recursal/Fanatic-Fandom huggingface dataset:
Adept Pyromancer (XCG Card)

Adept Pyromancer's Artwork in Base Set 1

### Rules Text

Fire Elementals you control gain 3ATK. 3G, T: This card deals 6 damage to target permanent.

### Printed in

BS1

Adept Pyromancer is one of the oldest cards in the game, with roots in the pre-alpha test deck and beyond.

It appeared as a planned Uncommon in several sets when the game was a Collectible Card Game, including the Bella Elementorum Starter Set, then known as "Rapturous Pyromancer"

Its artwork also appears on the front cover of the "Fire and Ice" starter deck.
Doppler Effect

## **Problem**

A police siren emits a frequency of 640 Hz. When the police car approaches you at a speed of 25 m/s, what frequency do you perceive when:

1) you are not moving

2) you are moving towards the police car at 5 m/s

3) you are moving away from the police car at 5 m/s.

Assume the speed of sound is 343 m/s.

# f

f
0

(

v
±

v
r

v
±

v
s

)

{\displaystyle f = f\_0 \left(\frac{v \pm v\_r}{v \pm v\_s}\right) }

where

f

{\displaystyle f}

 is the shifted frequency,

f
0

{\displaystyle f\_0}

 is the frequency of the source,

v

{\displaystyle v}

 is the speed of sound,

v
r

{\displaystyle v\_r}

 is the speed of the receiver, and

v
s

{\displaystyle v\_s}

 is the speed of the source.

## **Solution**

When you are not moving the speed of the receiver is

v
r

=
0

{\displaystyle v\_r = 0 }

. Then the Doppler shift formula becomes

# f

f
0

(

v

v
−

v
s

)

.

{\displaystyle f = f\_0 \left(\frac{v}{v - v\_s}\right). }

Substituting all the values into the formula yields:

f

=
640

(

343

343
−
25

)

=
690

H
z
.

{\displaystyle \begin{align}
f &= 640 \left(\frac{343}{343 - 25}\right) \\[5 pt]
&= 690 \, Hz.
\end{align} }

When you are moving towards the police car at 5 m/s, select the postive sign in the numerator. Then the Doppler shift formula becomes

# f

f
0

(

v

v
r

v
−

v
s

)

.

{\displaystyle f = f\_0 \left(\frac{v + v\_r}{v - v\_s}\right). }

Substituting all the values into the formula yields:

f

=
640

(

343
5

343
−
25

)

=
700.38

H
z
.

{\displaystyle \begin{align}
f &= 640 \left(\frac{343 + 5}{343 - 25}\right) \\[5 pt]
&= 700.38 \, Hz.
\end{align} }

When you are moving away from the police car at 5 m/s, select the negative sign in the numerator. Then the Doppler shift formula becomes

# f

f
0

(

v
−

v
r

v
−

v
s

)

.

{\displaystyle f = f\_0 \left(\frac{v - v\_r}{v - v\_s}\right). }

Substituting all the values into the formula yields:

f

=
640

(

343
−
5

343
−
25

)

=
680.25

H
z
.

{\displaystyle \begin{align}
f &= 640 \left(\frac{343 - 5}{343 - 25}\right) \\[5 pt]
&= 680.25 \, Hz.
\end{align} }
Liu Hui (劉徽)

**By: Tao Steven Zheng (鄭濤)**

### Notable works

Commentary on the Jiuzhang Suanshu (九章算術注 jiǔ zhāng suàn shù zhù), Haidao Suanjing (海島算經 hǎi dǎo suàn jīng)

**Liu Hui** was a Chinese mathematician who lived during the Three Kingdoms period (220 - 280 AD) and Wei-Jin dynasty (265 - 420 AD). Little is known about Liu Hui's life aside that he was a descendant of the Marquis of Zixiang, and that he completed his commentary on the *Jiuzhang Suanshu* (九章算術; *Nine Chapters on the Mathematical Art*) in the year 263 AD. Despite his relative obscurity, Liu Hui was known for his mathematical prowess and inventiveness. His contribution to Chinese mathematics was immense because he rigorously proved the validity of many algorithms and formulas known before his time. He also developed novel ideas in the study of geometry that are theoretically sound.

## **Discoveries**

### **Approximations of π**

Liu Hui devised a geometric algorithm for computing *π* called the *ge yuan shu* (割圓術; circle cutting method). Using a 96-sided polygon, Liu obtained the bound

3.141024
<
π
<
3.142708

{\displaystyle 3.141024 < \pi < 3.142708 }

. He later devised a quick method involving the use of geometric series to arrive at a more accurate approximation of

π
≈

3927
1250

{\displaystyle \pi \approx \frac{3927}{1250} }

. Liu Hui's quick method achieved the accuracy of a 3072-sided polygon using the result he obtained from a 96-sided polygon.

### **Arithmetic with Negative Numbers**

Liu Hui established the rules of arithmetic with negative numbers in his commentary of the *Jiuzhang Suanshu* (九章算術). However, it is likely that negative numbers were known in China during the Han dynasty (206 BC - 220 AD) or earlier. He then explained in great detail the use of negative numbers for solving *fangcheng* problems (方程). These problems are systems of simultaneous linear equations solved using the elimination method.

**See: Jiuzhang Suanshu (Fangcheng 8)**

### **The Cavalieri Principle**

Liu Hui discovered the Cavalieri principle (the method of indivisibles) 1400 years before Bonaventura Cavalieri (1598 - 1647). He applied this principle to derive precise formulas for calculating the volumes of prisms, pyramids, cylinders, wedges, and frustums. Liu Hui also applied the Cavalieri principle to determine the volumetric ratio of the sphere to the *mou he fang gai* (牟合方盖), a solid formed by the intersection of two perpendicular cylinders.

