recursal/Fanatic-Fandom huggingface dataset:
Alchemist Xu 方士许我

**《梦溪笔谈·卷18技艺1》**
贾魏公[1]为相日，有方士姓许，对人未尝称名，无贵贱皆称 “我 ”，时人谓之“许我 ”。言谈颇有可采，然傲诞，视公卿蔑如也。公欲见，使人邀召数四，卒不至。又使门人苦邀致之，许骑驴径欲造丞相厅事。门吏止之不可。吏曰 ：“此丞相厅门，虽丞郎亦须下 。” 许曰 ：“我无所求于丞相，丞相召我来，若如此，但须我去耳 。” 不下驴而去。门吏急追之不还， 以白丞相。魏公又使人谢而召之，终不至。公叹曰 ：“许市井人耳。惟其无所求于人，尚不可以势屈，况其以道义自任者乎。”
**【今译】**
贾昌朝担任丞相的那段日子，有一位姓许的方士，他对人从来没有说过自己的名字，不论对方是有身份的人，还是社会地位低下的人他都自称“我”，当时的人就叫他“许我”。言谈很有些可取之处，但性情高傲不合常理，对达官显贵常显露轻视的样子。贾昌朝想见他，派人邀请了多次，还是不来。又派手下门客苦苦邀请他来，许我骑着毛驴要直接进丞相客厅。守门的官吏阻止他不能这样。说：“这是丞相的厅门，即使是寺丞郎官也必须下马。”许我说：“我没什么求丞相的，丞相召我来。如果这样，只能让我走了。”他便没下驴而去。守门的官吏急忙追赶他，他也不肯回头。守门的官吏将实情向丞相禀告。贾昌朝又派人去道歉并邀请他，许我最终还是没有来。贾昌朝感叹地说：“许我只不过是个平民百姓罢了。只因为他无所求于人，尚不能以权势让他屈服，更何况他是以道义为己任的人呢?”

***Mengxi Bitan* (Volume 18 Jiyi 1)**
When Jia Changchao (贾昌朝) became prime minister, there was an alchemist surnamed Xu, who never made known to others his given name. As he always called himself “I” when talking to people of high social status and of low social position, he got the nickname Xu Wo (许我, “I am Xu”). Knowledgeable and talented, he was a man of peevish temper who had little regard for eminent officials. Jia longed to see him. He had invited him for several times, but Xu declined all his invitations. Jia went so far as to sending his men to invite again and again. Xu eventually consented to meet Jia in his official residence. He rode on a donkey and intended to go directly into the reception hall. However, the guard in front of the entrance stopped him, saying, “This is the prime minister’s official residence! Whoever you are, you will have to dismount from your donkey!”

Xu replied, “I am not here to as the prime minister for help. It is the prime minister who invited me here. If I cannot enter, I will have to leave.”

The alchemist left without dismounting from his donkey. The guard then chased Xu in a hurry, pleading him to stay, but he refused to come back. The guard had to report the event to Jia, who sent his men to apologize to the alchemist and invited him again, but he would not come. Jia sighed with mixed feeling, saying, “Xu is only a civilian, yet he does not yield to no influential officials because he seeks no profit for himself. What a virtuous man he is!”

**【注释】**

[1] 贾昌朝（公元997年－1065年），北宋宰相、文学家、书法家。

Jia Changchao (997 – 1065) was a prime minister, writer, and calligrapher of the Northern Song dynasty.

**【今译】**

贾昌朝担任丞相的那段日子，有一位姓许的方士，他对人从来没有说过自己的名字，不论对方是有身份的人，还是社会地位低下的人他都自称“我”，当时的人就叫他“许我”。言谈很有些可取之处，但性情高傲不合常理，对达官显贵常显露轻视的样子。贾昌朝想见他，派人邀请了多次，还是不来。又派手下门客苦苦邀请他来，许我骑着毛驴要直接进丞相客厅。守门的官吏阻止他不能这样。说：“这是丞相的厅门，即使是寺丞郎官也必须下马。”许我说：“我没什么求丞相的，丞相召我来。如果这样，只能让我走了。”他便没下驴而去。守门的官吏急忙追赶他，他也不肯回头。守门的官吏将实情向丞相禀告。贾昌朝又派人去道歉并邀请他，许我最终还是没有来。贾昌朝感叹地说：“许我只不过是个平民百姓罢了。只因为他无所求于人，尚不能以权势让他屈服，更何况他是以道义为己任的人呢?”
Divergence Theorem Reduction

# F
ϕ
a

{\displaystyle \boldsymbol{F} = \phi \boldsymbol{a} }

, where

ϕ

{\displaystyle \phi}

 is a scalar function, and

a

{\displaystyle \boldsymbol{a} }

 is a constant vector. Use the divergence theorem to prove that

∭
V

∇
ϕ

d
# V

∬

∂
V

ϕ

n
^

d
S
.

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

## **Solution**

The divergence theorem states that

∭
V

∇
⋅
F

d
# V

∬

∂
V

F
⋅

n
^

d
S
.

{\displaystyle \iiint\_V \nabla \cdot \boldsymbol{F} \,dV = \iint\_{\partial V} \boldsymbol{F} \cdot \hat{\boldsymbol{n}} \,dS. }

Calculate the divergence of

# F
ϕ
a

{\displaystyle \boldsymbol{F} = \phi \boldsymbol{a} }

.

∇
⋅
(
ϕ
a
# )
(
∇
ϕ
)
a
ϕ
(
∇
⋅
a
)

{\displaystyle \nabla \cdot (\phi \boldsymbol{a}) = (\nabla \phi)\boldsymbol{a} + \phi(\nabla \cdot \boldsymbol{a})}

Thus

∭
V

(
∇
ϕ
)
⋅
a
ϕ
(
∇
⋅
a
)

d
# V

∬

∂
V

ϕ
a
⋅

n
^

d
S

{\displaystyle \iiint\_V (\nabla \phi) \cdot \boldsymbol{a} + \phi(\nabla \cdot \boldsymbol{a}) \,dV = \iint\_{\partial V} \phi\boldsymbol{a} \cdot \hat{\boldsymbol{n}} \,dS }

Since

a

{\displaystyle \boldsymbol{a} }

 is a constant vector,

∇
⋅
# a
0

{\displaystyle \nabla \cdot \boldsymbol{a} =
 0 }

.

Consequently,

∭
V

(
∇
ϕ
)
⋅
a

d
# V

∬

∂
V

ϕ
a
⋅

n
^

d
S

{\displaystyle \iiint\_V (\nabla \phi) \cdot \boldsymbol{a} \,dV = \iint\_{\partial V} \phi\boldsymbol{a} \cdot \hat{\boldsymbol{n}} \,dS }

or

∭
V

(
∇
ϕ
)
⋅
a

d
V
−

∬

∂
V

ϕ
a
⋅

n
^

d
# S

{\displaystyle \iiint\_V (\nabla \phi) \cdot \boldsymbol{a} \,dV - \iint\_{\partial V} \phi\boldsymbol{a} \cdot \hat{\boldsymbol{n}}\, dS = 0. }

Since

a

{\displaystyle \boldsymbol{a} }

 is a constant vector, and the dot product is commutative, it can be factored out.

a
⋅

(

∭
V

∇
ϕ

d
V
−

∬

∂
V

ϕ

n
^

d
S

)

=

{\displaystyle \boldsymbol{a} \cdot \left(\iiint\_V \nabla \phi\,dV - \iint\_{\partial V} \phi \hat{\boldsymbol{n}} \,dS \right) = 0. }

Therefore,

∭
V

∇
ϕ

d
# V

∬

∂
V

ϕ

n
^

d
S
.

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

Click here for an application of this identity.
Angle Maximization Problem

## **Problem**

The following optimization problem was first posed and solved in 1471 by a late Medieval German astronomer Johannes Müller von Königsberg (1436 – 1476 AD), known more famously as Regiomontanus. Regiomontanus solved the problem without calculus, using clever tricks in elementary geometry. The solution presented below uses differentiation.

