recursal/Fanatic-Fandom huggingface dataset:
Atmospheric Thermodynamics

## **Problem**

Consider the following information.

**Solar constant:**

# S
1370

W

m
2

{\displaystyle S = 1370 \frac{W}{m^2} }

**Albedo of the Earth:**

# α
0.3

{\displaystyle \alpha = 0.3 }

**Emissivity of the Earth:**

# e
1

{\displaystyle e = 1}

 (Blackbody)

**Stefan-Boltzmann constant:**

# σ
5.67
×

10

−
8

{\displaystyle \sigma = 5.67 \times {10}^{-8} }

**Part 1:** Calculate the surface temperature of the Earth

T

{\displaystyle T}

 with the absence of the atmosphere. Note that the surface temperature is labelled

T
S

{\displaystyle T\_S}

 in the above diagram.

**Part 2:** Calculate the surface temperature of the Earth

T

{\displaystyle T}

 with the atmosphere present.

## **Solution**

**Part 1**

Without an atmosphere the power of solar radiation absorbed by the surface of the Earth is equal to the power radiated by the Earth.

(
1
−
α
)
S
π

R
2

=
e
σ
4
π

R
2

T
4

{\displaystyle (1-\alpha)S \pi R^2 = e \sigma 4 \pi R^2 T^4 }

# T

(
1
−
0.3
)
(
1370
)

4
(
5.67
×

10

−
8

)

4

=
255

K

{\displaystyle T = \sqrt[4]{\frac{(1-0.3)(1370)}{4(5.67 \times {10}^{-8})}} = 255\, K }

This temperature is

−

18

∘

C

{\displaystyle -{18}^{\circ} C }

, which is an ice world!

**Part 2**

Add an atmosphere. The power of solar radiation absorbed by the atmosphere of the Earth is equal to the power radiated by the atmosphere. Thus,

(
1
−
α
)
S
π

R
2

=
e
σ
4
π

R
2

T

A

4

.

{\displaystyle (1-\alpha)S \pi R^2 = e \sigma 4 \pi R^2 {{T}\_{A}}^{4}. }

This is the same calculation as in part 1, so

T
A

=
255
K

{\displaystyle T\_A = 255 K }

.

Now, the power of the solar radiation still reaches the surface of the Earth alongside and the power radiated by the atmosphere. Therefore, we tweak the equation from part 1 by adding an atmospheric radiation term.

(
1
−
α
)
S
π

R
2

e
σ
4
π

R
2

T

A

4

=
e
σ
4
π

R
2

T
4

{\displaystyle (1-\alpha)S \pi R^2 + e \sigma 4 \pi R^2 {{T}\_{A}}^{4} = e \sigma 4 \pi R^2 T^4 }

Since

(
1
−
α
)
S
π

R
2

=
e
σ
4
π

R
2

T

A

4

{\displaystyle (1-\alpha)S \pi R^2 = e \sigma 4 \pi R^2 {{T}\_{A}}^{4} }

,

2

(

e
σ
4
π

R
2

T

A

4

)

=
e
σ
4
π

R
2

T
4

{\displaystyle 2\left(e \sigma 4 \pi R^2 {{T}\_{A}}^{4}\right) = e \sigma 4 \pi R^2 T^4 }

# T

2
4

T

A

{\displaystyle T = \sqrt[4]{2} {T}\_{A} }

# T

2
4

(
255
# )
303

K
.

{\displaystyle T = \sqrt[4]{2}(255) = 303\, K. }

This temperature is around

30

∘

C

{\displaystyle {30}^{\circ} C }

, which is too hot! Sure, some places on Earth can be hotter than this temperature, but the average surface temperature is only around

14.5

∘

C

{\displaystyle {14.5}^{\circ} C }

. One possible source of error in our calculation in part 2 is the amount of solar radiation that reaches the surface of the Earth, but there could be a plethora of other causes such as other thermal transfer effects.
Keplar's Third Law

## **Problem**

Derive Keplar’s third law of planetary orbits for comparing the periods and orbital radius of two planets. This law is sometimes called the harmonic law.

(

T
1

T
2

)

2

=

(

r
1

r
2

)

3

{\displaystyle \left(\frac{T\_1}{T\_2}\right)^2=\left(\frac{r\_1}{r\_2}\right)^3 }

Assume the orbits are circles.

## **Solution**

The net force in orbital motion is the gravitational force

F
g

=

G
M
m

r
2

{\displaystyle F\_g=\frac{GMm}{r^2} }

.
Since the net force is the centripetal force for planetary orbits,

F
g

=

F
c

{\displaystyle F\_g=F\_c }

G
M
m

r
2

=

m
4

π
2

r

T
2

{\displaystyle \frac{GMm}{r^2} =\frac{m4 \pi^2 r}{T^2} }

G
M

4

π
2

T
2

=

r
3

{\displaystyle \frac{GM}{4 \pi^2} T^2=r^3 }

.
For a planet with an orbital radius

r
1

{\displaystyle r\_1 }

G
M

4

π
2

T

1

2

=

r

1

3

{\displaystyle \frac{GM}{4 \pi^2} {T}\_{1}^2={r}\_{1}^3 }

.
For a planet with an orbital radius

r
2

{\displaystyle r\_2 }

G
M

4

π
2

T

2

2

=

r

2

3

{\displaystyle \frac{GM}{4 \pi^2} {T}\_{2}^2={r}\_{2}^3 }

.
Divide the two equations to discover that

(

T
1

T
2

)

2

=

(

r
1

r
2

)

3

{\displaystyle \left(\frac{T\_1}{T\_2}\right)^2=\left(\frac{r\_1}{r\_2}\right)^3 }

.
## **Note**

Most planetary orbits are elliptical, not circular. Although the derivation uses more advanced mathematics, the final result is not that different.

G
(
M
m
)

4

π
2

T
2

=

a
3

{\displaystyle \frac{G(M + m)}{4 \pi^2} T^2=a^3 }

Here,

a

{\displaystyle a}

 is the semi-major axis of the ellipse. Thus the Harmonic law would be

(

T
1

T
2

)

2

=

(

a
1

a
2

)

3

{\displaystyle \left(\frac{T\_1}{T\_2}\right)^2=\left(\frac{a\_1}{a\_2}\right)^3 }

.
Elasticity of Demand

## **Economics Concept**

The **elasticity of demand**

ϵ

{\displaystyle \epsilon}

 is a measure of how sensitive the quantity demanded is to its price. When the price rises, quantity demanded falls for almost any good (hence negative), but it falls more for some than for others. The price elasticity gives the percentage change in quantity demanded when there is a one percent increase in price, holding everything else constant. For example, if the elasticity is −2, that means a one percent increase in the price leads to a two percent decrease in the quantity demanded.

## **Problem**

Let the price of a good be

p

{\displaystyle p}

, the quantity of a good be

# q
q
(
p
)

{\displaystyle q = q(p) }

, and the revenue be

# R
p
q

{\displaystyle R = pq }

. Show that revenue is maximized if

p
q

d
q

d
p

=
−
1

{\displaystyle \frac{p}{q}\frac{dq}{dp} = -1 }

.

## **Solution**

Begin by taking the derivative of the revenue with respect to price:

# R
p
q

{\displaystyle R = pq }

d
R

d
p

=
q
p

d
q

d
p

{\displaystyle \frac{dR}{dp} = q + p\frac{dq}{dp} }

d
R

d
p

=
q

(

1

p
q

d
q

d
p

)

{\displaystyle \frac{dR}{dp} = q\left( 1 + \frac{p}{q}\frac{dq}{dp} \right) }

.
Set the derivative equal to zero

# 0
q

(

1

p
q

d
q

d
p

)

{\displaystyle 0 = q\left( 1 + \frac{p}{q}\frac{dq}{dp} \right) }

and analyze the result.

When quantity equals zero (

# q
0

{\displaystyle q=0}

), the revenue is minimized because there are no sales; thus, revenue is maximized when

p
q

d
q

d
p

=
−
1

{\displaystyle \frac{p}{q}\frac{dq}{dp} = -1 }

.

