recursal/Fanatic-Fandom huggingface dataset:
Volume of Prismoids 2

## **Problem**

The solids introduced to schoolchildren are called prismoids. The general formula for computing the volume of a prismoid is

# V

h
6

[

A
T

4

A
M

A
B

]

{\displaystyle V = \frac{h}{6}\left[A\_T + 4A\_M + A\_B \right] }

where

h

{\displaystyle h}

 is the height,

A
T

{\displaystyle A\_T}

 i the area of the top,

A
M

{\displaystyle A\_M }

 is the area of the midsection, and

A
B

{\displaystyle A\_B }

 is the area of the base.

Use the prismoid formula to derive the formulas for computing the volume of the cylinder, the paraboloid, and the bicylinder.

CylinderParaboloidBicylinder
## **Solution**

**Cylinder**

The cylinder is a prism of a circle. Let the height be

h

{\displaystyle h}

. The top, midsection and base are circles with area

π

r
2

{\displaystyle \pi r^2}

.

A
T

=

A
M

=

A
B

=
π

r
2

{\displaystyle A\_T = A\_M = A\_B = \pi r^2 }

# V

h
6

[

π

r
2

4
π

r
2

π

r
2

]

{\displaystyle V = \frac{h}{6}\left[\pi r^2 + 4\pi r^2+ \pi r^2 \right] }

# V

h
6

[

6
π

r
2

]

{\displaystyle V = \frac{h}{6}\left[6\pi r^2\right] }

# V
π

r
2

h

{\displaystyle V= \pi r^2h}

**Paraboloid**

Let

h

{\displaystyle h}

 be the height of the paraboloid. The base of a paraboloid is a circle with radius

r

{\displaystyle r}

. The top of the paraboloid is a point with no area. The midsection is harder to determine. From volume of a paraboloid it is found that the radius of a paraboloid as a function of height is

r
(
x
# )

r

h

x

{\displaystyle r(x) = \frac{r}{\sqrt{h}} \sqrt{x} }

.
Thus at

# x

h
2

{\displaystyle x = \frac{h}{2}}

, the radius of the circle at the midsection is

r
M

=

r

2

{\displaystyle r\_M = \frac{r}{\sqrt{2}} }

 and the area of the midsection is

A
M

=
π

(

r
M

)

2

=

π

r
2

2

{\displaystyle A\_M = \pi{(r\_M)}^2 = \frac{\pi r^2}{2} }

.

# V

h
6

[

0
4

(

π

r
2

2

)

π

r
2

]

{\displaystyle V = \frac{h}{6}\left[0 + 4\left(\frac{\pi r^2}{2}\right)+ \pi r^2 \right] }

# V

h
6

[

3
π

r
2

]

{\displaystyle V = \frac{h}{6}\left[3\pi r^2 \right] }

# V

π

r
2

h

2

{\displaystyle V = \frac{\pi r^2 h}{2} }

**Bicylinder**

The the top and base of the bicylinder are points with no area, hence,

A
T

=

A
B

=
0

{\displaystyle A\_T = A\_B = 0 }

. The midsection is a square with area

d
2

{\displaystyle d^2}

.

# V

d
6

[

0
4

d
2

0

]

{\displaystyle V = \frac{d}{6}\left[0 + 4d^2+ 0 \right] }

# V

d
6

[

4

d
2

]

{\displaystyle V = \frac{d}{6}\left[4d^2 \right] }

# V

2

d
3

3

{\displaystyle V = \frac{2d^3}{3} }
Thoughts on Religions

**By: Tao Steven Zheng (郑涛)**

## **On Confucianism**

**Renaming Confucianism**

In the West, what is called *ru jia* 儒家 is called Confucianism, after the Latinized name the philosophy's founder Kong Fuzi 孔夫子. The character *ru* 儒, as a noun, literally translates to a well-read scholar. More interestingly, the original role of a *ru* 儒 is closer to a priest who is adept in conducting ritual ceremony and divination. So perhaps Confucianism should be called "Scholasticism". Of course, this should not be confused with the Scholasticism of monk-scholars in Medieval Europe!

## **On Christianity**

**The Making of Jesus Christ**

Our more historically examined Jesus portrays a less supernatural religious figure. Jesus of Nazareth was a carpenter by trade who was fed up by the dogmatic teachings of the Jewish rabbis of his time; a time of great social and political uncertainty under Roman rule. He and his disciples (later called Christians) became convenient scapegoats for heresy, so the Jewish religious leaders demanded Pontius Pilate to execute Jesus (via crucifixion, a common Roman execution method). Nearly 50 - 200 years later, we get a collection of inconsistent stories of Jesus (the Gospels) written by several biographers. The earlier Gospels written in the 1st century AD portrayed a less supernatural Jesus, and the later Gospels describe a progressively more supernatural Jesus. These spiritual embellishments to Jesus' life story transformed "Jesus of Nazareth" (the man) into "Jesus Christ" (the God-man).
The Fossil in Zezhou 泽州化石

**《梦溪笔谈·卷21异事18》**
治平中，泽州人家穿井，土中见一物，蜿蜿如龙蛇。畏之不敢角，久之见其不动，试扑之，乃石也。村民无知，遂碎之。时程伯纯为晋城令，求得一段，鳞甲皆如生物。盖蛇蜃（shèn）所化，如石蟹之类。

***Mengxi Bitan* Chapter 21 Yishi 18**
During Zhiping period of the reign of Emperor Yingzong, a family in Zezhou found an object in the earth when digging a well. It was winding like a snake or a dragon. At first they did not dare to touch it. Seeing that it remained motionless for a long while, they began to lay their hands on it and found that it was a stone-like object. As the villagers were ignorant, they broke it into pieces. At that time Cheng Bochun was the county magistrate of Jincheng, he got a small piece and found that its scales looked like those of a living creature. Most probably it was transformed from a snake-like animal in the same way as the fossil of the crab was.

**【今译】**

治平年间，泽州有家人家打井，在土里看到一样东西，形状象龙蛇那样弯弯曲曲。人们害怕，不敢触摸。过了很久，看见它不动，就试着拍打了它一下，原来是石头。村里的人无知，将它打碎了。当时程伯纯在晋城当县官，拿到一段，上面的鳞甲都象活的一样，大概是蛇蜃之类的生物变化而成的。就象我们看到的已经变成石头的螃蟹一样。
Harmonic Series

## **Problem**

The **harmonic series** is an infinite series given by the sum of reciprocals.

∑

# n
1

∞

1
n

=
1

1
2

1
3

1
4

1
5

⋯

{\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{n}}=1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots }

**Part 1**

Use the **comparison test** to prove that the harmonic series is divergent.

**Part 2**

Use the **integral test** to prove that the harmonic series is divergent.

## **Solution**

**Part 1**

Group the terms in the harmonic series as follows:

1

1
2

(

1
3

1
4

)

(

1
5

1
6

1
7

1
8

)

⋯

{\displaystyle 1+{\frac {1}{2}}+\left({\frac {1}{3}}+{\frac {1}{4}}\right)+\left({\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}\right)+\cdots }

It can be shown that each bracket grouped above is always greater than

1
2

{\displaystyle \frac{1}{2}}

, for example,

1
3

1
4

>

1
4

1
4

>

1
2

{\displaystyle {\begin{aligned}{\frac {1}{3}}+{\frac {1}{4}}&>{\frac {1}{4}}+{\frac {1}{4}}\\&>{\frac {1}{2}}\end{aligned}}}

and

1
5

1
6

1
7

1
8

>

1
8

1
8

1
8

1
8

>

1
2

{\displaystyle {\begin{aligned}{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}&>{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}\\&>{\frac {1}{2}}\end{aligned}}}

Therefore,

1

1
2

1
3

1
4

⋯

1

2

k

>
1

k
2

{\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots +{\frac {1}{2^{k}}}>1+{\frac {k}{2}}}

and

1

1
2

1
3

1
4

1
5

1
6

1
7

1
8

.
.
.
>
1

1
2

1
2

1
2

.
.
.