Refer to the diagram on the right (Figure 1). If the heights

a

{\displaystyle a}

 and

b

{\displaystyle b}

 are given constants, at what horizontal distance

x

{\displaystyle x}

 should one stand to maximize the viewing angle

θ

{\displaystyle \theta}

. Solve this problem using derivatives.

## **Hint**

(1) There are two other angles in Figure 1,

α

{\displaystyle \alpha}

 and

β

{\displaystyle \beta}

, where

# θ
α
−
β

{\displaystyle \theta = \alpha - \beta }

.

## **Solution**

Consider the angles

α

{\displaystyle \alpha}

 and

β

{\displaystyle \beta}

 such that

# θ
α
−
β

{\displaystyle \theta = \alpha - \beta }

. According to the above diagram,

tan
⁡

α

=

b
x

{\displaystyle \tan{\alpha} = \frac{b}{x} }

tan
⁡

β

=

a
x

{\displaystyle \tan{\beta} = \frac{a}{x} }

.
Thus, the viewing angle

θ

{\displaystyle \theta}

 can be expressed as

# θ
arctan
⁡

(

b
x

)

−
arctan
⁡

(

a
x

)

.

{\displaystyle \theta = \arctan{\left(\frac{b}{x}\right)} - \arctan{\left(\frac{a}{x}\right)}. }

To maximize the viewing angle

θ

{\displaystyle \theta}

, derive the above function with respect to

x

{\displaystyle x}

:

d
θ

d
x

=

1

1

(

b
x

)

2

⋅

−
b

x
2

−

1

1

(

a
x

)

2

⋅

−
a

x
2

{\displaystyle \frac{d\theta}{dx} = \frac{1}{1+{\left(\frac{b}{x} \right)}^{2}} \cdot \frac{-b}{x^2} - \frac{1}{1+{\left(\frac{a}{x} \right)}^{2}} \cdot \frac{-a}{x^2}}

d
θ

d
x

=

−
b

x
2

b
2

a

x
2

a
2

.

{\displaystyle \frac{d\theta}{dx} = \frac{-b}{x^2+b^2} + \frac{a}{x^2+a^2}. }

Set

d
θ

d
x

=
0

{\displaystyle \frac{d\theta}{dx} = 0 }

 and solve for

x

{\displaystyle x}

:

# 0

−
b

(

x
2

a
2

)

a

(

x
2

b
2

)

(

x
2

a
2

)

(

x
2

b
2

)

{\displaystyle 0 = \frac{-b\left(x^2+a^2\right) + a\left(x^2+b^2\right)}{\left(x^2+a^2\right) \left(x^2+b^2\right)} }

# 0
−
b

x
2

−

a
2

b
a

x
2

a

b
2

{\displaystyle 0 = -bx^2 - a^2 b + ax^2 + ab^2 }

(
b
−
a
)

x
2

=
a

b
2

−

a
2

b

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

(
b
−
a
)

x
2

=
a
b
(
b
−
a
)

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

x
2

=
a
b

{\displaystyle x^2 = ab }

# x
±

a
b

.

{\displaystyle x = \pm \sqrt{ab}. }

Testing the regions of the derivative function, we discover that

{

d
θ

d
x

>
0

if

0
≤
x
<

a
b

d
θ

d
x

=
0

if

# x

a
b

d
θ

d
x

<
0

if
x
>

a
b

.

{\displaystyle \begin{cases}
\frac{d \theta}{dx} > 0 & \text{if} \; 0 \le x < \sqrt{ab} \\[5 pt]
\frac{d \theta}{dx} = 0 & \text{if} \;x = \sqrt{ab} \\[5 pt]
\frac{d \theta}{dx} < 0 & \text{if } x > \sqrt{ab}. \end{cases} }

Hence, the angle increases on the interval

0
≤
x
<

a
b

{\displaystyle 0 \le x < \sqrt{ab} }

 and decreases when

x
>

a
b

{\displaystyle x > \sqrt{ab} }

. Therefore, the angle is maximized precisely at

x

a
b

{\displaystyle x \sqrt{ab}}

.
Chinese Saying – 雲開見月

**Translation:** After the clouds disperse, the moon is in view.

**English equivalent:** Things are brightening up.

**Source:** Biography of Yuan Shao from the *History of the Later Han*《後漢書·袁紹傳》

## **Historical Notes**

***History of the Later Han*** (Chinese: 後漢書; Pinyin: *Hòu hàn shū*), chronicles the history of the Eastern Han dynasty (25 – 220 AD). The book was compiled by Fan Ye (Chinese: 范曄; Pinyin: *Fàn Yè*; 398 – 446 AD) in the 5th century during the Northern and Southern dynasties period, using a number of earlier histories and documents as sources.

**Yuan Shao** (Chinese: 袁紹;  Pinyin: *Yuán Shào*; ? – 202 AD), was a Chinese military general, politician, and warlord who lived in the late Eastern Han dynasty. He occupied the northern territories of China during the civil wars that occurred prior to the Three Kingdoms period (220 – 280 AD).
2D Random Walks with Python

**By: Tao Steven Zheng (郑涛)**

```
import random
import matplotlib.pyplot as plt
import turtle

def randwalk(n):
    x = 0
    y = 0
    step\_x = [x]
    step\_y = [y]

    for i in range(1,n+1):
        move = random.randint(0,3)
        if move == 0:
            x += 1
        if move == 1:
            x += -1
        if move == 2:
            y += 1
        if move == 3:
            y += -1
        step\_x.append(x)
        step\_y.append(y)
    return [step\_x, step\_y]

'''Plotting the 2-D Random Walks'''
randwalk1 = randwalk(1000)
randwalk2 = randwalk(1000)
randwalk3 = randwalk(1000)

plt.plot(randwalk1[0], randwalk1[1], 'r-', label = "randwalk1")
plt.plot(randwalk2[0], randwalk2[1], 'g-', label = "randwalk2")
plt.plot(randwalk3[0], randwalk3[1], 'b-', label = "randwalk3")
plt.title("2-D Random Walks")
plt.legend(loc='upper center', bbox\_to\_anchor=(0.5,-0.1),
           fancybox=True, shadow=True)
plt.show()

'''Animate the 2-D Random walk'''
#import turtle library to animate the random walk
turtle.speed('fastest')

walk = randwalk(1000)

for x, y in zip(\*walk):
    #multiply by 5 to enlarge the random walk
    turtle.goto(x\*5,y\*5)
turtle.exitonclick() #neat feature of turtle

```
Jiuzhang Suanshu Gougu 5

This classic problem from the *Jiuzhang Suanshu* (九章算术) is called **Vine Winding up a Tree** (葛缠木).

## **Problem**

今有木長二丈，圍之三尺。葛生其下，纏木七周，上與木齊。問：葛長幾何？

Suppose there is a tree 2 *zhang* tall whose circumference is 3 *chi*. A kudzu vine grows from the bottom of the tree and coils around the tree 7 times, such that the top reaches the top of the tree. Question: How long is the kudzu vine?

**Note:** 1 *zhang* is 10 *chi*

## **Solution**

The problem requires the Pythagorean theorem to compute the length of a helical vine. As shown in the figure above, let

a

{\displaystyle a}

 be the height of the tree,

b

{\displaystyle b}

 be the circumference of the tree multiplied by the number of coils of the tree, and

c

{\displaystyle c}

 be the length of the kudzu vine. It is known that the height of the tree is 2 *zhang*, or 20 *chi*. The base of the right triangle is 7 times the circumference of the tree. Since the circumference of the tree is 3 *chi*, the base is 21 *chi*. When unwound, the length of the vine is the hypotenuse of the right triangle.

Hence,

# c

a

2

b

2

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

# c

20

2

21

2

=
29

{\displaystyle c={\sqrt {{20}^{2}+{21}^{2}}}=29}

The vine measures 29 *chi*, or 2 *zhang* 9 *chi*, in length.
De Morgan's Law

## **Problem**

Use sets to prove De Morgan's laws.