# ϵ

p
q

d
q

d
p

{\displaystyle \epsilon = \frac{p}{q}\frac{dq}{dp} }

.

When

ϵ

=
1

{\displaystyle |\epsilon| = 1 }

, the good is unit elastic. This means a 1% change in price will cause a 1% change in quantity demanded. Revenue is maximized.

When

ϵ

>
1

{\displaystyle |\epsilon| > 1 }

, the good is price elastic. This means a 1% change in price will cause more than 1% change in quantity demanded. Decrease price to increase revenue.

When

ϵ

<
1

{\displaystyle |\epsilon| < 1 }

, the good is price inelastic. This means a 1% change in price will cause less than 1% change in quantity demanded. Increase price to increase revenue.
Jiuzhang Suanshu Shanggong 10

## **Problem**

答曰：一十萬一千六百六十六尺、太半尺。

術曰：上、下方相乘，又各自乘，並之，以高乘之，三而一。

Answer:

101666

2
3

{\displaystyle 101666\frac{2}{3} }

 [cubic] *chi*.

Method: Multiply the edge of the upper square by the edge of the lower square, then multiply each of them by itself and add all the products, then multiply by the height, divide by 3.

1 Literally translates to “square pavilion”.

## **Derivation**

This problem from the *Jiuzhang Suanshu* introduces a solid called the *fang ting* (方亭, “square pavilion”), a pyramidal frustum, which is formed by truncating a square pyramid by a plane parallel to the base. Let

a
,
b
,
h

{\displaystyle a, b, h }

 denote the the width at the top base, the width of the bottom base, and the height respectively. The *Jiuzhang Suanshu* gives the correct volumetric formula:

V

f
a
n
g
t
i
n
g

=

1
3

(

a
2

a
b

b
2

)

h
.

{\displaystyle V\_{fangting} = \frac{1}{3} \left(a^2 + ab + b^2 \right)h. }

Here I present Liu Hui's derivation. The frustum of a square pyramid can be divided into nine sections of three types of volumes: one rectangular prism (方堡壔 *fang bao dao*, "square watchtower"), four triangular prisms (塹堵 *qian du*, "moat rampart"), and four square pyramids (陽馬 *yang ma*, "male horse").

V

f
a
n
g
b
a
o
d
a
o

=

a
2

h

{\displaystyle V\_{fangbaodao} = a^2 h }

V

q
i
a
n
d
u

=

1
2

a

(

b
−
a

2

)

h

{\displaystyle V\_{qiandu} = \frac{1}{2} a\left(\frac{b-a}{2}\right)h }

V

y
a
n
g
m
a

=

1
3

(

b
−
a

2

)

2

h

{\displaystyle V\_{yangma} = \frac{1}{3} \left(\frac{b-a}{2}\right)^2 h }

Subsequently, the total volume is

V

f
a
n
g
t
i
n
g

=

V

f
a
n
g
b
a
o
d
a
o

4
×

V

q
i
a
n
d
u

4
×

V

y
a
n
g
m
a

=

a
2

h
4
×

1
4

a
(
b
−
a
)
h
4
×

1
12

(
b
−
a

)
2

h

=

a
2

h
a
(
b
−
a
)
h

1
3

(
b
−
a

)
2

h

=

(

a
2

a
(
b
−
a
)

b
2

−
2
a
b

a
2

3

)

h

=

1
3

(

a
2

a
b

b
2

)

h
.

{\displaystyle \begin{align}
V\_{fangting} &= V\_{fangbaodao} + 4 \times V\_{qiandu} + 4 \times V\_{yangma} \\[5 pt]
&= a^2h + 4 \times \frac{1}{4}a(b-a)h + 4 \times \frac{1}{12}(b-a)^2 h \\[5 pt]
&= a^2h + a(b-a)h + \frac{1}{3}(b-a)^2h \\[5 pt]
&= \left(a^2 + a(b-a) +\frac{b^2 - 2ab +a^2}{3}\right)h \\[5 pt]
&= \frac{1}{3} \left(a^2 + ab + b^2 \right)h.
\end{align} }

## **Calculation**

Convert the edge of the lower square base, edge of the upper square base, and height of the pyramidal frustum (方亭*fang ting*) to *chi.*

**Lower edge:** 5 *zhang* = 50 *chi*

**Upper edge:** 4 *zhang* = 40 *chi*

**Height:** 5 *zhang* = 50 *chi*

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

V

=

1
3

(

a
2

a
b

b
2

)

h

=

1
3

×

(

40
2

40
×
50

50
2

)

×
50

=

1
3

×
6100
×
50

=

305000
3

=
101666

2
3

.

{\displaystyle \begin{align}
V &= \frac{1}{3} \left(a^2 + ab + b^2 \right)h \\[5 pt]
&= \frac{1}{3} \times \left(40^2 + 40 \times 50 + 50^2 \right) \times 50 \\[5 pt]
&= \frac{1}{3} \times 6100 \times 50 \\[5 pt]
&= \frac{305000}{3} \\[5 pt]
&= 101666\frac{2}{3}.
\end{align} }

The volume of the pyramidal frustum (方亭 *fang ting*) is

101666

2
3

{\displaystyle 101666\frac{2}{3} }

 *cubic chi*.
Jiuzhang Suanshu Fangtian 31

## **Problem**

答曰：七十五步。

Answer: 75 [square] *bu*.

## **Solution**

The *Jiuzhang Suanshu* states four methods for computing the area of a circle:

又術曰：周徑相乘，四而一。
又術曰：徑自相乘，三之，四而一。
又術曰：周自相乘，十二而一。
Another method: Multiply the circumference and the diameter, divide by 4.
Another method: Multiply the diameter by itself, multiply by 3, and divide by 4.
Another method: Multiply the circumference by itself, divide by 12.

In modern mathematical notation, the formulas are:

(1)

# A

C
2

⋅

d
2

{\displaystyle A = \frac{C}{2} \cdot \frac{d}{2} }

(2)

# A

C
d

4

{\displaystyle A = \frac{Cd}{4} }

(3)

# A

3

d
2

4

{\displaystyle A = \frac{3d^2}{4} }

(4)

# A

C
2

12

{\displaystyle A = \frac{C^2}{12} }

where

A
,
C
,
d

{\displaystyle A, C, d }

represent the area, circumference, and diameter of the circle respectively.

Methods (1) and (2) are identical formulas and give the exact area if the circumference is indeed

π

{\displaystyle \pi }

 times the diameter. However, a close inspection of the given measurements of the circumference and diameter in each problem indicates that the original authors of the *Jiuzhang Suanshu* took

# π
3

{\displaystyle \pi = 3 }

. This crude approximation for

π

{\displaystyle \pi }

 is then used to express the formulas for methods (3) and (4). Had they expressed the formulas in terms of

π

{\displaystyle \pi }

 where

π
≠
3

{\displaystyle \pi \ne 3 }

, the last two formulas would be exact:

(3)

# A

π

d
2

4

{\displaystyle A = \frac{\pi d^2}{4} }

(4)

# A

C
2

4
π

.

{\displaystyle A = \frac{C^2}{4\pi}. }

Substitute the given circumference and diameter into each area formula.