{\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+...>1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+...}

which culminates to the conclusion that

S

2

n

>
1

n
2

{\displaystyle {S}\_{2^{n}}>1+{\frac {n}{2}}}

.

Since

1

1
2

1
2

1
2

# ⋯

lim

n
→
∞

(

1

n
2

)

=
∞

{\displaystyle 1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+\cdots =\lim \_{n\rightarrow \infty }\left(1+{\frac {n}{2}}\right)=\infty }

it follows that

∑

# n
1

∞

1
n

=
∞
.

{\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{n}}=\infty .}

**Part 2**

∑

# n
1

∞

1
n

>

∫

1

∞

1
x

d
x

{\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{n}}>\int \_{1}^{\infty }{\frac {1}{x}}dx}

∫

1

∞

1
x

d
# x

lim

b
→
∞

(

ln
⁡

(
x
)

1

b

)

{\displaystyle \int \_{1}^{\infty }{\frac {1}{x}}dx=\lim \_{b\rightarrow \infty }\left(\ln {(x)}{\Biggr |}\_{1}^{b}\right)}

∫

1

∞

1
x

d
# x
∞

{\displaystyle \int \_{1}^{\infty }{\frac {1}{x}}dx=\infty }

Therefore,

∑

# n
1

∞

1
n

=
∞

{\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{n}}=\infty }

## **Historical Note**

The earliest proof that the harmonic series diverges can be found in the mathematics text *Quaestiones super geometriam Euclidis* (*Questions Concerning Euclid's Geometry*) by the Medieval French theologian, philosopher, and mathematician Nicholas Oresme (c. 1325 - 1382 AD). This proof dates to around 1350 AD. The arguments in Part 1 (using the comparison test) is based on Nicholas Oresme's proof.
Producing Aluminum Chloride

## **Problem**

Aluminum chloride

(

AlCl

3

)

{\displaystyle \ce{(AlCl3) }}

 is produced by reacting aluminum and chlorine gas.

**Part 1:** Determine the chemical formula.

**Part 2:** What is the **limiting reagent** if 10.0 grams of aluminum react with 15.0 grams of chlorine.

**Part 3:** Determine the mass of aluminum chloride

(

AlCl

3

)

{\displaystyle \ce{(AlCl3) }}

 that could be produced from part 2. This mass is called the **theoretical yield**.

## **Solution**

**Part 1**

Chlorine gas is diatomic, so begin with the guess:

Al

Cl

2

⟶

AlCl

3

{\displaystyle \ce{ Al + Cl2 -> AlCl3 }}

and then balance the equation

2

Al
3

Cl

2

⟶
2

AlCl

3

{\displaystyle \ce{ 2Al + 3Cl2 -> 2AlCl3 }}

The balanced equation shows the exact amount of reagents required to produce the exact amount of products. This is called the **stoichiometric amount**. When one reagent is in excess of this stochiometric amount, the excess will not react. The reagent present in the most limiting amount is called the **limiting reagent**. The limiting reagent dictates how much product will be formed.

**Part 2**

Perform a stoichiometric calculation with the given information for aluminum. Take all known quantities of each compound and convert to the number of moles.

10.0

g
×

1

m
o
l

A
l

27.0

g

A
l

=
0.370

m
o
l

A
l

{\displaystyle 10.0 \; g \times \frac{1 \; mol \; Al}{27.0 \; g \; Al} = 0.370 \; mol \; Al }

0.370

m
o
l

A
l
×

3

C

l
2

2

A
l

=
0.556

m
o
l

C

l
2

{\displaystyle 0.370 \; mol \; Al \times \frac{3 \; Cl\_2}{2 \; Al} = 0.556 \; mol \; Cl\_2 }

0.556

m
o
l

C

l
2

×

71.0

g

C

l
2

1

m
o
l

C

l
2

=
39.4

g

C

l
2

{\displaystyle 0.556 \; mol \; Cl\_2 \times \frac{71.0 \; g \; Cl\_2}{1 \; mol \; Cl\_2} = 39.4 \; g \; Cl\_2 }

We only have 15.0 grams of chlorine, so the chlorine will run out first; this means chlorine is the **limiting reagent**.

**Part 3**

How much aluminum chloride (in grams) is produced?

15.0

g

C

l
2

×

1

m
o
l

C

l
2

71.0

g

C

l
2

=
0.211

m
o
l

C

l
2

{\displaystyle 15.0 \; g \; Cl\_2 \times \frac{1 \; mol \; Cl\_2}{71.0 \; g \; Cl\_2} = 0.211 \; mol \; Cl\_2 }

0.211

m
o
l

C

l
2

×

2

A
l
C

l
3

3

C

l
2

=
0.141

m
o
l

A
l
C

l
3

{\displaystyle 0.211 \; mol \; Cl\_2 \times \frac{2 \; AlCl\_3}{3 \; Cl\_2} = 0.141 \; mol \; AlCl\_3 }

0.141

m
o
l

A
l
C

l
3

×

133.5

g

A
l
C

l
3

1

m
o
l

A
l
C

l
3

=
18.8

g

A
l
C

l
3

{\displaystyle 0.141 \; mol \; AlCl\_3 \times \frac{133.5 \; g \; AlCl\_3}{1 \; mol \; AlCl\_3} = 18.8 \; g \; AlCl\_3 }

The **theoretical yield** aluminum chloride produced from the above process is 18.8 grams.
Predicting Chemical Reactions

## **Problem**

Consider the chemical system consisting of the species carbon dioxide (

CO

2

{\displaystyle \ce{CO2}}

), water (

H

2

O

{\displaystyle \ce{H2O}}

), diatomic hydrogen (

H

2

{\displaystyle \ce{H2}}

), methane (

CH

4

{\displaystyle \ce{CH4}}

), and carbon monoxide (

CO

{\displaystyle \ce{CO}}

). What are the possible reactions for this system? Use linear algebra to solve this problem.

## **Solution**

Construct a matrix based on the atoms of each compound.

CO

2

{\displaystyle \ce{CO2}}

H

2

O

{\displaystyle \ce{H2O}}

H

2

{\displaystyle \ce{H2}}

CH

4

{\displaystyle \ce{CH4}}

CO

{\displaystyle \ce{CO}}

H

{\displaystyle \ce{H}}

C

{\displaystyle \ce{C}}

O

{\displaystyle \ce{O}}

Our matrix is

[

0

2

2

4

0

1

0

0

1

1

2

1

0

0

1

]

{\displaystyle \begin{bmatrix} 0 & 2 & 2 & 4 & 0 \\ 1 & 0 & 0 & 1 & 1 \\ 2 & 1 & 0 & 0 & 1 \end{bmatrix}}

The matrix reduced to row echelon form is

[

1

0

0

1

1

0

1

0

−
2

−
1

0

0

1

4

1

]

{\displaystyle \begin{bmatrix} 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & -2 & -1 \\ 0 & 0 & 1 & 4 & 1 \end{bmatrix}}

Which means we get the basis vectors

(

x
1

x
2

x
3

x
4

x
5

)

=

x
4

(

−
1

2

−
4

1

0

)

x
5

(

−
1

1

−
1

0

1

)

{\displaystyle \begin{pmatrix} x\_1 \\ x\_2 \\ x\_3 \\ x\_4 \\ x\_5 \end{pmatrix} =
x\_4\begin{pmatrix} -1 \\ 2 \\ -4 \\ 1 \\ 0 \end{pmatrix} + x\_5\begin{pmatrix} -1 \\ 1 \\ -1 \\ 0 \\ 1 \end{pmatrix}
}

Here, the basis vectors translate to two chemical reactions:

CO

2

4

H

2

↽

−

−

⇀

2

H

2

O

CH

4

{\displaystyle \ce{CO2 + 4H2 <=> 2H2O + CH4}}

CO

2

H

2

↽

−

−

⇀

H

2

O
CO

{\displaystyle \ce{CO2 + H2 <=> H2O + CO }}
Sangaku Three Tangent Circles Problem

## **Problem**

Sangaku (算額, lit. "calculation tablet") are Japanese geometry problems presented on wooden tablets. These tablets were hung in Buddhist temples and Shinto shrines during the Edo period (1603 - 1867) in the 18th to 19th centuries.