**Law 1**

∼
(
A
∪
B
# )
(
∼
A
)
∩
(
∼
B
)

{\displaystyle \sim (A \cup B) = (\sim A) \cap (\sim B) }

**Law 2**

∼
(
A
∩
B
# )
(
∼
A
)
∪
(
∼
B
)

{\displaystyle \sim (A \cap B) = (\sim A) \cup (\sim B) }

**Note:**

∼

{\displaystyle \sim}

 means **not**.

∪

{\displaystyle \cup}

 means **union**, which is equivalent to **or**

∨

{\displaystyle \lor}

.

∩

{\displaystyle \cap}

 means **intersection**, which is equivalent to **and**

∧

{\displaystyle \and }

.

In set theory, De Morgan's law is usually written with the complement of the set.

(
A
∪
B
)

C

=

A
C

∩

B
C

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

(
A
∩
B
)

C

=

A
C

∪

B
C

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

## **Solution**

**Law 1**

x
∈
[
∼
(
A
∪
B
)
]

{\displaystyle x \in [\sim (A \cup B)] }

 implies

x
∉
(
A
∪
B
)

{\displaystyle x \notin (A \cup B) }

x
∉
A
∧
x
∉
B

{\displaystyle x \notin A \and x \notin B }

x
∈
(
∼
A
)
∧
x
∈
(
∼
B
)

{\displaystyle x \in (\sim A) \and x \in (\sim B) }

x
∈
[
(
∼
A
)
∩
(
∼
B
)
]

{\displaystyle x \in [(\sim A) \cap (\sim B)] }

**Law 2**

x
∈
[
∼
(
A
∩
B
)
]

{\displaystyle x \in [\sim (A \cap B)] }

 implies

x
∉
(
A
∩
B
)

{\displaystyle x \notin (A \cap B) }

x
∉
A
∨
x
∉
B

{\displaystyle x \notin A \or x \notin B }

x
∈
(
∼
A
)
∨
x
∈
(
∼
B
)

{\displaystyle x \in (\sim A) \or x \in (\sim B) }

x
∈
[
(
∼
A
)
∪
(
∼
B
)
]

{\displaystyle x \in [(\sim A) \cup (\sim B)] }

## **Logic and De Morgan' laws**

In logic, De Morgan's laws are expressed as

¬
(
P
∨
Q
)

⟺

(
¬
P
)
∧
(
¬
Q
)
,

{\displaystyle \neg(P\or Q)\iff(\neg P)\and(\neg Q),}

and

¬
(
P
∧
Q
)

⟺

(
¬
P
)
∨
(
¬
Q
)

{\displaystyle \neg(P\and Q)\iff(\neg P)\or(\neg Q)}

for propositions

P
,
Q

{\displaystyle P, Q}

.
Fibonacci Series 1

## **Problem**

Consider the infinite series

∑

# k
1

∞

F
k

9
n

=

1
9

1
81

2
729

3
6561

5
59049

⋯

{\displaystyle \sum\_{k=1}^\infty \frac{F\_k}{9^n} = \frac{1}{9} + \frac{1}{81} + \frac{2}{729} + \frac{3}{6561} + \frac{5}{59049} + \cdots }

where

F
k

{\displaystyle F\_k }

 represents the Fibonacci sequence

F

n

=

F

n
−
1

F

n
−
2

{\displaystyle F\_{n} = F\_{n-1} + F\_{n-2} }

. Determine the total sum.

## **Solution**

Let

S

{\displaystyle S \; }

represent the infinite series:

# S

1
9

1

9
2

2

9
3

3

9
4

5

9
5

⋯

{\displaystyle S = \frac{1}{9} + \frac{1}{9^2} + \frac{2}{9^3} + \frac{3}{9^4} + \frac{5}{9^5} + \cdots }

Multiply both sides of the equation with

1
9

{\displaystyle \frac{1}{9}}

:

1
9

# S

1

9
2

1

9
3

2

9
4

3

9
5

5

9
6

⋯

{\displaystyle \frac{1}{9}S = \frac{1}{9^2} + \frac{1}{9^3} + \frac{2}{9^4} + \frac{3}{9^5} + \frac{5}{9^6} + \cdots }

Next, subtract

1
9

S

{\displaystyle \frac{1}{9}S }

 from

S

{\displaystyle S \; }

:

S
−

1
9

# S

1
9

0

1

9
3

1

9
4

2

9
5

3

9
6

5

9
7

⋯

{\displaystyle S - \frac{1}{9}S = \frac{1}{9} + 0 + \frac{1}{9^3} + \frac{1}{9^4} + \frac{2}{9^5} + \frac{3}{9^6} + \frac{5}{9^7} + \cdots }

8
9

# S

1
9

1

9
2

(

1
9

1

9
2

2

9
3

3

9
4

5

9
5

⋯

)

{\displaystyle \frac{8}{9}S = \frac{1}{9} + \frac{1}{9^2} \left(\frac{1}{9} + \frac{1}{9^2} + \frac{2}{9^3} + \frac{3}{9^4} + \frac{5}{9^5} + \cdots \right) }

8
9

# S

1
9

1

9
2

S

{\displaystyle \frac{8}{9}S = \frac{1}{9} + \frac{1}{9^2}S }

Lastly, solve for

S

{\displaystyle S \; }

:

8
9

# S

1
9

1
81

S

{\displaystyle \frac{8}{9}S = \frac{1}{9} + \frac{1}{81}S }

# S

9
71

{\displaystyle S = \frac{9}{71} }
Jiuzhang Suanshu Fangtian 37

## **Problem**

答曰：二畝五十五步。

術曰：并中外周而半之，以徑乘之為積步。

92

{\displaystyle 92}

 *bu*, and outer circumference of

122

{\displaystyle 122 }

 *bu*. The width (徑 *jing*) of the field is

5

{\displaystyle 5}

 *bu*. Question: What is the area of the field?

Answer: 2 *mu* 55 [square] *bu*.
Method: Add the inner and outer circumferences and halve the result, then multiply the width to obtain the [area] product in [square] *bu*.

## **Solution**

This problem exploits the ring-shaped field method [annulus]. This method involves three quantities: the outer circumference (外周 wai zhou), the inner circumference (中周 zhong zhou), and *jing* (徑). The word *jing* is typically defined as the diameter of a circle. However, for this type of problem, the *jing* of an annulus is defined as the distance between the outer and central circumference. Consider this the width of the annulus.

**Derivation**

The area of a circle given its diameter is

π

d
2

4

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

. The area of the annulus is given by the areal difference of two concentric circles

# A

π
4

(

D
2

−

d
2

)

{\displaystyle A = \frac{\pi}{4}(D^2-d^2) }

where

D

{\displaystyle D}

 is the outer diameter, and

d

{\displaystyle d}

 is the central diameter. The outer circumference and central circumference are

C
1

=
π
D

{\displaystyle C\_1 = \pi D }

 and

C
2

=
π
d

{\displaystyle C\_2 = \pi d }

 respectively. One can factorize the area of an annulus as follows

# A

π
4

(
D
d
)
(
D
−
d
)
.

{\displaystyle A = \frac{\pi}{4} (D+d)(D-d). }

This is equivalent to the prescribed method in the *Jiuzhang Suanshu*

# A

π
D
π
d

2

(

D
−
d

2

)

{\displaystyle A = \frac{\pi D + \pi d}{2}\left(\frac{D-d}{2}\right) }

where

D
−
d

2

{\displaystyle \frac{D-d}{2} }

 is the width (徑 *jing*).

Let

w

{\displaystyle w}

 represent the width (徑 *jing*),

C
1

{\displaystyle C\_1}

the outer circumference (外周 *wai zhou*), and

C
2

{\displaystyle C\_2}

 the inner circumference (中周 *zhong zhou*) . The area of he annulus can be alternatively expressed as

# A

(

C
1

C
2

2

)

w
.

{\displaystyle A = \left(\frac{C\_1 + C\_2}{2}\right)w. }

This formula shows that the area of the annulus is a circular analogon of the trapezoid!