**Formula 1**

A

=

30

b
u

2

×

10

b
u

2

=
15

b
u
×
5

b
u

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{30 \; bu}{2} \times \frac{10 \; bu}{2} \\
&= 15 \; bu \times 5 \; bu \\
& = 75 \; bu^2
\end{align}}

**Formula 2**

A

=

30

b
u
×
10

b
u

4

=

300

b

u
2

4

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{30 \; bu \times 10 \; bu}{4} \\
&= \frac{300 \; bu^2}{4} \\
& = 75 \; bu^2
\end{align}}

**Formula 3**

A

=

3
×
(
10

b
u

)
2

4

=

3
×
100

b

u
2

4

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{3\times (10 \; bu)^2}{4} \\
&= \frac{3 \times 100 \; bu^2}{4} \\
& = 75 \; bu^2
\end{align}}

**Formula 4**

A

=

(
30

b
u

)
2

12

=

900

b

u
2

12

=
75

b

u
2

{\displaystyle \begin{align}
A &= \frac{(30 \; bu)^2}{12} \\
&= \frac{900 \; bu^2}{12} \\
& = 75 \; bu^2
\end{align}}

Since the formulas equivalent, each method gives the same answer. If only the given circumference is taken to calculate the precise area of this circle, its true area would be, to 2 decimal places,

71.62

b

u
2

{\displaystyle 71.62 \; bu^2 }

. If only the given diameter is used to calculate the precise area of this circle, its true area would be, to 2 decimal places,

78.54

b

u
2

{\displaystyle 78.54 \; bu^2 }

.
Jiuzhang Suanshu Gougu 16

## **Problem**

This classic problem from the *Jiuzhang Suanshu* (九章算術) is given the name "circle in a right triangle" (勾股容圓 gōu gǔ róng yuán).

答曰：六步。

術曰：八步為勾，十五步為股，為之求弦。三位並之為法，以勾乘股，倍之為實。實如法得徑一步。

Answer: 6 *bu*.

Method: Let the *gou* be 8 *bu*, the *gu* be 15 *bu*, and use these to find the *xian*. Sum all three and let this result be the divisor. Multiply the *gou* with the *gu*, then double, and let this result be the dividend. Divide the dividend by the divisor to obtain the diameter in *bu*.

## **Solution**

Here I will present Liu Hui's derivations of three formulas for computing the diameter of the inscribed circle. Liu Hui's derivation of the first two formulas uses this diagram.

**Formula 1**

Let

a

{\displaystyle a}

 be the shorter leg (勾 *gou*),

b

{\displaystyle b}

 be the longer leg (股 *gu*), and

c

{\displaystyle c}

 be the hypotenuse (弦 *xian*). Let

r

{\displaystyle r}

 be the radius and

d

{\displaystyle d}

 be the diameter of the inscribed circle.

Dissect the right triangle into 6 pieces: 2 red triangles, 2 blue triangles, and 2 yellow triangles. Then combine each pair of triangles of the same color to form rectangles (the yellow triangles form a square). Copy each rectangle four times and then assemble them to form one large rectangle. Use this large rectangle to derive three different formulas for calculating the diameter of the inscribed circle.
The area of a large rectangle is equal to the area of four right triangles:

d
×
(
a
b
c
# )
4
×

1
2

a
b

{\displaystyle
d \times (a+b+c) = 4\times \frac{1}{2} ab
}

d
(
a
b
c
# )
2
a
b

{\displaystyle
d(a+b+c) = 2ab
}

# d

2
a
b

a
b
c

.

{\displaystyle
d = \frac{2ab}{a+b+c}.
}

**Calculation**

First, compute the length of the hypotenuse.

# c

8

2

15

2

=
17

{\displaystyle
c = \sqrt{{8}^{2} + {15}^{2}} = 17
}

# d

2
×
8
×
15

8
15
17

{\displaystyle
d = \frac{2 \times 8 \times 15}{8+15+17}
}

# d
6

{\displaystyle d = 6 }

The diameter of the inscribed circle is 6 *bu*.

**Formula 2**

It can be seen immediately from that:

# c
(
a
−
r
)
(
b
−
r
)

{\displaystyle c = (a-r)+(b-r) }

c
2
# r
a
b

{\displaystyle c + 2r = a+b }

c
# d
a
b

{\displaystyle c+d = a+b }

# d
a
b
−
c
.

{\displaystyle d = a + b - c. }

**Calculation**

# d
a
b
−
c

{\displaystyle d = a + b -c }

# d
8
15
−
17

{\displaystyle d = 8 + 15 - 17 }

# d

{\displaystyle d = 6. }

**Formula 3**

From the Door and Pole problem (Jiuzhang Suanshu Gougu 12), it was derived that

a
b
−
# c

2
(
c
−
a
)
(
c
−
b
)

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

.

Thus,

# d

2
(
c
−
a
)
(
c
−
b
)

.

{\displaystyle d = \sqrt{2(c-a)(c-b)}. }

**Calculation**

# d

2
(
17
−
8
)
(
17
−
15
)

{\displaystyle d = \sqrt{2(17-8)(17-15)} }

# d
6

{\displaystyle d = 6 }

For a derivation of the Pythagorean theorem (gougu theorem), check out this page!
Forms of Formation

## **Problem**

Identify the energetic chemical phenomena behind each equation.

(1)

Br

2

(
g
)

⟶
2

Br

(
g
)

{\displaystyle \ce{{Br2}\_{(g)} -> 2{Br}\_{(g)}}}

(2)

K

(
s
)

⟶

K

(
g
)

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

(3)

K

(
g
)

Br

−

(
g
)

⟶

KBr

(
s
)

{\displaystyle \ce{{K^+}\_{(g)} + {Br^-}\_{(g)} -> {KBr}\_{(s)}}}

(4)

K

(
g
)

⟶

K

(
g
)

e

−

{\displaystyle \ce{{K}\_{(g)} -> {K^+}\_{(g)} + e^-}}

(5)

KBr

(
s
)

⟶

K

(
aq
)

Br

−

(
aq
)

{\displaystyle \ce{{KBr}\_{(s)} -> {K^+}\_{(aq)} + {Br^-}\_{(aq)}}}

(6)

Br

(
g
)

e

−

⟶

Br

−

(
g
)

{\displaystyle \ce{{Br}\_{(g)} + e^- -> {Br^-}\_{(g)}}}

(7)

K

(
s
)

1

2

Br

2

(
g
)

⟶

KBr

(
s
)

{\displaystyle \ce{{K}\_{(s)} + \frac{1}{2} {Br\_2}\_{(g)} -> {KBr}\_{(s)}}}

## **Solution**

(1) Heat of dissociation

(2) Heat of sublimation (enthalphy of sublimation)

(3) Lattice energy

(4) Ionization energy

(5) Heat of solution (enthalphy of solution)

(6) Electron affinity

(7) Heat of formation (enthalphy of formation)
Graphing Sinusoidal Functions with Python

**By: Tao Steven Zheng (郑涛)**

## **Code**

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

x = np.linspace(-3\*np.pi,3\*np.pi,200)
y = np.sin(x)
x1 = np. linspace(-3\*np.pi,3\*np.pi,200)
y1 = 2 \* np.cos(np.pi\*x)