The following sangaku problem is from a tablet in Gunma Prefecture that dates to 1824. There are three mutually tangent circles resting on a line. Find the radius of the middle circle

R

{\displaystyle R}

, given the radii of the adjacent circles

R
1

{\displaystyle R\_1}

 and

R
2

{\displaystyle R\_2}

.

## **Solution**

Consider the images on the right (Figure 1 and Figure 2). Let the radius of the middle circle be

R

{\displaystyle R}

, the radii of the left circle be

R
1

{\displaystyle R\_1}

, and the radii of the right circle be

R
2

{\displaystyle R\_2}

. In Figure 3, we have

R
1

<

R
2

{\displaystyle R\_1 < R\_2 }

. Introduce two lengths

x

{\displaystyle x}

 and

y

{\displaystyle y}

 that forms the base of three right triangles (Figure 2).

By the Pythagorean theorem,

x

2

(

R
1

−
R

)

2

=

(

R
1

R

)

2

{\displaystyle {x}^{2} + {\left(R\_1 - R\right)}^{2} = {\left(R\_1 + R\right)}^{2} }

y

2

(

R
2

−
R

)

2

=

(

R
2

R

)

2

{\displaystyle {y}^{2} + {\left(R\_2 - R\right)}^{2} = {\left(R\_2 + R\right)}^{2} }

(

x
y

)

2

(

R
2

−

R
1

)

2

=

(

R
1

R
2

)

2

{\displaystyle {\left(x+y \right)}^{2} + {\left(R\_2 - R\_1\right)}^{2} = {\left(R\_1 + R\_2\right)}^{2} }

Simplify the above three expressions.

x

2

=
4

R
1

R

{\displaystyle {x}^{2} = 4 R\_1 R }

y

2

=
4

R
2

R

{\displaystyle {y}^{2} = 4 R\_2 R }

(

x
y

)

2

=
4

R
1

R
2

{\displaystyle {\left(x+y \right)}^{2} = 4 R\_1 R\_2 }

Take the square roots.

# x
2

R
1

R

{\displaystyle x = 2 \sqrt{R\_1 R} }

# y
2

R
2

R

{\displaystyle y = 2 \sqrt{R\_2 R} }

x
# y
2

R
1

R
2

{\displaystyle x+y = 2 \sqrt{R\_1 R\_2} }

Therefore,

2

R
1

R

2

R
2

R

=
2

R
1

R
2

{\displaystyle 2 \sqrt{R\_1 R} + 2 \sqrt{R\_2 R} = 2 \sqrt{R\_1 R\_2} }

and

# R

R
1

R
2

(

R
1

R
2

)

2

.

{\displaystyle R = \frac{R\_1 R\_2}{{\left(\sqrt{R\_1} + \sqrt{R\_2}\right)}^{2}}. }
Volume of the Bicylinder

## **Problem**

The **bicylinder** is a solid formed by the intersection of perpendicular two cylinders of equal cross-sections (Figure 1).

{

x
2

y
2

=

r
2

x
2

z
2

=

r
2

{\displaystyle \begin{cases} x^2 + y^2 = r^2 \\ x^2 + z^2 = r^2 \end{cases} }

Use integration to determine the volume of the bicylinder if the cross-sections of the intersecting cylinders are circles of radius

r

{\displaystyle r}

.

## **Solution**

### **Single Integral Solution**

The first thing to consider are the **level-curves** in the direction of integration. Let the integration be carried out in the y-axis. The cross-sections in the y-axis are squares that vary under the constraint of a circle with radius

r

{\displaystyle r}

, which is the circle

x
2

y
2

=

r
2

{\displaystyle x^2 + y^2 = r^2 }

. The area of the cross-sectional squares is determined by the formula

# y
2

r
2

−

x
2

.

{\displaystyle b = 2y = 2\sqrt{r^2-x^2}. }

Subsequently, the cross-sectional areas are

A
(
x
# )
4
(

r
2

−

x
2

)

{\displaystyle A(x) = 4(r^2 - x^2) }

. To determine the volume, integrate in the y-axis from one end of the circle to the other end:

V

=

∫

−
r

r

4
(

r
2

−

x
2

)
d
x

=
4

[

r
2

x
−

x
3

3

−
r

r

]

=

16

r
3

3

.

{\displaystyle \begin{align}
V &= \int^{r}\_{-r} 4(r^2 - x^2) dx \\[5 pt]
&= 4 \left[r^2x - \frac{x^3}{3}\Biggr|\_{-r}^{r}\right] \\[5 pt]
&= \frac{16r^3}{3}.
\end{align} }

### **Double Integral Solution**

The required volume is equal to eight times the octant of the bicylinder. First set up the integral

V

=
8

∫

0

r

∫

0

r
2

−

x
2

z

d
y

d
x

=
8

∫

0

r

∫

0

r
2

−

x
2

r
2

−

x
2

d
y

d
x
.

{\displaystyle \begin{align}
V &= 8\int\_{0}^{r}\int\_{0}^{\sqrt{r^2-x^2}}z\,dy\, dx \\[5 pt]
&= 8 \int\_{0}^{r}\int\_{0}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2}\,dy\, dx.
\end{align} }

Integrate with respect to

y

{\displaystyle y}

 and evaluate:

V

=
8

∫

0

r

r
2

−

x
2

[

y

0

r
2

−

x
2

]

d
x

=
8

∫

0

r

r
2

−

x
2

d
x
.

{\displaystyle \begin{align}
V &= 8 \int\_{0}^{r} \sqrt{r^2-x^2} \left[y\, \Biggr|\_{0}^{\sqrt{r^2-x^2}}\right]\, dx \\[5 pt]
&= 8 \int\_{0}^{r}r^2-x^2\, dx.
\end{align} }

Integrate with respect to

x

{\displaystyle x}

 and evaluate:

V

=
8

∫

0

r

r
2

−

x
2

d
x

=
8

[

r
2

x
−

x
3

3

0

r

]

=

16

r
3

3

.

{\displaystyle \begin{align}
V &= 8 \int\_{0}^{r}r^2-x^2\, dx \\[5 pt]
&= 8 \left[r^2x - \frac{x^3}{3} \, \Biggr|\_{0}^{r}\right] \\[5 pt]
&= \frac{16r^3}{3}.
\end{align} }
Phone Numbers

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

## **Problem**

Suppose every telephone number consists of seven digits such that the first digit of all telephone numbers is an odd number, and the last digit of all telephone numbers is a non-zero even number. Question: What is the maximum number of seven-digit telephone numbers that can be issued?

## **Solution**

Each telephone number has seven digits, so imagine there are 7 cells. The first digit must be odd; thus, there are only 5 choices for the first cell: 1, 3, 5, 7, or 9. The last digit must be a non-zero even number; thus, there are only 4 choices for the first cell: 2, 4, 6, or 8. The the remaining 5 cells can be filled by any digit from 0 to 9; hence, there are 10 choices each.