**Calculation**

# A

1
2

(

92
122

)

(
5
)

=
535

b
u

2

{\displaystyle A = \frac{1}{2} \left(92 + 122\right)\left(5\right)= 535\, {bu}^{2}}

This answer is correct, but the solution from the *Jiuzhang Suanshu* states

2

m
u

55

b
u

2

{\displaystyle 2 \, mu \, 55 \, {bu}^{2} }

. This is because

1

m
# u
240

b
u

2

{\displaystyle 1 \, mu = 240 \, {bu}^{2} }

.

# w

D
−
d

2

{\displaystyle w = \frac{D-d}{2} }

, it follows that

# w

1
2

(

C
1

π

−

C
2

π

)

=

C
1

−

C
2

2
π

.

{\displaystyle w = \frac{1}{2} \left(\frac{C\_1}{\pi} - \frac{C\_2}{\pi} \right) = \frac{C\_1 - C\_2}{2 \pi}. }

From the given measurements of the central circumference and outer circumference suggests that the *jing* of this annular field should be

# w

122
−
92

2
π

=
4.77.

{\displaystyle w = \frac{122 - 92}{2 \pi} = 4.77. }

The value given in the *Jiuzhang Suanshu* assumes

# π
3

{\displaystyle \pi = 3 }

.
Hassan bin al-Haytham (الحسن بن الهيثم)

**By: Tao Steven Zheng (郑涛)**

### Known for

Alhazen's problem, experimental science, scientific methodology, animal psychology

### Fields

Catoptrics, visual perception, moon illusion

### Notable works

*Book of Optics* (كتاب المناظر Kitab al-Manazir), *Doubts Concerning Ptolemy* (Al-Shukūk ‛alā Batlamyūs),

**Hassan bin al-Haytham** was an Arab mathematician, astronomer, physicist, and theologian. As an influential scientist of the Islamic world, his work, attitude, and approach to scientific research was ahead of his time. As an astronomer, he was skeptical of the overly-elaborate Ptolemaic model of planetary motion. As a physicist, he applied reason and controlled experiments to the investigation of optics. By uniting mathematics and empiricism centuries before the experiments of Galileo and the philosophy of Francis Bacon, al-Haytham may be considered the first true scientist in the modern sense. In the West, his name is Latinized as Alhazen. This Latinized name will be used in this biography.

## **Discoveries**

**Optics of the Eye**

Alhazen contested the ancient theory of light rays being emitted from the eyes. Through careful observation and experiments, Alhazen demonstrated that that light enters the eye, in which images are projected within. He also mapped the internal anatomy of the eye, and explained how the brain and eye are connected to produce the sense of vision. This is the first accurate explanation of vision from a physical and biological standpoint.

**Atmospheric Optics**

Alhazen studied optics extensively, and recorded several observations and experiments in his seminal work *Book of Optics* (كتاب المناظر Kitab al-Manazir). Aside from the general theory of optics, Alhazen also investigated the optics of atmospheric and meteorological phenomena. He made several observations on refraction, the bending of light through materials with different densities. Rainbows and the afterglow at the horizon during twilight were some of the atmospheric phenomena that Alhazen explained with the principle of refraction.

**Lunes of Alhazen**

In attempt to square the circle using a compass and straightedge (proven to be impossible by Lindemann in 1882), Alhazen proved that the sum of the areas of the two lunes (crescent shape) is equal to the area of that the right triangle in which the lunes subtend.

Lunes of Alhazen

**Primality Tests**

Eratosthenes was credited to be the first mathematician to develop a sieve for finding prime numbers. However, the ancient Greeks did not consider tests for primality. Alhazen stated two primality tests but did not prove them. The first primality test is more of a property of perfect numbers (a whole number that is equal to the sum of its positive divisors). Alhazen stated that every even perfect number is of the form

2

n
−
1

(

2
n

−
1
)

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

where

2
n

−
1

{\displaystyle 2^n-1}

 is prime. Alhazen also stated what is known in the West as Wilson's theorem, which states

(
p
−
1
)
!
≡
−
1

(
mod

p
)

{\displaystyle (p-1)! \equiv -1 \pmod p}

where

p

{\displaystyle p}

 is prime.
Convex Lens

## **Problem**

A 3-cm high arrowhead oriented upwards is placed 75 cm from a thin convex lens (Figure 1). The focal length of the convex lens is 25 cm.

**Part 1:** Where is the image located?

**Part 2:** What is the lateral magnification? Is the image real or virtual?

**Part 3:** What happens to the image if the object is placed at the focus?

## **Hint**

(1) Thin lens formula:

1

d
o

1

d
i

=

1
f

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

.

(2) Magnification formula:

# M
−

d
i

d
o

=

h
i

h
o

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

.

## **Solution**

**Part 1**

For thin lenses, we can use the approximate formula

1

d
o

1

d
i

=

1
f

.

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

By rearranging this formula for the image distance, we get

d

i

=

d

o

f

d

o

−
f

{\displaystyle d\_{i}={\frac {d\_{o}f}{d\_{o}-f}}}

d
i

=

(
75

c
m
)
(
25

c
m
)

75

c
m
−
25

c
m

=
37.5

c
m
.

{\displaystyle d\_i = \frac{(75 \; cm)(25 \; cm)}{75 \; cm - 25 \; cm} = 37.5 \; cm .}

This result is positive, which means the image is real and located beyond the lens.

**Part 2**

The magnification is calculated by the fraction

# M
−

d
i

d
o

=

h
i

h
o

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

thus

# M
−

37.5

c
m

75

c
m

=
−
0.5.

{\displaystyle M=-{\frac {37.5\;cm}{75\;cm}}=-0.5.}

−
0.5

=

h

i

3

c
m

h

i

=
−
1.5

c
m

{\displaystyle {\begin{aligned}-0.5&={\frac {h\_{i}}{3\;cm}}\\[5pt]h\_{i}&=-1.5\;cm\end{aligned}}}

The magnification is

−
0.5

{\displaystyle -0.5}

, which means the image is smaller than the object. The height of the image is therefore

h

i

=
−
1.5

c
m

{\displaystyle h\_{i}=-1.5\;cm}

. Since the magnification is negative, we get an inverted image. The inverted image makes it a virtual image.

**Part 3**

No image is formed along the principle axis and the image is located very far away (the image will form at infinity). This can be verified by calculation. Make the substitution

d

o

=
f

{\displaystyle d\_{o}=f}

1

d

o

1

d

i

=

1
f

1
f

1

d

i

=

1
f

1

d

i

=
0

d

i

→
∞

{\displaystyle {\begin{aligned}{\frac {1}{d\_{o}}}+{\frac {1}{d\_{i}}}&={\frac {1}{f}}\\[5pt]{\frac {1}{f}}+{\frac {1}{d\_{i}}}&={\frac {1}{f}}\\[5pt]{\frac {1}{d\_{i}}}&=0\\[5pt]d\_{i}&\rightarrow \infty \\[5pt]\end{aligned}}}
Infinite Arithmetic-Geometric Series

This is one of my favourite problems recorded in one of my high school mathematical notebooks.

## **Problem**

Show that if

r

<
1

{\displaystyle |r|<1}

, and

a

{\displaystyle a}

 and

b

{\displaystyle b}

 are constants, then the series

∑

# n
0

∞

(
a
n
b
)

r

n

=

a

1
−
r

b
r

(
1
−
r

)
2

{\displaystyle \sum\_{n=0}^{\infty} (a+nb){r}^{n} = \frac{a}{1-r} + \frac{br}{(1-r)^2} }

.
**Hint**

(1) Differentiation will help save a lot of work!

## **Solution**

It is strategic to break the sum into two parts

∑

# n
0

∞

(
a
n
b
)

r

n

=

∑

# n
0

∞

a

r

n

∑

# n
0

∞

n
b

r

n

{\displaystyle \sum\_{n=0}^{\infty} (a+nb){r}^{n} = \sum\_{n=0}^{\infty} a{r}^{n} + \sum\_{n=0}^{\infty} nb{r}^{n} }

.
When

r

<
1

{\displaystyle |r|<1}

, the first part of the sum is a convergent infinite geometric series. The formula for calculating this series is well known:

∑

# n
0

∞

a

r

n

=
a
(
1
r

r
2

.
.
.
# )

a

1
−
r

{\displaystyle \sum\_{n=0}^{\infty} a{r}^{n} = a(1+r+r^2+...) = \frac{a}{1-r}}

.
Now consider the second part of the sum and factor our

b
r

{\displaystyle br }

:

∑

# n
0

∞

n
b

r

n

=
b
r
2
b

r
2

3
b

r
3

.
.
.