plt.plot(x, y, 'r-')
plt.plot(x1, y1,'b-')
plt.axis([-3\*np.pi,3\*np.pi, -3, 3]
plt.xlabel('x') #x-axis label
plt.ylabel('y') #y-axis label
plt.title('Plot of two sinusoidal functions')
plt.show()

```
Dispersion Relation

## **Problem**

Use the de Broglie relations to obtain the dispersion relation of matter waves. The result should show that

ω

{\displaystyle \omega}

 is proportional to

k

2

{\displaystyle {k}^{2} }

.

## **Solution**

Elementary particles, atomic nuclei, atoms, and even molecules behave in some contexts as matter waves. According to the de Broglie relations, their kinetic energy

E

{\displaystyle E}

 can be expressed in terms of the angular frequency

ω

{\displaystyle \omega}

, and their momentum

p

{\displaystyle p}

 in terms of the wavenumber

k

{\displaystyle k}

:

E

=
ℏ
ω

p

=
ℏ
k

{\displaystyle \begin{align}
E &= \hbar\omega \\[5 pt]
p &= \hbar k \end{align} }

Since

# E

p
2

2
m

{\displaystyle E=\frac{p^2}{2m}}

, we get

ℏ
# ω

(
ℏ
k

)
2

2
m

{\displaystyle \hbar\omega = \frac{(\hbar k)^2}{2m} }

ℏ
# ω

ℏ
2

k
2

2
m

{\displaystyle \hbar\omega = \frac{\hbar^2 k^2}{2m} }

# ω

ℏ

k
2

2
m

.

{\displaystyle \omega = \frac{\hbar k^2}{2m}. }

Evidently, the angular frequency is proportional to the wavenumber squared.
Jiuzhang Suanshu Gougu 6

This classic problem from the *Jiuzhang Suanshu* (九章算术) is called **Reed in a Pond** (池中葭).

## **Problem**

今有池方一丈，葭生其中央，出水一尺。引葭赴岸，適與岸齊。問：水深、葭長各幾何？

Suppose there is a square pond measuring 1 *zhang* wide. A reed grows at the center of the pond with 1 *chi* extending above the water. When the reed is drawn towards the edge of the pond, the tip just reaches the edge. Question: What is the depth of the pond and the length of the reed?

## **Solution**

Draw a right triangle such that the reed drawn towards the edge of then pond is the hypotenuse. Let

# a
5

{\displaystyle a = 5}

 be the distance from the centre of the pond to the edge of the pond. Let

c

{\displaystyle c}

 be the length of the entire reed, and

b

{\displaystyle b}

 be the segment of the reed submerged underwater (this is also the depth of the pond). The length of the reed is 1 chi above the surface of the pond, thus it follows that

c
−
# b
1

{\displaystyle c - b = 1 }

.

Substitute

# c
b
1

{\displaystyle c = b + 1 }

 into

a

2

b

2

=

c

2

{\displaystyle {a}^{2} + {b}^{2} = {c}^{2} }

.

a

2

b

2

=

(
b
1
)

2

{\displaystyle
{a}^{2} + {b}^{2} = {(b + 1)}^{2}
}

a

2

b

2

=

b

2

2
b
1

{\displaystyle
{a}^{2} + {b}^{2} = {b}^{2} + 2b + 1
}

# b

a

2

−
1

2

{\displaystyle
b = \frac{{a}^{2} - 1}{2}
}

**Calculation**

Since

# a
5

{\displaystyle a = 5}

,

# b

5

2

−
1

2

=
12

{\displaystyle
b = \frac{{5}^{2} - 1}{2} = 12
}

# c
12
# 1
13

{\displaystyle
c = 12 + 1 = 13
}

The depth of the pool is 12 chi, and the height of the reed is 13 chi.

**Traditional Chinese Solution**

As shown in Figure 5-5, let

a

{\displaystyle a}

 represent half the width of the pond,

b

{\displaystyle b}

 the depth of the pool,

c

{\displaystyle c}

 the length of reed, and

c
−
b

{\displaystyle c-b}

 the length of reed extending above the water. The formula given in *Jiuzhang Suanshu* is:

# b

a
2

−

(
c
−
b
)

2

2
(
c
−
b
)

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

To derive the above formula, rewrite the Pythagorean theorem as

a
2

=

c
2

−

b
2

{\displaystyle a^2 = c^2 - b^2 }

a
2

=
(
c
b
)
(
c
−
b
)

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

c
# b

a
2

(
c
−
b
)

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

Substituting this formula into

# b

1
2

(
c
b
)
−

1
2

(
c
−
b
)

{\displaystyle b = \frac{1}{2}(c+b) - \frac{1}{2}(c-b)}

gives

# b

1
2

(

a
2

(
c
−
b
)

)

−

1
2

(
c
−
b
)

{\displaystyle b =\frac{1}{2}\left(\frac{a^2}{(c-b)} \right) - \frac{1}{2}(c-b) }

# b

a
2

−

(
c
−
b
)

2

2
(
c
−
b
)

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

**Calculation**

# b

a
2

−

(
c
−
b
)

2

2
(
c
−
b
)

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

# b

5
2

−

1

2

2
×
1

{\displaystyle b = \frac{5^2 - {1}^{2}}{2\times 1} }

# b
12

{\displaystyle b = 12 }

# c
b
(
c
−
b
)

{\displaystyle c = b + (c-b) }

# c
12
1

{\displaystyle c = 12 + 1 }

# c
13

{\displaystyle c=13}
Ground-state Electron Configuration

## **Problem**

What is the ground-state configuration of

Fe

2

{\displaystyle \ce{{Fe}^{2+}}}

?

## **Solution**

The electron configuration for iron is

1

s
2

2

s
2

2

p
6

3

s
2

3

p
6

3

d
6

4

s
2

{\displaystyle 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2 }

 or

[
Ar
]
3

d
6

4

s
2

{\displaystyle [\text{Ar}]3d^6 4s^2 }

. Iron(II) ion removes 2 electrons from the outermost shell (valence shell), which corresponds to the

4
s

{\displaystyle 4s}

 orbital. Thus the configuration for Iron(II) is

1

s
2

2

s
2

2

p
6

3

s
2

3

p
6

3

d
6

{\displaystyle 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 }

 or

[
Ar
]
3

d
6

.

{\displaystyle [\text{Ar}]3d^6. }
Cylindrical Capacitor

## **Problem**

A cylindrical capacitor consists of a cylindrical wire of radius

a

{\displaystyle a}

, and a coaxial cylindrical shell of radius

b

{\displaystyle b}

. Both the wire and cylindrical shell have length

L

{\displaystyle L}

. The wire has a total charge of

Q

{\displaystyle Q}

 distributed on its surface. The cylindrical shell has a total charge of

−
Q

{\displaystyle -Q }

 distributed on its surface.

Calculate the following:

**Part 1:** Electric field in the region

a
<
r
<
b

{\displaystyle a < r < b }

**Part 2:** Electric potential in the region

a
<
r
<
b

{\displaystyle a < r < b }

**Part 3:** Capacitance of the spherical capacitor.

## **Solution**

**Part 1: Electric field**

The first thing to calculate is the electric field between the spherical shells. Use Gauss' law

∮
E
⋅
d
# A

Q

e
n

ϵ
0

{\displaystyle \oint \boldsymbol{E} \cdot d\boldsymbol{A} = \frac{Q\_{en}}{\epsilon\_0}}

.
The total charge enclosed in a Gaussian surface between the wire and the cylindrical shell is

Q

{\displaystyle Q}

. For cylindrical geometry

∮
E
⋅
d
# A

E

(
2
π
r
L
)

{\displaystyle \oint \boldsymbol{E} \cdot d\boldsymbol{A} = |\boldsymbol{E}| (2\pi rL) }

E

(
2
π
r
L
# )

Q

ϵ
0

{\displaystyle |\boldsymbol{E}| (2\pi rL) = \frac{Q}{\epsilon\_0} }

E

=

Q

2
π

ϵ
0

L
r

{\displaystyle |\boldsymbol{E}| = \frac{Q}{2\pi \epsilon\_0 Lr} }

.
The electric field from the inner spherical shell emanates radially outward, so

# E

Q

2
π

ϵ
0

L
r

r

{\displaystyle \boldsymbol{E} = \frac{Q}{2\pi \epsilon\_0 Lr} \boldsymbol{r}}

.
**Part 2: Electric potential**

# V
−

∫

b

a

E
⋅
d
l

{\displaystyle V = -\int\_{b}^{a} \boldsymbol{E} \cdot d\boldsymbol{l} }

# V
−

∫

b

a

Q

2
π

ϵ
0

L
r

d
r

{\displaystyle V = -\int\_{b}^{a} \frac{Q}{2\pi \epsilon\_0 Lr} dr }

# V
−

Q

2
π

ϵ
0

L

(

ln
⁡
(
a
)
−
ln
⁡
(
b
)

)

{\displaystyle V = -\frac{Q}{2\pi \epsilon\_0 L} \left(\ln(a) - \ln(b) \right) }

# V

Q

2
π

ϵ
0

L

ln
⁡

(

b
a

)

{\displaystyle V = \frac{Q}{2\pi \epsilon\_0 L} \ln\left(\frac{b}{a}\right) }

**Part 3: Capacitance**

# C

Q
V

{\displaystyle C = \frac{Q}{V} }

# C

2
π

ϵ
0

L

ln
⁡

(

b
a

)

{\displaystyle C = \frac{2\pi \epsilon\_0 L}{\ln\left(\frac{b}{a}\right)} }
Longest Collatz Sequence with Python

## **Task**

Here is the problem statement for Problem 14 of Project Euler (link here).

The following iterative sequence is defined for the set of positive integers:

n → n/2 (n is even)

n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

## **Code**