Subsequently, the maximum number of telephone numbers is

5
×
10
×
10
×
10
×
10
×
10
×
4

=
5
×
4
×

10

5

=
2000000

=
2
×

10

6

{\displaystyle \begin{align} 5 \times 10 \times 10 \times 10 \times 10 \times 10 \times 4 &= 5 \times 4 \times {10}^{5} \\
&= 2000000 \\
&= 2\times {10}^{6}
\end{align} }
Primal and Dual

## **Problem**

Consider the primal linear programming problem

M
a
x

6

x
1

7

x
2

2

x
1

5

x
2

≤
1

x
1

3

x
2

≤
4

x
1

,

x
2

≥
0

{\displaystyle \begin{align}
Max \quad 6x\_1 + 7x\_2 \\
2x\_1 + 5x\_2 &\le 1 \\
x\_1 + 3x\_2 &\le 4 \\
x\_1, x\_2 &\ge 0
\end{align}}

If the optimal solution is

x
∗

=
(

x
1

,

x
2

# )
(
0.5
,
0
)

{\displaystyle x^\* = (x\_1,x\_2) = (0.5, 0) }

, what is the optimal solution of its dual problem if

y
∗

=
(

y
1

,

y
2

# )
(
?
,
0
)

{\displaystyle y^\* = (y\_1,y\_2) = (?, 0) }

?

## **Solution**

For a primal problem

M
a
x

c

T

x

A
x
≤
b

x
≥
0

{\displaystyle \begin{align}
Max \quad {c}^{T} x \\
Ax \le b \\
x \ge 0
\end{align}}

the dual problem is

M
a
x

b

T

y

A

T

y
≥
c

y
≥
0

{\displaystyle \begin{align}
Max \quad {b}^{T} y \\
{A}^{T}y \ge c \\
y \ge 0
\end{align}}

where

x
,
y
,
b
,
c

{\displaystyle x,y,b,c }

 are vectors and

A

{\displaystyle A}

 is a matrix.

Thus the dual problem of the above primal problem is:

M
i
n

y
1

4

y
2

2

y
1

y
2

≥
6

5

y
1

3

y
2

≥
7

y
1

,

y
2

≥
0

{\displaystyle \begin{align}
Min \quad y\_1 + 4y\_2 \\
2y\_1 + y\_2 &\ge 6 \\
5y\_1 + 3y\_2 &\ge 7 \\
y\_1, y\_2 &\ge 0
\end{align}}

If

x
∗

=
(
0.5
,
0
)

{\displaystyle x^\* = (0.5,0) }

, the optimal value of the objective function is

# Z
6
(
0.5
)
7
(
0
# )
3

{\displaystyle Z = 6(0.5) + 7(0) = 3 }

.
The **strong duality theorem** states that the optimal value of the primal equals the optimal value of the dual if there is no duality gap.

c

T

x

∗

=

b

T

y

∗

{\displaystyle {c}^{T} {x}^{\*} = {b}^{T} {y}^{\*} }

It should be noted that if the primal problem has an optimal solution, then the dual problem also has an optimal solution.

Thus

# W

y
1

4
(
0
# )
3

{\displaystyle W = y\_1 + 4(0) = 3 }

and

y
1

=
3

{\displaystyle y\_1 = 3 }

.
Thus the solution of

y
∗

=
(

y
1

,

y
2

# )
(
3
,
0
)

{\displaystyle y^\* = (y\_1,y\_2) = (3, 0) }

.
Taylor Series of Inverse Tangent

## **Problem**

Begin with the **geometric series**

1

1
−
x

=

∑

# n
0

∞

x
n

{\displaystyle \frac{1}{1-x} = \sum^{\infin}\_{n=0} x^n}

and use this summation formula to derive the Taylor series of

# y
arctan
⁡
(
x
)

{\displaystyle y = \arctan(x) }

 centered at

# x
0

{\displaystyle x = 0}

.

## **Solution**

Note that

∫

1

1

x
2

d
# x
arctan
⁡
(
x
)
.

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

Since the geometric series centered at zero is

1

1
−
x

=
1
x

x
2

x
3

⋯

{\displaystyle \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots}

then

1

1
x

=

1

1
−
(
−
x
)

=
1
−
x

x
2

−

x
3

⋯

{\displaystyle \frac{1}{1+x} = \frac{1}{1-(-x)} = 1 - x + x^2-x^3 + \cdots }

and

1

1

x
2

=
1
−

x
2

x
4

−

x
6

⋯

{\displaystyle \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + \cdots }

Subsequently,

∫

1

1

x
2

d
# x
x
−

x
3

3

x
5

5

−

x
7

7

⋯

{\displaystyle \int \frac{1}{1+x^2} \;dx = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots }

and

arctan
⁡
(
x

# n
1

∞

(
−
1
)

n
−
1

x

2
n
1

2
n
1

.

{\displaystyle \arctan(x) = \sum\_{n=1}^{\infty} {(-1)}^{n-1}\frac{x^{2n+1}}{2n+1}.}

Note that the series can be alternatively written as

arctan
⁡
(
x

# n
1

∞

(
−
1
)

n
1

x

2
n
1

2
n
1

.

{\displaystyle \arctan(x) = \sum\_{n=1}^{\infty} {(-1)}^{n+1}\frac{x^{2n+1}}{2n+1}.}
Moscow Papyrus Problem 19

## **Problem**

There are a number of problems recorded on Egyptian papyri that involve linear equations. Among the earliest and most basic linear equations problems were called **Ahau problems**. Problem 19 of the *Moscow Mathematical Papyrus* (c. 1850 BC) asks:

A number taken 1 and 1/2 times and added to 4 to make 10. What is the number?

## **Solution**

The scribe who wrote the *Moscow Mathematical Papyrus* gives a solution:

Calculate the excess of this 10 over 4. The result is 6. You operate on 1 and 1/2 to find 1. The result is 2/3. You take 2/3 of this 6. The result is 4. Behold, 4 says it. You will find that this is correct.

In modern notation, the equation is

x

1
2

x
# 4

{\displaystyle x + \frac{1}{2}x + 4 = 10. }

The ancient Egyptian scribe instructs the reader to subtract 4 from 10

x

1
2

# x

{\displaystyle x + \frac{1}{2}x = 6. }

After subtracting 4, the scribe noted that the reciprocal of

1

1
2

{\displaystyle 1 \frac{1}{2}}

is equal to

2
3

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

. The scribe then multiplies 6 by

2
3

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

to obtain the answer:

2
3

×
# 6

{\displaystyle \frac{2}{3} \times 6 = 4. }
Volume of Prismoids

## **Problem**

The solids introduced to schoolchildren are called prismoids. The general formula for computing the volume of a prismoid is

# V

h
6

[

A
T

4

A
M

A
B

]

{\displaystyle V = \frac{h}{6}\left[A\_T + 4A\_M + A\_B \right] }

where

h

{\displaystyle h}

 is the height,

A
T

{\displaystyle A\_T}

 i the area of the top,

A
M

{\displaystyle A\_M }

 is the area of the midsection, and

A
B

{\displaystyle A\_B }

 is the area of the base.

Use the prismoid formula to derive the formulas for computing the volume of the cube, the cone, and the sphere.

CubeConeSphere
## **Solution**

**Cube**

The cube is a square prism. Let the length of each edge be

d

{\displaystyle d}

. The area of the top, midsection and base must be the area of the same square.