{\displaystyle \sum\_{n=0}^{\infty} nb{r}^{n} = br+2br^2+3br^3+...}

∑

# n
0

∞

n
b

r

n

=
b
r
(
1
2
r
3

r
2

.
.
.
)

{\displaystyle \sum\_{n=0}^{\infty} nb{r}^{n} = br(1+2r+3r^2+...)}

.
It is known that from the convergent infinite geometric series that

1
r

r
2

.
.
# .

1

1
−
r

{\displaystyle 1+r+r^2... = \frac{1}{1-r} }

Notice that by differentiating the left-hand side of the above series, we get

1
2
r
3

r
2

.
.
# .

d

d
r

(

1

1
−
r

)

{\displaystyle 1+2r+3r^2+... = \frac{d}{dr}\left(\frac{1}{1-r}\right)}

.
Consequently,

1
2
r
3

r
2

.
.
# .

1

(
1
−
r

)
2

{\displaystyle 1+2r+3r^2+... = \frac{1}{(1-r)^2}}

and

∑

# n
0

∞

n
b

r

n

=

b
r

(
1
−
r

)
2

{\displaystyle \sum\_{n=0}^{\infty} nb{r}^{n} = \frac{br}{(1-r)^2}}

.
Therefore, combining both parts gives the result

∑

# n
0

∞

(
a
n
b
)

r

n

=

a

1
−
r

b
r

(
1
−
r

)
2

{\displaystyle \sum\_{n=0}^{\infty} (a+nb){r}^{n} = \frac{a}{1-r} + \frac{br}{(1-r)^2} }

.
Diffraction Grating

## **Problem**

**Part 1:** Light of wavelength 700 nm is directed onto a diffraction grating with 5000 lines/cm. What are the angular deviations of the first and second order maxima from the central maxima?

**Part 2:** Explain why X rays, rather than visible light, are used to study crystal structure?

## **Hint**

(1) The nth-order maxima is calculated using the formula

d
sin
⁡
# θ
n
λ
.

{\displaystyle d\sin \theta = n\lambda.}

## **Solution**

**Part 1**

The first-order maxima occur when

d
sin
⁡

θ
1

=
1
λ

1
×

10

−
6

5000

sin
⁡

θ
1

=
1
(
700
×

10

−
9

)

2
×

10

−
6

sin
⁡

θ
1

=
7
×

10

−
6

sin
⁡

θ
1

=
0.35

θ
1

=

20.49
∘

.

{\displaystyle \begin{align}
d\sin \theta\_1 &= 1\lambda \\[5 pt]
\frac{1 \times 10^{-6}}{5000} \sin \theta\_1 &= 1 (700 \times 10^{-9}) \\[5 pt]
2 \times 10^{-6}\sin \theta\_1 &= 7 \times 10^{-6} \\[5 pt]
\sin \theta\_1 &= 0.35 \\[5 pt]
\theta\_1 &= 20.49^\circ.
\end{align}}

The second-order maxima occur when

d
sin
⁡

θ
2

=
2
λ

1
×

10

−
6

5000

sin
⁡

θ
2

=
2
(
700
×

10

−
9

)

2
×

10

−
6

sin
⁡

θ
2

=
1.4
×

10

−
6

sin
⁡

θ
2

=
0.7

θ
2

=

44.43
∘

.

{\displaystyle \begin{align}
d\sin \theta\_2 &= 2\lambda \\[5 pt]
\frac{1 \times 10^{-6}}{5000} \sin \theta\_2 &= 2 (700 \times 10^{-9}) \\[5 pt]
2 \times 10^{-6}\sin \theta\_2 &= 1.4 \times 10^{-6} \\[5 pt]
\sin \theta\_2 &= 0.7 \\[5 pt]
\theta\_2 &= 44.43^\circ.
\end{align}}

**Part 2**

For diffraction to occur, the distance between the lines in a diffraction grating has to be comparable to the wavelength of the incident light. The distance between atoms in a crystal structure is of the order of

10

−
10

m

{\displaystyle 10^{-10} \; m}

. If we look at a crystal structure as a diffraction grating, the wavelength of the electromagnetic waved used has to be comparable to this distance, which is X ray in the electromagnetic radiation.
Measurement of the Torus

## **Problem**

This problem is an investigation of two measurements of the **torus**: surface area

S

{\displaystyle S}

 and volume

V

{\displaystyle V}

.

The **torus** is a shape generated by revolving a circle of radius

r

{\displaystyle r}

 around an axis a distance

R

{\displaystyle R}

 away from the axis.

**Part 1:** Determine the surface area of the torus.

**Part 2:** Determine the volume of the torus.

## **Solution**

**Part 1**

The surface area

S

{\displaystyle S}

 of the torus is generated by extruding the circumference of the circle with radius

r

{\displaystyle r}

 around the circular path that passes through its center. The length of the path is the circumference of the circle with radius

R

{\displaystyle R}

:

S

=
2
π
r
×
2
π
R

[
5
p
t
]

=
4

π
2

R
r

{\displaystyle \begin{align}
S &= 2\pi r \times 2 \pi R \\ [5 pt]
&= 4 \pi^2 Rr\\
\end{align} }

**Part 2**

The volume

V

{\displaystyle V}

 of the torus is generated by extruding the area of the circle with radius

r

{\displaystyle r}

 around the circular path that passes through its center. The length of the path is the circumference of the circle with radius

R

{\displaystyle R}

:

V

=
π

r
2

×
2
π
R

[
5
p
t
]

=
2

π
2

R

r
2

{\displaystyle \begin{align}
V &= \pi r^2 \times 2 \pi R \\ [5 pt]
&= 2 \pi^2 Rr^2
\end{align} }
Petroleum 石油

**《梦溪笔谈·卷24杂志（一）2》**
鄜延境内有石油，旧说高奴县出“脂水”，即此也。生于水际，沙石与泉水相杂，惘惘而出，土人以雉尾裛[1]之，乃采 入缶中，颇似淳漆，然之如麻，但烟甚浓，所沾幄幕皆黑。予疑其烟可用，试扫其煤以为墨，黑光如漆，松墨不及也，遂大为之，其识文为“延川石液”者是也。此物后必大行于世，自予始为之。盖石油至多，生于地中无穷，不若松木有时而竭。今齐、鲁间松林尽矣，渐至太行、京西、江南，松山大半皆童矣。造煤人盖知石烟之利也。石炭烟亦大，墨人衣。予戏为《延州》诗云：“二郎山下雪纷纷，旋卓穹庐学塞人。化尽素衣冬未老，石烟多似洛阳尘。”

***Mengxi Bitan* Chapter 24 Zazhi (Part I) 2**

There is petroleum in Fuyan region, which was called “oily water” that was produced in Gaonu County in the past. The petroleum emerges among the sand and pebbles beside water. Being mixed with springs, it often wells up slowly from underground. Local people use the tail feathers of a pheasant to absorb the oily water and then collect it in earthen jars. The petroleum looks like pure paint and burns like hemp stalks. Its smoke is very dark and curtains will be darkened. I believe that the soot can be used, so I collect some of it and make an ink stick which is as dark and bright as black paint. The quality of this kind of ink stick is even better than the one made of pine soot. This type of ink stick is manufactured in large quantities and “Yanzhou Stone Liquid” is the label carved on its surface. I am the first manufacturer of this kind of ink stick and I know that it will certainly become popular, because there is plenty of petroleum in the world which can spring up from the underground inexhaustibly. On the contrary pine trees will be exhausted one day. Nowadays all the pine trees in Qi and Lu regions have already been cut down. Meanwhile mountains covered with pine trees in Taihang, Jingxi and Jiangnan regions have become half bare. People making ink sticks probably still do not know the advantages of the smoke from burning petroleum. Smoke from the burning coal is also very dark and can blacken people’s clothes. I once wrote a poem called “Yanzhou” in a jocular manner, which went as follows:
“It is snowing heavily under Erlang Mountain, People set up tents like those nomadic tribes. The winter is not over though the light-colored clothes have all been blackened,
The smoke of burning petroleum looks like the dust in Luoyang.”