```
def collatz\_sequence(n):
    chain\_length = 1
    n\_i = n
            n\_i = n\_i/2
            chain\_length += 1
        elif n\_i % 2 == 1:
            n\_i = 3 \* n\_i + 1
            chain\_length += 1
    return [chain\_length, n]

def longest\_collatz\_chain(n):
    fullList = []
    for i in range(2, n):
        fullList.append(collatz\_sequence(i))
        sortedList = sorted(fullList, reverse=True)
    print(sortedList[:1])

```

## **Experiments and Results**

```
'''Experiments'''
longest\_collatz\_chain(10)
longest\_collatz\_chain(100)
longest\_collatz\_chain(1000)

'''Results'''
[[20, 9]]
[[119, 97]]
[[179, 871]]

```
Gougu Problem

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

## **Problem**

This problem was inspired by the **gougu diagram** (勾股图) and problem 9.15 of the *Jiuzhang Suanshu* called **square in a right triangle** (勾股容方).

Suppose the area of the large square is 50 and the area of the blue square is 29. Question: What is the area of the red square?

## **Solution**

**Algebraic Solution**

Each red square is inscribed inside a right triangle. Let

a
,
b
,
c

{\displaystyle a,b,c}

 denote the base, height, and hypotenuse of each right triangle. The hypotenuse of the right traingles form the perimeter of the blue square. Let

x

{\displaystyle x}

 denote the side of the red square.

From the diagram, we get the area of the large square being

(
a
b

)
2

=
50

{\displaystyle (a + b)^2 = 50}

and the area of the blue squre being

a
2

b
2

=

{\displaystyle a^2 + b^2 = 29.}

Expanding the first equation and then subtracting the second equation yields:

(
a
b

)
2

−
(

a
2

b
2

)

=
50
−
29

(

a
2

2
a
b

b
2

)
−
(

a
2

b
2

)

=
21

2
a
b

=

{\displaystyle \begin{align}
(a + b)^2 - (a^2 + b^2) &= 50 - 29 \\[5 pt]
(a^2 + 2ab + b^2) - (a^2 + b^2) &= 21 \\[5 pt]
2ab &= 21.
\end{align}}

From Problem 9.15 of the *Jiuzhang Suanshu*, the sidelength of the inscribed square is equal to

a
b

a
b

.

{\displaystyle \frac{ab}{a+b}.}

Therefore, the area of the red square is

(

a
b

a
b

)

2

=

(

21
2

)

2

⋅

1
50

=

441
200

.

{\displaystyle \begin{align}
\left(\frac{ab}{a+b}\right)^2 &= \left(\frac{21}{2}\right)^2\cdot \frac{1}{50} \\[5 pt]
&= \frac{441}{200}.
\end{align}}

**Visual Solution**

   Consider the dissection diagram on the right (Figure 1). The blue square can be visualized to be made up of a central square and 4 right triangles. Each triangle is half the area of each grey rectangle between the sides of two adjacent red squares. Let

x

{\displaystyle x}

 be the side length of the red square.

Cut up and rearrange the blue square to fill up the rectangular section of the big square (rectangle PQRS), a rectangle whose width is

50

{\displaystyle \sqrt{50}}

, and height is

50

−
2
x

{\displaystyle \sqrt{50} - 2x}

. By the out-in complementary method, the area of rectangle PQRS is 29. If you remove that middle strip you're left with two identical side rectangles (CDGH and ABEF) whose combined area must be

50
−
# 29
21

{\displaystyle 50 - 29 = 21}

, so each side rectangle is is

21
2

=
10.5

{\displaystyle \frac{21}{2} = 10.5}

 in area. Since each side rectangle is

50

{\displaystyle \sqrt{50}}

 in width, and its height is equal to the side length of a red square, we get

x

50

=

21
2

{\displaystyle x\sqrt{50} = \frac{21}{2}}

and

# x

21

2

50

{\displaystyle x = \frac{21}{2\sqrt{50} }}

Therefore, the area of one red square is

x
2

=

441
200

.

{\displaystyle x^2 = \frac{441}{200}.}
Bypassing Cyclic Integration by Parts

## **Problem**

Use complex integration to compute

∫

e

a
x

sin
⁡
(
b
x
)
d
x

{\displaystyle \int {e}^{ax}\sin(bx) dx }

 and

∫

e

a
x

cos
⁡
(
b
x
)
d
x

{\displaystyle \int {e}^{ax}\cos(bx) dx }

 simultaneously.

**Remark:** This exercise saves a lot of work!

## **Solution**

Consider the integral

∫

e

(
a
i
b
)
x

d
# x

e

(
a
i
b
)
x

a
i
b

.

{\displaystyle \int {e}^{(a+ib)x} dx = \frac{{e}^{(a+ib)x}}{a+ib}.}

Since

e

(
a
i
b
)
x

=

e

a
x

[
cos
⁡
(
b
x
)
i
sin
⁡
(
b
x
)
]

{\displaystyle {e}^{(a+ib)x} = {e}^{ax} [\cos(bx) + i\sin(bx)] }

,

∫

e

a
x

[
cos
⁡
(
b
x
)
i
sin
⁡
(
b
x
)
]
d
# x

e

(
a
i
b
)
x

a
i
b

{\displaystyle \int {e}^{ax} [\cos(bx) + i\sin(bx)] dx = \frac{{e}^{(a+ib)x}}{a+ib}}

∫

e

a
x

cos
⁡
(
b
x
)
d
x
i
∫

e

a
x

sin
⁡
(
b
x
)
d
# x

e

a
x

[
cos
⁡
(
b
x
)
i
sin
⁡
(
b
x
)
]

a
i
b

.

{\displaystyle \int e^{ax}\cos(bx)dx + i \int e^{ax}\sin(bx) dx = \frac{{e}^{ax} [\cos(bx) + i\sin(bx)]}{a+ib}.}

Multiply the numerator and denominator by the complex conjugate on the right hand side.

∫

e

a
x

cos
⁡
(
b
x
)
d
x
i
∫

e

a
x

sin
⁡
(
b
x
)
d
# x

e

a
x

[
cos
⁡
(
b
x
)
i
sin
⁡
(
b
x
)
]
(
a
−
i
b
)

(
a
i
b
)
(
a
−
i
b
)

{\displaystyle \int e^{ax}\cos(bx)dx + i \int e^{ax}\sin(bx) dx = \frac{{e}^{ax} [\cos(bx) + i\sin(bx)](a-ib)}{(a+ib)(a-ib)}}

=

e

a
x

a
2

b
2

[
a
cos
⁡
(
b
x
)
i
a
sin
⁡
(
b
x
)
−
i
b
cos
⁡
(
b
x
)
b
sin
⁡
(
b
x
)
]

{\displaystyle = \frac{{e}^{ax}}{a^2 + b^2} [a\cos(bx) + ia\sin(bx)-ib\cos(bx) + b\sin(bx)] }

By matching real terms and complex terms, and tacking on the arbitrary constant from integrating, we can conclude that

∫

e

a
x

sin
⁡
(
b
x
)
d
# x

e

a
x

[
a
sin
⁡
(
b
x
)
−
b
cos
⁡
(
b
x
)
]

a
2

b
2

C

{\displaystyle \int {e}^{ax}\sin(bx) dx = \frac{{e}^{ax} [a\sin(bx) - b\cos(bx)]}{a^2 + b^2}
 + C }

and

∫

e

a
x

cos
⁡
(
b
x
)
d
# x

e

a
x

[
a
cos
⁡
(
b
x
)
b
sin
⁡
(
b
x
)
]

a
2

b
2

C
.

{\displaystyle \int {e}^{ax}\cos(bx) dx = \frac{{e}^{ax} [a\cos(bx) + b\sin(bx)]}{a^2 + b^2}
 + C. }
Math Book Reviews by Steven Zheng

## ***Elementary Number Theory in Nine Chapters*** **(Second Ed.)**

**Author:** James J. Tattersall

Rating: 10/10

This is an excellent (and readable) intro text to number theory with proofs to many important number-theoretic formulas and theorems. Aside from theory, there is also an emphasis on applications such as calendrics, representations, and cryptography. From front to back, there are many interesting historical notes that connect the subject to several cultures (including Chinese, Muslim and Indian). I highly recommend this text to anyone who enjoys challenging and enlightening problems.

## ***Explorations in Geometry***

   **Author:** Bruce Shawyer

Rating: 8/10

This book draws many problems from the IMO and other math contests, with supplementary information and commentary. Certainly, this book is great for expanding one's horizons to solving challenging geometry problems in the spirit of competitions.

## ***Ancient Puzzles: Classic Brainteasers and Other Timeless Mathematical Games of the Last 10 Centuries***

   **Author:** Dominic Olivastro

Rating 10/10

This book is great for three reasons:

A) You learn a lot about problem solving in discrete math

B) It is a fun read because the problems are presented in recreational and novel ways

C) The book introduces tons of ancient and medieval math outside the Greco-Roman tradition (which is a breath of fresh air)

## ***How to Solve It***

    **Author:** G.Polya

Rating 10/10

Anybody interested in the teaching of mathematics, or teaching in general should read this book! This classic book teaches people that anyone can do mathematics and problem solve, all you need is clear thinking, patience and the right type of guidance (nature and/or nuture).

## ***A History of Chinese Mathematics***

    **Author:** Jean-Claude Martzloff

Rating 10/10

For those interested in mathematics in other cultures, this book is a perfect window into Ancient Chinese Mathematics. Concise but very informative, this book will open new eyes to mathematics from the layman to the expert.