A
T

=

A
M

=

A
B

=

d
2

{\displaystyle A\_T = A\_M = A\_B = d^2 }

# V

d
6

[

d
2

4

d
2

d
2

]

{\displaystyle V = \frac{d}{6}\left[d^2 + 4d^2+ d^2 \right] }

# V

d
6

[

6

d
2

]

{\displaystyle V = \frac{d}{6}\left[6d^2\right] }

# V

d
3

{\displaystyle V = d^3 }

**Cone**

The base of a cone is a circle with radius

r

{\displaystyle r}

. By proportion (similar triangles), the midsection is a circle with radius

r
2

{\displaystyle \frac{r}{2}}

. The top of the cone is a point with no area. Let

h

{\displaystyle h}

 be the height of the cone.

# V

h
6

[

0
4
π

(

r
2

)

2

π

r
2

]

{\displaystyle V = \frac{h}{6}\left[0 + 4\pi {\left(\frac{r}{2} \right)}^2 + \pi r^2 \right] }

# V

h
6

[

2
π

r
2

]

{\displaystyle V = \frac{h}{6}\left[2\pi r^2 \right] }

# V

π

r
2

h

3

{\displaystyle V = \frac{\pi r^2 h}{3} }

**Sphere**

Let

r

{\displaystyle r}

 be the radius of the sphere, and

d

{\displaystyle d}

 be the diameter of the sphere. The the top and base of a sphere are points with no area, hence,

A
T

=

A
B

=
0

{\displaystyle A\_T = A\_B = 0 }

. The midsection is a circle with area

π

r
2

{\displaystyle \pi r^2}

 or

π

d
2

4

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

.

# V

2
r

6

[

0
4
π

r
2

0

]

{\displaystyle V = \frac{2r}{6}\left[0 + 4\pi r^2+ 0 \right] }

# V

2
r

6

[

4
π

r
2

]

{\displaystyle V = \frac{2r}{6}\left[4\pi r^2 \right] }

# V

4
π

r
3

3

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

or

# V

d
6

[

0
4
π

(

d
2

)

2

0

]

{\displaystyle V = \frac{d}{6}\left[0 + 4\pi {\left(\frac{d}{2} \right)}^2 + 0 \right] }

# V

d
6

[

π

d
2

]

{\displaystyle V = \frac{d}{6}\left[\pi d^2 \right] }

# V

π

d
3

6

{\displaystyle V = \frac{\pi d^3}{6} }
Synthetic Division

## **Problem**

Use synthetic division to calculate

x

3

−
6

x

2

11
x
−
6

{\displaystyle x^{3}-6x^{2}+11x-6}

 divided by

x
−
4

{\displaystyle x-4}

.

## **Solution**

Arrange the coefficients of the polynomial as follows, with the zero of the divisor on the left.

4

1

−
6

11

−
6

{\displaystyle {\begin{array}{cc}{\begin{array}{r}4\\\\\end{array}}&{\begin{array}{|rrrr}\ 1&-6&11&-6\\&&&\\\hline \end{array}}\end{array}}}

The first coefficient after the bar is "dropped" to the bottom row.

4

1

−
6

11

−
6

1

{\displaystyle {\begin{array}{cc}{\begin{array}{r}4\\\\\\\end{array}}&{\begin{array}{|rrrr}\color {blue}1&-6&11&-6\\&&&\\\hline \color {blue}1&&&\\\end{array}}\end{array}}}

The dropped number is multiplied by the zero of the divisor, and placed in the next column.

4

1

−
6

11

−
6

4

1

{\displaystyle {\begin{array}{cc}{\begin{array}{r}4\\\\\\\end{array}}&{\begin{array}{|rrrr}1&-6&11&-6\\&\color {red}4&&\\\hline \color {blue}1&&&\\\end{array}}\end{array}}}

Add the numbers in the next column.

4

1

−
6

11

−
6

4

1

−
2

{\displaystyle {\begin{array}{cc}{\begin{array}{c}4\\\\\\\end{array}}&{\begin{array}{|rrrr}1&\color {green}-6&11&-6\\&\color {green}4&&\\\hline 1&\color {green}-2&&\\\end{array}}\end{array}}}

Repeat above procedure for the rest of the polynomial.

4

1

−
6

11

−
6

4

−
8

12

1

−
2

3

6

{\displaystyle {\begin{array}{cc}{\begin{array}{c}4\\\\\\\end{array}}&{\begin{array}{|rrrr}1&-6&11&-6\\&4&-8&12\\\hline 1&-2&3&6\end{array}}\end{array}}}

Here, the last term -6 is the remainder while the rest correspond to the coefficients of the quotient; hence, the quotient is

x

2

−
2
x
3

{\displaystyle x^{2}-2x+3}

 and the remainder is 6.
Molar Mass of Theobromine

## **Problem**

Theobromine, also known as 3,7-dimethylxanthine, is a compound found in chocolate. When caffeine is metabolized in the liver, around 12% becomes theobromine. The chemical formula of theobromine is

C

7

H

8

N

4

O

2

{\displaystyle \ce{ C7 H8 N4 O2 }}

. Determine the molar mass of theobromine.

Interestingly, the element bromine is not found in theobromine; rather, the name derives from the Greek words *'theo'* (god) and *'broma'* (food).

## **Solution**

Construct a table of the elements and the subscripts (number of respective elements) found in theobromine. Search the molar masses of each element from the periodic table.

C

7

H

8

N

4

O

2

{\displaystyle \ce{ C\_7 H\_8 N\_4 O\_2 }}

Multiply each the subscripts of each element with their respective molar mass.

 Hydrogen
 Nitrogen
 Oxygen

Add the net molar mass of each element to obtain the net molar mass of theobromine.

84.07

g

m
o
l

8.08

g

m
o
l

56.04

g

m
o
l

32.00

g

m
o
l

=
180.19

g

m
o
l

{\displaystyle 84.07 \; \frac{g}{mol} + 8.08 \; \frac{g}{mol} + 56.04 \; \frac{g}{mol} + 32.00 \; \frac{g}{mol} = 180.19 \; \frac{g}{mol} }
Quantum Free Particle

## **Problem**

Consider a quantum free particle characterized by the complex plane wave

Ψ
(
x
,
t
# )
A

e

i
(
k
x
−
ω
t
)

{\displaystyle \Psi (x,t) = A{e}^{i(kx-\omega t)}}

. Show that the free particle satisfies the Schrodinger equation

i
ℏ

∂
Ψ

∂
t

=

−

ℏ

2

2
m

∂

2

Ψ

∂

x

2

V
(
x
)
Ψ
(
x
,
t
)

{\displaystyle i\hbar \frac{\partial \Psi}{\partial t} = \frac{-{\hbar}^{2}}{2m} \frac{{\partial}^{2} \Psi}{\partial {x}^{2}} + V(x)\Psi (x,t) }

 for some potential function

V
(
x
)

{\displaystyle V(x)}

.

# H
T
V

{\displaystyle H = T + V }

 where

T

{\displaystyle T}

 is the kinetic energy and

V

{\displaystyle V}

 is the potential energy. Since the quantity

H

{\displaystyle H}

 is the total energy, let us rewrite the Hamiltonian as

# E

p

2

2
m

V
(
x
)

{\displaystyle E = \frac{{p}^{2}}{2m} + V(x) }

.