**【注释】**

[1] 裛yì：古同“浥”，沾湿。

**【今译】**

鄜延路一带有石油，过去所说的高奴县出“脂水”就是这种东西。石油出产在水边，在沙石与泉水相杂的地方，慢慢地溢出，当地人用野鸡尾部的羽毛把油沾上来，收集在瓦罐里。这种油看上去挺象淳漆，燃烧起来象麻杆，但冒出来的烟很浓，把帐篷都熏黑了。我猜想这种烟可以利用，就试着扫了一些烟灰来做墨，结果写出来的字又黑又亮。象漆一般，即使松墨也比不上它。于是，就大量制作，并给它定名叫“延州石液”。这种东西以后一定会在社会上广泛流传，但是我最初制作使用它。石油非常多，地下的蕴藏无穷无尽，不比松木那样有时会用完。现在齐、鲁一带的松木已经砍光，慢慢地连太行山区、京西、江南一带的松山也大部分砍伐得成为秃山了。造墨的人必定不知道用石油 烟的好处。煤的烟也大，将人们的衣服熏黑。我戏作了一首《延州》诗说：“二郎山下雪纷纷，旋卓穹庐学塞人。化尽素衣冬未老，石烟多似洛阳尘。”
Atwood Machine

## **Problem**

The Atwood machine is a simple contraption used to demonstrate motion with constant acceleration involving two masses

m
1

{\displaystyle m\_1 }

 and

m
2

{\displaystyle m\_2 }

 attached to a pulley. Consider the free body diagram on the right.

**Part 1:** Suppose

m
1

>

m
2

{\displaystyle m\_1 > m\_2 }

, determine the acceleration of the two-mass system for a frictionless, ideal pulley.

**Part 2:** Suppose

m
1

>

m
2

{\displaystyle m\_1 > m\_2 }

, determine the tension on the string for a frictionless, ideal pulley.

## **Solution**

**Part 1**

Based on the above free-body diagram of the two masses of the Atwood machine, where

m
1

>

m
2

{\displaystyle m\_1 > m\_2 }

, the weight of

m
1

{\displaystyle m\_1 }

 is the dominant force driving the machine. Knowing this, we can write the equations of motion

(

F

n
e
t

=
m
a

)

{\displaystyle \left({F}\_{net} = ma \right) }

 for both masses.

**Mass 1**

m
1

g
−
# T

m
1

a

{\displaystyle m\_1 g - T = m\_1 a}

**Mass 2**

T
−

m
2

# g

m
2

a

{\displaystyle T - m\_2 g = m\_2 a }

By adding the two equations we get

m
1

g
−

m
2

# g

m
1

a

m
2

a

{\displaystyle m\_1 g - m\_2 g = m\_1 a + m\_2 a }

g
(

m
1

−

m
2

# )
a
(

m
1

m
2

)
.

{\displaystyle g(m\_1 - m\_2) = a(m\_1 + m\_2). }

Thus, the acceleration is

# a

g
(

m
1

−

m
2

)

m
1

m
2

.

{\displaystyle a = \frac{g(m\_1 - m\_2)}{m\_1 + m\_2}. }

**Part 2**

Substitute the acceleration from part 1 into the either equation of motion. The results will be identical. Here we will use the equation for

m
2

{\displaystyle m\_2 }

.

T
−

m
2

# g

m
2

a

{\displaystyle T - m\_2 g = m\_2 a }

# T

m
2

(
a
g
)

{\displaystyle T = m\_2 (a + g) }

# T
g

m
2

(

m
1

−

m
2

m
1

m
2

1

)

{\displaystyle T = g m\_2 \left(\frac{m\_1 - m\_2}{m\_1 + m\_2} + 1 \right) }

# T

2
g

m
1

m
2

m
1

m
2

{\displaystyle T = \frac{2g \, m\_1 m\_2}{m\_1 + m\_2} }
Jiuzhang Suanshu Yingbuzu 1

## **Problem**

今有共買物，人出八，盈三；人出七，不足四。問：人數、物價各幾何？

Suppose there is an item bought in common. If each person pays 8 coins, there is an excess of 3 coins; if each person pays 7 coins, there is a deficiency of 4 coins. Question: What is the number of people, and what is the price of the item?

## **Solution**

Let

x

{\displaystyle x}

 be the number of people, and

y

{\displaystyle y}

 the price of the items in coins.

Write the problem as a system of linear equations:

{

8
x
−
y

=
3

7
x
−
y

=
−
4

{\displaystyle \begin{cases}
8x - y &= 3 \\
7x - y &= -4
\end{cases}
}

Use Cramer's rule

# x

(
3
)
(
−
1
)
−
(
−
1
)
(
−
4
)

(
8
)
(
−
1
)
−
(
−
1
)
(
7
)

=
7

{\displaystyle
x = \frac{(3)(-1) - (-1)(-4)}{(8)(-1) - (-1)(7)} = 7
}

# y

(
8
)
(
−
4
)
−
(
3
)
(
7
)

(
8
)
(
−
1
)
−
(
−
1
)
(
7
)

=
53

{\displaystyle
y = \frac{(8)(-4) - (3)(7)}{(8)(-1) - (-1)(7)} = 53
}

There are 7 people, and the items cost 53 coins.

**Note:** The method employed in the *Jiuzhang Suanshu* is actually the method of Double False-Position, which turns out to be equivalent to Cramer's rule for 2 x 2 systems of linear equations.

**Excess:**

a
1

x
−
# y

c
1

{\displaystyle a\_1 x - y = c\_1 }

**Deficit:**

a
2

x
−
# y

c
2

{\displaystyle a\_2 x - y = c\_2 }

# x

c
1

−

c
2

a
1

−

a
2

{\displaystyle x = \frac{c\_1 - c\_2}{a\_1 - a\_2} }

# y

a
2

c
1

−

a
1

c
2

a
1

−

a
2

{\displaystyle y = \frac{a\_2 c\_1 - a\_1 c\_2}{a\_1 - a\_2} }

8
x
−
# y
3

{\displaystyle 8x - y = 3 }

7
x
−
# y
−
4

{\displaystyle 7x - y = -4 }

# x

3
−
(
−
4
)

8
−
7

=
7

{\displaystyle x = \frac{3-(-4)}{8-7} = 7 }

# y

7
(
3
)
−
8
(
−
4
)

8
−
7

=
53

{\displaystyle y = \frac{7(3)-8(-4)}{8-7} = 53 }
Mathematical Terms in Greek

**Compiled by Tao Steven Zheng (郑涛)**

## **Αριθμητική Arithmitikí**

Πρόσθεση Prósthesi - Addition

Αφαίρεση Aphaíresi - Subtraction

Πολλαπλασιασμός Pollaplasiasmós - Multiplication

Διαίρεση Diaíresi - Division

Παραγοντοποίηση Paragontopoíisi - Factorization

Διαιρέτης Diairétis - Divisor

Μέρισμα Mérisma - Dividend

Πηλίκο Pilíko - Quotient

Προϊόν Proïón - Product
Derivative of Exponential Functions

## **Problem**

**Part 1:** Use the limit definition of the derivative to show that

d

d
x

a
x

=

a
x

ln
⁡
(
a
)

{\displaystyle \frac{d}{dx}a^x = a^x \ln(a) }

.

**Part 2:** Use logarithmic differentiation to show that

d

d
x

a
x

=

a
x

ln
⁡
(
a
)

{\displaystyle \frac{d}{dx}a^x = a^x \ln(a) }

.