## ***Janos Bolyai: Non-Euclidean Geometry and the Nature of Space***

    **Author:** Jeremy J. Gray

Rating 10/10

This book goes into the details of how Janos Bolyai obsession with Euclid's Fifth Postulate ushered a new age of geometry. The book also comes with Bolyai's groundbreaking dissertation on the Fifth Postulate.
Gamma-Zeta Product

# s
n
u

{\displaystyle s = nu }

 and show that

Γ
(
x
)
ζ
(
x
# )

∫

0

∞

e

−
s

s

x
−
1

d
s

(

∑

# n
1

∞

n

−
x

)

{\displaystyle \Gamma(x)\zeta(x) = \int \_{ 0 }^{ \infty }{ { e }^{ -s } } { s }^{ x-1 }ds\left( \sum \_{ n=1 }^{ \infty }{ { n }^{ -x } } \right) }

is equivalent to the integral

Γ
(
x
)
ζ
(
x
# )

∫

0

∞

u

x
−
1

e

u

−
1

d
u

.

{\displaystyle \Gamma(x)\zeta(x) = \int \_{ 0 }^{ \infty }{ \frac { { u }^{ x-1 } }{ { e }^{ u }-1 } \,du }. }

## **Solution**

Since the gamma function is a real number, we may treat the product as

Γ
(
x
)
ζ
(
x

# n
1

∞

∫

0

∞

e

−
s

(

s
n

)

x
−
1

1
n

d
s

.

{\displaystyle \Gamma(x)\zeta(x) = \sum \_{ n=1 }^{ \infty }{ \int \_{ 0 }^{ \infty }{ { e }^{ -s } } {\left( \frac{s}{n} \right)}^{ x-1 }\frac{1}{n}\,ds }. }

We let

# s
n
u

{\displaystyle s = nu }

 and

d
# s
n
d
u

{\displaystyle ds=ndu }

, thus

Γ
(
x
)
ζ
(
x
)

=

∑

# n
1

∞

∫

0

∞

e

−
n
u

u

x
−
1

d
u

[
5
p
t
]

=

∫

0

∞

(

e

−
u

e

−
2
u

e

−
3
u

.
.
.
)

u

x
−
1

d
u

[
5
p
t
]

=

∫

0

∞

u

x
−
1

e

−
u

(

e

u

e

u

−
1

)

d
u

[
5
p
t
]

=

∫

0

∞

u

x
−
1

e

u

−
1

d
u
.

{\displaystyle \begin{align}
\Gamma(x)\zeta(x) &= \sum \_{ n=1 }^{ \infty }{ \int \_{ 0 }^{ \infty }{ { e }^{ -nu } } {u}^{ x-1 }du } \\ [5 pt]
 &= \int \_{ 0 }^{ \infty }{ ({ e }^{ -u} +{ e }^{ -2u} +{ e }^{ -3u} +... ) } {u}^{ x-1 }du \\ [5 pt]
 &= \int \_{ 0 }^{ \infty }{ { u }^{ x-1 }{ e }^{ -u } } \left( \frac { { e }^{ u } }{ { e }^{ u }-1 } \right) du \\ [5 pt]
 &= \int \_{ 0 }^{ \infty }{ \frac { { u }^{ x-1 } }{ { e }^{ u }-1 } } \,du.
\end{align} }
Archimedes (Aρχιμήδης)

**By: Tao Steven Zheng (郑涛)**

**Born:** c. 287 BC in Syracuse

**Died:** c. 212 BC in Syracuse

**Archimedes** was a Greek mathematician and inventor. Heralded by some to be the greatest scientist of classical antiquity, Archimedes uncovered several mathematical truths with great ingenuity. Many of his discoveries were made by implementing the "Method of Exhaustion", an nascent form of geometric integration, and the principle of levers. Archimedes used the principle of levers for solving mechanical problem when it comes to weights, and purely mathematical problems when it comes to volumes. Archimedes was well-known for inventing military machines, especially machines designed for city defense.

## **Discoveries**

**Archimedes' Principle**

In his work *On Floating Bodies*, Archimedes discussed how the volume of the displaced liquid is the same as the volume of the immersed object. According to legend, Archimedes leaped out of his bath shouting "Eureka!" after making this observation.

**Approximation of pi**

Archimedes titled his work on the approximation of pi *On the Measurement of the Circle*. He applied the method of exhaustion up to a 96-sided polygon to arrive at the bound

3

1
7

<
π
<
3

10
71

{\displaystyle 3\frac{1}{7} < \pi < 3\frac{10}{71} }

.

**Principle of Levers**

Archimedes mathematically described the relationship between the fulcrum, effort, and load.

**Volume and Surface Area of the Sphere**

In his two-part work *On the Sphere and the Cylinder*, Archimedes determined that the volume of a sphere is two-thirds the volume of its circumscribed cylinder. He also discovered the formulas for computing the volume and surface area of a sphere.

V

s
p
h
e
r
e

=

4
π

r
3

3

{\displaystyle V\_{sphere} = \frac{4\pi r^3}{3} }

S

A

s
p
h
e
r
e

=
4
π

r
2

{\displaystyle SA\_{sphere} = 4\pi r^2 }

Archimedes' Triumph

**Quadrature of the Parabola**

In his work *Quadrature of the Parabola*, Archimedes applied the method of exhaustion to prove that the area of a parabolic segment is four-thirds the area of its inscribed triangle.

## **Inventions**

**Archimedean Screw**

An irrigation device that uses a rotating screw to displace water from a lower elevation to a higher elevation.

**Archimedes' Claw**

A large mechanical claw (more likely a hook) attached to a lever to tip over enemy ships.
Ideal Gas Law at STP

## **Problem**

The ideal gas law states:

P
# V
n
R
T

{\displaystyle PV = nRT}

where

P

{\displaystyle P}

 is the pressure measured in pascals (Pa).

V

{\displaystyle V}

 is the volume measured in cubic meters (m3).

n

{\displaystyle n}

 moles of a gas.

R

{\displaystyle R}

 the gas constant 8.31 J/K/mol.

T

{\displaystyle T}

 is the temperature measured in kelvins (K).
**Part 1:** Before 1982, STP (standard temperature and pressure) is defined as the condition when temperature is 0 degrees Celsius and the pressure is 1 atmosphere. Calculate the volume of 1 mole of any gas in STP in cubic meters and in liters.

**Part 2:** After 1982, STP (standard temperature and pressure) is defined as the condition when temperature is 0 degrees Celsius and the pressure is 1 bar. Calculate the volume of 1 mole of any gas in STP in cubic meters and in liters.

## **Solution**

**Part 1**

Convert the pressure from atmospheres (atm) to pascals (Pa).

# P
1

a
t
# m
101325

P
a

{\displaystyle P = 1 \; atm = 101325 \; Pa }

Convert the temperature from degrees Celsius (degs C) to kelvins (K).

# T

0

∘

C
# 273.15
273.15

K

{\displaystyle T={0}^{\circ }C+273.15=273.15\;K}

Use the ideal gas law to determine the volume.

P
# V
n
R
T

{\displaystyle PV = nRT}

(
101325

P
a
)
# V
(
1

m
o
l
)

(

8.31

J

K
⋅
m
o
l

)

(
273.15

K
)

{\displaystyle (101325\;Pa)V=(1\;mol)\left(8.31\;{\frac {J}{K\cdot mol}}\right)(273.15\;K)}

# V
0.0224

m

3

{\displaystyle V=0.0224\,m^{3}}

# V
22.4

L

{\displaystyle V=22.4\,L}

**Part 2**

Convert the pressure from bars (bar) to pascals (Pa).

# P
1

b
a
# r
100000

P
a

{\displaystyle P=1\;bar=100000\;Pa}

Convert the temperature from degrees Celsius (degs C) to kelvins (K).

# T

0

∘

C
# 273.15
273.15

K

{\displaystyle T={0}^{\circ }C+273.15=273.15\,K}

Use the ideal gas law to determine the volume.

P
# V
n
R
T

{\displaystyle PV = nRT}

(
100000

P
a
)
# V
(
1

m
o
l
)

(

8.31

J

K
⋅
m
o
l

)

(
273.15

K
)

{\displaystyle (100000\;Pa)V=(1\;mol)\left(8.31\;{\frac {J}{K\cdot mol}}\right)(273.15\;K)}

# V
0.0227

m

3

{\displaystyle V=0.0227\;m^{3}}

# V
22.7

L

{\displaystyle V=22.7\;L}
Chinese Remainder Theorem with Python

**By: Tao Steven Zheng (郑涛)**

## **Code**

**Description**

Chinese Remainder Theorem:
GCD ( Greatest Common Divisor)
If

G
C
D
(
a
,
b
# )
1

{\displaystyle GCD(a,b) = 1 }

, then for any remainder

r
⋅
a
(
m
o
d

a
)

{\displaystyle r\cdot a (mod \; a) }

 and any remainder

r
⋅
b
(
m
o
d

b
)

{\displaystyle r\cdot b (mod \; b) }

 there exists integer

n

{\displaystyle n}

, such that

# n
r
⋅
a
(
m
o
d

a
)

{\displaystyle n = r \cdot a (mod \; a) }

 and

# n
r
⋅
b
(
m
o
d

b
)

{\displaystyle n = r\cdot b (mod \; b) }

. If

n

1

{\displaystyle {n}\_{1} }

 and

n

2

{\displaystyle {n}\_{2} }

 are
two such integers, then

n

1

=

n

2

(
m
o
d

a
⋅
b
)

{\displaystyle {n}\_{1} = {n}\_{2} (mod \; a\cdot b) }

.

Qin Jiushao's Algorithm :
1. Use extended Euclidean Algorithm to find

x
,
y

{\displaystyle x, y}

 such that

a
⋅
x
b
⋅
# y
1

{\displaystyle a\cdot x + b\cdot y = 1 }

2. Take

# n
r
⋅
a
⋅
b
⋅
y
r
⋅
b
⋅
a
⋅
x

{\displaystyle n = r\cdot a\cdot b\cdot y + r\cdot b\cdot a\cdot x }