Now, we take the derivatives:

∂
Ψ

∂
t

=
−
i
ω
A

e

i
(
k
x
−
ω
t
)

=
−
i
ω
Ψ
(
x
,
t
)

{\displaystyle \frac{\partial \Psi}{\partial t} = -i \omega A{e}^{i(kx-\omega t)} = -i \omega \Psi (x,t) }

∂

2

Ψ

∂

x

2

=
−

k

2

A

e

i
(
k
x
−
ω
t
)

=
−

k

2

Ψ
(
x
,
t
)

{\displaystyle \frac{{\partial}^{2} \Psi}{\partial {x}^{2}} = -{k}^{2}A{e}^{i(kx-\omega t)} = -{k}^{2}\Psi (x,t) }

Since

# p

2
π
ℏ

λ

{\displaystyle p = \frac{2\pi \hbar}{\lambda} }

 and

# k

2
π

λ

{\displaystyle k = \frac{2\pi}{\lambda}}

, where

k

{\displaystyle k}

 is the wavenumber and

λ

{\displaystyle \lambda}

 is the wavelength, we have

# k

p
ℏ

{\displaystyle k = \frac{p}{\hbar} }

.

Therefore,

∂

2

Ψ

∂

x

2

=
−

p

2

ℏ

2

Ψ
(
x
,
t
)

{\displaystyle \frac{{\partial}^{2} \Psi}{\partial {x}^{2}} = -\frac{{p}^{2}}{{\hbar}^{2}}\Psi (x,t) }

Next, we multiply

Ψ
(
x
,
t
)

{\displaystyle \Psi(x,t)}

 to the Hamiltonian:

E
Ψ
(
x
,
t
# )

p

2

2
m

Ψ
(
x
,
t
)
V
(
x
)
Ψ
(
x
,
t
)

{\displaystyle E\Psi (x,t) = \frac{{p}^{2}}{2m}\Psi (x,t) + V(x)\Psi (x,t) }

Notice that the above equation can be expressed as

E
Ψ
(
x
,
t
# )

−

ℏ

2

2
m

∂

2

Ψ

∂

x

2

V
(
x
)
Ψ
(
x
,
t
)

{\displaystyle E\Psi (x,t) = \frac{{-{\hbar}^{2}}}{2m}\frac{{\partial}^{2} \Psi}{\partial {x}^{2}} + V(x)\Psi (x,t) }

Since the energy of matter waves is given by

# E
ℏ
ω

{\displaystyle E = \hbar \omega}

, we can show that

E
Ψ
(
x
,
t
# )

ℏ
ω

−
i
ω

Ψ
(
x
,
t
)

{\displaystyle E\Psi (x,t) = \frac{\hbar \omega}{-i \omega} \Psi (x,t) }

Combining all the right parts, we assemble the Schrödinger equation:

i
ℏ

∂
Ψ

∂
t

=

−

ℏ

2

2
m

∂

2

Ψ

∂

x

2

V
(
x
)
Ψ
(
x
,
t
)

{\displaystyle i\hbar \frac{\partial \Psi}{\partial t} = \frac{-{\hbar}^{2}}{2m} \frac{{\partial}^{2} \Psi}{\partial {x}^{2}} + V(x)\Psi (x,t) }
Nine Islands

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

This is a problem I created when studying polygon networks and combinatorial numbers.

## **Problem**

Suppose there are nine islands mutually connected by bridges, such that no two bridges intersect. If each island can be visited only once, how many possible paths are there to traverse between any selected pair of islands?

## **Solution**

Label the nine islands 0 to 8. Isolate our designated island of origin as island number 8. Enumerate the remaining islands numbers 0 to 7. To travel from our island of origin to another island, we may choose to visit any number of islands in any order (but only once). Remember that each island is interconnected.

Suppose we visit

k

{\displaystyle k}

 islands before reaching the island of destination. Then there are

n
!

(
n
−
k
)
!

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

 possible paths, for

# k
0
,
1
,
2
,
3
,
.
.
.
,
7

{\displaystyle k = 0,1,2,3,...,7 }

, and where

# n
7

{\displaystyle n = 7 }

. Therefore, the total number of paths is the sum of permutations:

∑

# k
0

7

7
!

(
7
−
k
)
!

=

{\displaystyle \sum\_{k=0}^{7} \frac{7!}{(7-k)!} = 13700. }
Molar Conversions

## **Problem**

**Part 1**

Calculate the mass of 0.550 moles of

NaOH

{\displaystyle \ce{NaOH}}

(sodium hydroxide). How many

NaOH

{\displaystyle \ce{NaOH}}

 molecules are there?

**Part 2**

Calculate the number of moles of 34.5 grams of

HCl

{\displaystyle \ce{HCl}}

(hydrochloric acid). How many

HCl

{\displaystyle \ce{HCl}}

 molecules are there?

## **Solution**

**Part 1**

First calculate the molar mass of

NaOH

{\displaystyle \ce{NaOH}}

.

22.99

g

m
o
l

16.00

g

m
o
l

1.01

g

m
o
l

=
40.00

g

m
o
l

{\displaystyle 22.99 \; \frac{g}{mol} + 16.00 \; \frac{g}{mol} + 1.01 \; \frac{g}{mol} = 40.00 \; \frac{g}{mol} }

0.550

m
o
l
×

40

g

m
o
l

=
22.0

g

{\displaystyle 0.550 \; mol \times \frac{40 \; g}{mol} = 22.0 \; g}

To calculate the number of

NaOH

{\displaystyle \ce{NaOH}}

molecules, simply multiply the number of moles of

NaOH

{\displaystyle \ce{NaOH}}

 with Avogadro's number.

0.550
×

(

6.02
×

10

23

)

=
3.31
×

10

23

{\displaystyle 0.550 \times \left(6.02 \times {10}^{23} \right)= 3.31 \times {10}^{23} }

**Part 2**

First calculate the molar mass of

HCl

{\displaystyle \ce{HCl}}

.

1.01

g

m
o
l

35.45

g

m
o
l

=
36.46

g

m
o
l

{\displaystyle 1.01 \; \frac{g}{mol} + 35.45 \; \frac{g}{mol} = 36.46 \; \frac{g}{mol} }

34.5

g
×

1

m
o
l

36.46

g

=
0.945

m
o
l

{\displaystyle 34.5 \; g \times \frac{1 \; mol}{36.46 \; g} = 0.945 \; mol }

To calculate the number of

HCl

{\displaystyle \ce{HCl}}

molecules, simply multiply the number of moles of

HCl

{\displaystyle \ce{HCl}}

 with Avogadro's number.

0.945
×

(

6.02
×

10

23

)

=
5.69
×

10

23

{\displaystyle 0.945 \times \left(6.02 \times {10}^{23}\right) = 5.69 \times {10}^{23} }
Rhind Papyrus Problem 67

## **Problem**

The following problem is from the Rhind Papyrus (1650 BC):

A shepherd had a flock of sheep. He had to give a portion of his flock to a lord as tribute. The shepherd was told to give two-thirds of one-third of his original flock. The shepherd gave 70 sheep. How many sheep did the shepherd original have?

## **Solution**

Let

n

{\displaystyle n}

 be the size of the original flock. Two-thirds of one-third of the original flock translates to

2
3

(

1
3

n

)

.

{\displaystyle \frac{2}{3} \left(\frac{1}{3} n \right). }

Since 70 sheep were given as tribute, we get:

2
3

(

1
3

n

)

=
70

{\displaystyle \frac{2}{3} \left(\frac{1}{3} n \right) = 70 }

2
9

# n
70

{\displaystyle \frac{2}{9} n = 70 }

# n

9
(
70
)

2

{\displaystyle n = \frac{9(70)}{2} }

# n

{\displaystyle n = 315. }

The shepherd had 315 sheep in his original flock.
Middle Chinese Phonology

The traditional analysis of the Chinese syllable, derived from the *fanqie* method, is into an initial consonant, or "initial", (Template:Transl Template:Lang) and a final (Template:Transl Template:Lang).
Modern linguists subdivide the final into an optional "medial" glide (Template:Transl Template:Lang), a main vowel or "nucleus" (Template:Transl Template:Lang) and an optional final consonant or "coda" (Template:Transl Template:Lang).
Most reconstructions of Middle Chinese include the glides Template:IPA and Template:IPA, as well as a combination Template:IPA, but many also include vocalic "glides" such as Template:IPA in a diphthong Template:IPA.
Final consonants Template:IPA, Template:IPA, Template:IPA, Template:IPA, Template:IPA, Template:IPA, Template:IPA and Template:IPA are widely accepted, sometimes with additional codas such as Template:IPA or Template:IPA.Template:Sfnp
Rhyming syllables in the *Qieyun* are assumed to have the same nuclear vowel and coda, but often have different medials.Template:Sfnp

Middle Chinese reconstructions by different modern linguists vary.Template:Sfnp These differences are minor and fairly uncontroversial in terms of consonants; however, there is a more significant difference as to the vowels.
The most widely used transcriptions are Li Fang-Kuei's modification of Karlgren's reconstruction and William Baxter's typeable notation.