### **Hint**

(1) Use the limit

lim

h
→
0

a
h

−
1

h

=
ln
⁡
(
a
)

{\displaystyle \lim\_{h \to 0} \frac{a^h - 1}{h} = \ln(a) }

.

## **Solution**

**Part 1**

The limit definition of the derivative states:

f
′

(
x
# )

lim

h
→
0

f
(
x
h
)
−
f
(
x
)

h

.

{\displaystyle f'(x) = \lim\_{h \to 0}\frac{f(x+h) - f(x)}{h}.}

For the function

f
(
x
# )

a
x

{\displaystyle f(x) = a^x }

f
′

(
x
# )

lim

h
→
0

a

(
x
h
)

−

a
x

h

{\displaystyle f'(x) = \lim\_{h \to 0} \frac{a^{(x+h)} - a^x}{h} }

f
′

(
x
# )

lim

h
→
0

a
x

a
h

−

a
x

h

{\displaystyle f'(x) = \lim\_{h \to 0} \frac{a^x a^h - a^x}{h} }

f
′

(
x
# )

a
x

lim

h
→
0

a
h

−
1

h

.

{\displaystyle f'(x) = a^x \lim\_{h \to 0} \frac{a^h - 1}{h}. }

Since

lim

h
→
0

a
h

−
1

h

=
ln
⁡
(
a
)
,

{\displaystyle \lim\_{h \to 0} \frac{a^h - 1}{h} = \ln(a), }

 we get

f
′

(
x
# )

a
x

ln
⁡
(
a
)

{\displaystyle f'(x) = a^x \ln(a) }

d

d
x

a
x

=

a
x

ln
⁡
(
a
)
.

{\displaystyle \frac{d}{dx}a^x = a^x \ln(a). }

**Part 2**

Let

# y

a
x

{\displaystyle y=a^x}

, then

ln
⁡
(
y
# )
x
ln
⁡
(
a
)

{\displaystyle \ln(y) = x \ln(a) }

. Differentiating with respect to

x

{\displaystyle x}

 yields:

d

d
x

ln
⁡
(
y
# )

d

d
x

x
ln
⁡
(
a
)

{\displaystyle \frac{d}{dx} \ln(y) = \frac{d}{dx} x \ln(a) }

1
y

d
y

d
x

=
ln
⁡
(
a
)

{\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln(a) }

d
y

d
x

=
y
ln
⁡
(
a
)
.

{\displaystyle \frac{dy}{dx} = y\ln(a). }

Since

# y

a
x

,

{\displaystyle y = a^x, }

d
y

d
x

=

a
x

ln
⁡
(
a
)

{\displaystyle \frac{dy}{dx} = a^x\ln(a) }

d

d
x

a
x

=

a
x

ln
⁡
(
a
)
.

{\displaystyle \frac{d}{dx}a^x = a^x \ln(a). }
Capicitors in Series and Parallel

## **Problem**

Consider two capacitors with different capacitance

C
1

{\displaystyle C\_1}

 and

C
2

{\displaystyle C\_2}

. The formula for calculating the stored charge across a capacitor is

# Q
C
V

{\displaystyle Q=CV}

.

**Part 1:** Determine the equivalent capacitance when the two capacitors are connected in series.

**Part 2:** Determine the equivalent capacitance when the two capacitors are connected in parallel.

## **Solution**

**Part 1: Capacitors connected in series (same charge, different voltage)**

When two capacitors are connected in series, the stored charge across each capacitor is the same, and the voltage of each capacitor is inversely-proportional to their respective capacitance.

# V

V

1

V

2

{\displaystyle V=V\_{1}+V\_{2}}

Q

C

e
q

=

Q

1

C

1

Q

1

C

2

.

{\displaystyle {\frac {Q}{C\_{eq}}}={\frac {Q\_{1}}{C\_{1}}}+{\frac {Q\_{1}}{C\_{2}}}.}

Since

Q

1

=

Q

2

=
Q

{\displaystyle Q\_{1}=Q\_{2}=Q}

,

Q

C

e
q

=

Q

C

1

Q

C

2

{\displaystyle {\frac {Q}{C\_{eq}}}={\frac {Q}{C\_{1}}}+{\frac {Q}{C\_{2}}}}

1

C

e
q

=

1

C

1

1

C

2

.

{\displaystyle {\frac {1}{C\_{eq}}}={\frac {1}{C\_{1}}}+{\frac {1}{C\_{2}}}.}

In general, given

n

{\displaystyle n}

 capacitors added in series

1

C

e
q

=

1

C

1

1

C

2

.
.
.

1

C

n

.

{\displaystyle {\frac {1}{C\_{eq}}}={\frac {1}{C\_{1}}}+{\frac {1}{C\_{2}}}+...+{\frac {1}{C\_{n}}}.}

**Part 2: Capacitors connected in parallel (same voltage, different charge)**

When two capacitors are connected in parallel, the voltage across each capacitor is the same, and the stored charge of each capacitor is proportional to their respective capacitance.

# Q

Q

1

Q

2

{\displaystyle Q=Q\_{1}+Q\_{2}}

C

e
q

# V

C

1

V

1

C

2

V

2

.

{\displaystyle C\_{eq}V=C\_{1}V\_{1}+C\_{2}V\_{2}.}

Since

V

1

=

V

2

=
V

{\displaystyle V\_{1}=V\_{2}=V}

,

C

e
q

# V

C

1

V

C

2

V

{\displaystyle C\_{eq}V=C\_{1}V+C\_{2}V}

C

e
q

=

C

1

C

2

.

{\displaystyle C\_{eq}=C\_{1}+C\_{2}.}

In general, given

n

{\displaystyle n}

 capacitors added in parallel

C

e
q

=

C

1

C

2

.
.
.

C

n

.

{\displaystyle C\_{eq}=C\_{1}+C\_{2}+...+C\_{n}.}
Zheng's Problem Solving Process

There are five steps to problem solving:

1) understanding

2) organization

3) strategy

4) execution

5) reflection

**Step 1: Picture the problem (Understanding)**

Read the problem carefully to know what is known and what is not known. Understand the objective of the problem. Write down what is known and what needs to be solved. Translate words into the mathematical language.

**Step 2: Get organized (Organization)**

Separate the relevant and irrelevant parts of the presented information, and eliminate the irrelevant information. Introduce variables for the unknown quantities discussed. Make a table of what is known and unknown. If a diagram will help, draw the diagram (this is especially important when solving physics or geometry problems). Be sure the diagrams are labeled with the information.

**Step 3: Develop a plan (Strategy)**

Attempt to obtain more information from prior knowledge, or by dabbling with the information given in the problem. Make correct connections between the gathered information in order to develop testable assumptions. Conjure up a process for solving the problem with well thought out steps. These steps will vary depending on the problem. Strategies may vary from starting with a simpler problem to working backwards.

**Step 4: Solve the problem (Execution)**

Carry out the strategy. Work out the steps detailed in the plan. Methods will vary.

**Step 5: Check and reflect (Reflection)**

Verify the correctness of the answer from the steps that led to the answer. Were mistakes made along the way? Do the results make sense? For the sake of curiosity, think about other ways of solving the problem. Are there any extensions to the problem?
A Puzzling Sequence

## **Problem**

Consider the sequence

a
n

=

n
2

n
1

−
n

{\displaystyle a\_n = {\sqrt{n^2+n+1} - n} }

 for

n
≥
0

{\displaystyle n\geq 0}

.

Calculate

lim

n
→
∞

a
n

{\displaystyle \lim\_{n\to\infty}a\_n}

.

