
4
×
7

=

70
3

=
23

1
3

{\displaystyle \begin{align}
V &= \frac{1}{6}abc \\[5 pt]
& = \frac{1}{6} \times 5 \times 4 \times 7 \\[5 pt]
&= \frac{70}{3} \\[5 pt]
&= 23 \frac{1}{3}
\end{align}}

The volume of the tetrahedron (鱉臑 *bie nao*) is

23

1
3

{\displaystyle 23 \frac{1}{3}}

 cubic *chi*.
1D Random Walks with Python

**By: Tao Steven Zheng (郑涛)**

```
import numpy as np
import matplotlib.pyplot as plt

 def Randwalk(n):

    x = 0
    y = 0

    time = [x]
    position = [y]

    for i in range (1,n+1):
        move = np.random.uniform(0,1)

        if move < 0.5:
            x += 1
            y += 1

        if move > 0.5:
            x += 1
            y += -1

        time.append(x)
        position.append(y)

    return [time,position]

'''Trials and plots'''
 Randwalk1 = Randwalk(1000)
 Randwalk2 = Randwalk(1000)
 Randwalk3 = Randwalk(1000)
 plt.plot(Randwalk1[0],Randwalk1[1],'r-', label = "Randwalk1")
 plt.plot(Randwalk2[0],Randwalk2[1],'g-', label = "Randwalk2")
 plt.plot(Randwalk3[0],Randwalk3[1],'b-', label = "Randwalk3")
 plt.title("1-D Random Walks")
 plt.legend(loc='upper center', bbox\_to\_anchor=(0.5,-0.1), fancybox=True, shadow=True)
 plt.show()

```
Haidao Suanjing 1

## **Problem**

   今有望海島，立兩表，齊高三丈，前後相去千步，令後表與前表參相直。從前表卻行一百二十三步，人目著地取望島峯，與表末參合。從後表卻行一百二十七步，人目著地取望島峯，亦與表末參合。問：島高及去表各幾何？

When surveying a sea island, erect two poles, each 3 zhang tall, so that the front and rear poles are 1000 bu apart. They are aligned with the summit of the island. Move back 123 bu from the front pole, sighting at ground level, and find that the summit of the island coincides with the tip of the pole. Move back 127 bu from the rear pole, sighting at ground level, and find that the summit of the island also coincides with the tip of the pole. What is the height of the island, and its distance from the front pole?

[Note: 3 zhang =  5 bu]

## **Solution**

Let

h

{\displaystyle h}

 be the height of each pole.

Let

d

{\displaystyle d}

 be the distance between the two poles.

Let

s
1

{\displaystyle s\_1}

 be the length of the shadow of pole 1.

Let

s
2

{\displaystyle s\_2}

 be the length of the shadow of pole 2.

Let

H

{\displaystyle H}

 be the height of the island summit.

Let

D

{\displaystyle D}

 be the distance of the island to pole 1.

By proportion

H
−
h

h

=

D

s

1

=

D
d

s

2

{\displaystyle {\frac {H-h}{h}}={\frac {D}{s\_{1}}}={\frac {D+d}{s\_{2}}}}

.
Subsequently

h
# d

s

1

(
H
−
h
)

{\displaystyle hd=s\_{1}(H-h)}

and

h
(
D
d
# )

s

2

(
H
−
h
)

{\displaystyle h(D+d)=s\_{2}(H-h)}

.
Subtracting the two equations yields

h
# d
(

s

2

−

s

1

)
(
H
−
h
)

{\displaystyle hd=(s\_{2}-s\_{1})(H-h)}

,
or

H
−
h

h

=

d

s

2

−

s

1

{\displaystyle {\frac {H-h}{h}}={\frac {d}{s\_{2}-s\_{1}}}}

.
Consequently,

H
−
h

h

=

D

s

1

=

d

s

2

−

s

1

{\displaystyle {\frac {H-h}{h}}={\frac {D}{s\_{1}}}={\frac {d}{s\_{2}-s\_{1}}}}

.
The height of the island is

# H

h
d

s

2

−

s

1

h

{\displaystyle H={\frac {hd}{s\_{2}-s\_{1}}}+h}

,
and the distance of the island to pole 1 is

# D

s

1

d

s

2

−

s

1

{\displaystyle D={\frac {s\_{1}d}{s\_{2}-s\_{1}}}}

.