### Initials

The preface of the *Yunjing* identifies a traditional set of 36 initials, each named with an exemplary character.
An earlier version comprising 30 initials is known from fragments among the Dunhuang manuscripts.
In contrast, identifying the initials of the *Qieyun* required a painstaking analysis of *fanqie* relationships across the whole dictionary, a task first undertaken by the Cantonese scholar Chen Li in 1842 and refined by others since.
This analysis revealed a slightly different set of initials from the traditional set.
Moreover, most scholars believe that some distinctions among the 36 initials were no longer current at the time of the rime tables, but were retained under the influence of the earlier dictionaries.Template:Sfnp

Early Middle Chinese (EMC) had three types of stops: voiced, voiceless, and voiceless aspirated. There were five series of coronal obstruents, with a three-way distinction between dental (or alveolar), retroflex and palatal among fricatives and affricates, and a two-way dental/retroflex distinction among stop consonants.
The following table shows the initials of Early Middle Chinese, with their traditional names and approximate values:
Spherical Capacitor

## **Problem**

A spherical capacitor consists of two concentric spherical shells. The spherical shells are conductors with radius

a

{\displaystyle a}

 and

b

{\displaystyle b}

. The inner shell has a total charge of

Q

{\displaystyle Q}

 distributed on its surface. The outer 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 two spherical shells is

Q

{\displaystyle Q}

. For spherical geometry

∮
E
⋅
d
# A

E

(
4
π

r
2

)

{\displaystyle \oint \boldsymbol{E} \cdot d\boldsymbol{A} = |\boldsymbol{E}| (4\pi r^2) }

E

(
4
π

r
2

# )

Q

ϵ
0

{\displaystyle |\boldsymbol{E}| (4\pi r^2) = \frac{Q}{\epsilon\_0} }

E

=

Q

4
π

ϵ
0

r
2

{\displaystyle |\boldsymbol{E}| = \frac{Q}{4\pi \epsilon\_0 r^2} }

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

# E

Q

4
π

ϵ
0

r
2

r

{\displaystyle \boldsymbol{E} = \frac{Q}{4\pi \epsilon\_0 r^2} \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

4
π

ϵ
0

r
2

d
r

{\displaystyle V = -\int\_{b}^{a} \frac{Q}{4\pi \epsilon\_0 r^2} dr }

# V

Q

4
π

ϵ
0

(

1
a

−

1
b

)

{\displaystyle V = \frac{Q}{4\pi \epsilon\_0} \left(\frac{1}{a} - \frac{1}{b} \right) }

# V

Q

4
π

ϵ
0

(

b
−
a

a
b

)

{\displaystyle V = \frac{Q}{4\pi \epsilon\_0} \left(\frac{b-a}{ab} \right) }

**Part 3: Capacitance**

# C

Q
V

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

# C
4
π

ϵ
0

(

a
b

b
−
a

)

{\displaystyle C = 4\pi \epsilon\_0 \left(\frac{ab}{b-a} \right) }
The Commutator

## **Problem**

Show that

x
^

,

p
^

{\displaystyle {\hat {x}},{\hat {p}}}

 are incompatible observables.

Recall that

x
^

=
x

{\displaystyle \hat{x} = x }

 and

p
^

=

ℏ
i

∂

∂
x

{\displaystyle {\hat {p}}={\frac {\hbar }{i}}{\frac {\partial }{\partial x}}}

.

## **Solution**

When working with the commutator, we must operate the Hermitian operators on a wavefunction.

[

x
^

,

p
^

]
# Ψ

x
^

(

p
^

Ψ
)
−

p
^

(
x
Ψ
)

{\displaystyle [{\hat {x}},{\hat {p}}]\Psi ={\hat {x}}({\hat {p}}\Psi )-{\hat {p}}(x\Psi )}

[

x
^

,

p
^

]
# Ψ

x
^

(

ℏ
i

∂
Ψ

∂
x

)

−

ℏ
i

∂
(
x
Ψ
)

∂
x

{\displaystyle [{\hat {x}},{\hat {p}}]\Psi ={\hat {x}}\left({\frac {\hbar }{i}}{\frac {\partial \Psi }{\partial x}}\right)-{\frac {\hbar }{i}}{\frac {\partial (x\Psi )}{\partial x}}}

[

x
^

,

p
^

]
# Ψ
x

ℏ
i

∂
Ψ

∂
x

−
x

ℏ
i

∂
Ψ

∂
x

−

ℏ
i

Ψ

{\displaystyle [{\hat {x}},{\hat {p}}]\Psi =x{\frac {\hbar }{i}}{\frac {\partial \Psi }{\partial x}}-x{\frac {\hbar }{i}}{\frac {\partial \Psi }{\partial x}}-{\frac {\hbar }{i}}\Psi }

[

x
^

,

p
^

]
# Ψ
−

ℏ
i

Ψ

{\displaystyle [{\hat {x}},{\hat {p}}]\Psi =-{\frac {\hbar }{i}}\Psi }

Drop the test wavefunction and simplify.

[

x
^

,

p
^

]
# Ψ
−

ℏ
i

⋅

i
i

{\displaystyle [{\hat {x}},{\hat {p}}]\Psi =-{\frac {\hbar }{i}}\cdot {\frac {i}{i}}}

[

x
^

,

p
^

# ]
i
ℏ

{\displaystyle [\hat{x},\hat{p}]=i\hbar }

The commutator of position and momentum is clearly not zero; hence, incompatible observables.
Property of Four Consecutive Integers

## **Problem**

Let

a
,
b
,
c
,
d

{\displaystyle a,b,c,d}

 be four consecutive integers.

**Part 1:** Show that

a
b
c
d
1

{\displaystyle abcd + 1 }

 is a perfect square.

**Part 2:** prove that the square root of

a
b
c
d
1

{\displaystyle abcd + 1 }

 is equal to the average of

a
d

{\displaystyle ad }

 and

b
c

{\displaystyle bc}

.

## **Solution**

**Part 1**

Let

a
b
c
d
1

{\displaystyle abcd + 1 }

 be represented as

(
n
−
1
)
(
n
)
(
n
1
)
(
n
2
)
1

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

, where

n
−
1
,
n
,
n
1
,
n
2

{\displaystyle n-1, n, n+1, n+2 }

 are integers. Thus,

a
b
c
d
# 1

n

4

2
n

3

−

n

2

−
2
n

{\displaystyle abcd + 1 = {n}^{4} + {2n}^{3} - {n}^{2} - 2n + 1. }

The right hand side can be factored as

(

n

2

n
−
1
)

2

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

, proving that it is a perfect square.

```
**Note**

```

A smart way to factor

a
b
c
d
1

{\displaystyle abcd + 1 }

 is to make a substitution from

(
n
−
1
)
(
n
)
(
n
1
)
(
n
2
)
1

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

.