```
from functools import reduce

# Sunzi's algorithm (中国剩余定理 Chinese remainder theorem )
def CRT(n, r):
    total = 0
    prod = reduce(lambda r, b: r\*b, n)

    for n\_i, r\_i in zip(n, r):
        p = prod // n\_i
        total += r\_i \* mod\_inv(p, n\_i) \* p
    return [total % prod, prod]

# Qin Jiushao's algorithm (大衍求一术 Dayan qiuyi method)
def mod\_inv(a, b):
    b0 = b
    x0, x1 = 0, 1
    if b == 1:
        return 1
    while a > 1:
        q = a // b
        a, b = b, a%b
        x0, x1 = x1 - q \* x0, x0
    if x1 < 0:
        x1 += b0
    return x1

'''Experiments'''
n\_1 = [3, 5, 7]
r\_1 = [2, 3, 2]
soln\_1 = CRT(n\_1, r\_1)
print(soln\_1)

n\_2 = [19, 17, 12]
r\_2 = [1, 14, 1]
soln\_2 = CRT(n\_2, r\_2)
print(soln\_2)

```

## **Results**

soln\_1 = [23, 105]

N
≡
23

(
mod

105
)

{\displaystyle N \equiv 23 \pmod{105} }

soln\_2 = [3193, 3876]

N
≡
3193

(
mod

3876
)

{\displaystyle N \equiv 3193 \pmod{3876} }
Carbon Atoms in the Human Body

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

## **Problem**

Estimate the number of carbon atoms in the human body. Give your answer as an *order of magnitude*.

## **Solution**

Molar mass of carbon: 12 g/mol

Mass of average person: 60 - 80 kg

Percentage of carbon mass of a human body: 18% - 18.5%

To calculate a rough estimate of the number of carbon atoms in the human body, use 70000 g (70 kg) for the mass and 20% for the percentage of carbon mass in the human body. Make sure to use Avogadro's number, which is roughly

6
×

10

23

m
o
l

−
1

{\displaystyle 6 \times {10}^{23} \; {mol}^{-1} }

. Thus, the number of carbon atoms is roughly

(
70000

g
)
(
0.20
)

12

g
⋅

m
o
l

−
1

(
6
×

10

23

m
o
l

−
1

# )
7
×

10

26

{\displaystyle \frac{(70000 \; g)(0.20)}{12 \; g\cdot {mol}^{-1}}(6 \times {10}^{23} \; {mol}^{-1}) = 7 \times {10}^{26} }

The order of magnitude is 26.
Area Under the Bell Curve

## **Problem**

The equation of the **bell curve** is

f
(
x
# )

e

−

x
2

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

. Show that the area under the bell curve is

∫

−
∞

∞

e

−

x
2

d
# x

π

.

{\displaystyle \int\_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.}

## **Solution**

This problem is solved by making use of the property:

(

∫

−
∞

∞

e

−

x
2

d
x

)

2

=

∫

−
∞

∞

e

−

x
2

d
x

∫

−
∞

∞

e

−

y
2

d
# y

∫

−
∞

∞

∫

−
∞

∞

e

−

(

x
2

y
2

)

d
x

d
y
.

{\displaystyle \left(\int\_{-\infty}^{\infty} e^{-x^2}\,dx\right)^2 = \int\_{-\infty}^{\infty} e^{-x^2}\,dx \int\_{-\infty}^{\infty} e^{-y^2}\,dy = \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} e^{-\left(x^2+y^2\right)}\, dx\,dy. }

Let

I
M

=

∫

−
M

M

e

−

x
2

d
# x

∫

−
M

M

e

−

y
2

d
y

{\displaystyle I\_M = \int\_{-M}^{M} e^{-x^2}\,dx = \int\_{-M}^{M} e^{-y^2}\,dy }

 and

lim

M
→
∞

I
M

=
I

{\displaystyle \lim\_{M \to \infty} I\_M = I }

 be the required value of the integral. Then

I
M
2

=

∫

−
M

M

e

−

x
2

d
x

∫

−
M

M

e

−

y
2

d
y

=

∫

−
M

M

∫

−
M

M

e

−

(

x
2

y
2

)

d
x

d
y
.

{\displaystyle \begin{align}
I^2\_M &= \int\_{-M}^{M} e^{-x^2}\,dx \int\_{-M}^{M} e^{-y^2}\,dy \\[5 pt]
&= \int\_{-M}^{M} \int\_{-M}^{M} e^{-\left(x^2+y^2\right)}\,dx\,dy.
\end{align} }

Taking the limit to infinity gives

I
2

=

∫

−
∞

∞

∫

−
∞

∞

e

−

(

x
2

y
2

)

d
x

d
y
.

{\displaystyle I^2 = \int\_{-\infty}^{\infty} \int\_{-\infty}^{\infty} e^{-\left(x^2+y^2\right)}\,dx\,dy.}

To evaluate the above integral, consider the function

e

−

(

x
2

y
2

)

=

e

−

r

2

{\displaystyle e^{-\left(x^2 + y^2\right)} = e^{-r^{2}}}

on the plane

R

2

{\displaystyle \mathbb R^{2}}

. Converting to polar coordinates, we have

∬

R

2

e

−

(

x
2

y
2

)

d
x

d
y

=

∫
0

2
π

∫
0

∞

e

−

r
2

r

d
r

d
θ

=
2
π

∫
0
∞

r

e

−

r
2

d
r

=
2
π

∫

−
∞

0

1
2

e
s

d
s
,

# s
−

r
2

=
π

∫

−
∞

0

e
s

d
s

=
π

(

e
0

−

e

−
∞

)

=
π
.

{\displaystyle \begin{align}
\iint\_{\R^2} e^{-\left(x^2 + y^2\right)}dx\,dy &= \int\_0^{2\pi} \int\_0^{\infty} e^{-r^2}r\,dr\,d\theta\\[6pt]
&= 2\pi \int\_0^\infty re^{-r^2}\,dr\\[6pt]
&= 2\pi \int\_{-\infty}^0 \tfrac{1}{2} e^s\,ds, && s = -r^2\\[6pt]