# H

5
(
1000
)

127
−
123

# 5
1255

{\displaystyle H={\frac {5(1000)}{127-123}}+5=1255}

# D

123
(
1000
)

127
−
123

=
30750

{\displaystyle D={\frac {123(1000)}{127-123}}=30750}

The height of the island is 1255 bu; the distance from the front pole is 30750 bu.
Arithmetic Triangle

## **Problem**

The **Arithmetic Triangle** is a diagram that illustrates the binomial coefficients of the expansion of

(
a
b
)

n

{\displaystyle {(a+b)}^{n} }

. Jia Xian (fl. 1050 AD) is known to have discussed the arithmetic triangle in his works. The combinatorial formula for calculating the binomial coefficients in the arithmetic triangle is

(

n
k

)

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \binom n k =\frac{n!}{k!(n-k)!}}

where

n

{\displaystyle n}

 is the row and

k

{\displaystyle k}

 is the column. Notice that by adding two adjacent coefficients in a row results in the coefficient below the two:

(

n
−
1

k

)

(

n
−
1

k
−
1

)

=

(

n
k

)

{\displaystyle \binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k} }

Prove this fact.

## **Solution**

(
n
−
1
)
!

k
!
(
n
−
1
−
k
)
!

(
n
−
1
)
!

(
k
−
1
)
!
(
n
−
k
)
!

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \frac{(n-1)!}{k!(n-1-k)!} + \frac{(n-1)!}{(k-1)!(n-k)!} = \frac{n!}{k!(n-k)!} }

(
n
−
1
)
!

k
(
k
−
1
)
!
(
n
−
1
−
k
)
!

(
n
−
1
)
!

(
k
−
1
)
!
(
n
−
k
)
(
n
−
1
−
k
)
!

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \frac{(n-1)!}{k(k-1)!(n-1-k)!} + \frac{(n-1)!}{(k-1)!(n-k)(n-1-k)!} = \frac{n!}{k!(n-k)!} }

(
n
−
1
)
!

(
k
−
1
)
!
(
n
−
1
−
k
)
!

[

1
k

1

n
−
k

]

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \frac{(n-1)!}{(k-1)!(n-1-k)!} \left[\frac{1}{k} + \frac{1}{n-k}\right] = \frac{n!}{k!(n-k)!} }

(
n
−
1
)
!

(
k
−
1
)
!
(
n
−
1
−
k
)
!

[

n

k
(
n
−
k
)

]

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \frac{(n-1)!}{(k-1)!(n-1-k)!} \left[\frac{n}{k(n-k)} \right] = \frac{n!}{k!(n-k)!} }

n
(
n
−
1
)
!

k
(
k
−
1
)
!
(
n
−
k
)
(
n
−
1
−
k
)
!

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \frac{n(n-1)!}{k(k-1)!(n-k)(n-1-k)!} = \frac{n!}{k!(n-k)!} }

n
!

k
!
(
n
−
k
)
!

=

n
!

k
!
(
n
−
k
)
!

{\displaystyle \frac{n!}{k!(n-k)!} = \frac{n!}{k!(n-k)!} }

Therefore,

(

n
−
1

k

)

(

n
−
1

k
−
1

)

=

(

n
k

)

{\displaystyle \binom{n-1}{k} + \binom{n-1}{k-1} = \binom{n}{k} }

## **Historical Note**

The Arithmetic Triangle has many names. In China, it is called Jia Xian's Triangle or Yang Hui's Triangle, named after the mathematicians Jia Xian (c. 1010 - 1070 AD), who first discovered it, and Yang Hui (1238 - 1298 AD), who commented on Jia Xian's lost works. In Iran, it is called the Khayyam's Triangle, named after the Persian astronomer and poet Omar Khayyam (1048 - 1131 AD). In Europe, it is called the Pascal's Triangle, named after the French mathematician, inventor, and philosopher Blaise Pascal (1623 - 1662 AD).