(
n
−
1
)
(
n
)
(
n
1
)
(
n
2
)
# 1
(
n
−
1
)
(
n
2
)
(
n
)
(
n
1
)
1

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

(
n
−
1
)
(
n
)
(
n
1
)
(
n
2
)
# 1
(

n

2

n
−
2
)
(

n

2

n
)
1

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

Let

# u

n

2

n

{\displaystyle u = {n}^{2} + n }

, then

(

n

2

n
−
2
)
(

n

2

n
)
# 1
(
u
−
2
)
(
u
)
1

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

u

2

−
2
u
# 1

(
u
−
1
)

2

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

(
u
−
1
)

2

=

n

2

n
−
1

2

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

Therefore,

(
n
−
1
)
(
n
)
(
n
1
)
(
n
2
)
# 1

n

2

n
−
1

2

.

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

```
**End Note**

```

**Part 2**

The average of

a
d

{\displaystyle ad }

 and

b
c

{\displaystyle bc}

 is

1
2

[
(
n
−
1
)
(
n
2
)
n
(
n
−
1
)
# ]

1
2

(
2

n

2

2
n
−
2
# )

n

2

n
−
1

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

which is the square-root of

a
b
c
d
1

{\displaystyle abcd + 1 }

.
Optimal Timing of Art Collection

## **Problem**

The art collection of a recently deceased painter has an estimated value of

V
(
t
# )
200000
(
1.25

)

t

2

/

3

{\displaystyle V(t) = 200000(1.25)^{t^{2/3}}}

How long should the executor of the estate hold on to the collection before putting it up for sale if the discount rate under continuous compounding is 6 percent?

### **Hint**

Model the **present value function** as

P
(
t
# )
V
(
t
)
⋅

e

−
r
t

{\displaystyle P(t) = V(t)\cdot e^{-rt} }

, where

r

{\displaystyle r}

is the discount rate.

## **Solution**

Substituting the value

V

{\displaystyle V}

 into the depreciating present value

P
(
t
# )
V
(
t
)
⋅

e

−
r
t

{\displaystyle P(t) = V(t)\cdot e^{-rt} }

 gives

P
(
t
# )
200000
(
1.25

)

t

2

/

3

e

−
0.06
t

{\displaystyle P(t) = 200000(1.25)^{t^{2/3}}e^{-0.06t}}

Use logarithmic differentiation to solve for the critical value of time.

ln
⁡

P

=
ln
⁡

200000

t

2

/

3

ln
⁡

1.25

−
0.06
t

{\displaystyle \ln{P} = \ln{200000} + t^{2/3} \ln{1.25} -0.06t }

d

d
t

ln
⁡

P

=

d

d
t

ln
⁡

200000

d

d
t

(

t

2

/

3

ln
⁡

1.25

)

−

d

d
t

0.06
t

{\displaystyle {d \over dt}\ln{P} = {d \over dt} \ln{200000} + {d \over dt} \left(t^{2/3} \ln{1.25}\right) -{d \over dt}0.06t }

1
P

d
P

d
t

=
0

(

2
3

ln
⁡

1.25

)

t

−
1

/

3

−
0.06

{\displaystyle \frac{1}{P}{dP \over dt} = 0 + \left(\frac{2}{3}\ln{1.25}\right) t^{-1/3} -0.06 }

d
P

d
t

=
P

[

(

2
3

ln
⁡

1.25

)

t

−
1

/

3

−
0.06

]

{\displaystyle {dP \over dt} = P\left[\left(\frac{2}{3}\ln{1.25}\right) t^{-1/3} -0.06 \right] }

Set

d
P

d
t

=
0

{\displaystyle {dP \over dt} = 0 }

and solve for time

t
c

{\displaystyle t\_c }

. For this problem, solve

(

2
3

ln
⁡

1.25

)

t

−
1

/

3

−
# 0.06
0

{\displaystyle \left(\frac{2}{3}\ln{1.25}\right) t^{-1/3} -0.06 = 0 }

or equivalently

t

−
1

/

3

=

3
(
0.06
)

2
ln
⁡
1.25

{\displaystyle t^{-1/3} = \frac{3(0.06)}{2 \ln 1.25} }

t
≈
15.3

{\displaystyle t \approx 15.3 }

Therefore, the executor should hold on to the art collection for about 15.4 years before putting it up for sale.
One Sheet, Two Cylinders

## **Problem**

Consider a sheet of paper where its length is longer than its width. There are two ways to roll up the sheet of paper into a cylinder: along its length, and along its width. Which cylinder has the greater volume?

## **Solution**

Let

L

{\displaystyle L}

 be the length and

W

{\displaystyle W}

 be the width. The formula for calculating the volume of a cylinder is

# V
π

r
2

h

{\displaystyle V= \pi r^2h}

. Since

# C
2
π
r

{\displaystyle C = 2 \pi r}

, the volume of the cylinder can be expressed as

# V

C
2

h

4
π

{\displaystyle V = \frac{C^2 h}{4\pi} }

.

When rolled along its length, the volume of the cylinder is

# V

L
2

W

4
π

{\displaystyle V = \frac{L^2 W}{4\pi} }

.
When rolled along its width, the volume of the cylinder is

# V

W
2

L

4
π

{\displaystyle V = \frac{W^2 L}{4\pi} }

.

Since

L
>
W

{\displaystyle L > W }

, it follows that

L
2

W
>

W
2

L

{\displaystyle L^2 W > W^2 L }

. Therefore, the cylinder of greater volume is formed by rolling the paper along its length.
Stairway of Pigeons

## **Problem**

The following problem is from the medieval Latin manuscript *Propositiones ad Acuendos Juvenes* (Problems to Sharpen the Young), attributed to **Alcuin of York** (c. 735 – 804 AD):

Suppose there is a stairway which has 100 steps. One pigeon sat on the first step, two pigeons on the second step, three pigeons on the third step, four on the fourth step, five on the fifth step, and so on up to the hundredth step. How many pigeons are there altogether?

## **Solution**

Pair the pigeon on the first step with the 100 pigeons on the last step. The total is 101. Pair the 2 pigeons on the second step with the the 99 pigeons on the 99th step. The toal is also 101. Continue this pairing process until we reach the 50th and 51st step. There are 50 pairs of 101 pigeons. The total is therefore

50
×
# 101

{\displaystyle 50 \times 101 = 5050.}

This problem can also be thought of the sum of consecutive positive integers

∑

# k
1

n

# k
1
2
3
⋯
n
.

{\displaystyle \sum\_{k=1}^{n} k = 1 + 2 + 3 + \cdots + n.}

Determine the formula for calculating this sum using the arithmetic series formula. For this particular problem, the first term is

a
1

=
1

{\displaystyle a\_1=1}

 and the final term is

a
n

=
100

{\displaystyle a\_n = 100}

; therefore,

∑

# k
1

n

# k

n
(
n
1
)

2

.

{\displaystyle \sum\_{k=1}^{n} k = \frac{n(n+1)}{2}.}

The total number of pigeons is the sum of the first 100 positive integers:

∑

# k
1

100

# k

100
(
100
1
)

2

=

{\displaystyle \sum\_{k=1}^{100} k = \frac{100(100+1)}{2} = 5050.}

## **Historical Note**

There is a famous story about Carl Friedrich Gauss (1777 – 1855 AD) solving this problem at the tender age of 10. The story goes that one day at school, Gauss’s teacher wanted to take a rest and asked the students in his class to add the integers from 1 to 100. The teacher saw Gauss sitting idle, and when asked why he was not frantically doing addition, Gauss quickly replied that the sum was 5050. His classmates and teacher were astonished, and Gauss ended up being the only pupil to calculate the correct answer.