## **Solution**

This sequence is actually convergent. This may not be obvious at first glance, but can be determined with the conjugate.

a
n

=

(

n
2

n
1

−
n

)

(

n
2

n
1

n

n
2

n
1

n

)

{\displaystyle a\_n = {(\sqrt{n^2+n+1} - n})\left(\frac{{\sqrt{n^2+n+1} + n}}{{\sqrt{n^2+n+1} + n}}\right) }

a
n

=

n
2

n
1
−

n
2

n
2

n
1

n

{\displaystyle a\_n = \frac{n^2+n+1 - n^2}{{\sqrt{n^2+n+1} + n}} }

a
n

=

n
1

n
2

n
1

n

{\displaystyle a\_n = \frac{n+1}{{\sqrt{n^2+n+1} + n}} }

Evaluate the limit as

n
→
∞

{\displaystyle n \to \infty}

.

lim

n
→
∞

a
n

=

lim

n
→
∞

n
1

n
2

n
1

n

{\displaystyle \lim\_{n \to \infty} a\_n = \lim\_{n \to \infty} \frac{n+1}{{\sqrt{n^2+n+1} + n}} }

lim

n
→
∞

a
n

=

lim

n
→
∞

n

(

1

1
n

)

n
2

(

1

1
n

1

n
2

)

n

{\displaystyle \lim\_{n \to \infty} a\_n =\lim\_{n \to \infty} \frac{n\left(1+\frac{1}{n}\right)}{{\sqrt{n^2 \left(1+\frac{1}{n}+\frac{1}{n^2}\right)} + n}} }

lim

n
→
∞

a
n

=

lim

n
→
∞

n

(

1

1
n

)

n

(

1

1
n

1

n
2

1

)

{\displaystyle \lim\_{n \to \infty} a\_n = \lim\_{n \to \infty} \frac{n\left(1+\frac{1}{n}\right)}{n\left({\sqrt{1+\frac{1}{n}+\frac{1}{n^2}} + 1}\right)} }

lim

n
→
∞

a
n

=

lim

n
→
∞

1

1
n

1

1
n

1

n
2

1

{\displaystyle \lim\_{n \to \infty} a\_n = \lim\_{n \to \infty} \frac{1+\frac{1}{n}}{{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}} + 1}} }

lim

n
→
∞

a
n

=

1

1

1

{\displaystyle \lim\_{n \to \infty} a\_n = \frac{1}{\sqrt{1} + 1} }

lim

n
→
∞

a
n

=

1
2

{\displaystyle \lim\_{n \to \infty} a\_n = \frac{1}{2} }

.
Jiuzhang Suanshu Gougu 15

## **Problem**

This classic problem from the *Jiuzhang Suanshu* (九章算術) is called "square in a right triangle" (勾股容方 gōu gǔ róng fāng).

答曰：方三步、十七分步之九。

術曰：並勾、股為法，勾、股相乘為實。實如法而一，得方一步。

Answer: The width of the square is

3

9
17

{\displaystyle 3 \frac{9}{17} }

 *bu*.

Method: Sum the *gou* and the *gu* and let this result be the divisor. Multiply the *gou* and the *gu* and let this result be the dividend. Divide the dividend by the divisor to obtain the width in *bu*.

## **Solution**

**Derivation**

Let

a

{\displaystyle a}

 be the shorter leg (勾 *gou*),

b

{\displaystyle b}

 be the longer leg (股 *gu*), and

x

{\displaystyle x}

 be the width of the inscribed square (Figure 1).

Dissect the right triangle into 3 pieces: 1 red triangle, 1 blue triangle, and 1 yellow square. Copy each shape two times and then assemble them to form one large rectangle. Use this rectangle to derive the formula for calculating the width of the inscribed square.

x
(
a
b
# )
a
b

{\displaystyle x(a+b) = ab }

# x

a
b

a
b

{\displaystyle x = \frac{ab}{a+b} }

**Calculation**

Substitute the known values

# a
5

{\displaystyle a = 5}

and

# b
12

{\displaystyle b = 12 }

into the derived formula:

# x

5
×
12

5
12

{\displaystyle x = \frac{5\times 12}{5+12} }

# x

60
17

{\displaystyle x = \frac{60}{17} }

# x
3

9
17

{\displaystyle x = 3 \frac{9}{17} }

Therefore, the width of the inscribed square is

3

9
17

{\displaystyle 3 \frac{9}{17} }

 *bu*.
Heat Conduction Through Two Slabs

## **Problem**

Consider two slabs of lengths

L
1

{\displaystyle L\_1}

 and

L
2

{\displaystyle L\_2}

, and cross-sectional area

A

{\displaystyle A}

. The respective thermal conductivity constants are

k
1

{\displaystyle k\_1 }

 and

k
2

{\displaystyle k\_2 }

.

**Part 1:** Based on the below diagram, calculate the temperature between the two slabs

T
m

{\displaystyle T\_m}

.

**Part 2:** Calculate

T
m

{\displaystyle T\_m}

 if

k
1

=

k
2

=
k

{\displaystyle k\_1 = k\_2 = k }

 and

L
1

=

L
2

=
L

{\displaystyle L\_1 = L\_2 = L }

.

## **Solution**

**Part 1**

Heat naturally flows from higher temperature to lower temperature. Based on the diagram, the heat flows at a rate from left to right.

d
Q

d
t

=

A

k
2

L
2

(

T
h

−

T
m

)

{\displaystyle \frac{dQ}{dt} = \frac{A k\_2}{L\_2}(T\_h - T\_m) }

d
Q

d
t

=

A

k
1

L
1

(

T
m

−

T
c

)

{\displaystyle \frac{dQ}{dt} = \frac{A k\_1}{L\_1}(T\_m - T\_c) }

Thus,

A

k
2

L
2

(

T
h

−

T
m

# )

A

k
1

L
1

(

T
m

−

T
c

)

{\displaystyle \frac{A k\_2}{L\_2}(T\_h - T\_m) = \frac{A k\_1}{L\_1}(T\_m - T\_c) }

k
2

L
2

T
h

−

k
2

L
2

T
m

=

k
1

L
1

T
m

−

k
1

L
1

T
c

{\displaystyle \frac{k\_2}{L\_2}T\_h - \frac{k\_2}{L\_2}T\_m = \frac{k\_1}{L\_1}T\_m - \frac{k\_1}{L\_1}T\_c }

k
1

L
1

T
m

k
2

L
2

T
m

=

k
2

L
2

T
h

k
1

L
1

T
c

{\displaystyle \frac{k\_1}{L\_1}T\_m + \frac{k\_2}{L\_2}T\_m = \frac{k\_2}{L\_2}T\_h + \frac{k\_1}{L\_1}T\_c }

(

k
1

L
1

k
2

L
2

)

T
m

=

k
2

L
2

T
h

k
1

L
1

T
c

{\displaystyle \left(\frac{k\_1}{L\_1} + \frac{k\_2}{L\_2}\right)T\_m = \frac{k\_2}{L\_2}T\_h + \frac{k\_1}{L\_1}T\_c }

Consequently,

T
m

=

k
2

L
2

T
h

k
1

L
1

T
c

k
1

L
1

k
2

L
2

{\displaystyle T\_m = \frac{\frac{k\_2}{L\_2}T\_h + \frac{k\_1}{L\_1}T\_c}{\frac{k\_1}{L\_1} + \frac{k\_2}{L\_2}} }

.
**Part 2**
If

k
1

=

k
2

=
k

{\displaystyle k\_1 = k\_2 = k }

 and

L
1

=

L
2

=
L

{\displaystyle L\_1 = L\_2 = L }

,

T
m

=

k
2

L
2

T
h

k
1

L
1

T
c

k
1

L
1

k
2

L
2

{\displaystyle T\_m = \frac{\frac{k\_2}{L\_2}T\_h + \frac{k\_1}{L\_1}T\_c}{\frac{k\_1}{L\_1} + \frac{k\_2}{L\_2}} }

becomes

T
m

=

k
L

T
h

k
L

T
c

k
L

k
L

{\displaystyle T\_m = \frac{\frac{k}{L}T\_h + \frac{k}{L}T\_c}{\frac{k}{L} + \frac{k}{L}} }

.
Factoring out

k
L

{\displaystyle \frac{k}{L} }

 yields

T
m

=

k
L

(

T
h

T
c

)

k
L

(
1
1
)

{\displaystyle T\_m = \frac{\frac{k}{L}(T\_h + T\_c)}{\frac{k}{L}(1+1)} }

T
m

=

T
h

T
c

2

{\displaystyle T\_m = \frac{T\_h + T\_c}{2} }

.