&= \pi \int\_{-\infty}^0 e^s\,ds \\[6pt]
&= \pi \left(e^0 - e^{-\infty}\right) \\[6pt]
&=\pi.
\end{align}}

Therefore,

(

∫

−
∞

∞

e

−

x
2

d
x

)

2

=
π

{\displaystyle \left ( \int\_{-\infty}^\infty e^{-x^2}\,dx \right )^2=\pi}

and so

∫

−
∞

∞

e

−

x
2

d
# x

π

.

{\displaystyle \int\_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}.}
2x2 Matrix Game Formula

## **Problem**

Let the payoff matrix of a 2 x 2 game be characterized by the matrix

A

=

[

a

b

c

d

]

{\displaystyle \mathbf{A} =
 \begin{bmatrix}
 a & b \\
 c & d \\
 \end{bmatrix}}

.
All entries are positive real numbers.

**Part 1:** Use the simplex method to show that the value of the matrix game is

# v

det

(
A
)

S

{\displaystyle v = \frac{\det{(A)}}{S} }

, where

# S
(
a
d
)
−
(
b
c
)

{\displaystyle S = (a+d) - (b+c) }

 and

det

(
A
)

=
a
d
−
b
c

{\displaystyle \det{(A)} = ad - bc }

.

**Part 2:** Determine the strategy of the row player and the column player.

## **Solution**

**Part 1**

Construct the initial tableau.

**Tableau 1**

Leaving variable is

x

{\displaystyle x}

.

Entering variable is

s
1

{\displaystyle s\_1}

.

Divide row 1 by

a

{\displaystyle a}

:

1
a

R

1

{\displaystyle {\frac {1}{a}}R\_{1}}

.

Perform the following row operations to get tableau 2:

R

3

R

1

{\displaystyle R\_{3}+R\_{1}}

 and

R

2

−
c

R

1

{\displaystyle R\_{2}-cR\_{1}}

.

**Tableau 2**

Leaving variable is

y

{\displaystyle y}

.

Entering variable is

s
2

{\displaystyle s\_2}

.

Divide row 1 by

a
d
−
b
c

a

{\displaystyle {\frac {ad-bc}{a}}}

:

(

a

a
d
−
b
c

)

R

1

{\displaystyle \left({\frac {a}{ad-bc}}\right)R\_{1}}

.

Perform the following row operations to get tableau 3:

R

1

−

b
a

R

2

{\displaystyle R\_{1}-{\frac {b}{a}}R\_{2}}

 and

R

3

−

(

b
a

−
1

)

R
2

{\displaystyle R\_{3}-\left({\frac {b}{a}}-1\right)R2}

.

The value of the game is the reciprocal of

Z

{\displaystyle Z}

 (the entry in the lower right corner). Therefore,

# v

a
d
−
b
c

(
a
d
)
−
(
b
c
)

{\displaystyle v={\frac {ad-bc}{(a+d)-(b+c)}}}

# v

det

(
A
)

S

{\displaystyle v = \frac{\det{(A)}}{S} }

.
**Part 2**

To obtain the strategy for the row player, take the entries of the bottom row under

s
1

{\displaystyle s1}

 and

s
2

{\displaystyle s2}

 and divide each by the value of

Z

{\displaystyle Z}

.

# R

[

r

1

r

2

]

{\displaystyle R={\begin{bmatrix}r\_{1}&r\_{2}\\\end{bmatrix}}}

where

r

1

=

d
−
c

S

{\displaystyle r\_{1}={\frac {d-c}{S}}}

 and

r

2

=

a
−
b

S

{\displaystyle r\_{2}={\frac {a-b}{S}}}

.

To obtain the strategy for the column player, take the entries of the right-most column on row

x

{\displaystyle x}

 and

y

{\displaystyle y}

 and divide each by the value of

Z

{\displaystyle Z}

.

# C

[

c

1

c

2

]

{\displaystyle C={\begin{bmatrix}c\_{1}\\c\_{2}\\\end{bmatrix}}}

where

c

1

=

d
−
b

S

{\displaystyle c\_{1}={\frac {d-b}{S}}}

 and

c

2

=

a
−
c

S

{\displaystyle c\_{2}={\frac {a-c}{S}}}

.
Logarithmic Function

# y
ln
⁡

1
ln
⁡

x

{\displaystyle y = \ln{1+\ln{x}} }

.

**Part 1:** What is the domain of the function?

**Part 2:** What is the inverse function?

## **Solution**

**Part 1**

The domain of a logarithmic function

# y
ln
⁡

x

{\displaystyle y = \ln{x} }

 is

{
x
∈

R

:
x
>
0
}

{\displaystyle \lbrace x \in \mathbb{R}: x > 0 \rbrace }

Thus, to solve for the domain of the function

# y
ln
⁡

1
ln
⁡

x

{\displaystyle y = \ln{1+\ln{x}} }

 set

1
ln
⁡

x

>
0

{\displaystyle 1+\ln{x} > 0 }

ln
⁡

x

>
−
1

{\displaystyle \ln{x} > -1 }

e

ln
⁡

x

>

e

−
1

{\displaystyle {e}^{\ln{x}} > {e}^{-1} }

Hence, the domain is

{
x
∈

R

:
x
>

e

−
1

}

{\displaystyle \lbrace x \in \mathbb{R}: x > {e}^{-1} \rbrace }

.
**Part 2**

# y
ln
⁡

1
ln
⁡

x

{\displaystyle y = \ln{1+\ln{x}} }

# x
ln
⁡

1
ln
⁡

y

{\displaystyle x = \ln{1+\ln{y}} }

e

x

=
1
ln
⁡

y

{\displaystyle {e}^{x} = 1+\ln{y} }

e

x

−
# 1
ln
⁡

y

{\displaystyle {e}^{x} - 1 = \ln{y} }

ln
⁡

y

=

e

x

−
1

{\displaystyle \ln{y} = {e}^{x} - 1 }

# y

e

e

x

−
1

{\displaystyle y = {e}^{{e}^{x} - 1} }

f

−
1

(
x
# )

e

e

x

−
1

{\displaystyle {f}^{-1}(x) = {e}^{{e}^{x} - 1} }