**See: Jiuzhang Suanshu Shanggong 10, Liu Hui's Puzzle**

### **Out-In Complementary Principle**

In his book *Haidao Suanjing* (海島算經), Liu Hui created and solved several surveying problems by exploiting the *chong cha* method (重差術), a nascent form of trigonometry.

Haidao Suanjing 1
Chinese Science and Technology

## **Paper**

Although coarse forms of paper already existed as early as the 2nd century BC, the invention of industrial papermaking is attributed to Cai Lun (蔡倫; c. 50 – 121 AD), an eunuch who lived during the Han dynasty (206 BC - 220 AD). As an inspector in charge of the imperial workshops, he worked with the elite artisans and craftsmen of the Han empire, and oversaw the production of various instruments, tools, and weapons. In 105 AD, Cai Lun presented his innovative technique of papermaking to the Emperor He (漢和帝; 79 – 106 AD). His method begins by turning fibrous materials into loose pulp. Materials such as tree bark, remnants of hemp, linen rags, and fishnets were boiled until they are completely disintegrated. Then the pulp is pounded using hammers before it is mixed into water to form a solution. The pulp-water solution was then strained through a fine sieve, separating excess water with the thin and evenly distributed layer of pulp. Paper is formed after the pulp layer is dried.
Guo Shoujing 郭守敬

**By: Tao Steven Zheng (郑涛)**

**Born:** 1231 AD in Hebei province

**Died:** 1316 AD in Khanbaliq (modern day Beijing)

**Guo Shoujing** was an eminent inventor, astronomer, and mathematician famous for developing the "Shou shi li" (授时历 lit. time-granting calendar) during the Yuan dynasty (1279 - 1368 AD). Historical records indicate that he demonstrated great aptitude for hydraulic engineering at a young age. His grandfather, Guo Rong (郭荣), was knowledgeable in the Confucian classics, in mathematics, and hydraulic engineering, and taught the young boy such subjects. At age 14 Guo Shoujing was able to construct an advanced type of clepsydra, or water clock, called the *lian hua lou* (莲花漏 lit. lotus clepsydra). In 1251, he restored an old bridge across the Dahuoquan River in Hebei province.

## **Discoveries**

**Spherical Trigonometry**

Guo Shoujing revived the study of spherical trigonometry in China, Likely sources of inspiration might have come from Shen Kuo's work in the 11th century, as well as Islamic influences during his time.

**Cubic Interpolation**

Along with the astronomer-mathematician Wang Xun (王恂, 1235 - 1281 AD), Guo Shoujing tabulated accurate records of celestial motions and designed a new calendar called the "Shou shi li" (授时历). They utilized cubic interpolation in their calculations and determined the length of a year to be 365.2425 days (26 seconds longer than the modern value).

## **Inventions**

**Astronomical Instruments**

Guo Shoujing invented the *jian yi* (简仪 lit. simplified armilla), which is a torquetem set in equatorial coordinates. He also improved the design of various astronomical instruments, such as the sundial, gnomon, and clepsydra.
Concentration Dilution of Solution

## **Problem**

Determine the volume of water required to dilute 1.25 L of 0.0100 M hydrochloric acid to 0.00340 M hydrochloric acid.

## **Solution**

Use the dilution formula, and solve for the final volume

V
f

{\displaystyle V\_f }

:

c
i

V
i

=

c
f

V
f

{\displaystyle c\_i V\_i = c\_f V\_f }

V
f

=

c
i

V
i

c
f

.

{\displaystyle V\_f = \frac{c\_i V\_i}{c\_f}. }

Substitute the known values into the formula:

V
f

=

0.0100

M
×
1.25

L

0.00340

M

:

{\displaystyle V\_f = \frac{0.0100 \; M \times 1.25 \; L }{0.00340 \; M}
:}

V
f

=
3.68

L
.

{\displaystyle V\_f = 3.68 \; L. }

This is the final volume. To calculate the volume change, subtract the initial fluid volume from the final volume.

Δ
# V

V
f

−

V
i

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

Δ
# V
3.68

L
−
1.25

# L
2.43

L
.

{\displaystyle \Delta V = 3.68 \; L - 1.25 \; L = 2.43 \; L. }
Chinese Quotations

**Compiled by Tao Steven Zheng (郑涛)**

## **Yijing (11th - 7th century BC)**

上古結繩而治，後世聖人易之以書契，百官以治，萬民以察，蓋取諸夬。《周易· 繫辭下》

In the highest antiquity, government was carried on successfully by the use of knotted cords (to preserve the memory of things). In subsequent ages the sages substituted for these written characters and bonds. By means of these (the doings of) all the officers could be regulated, and (the affairs of) all the people accurately examined. The idea of this was taken, probably, from Guai (hexagram 43). --- *Yijing* (Xici II)

## **Xu Yue (180 - 220)**

刻板為三分，其上下二分以停遊珠，中間一分以定算位，位各五珠，上一珠與下四珠色別，其上別色之珠當五，其下四珠，珠各當一。《算數記遺》

A grooved board is split into three sections, with moveable beads placed in the top and bottom sections, and a middle section for setting the positions of computation. Each position [column] contains five beads, with one bead in the upper section and four beads of a different colour in the bottom section: the top beads of one colour are worth five; the four bottom beads each being worth one. --- *Suanshu Jiyi* (Notes on Traditions of Arithmetic Methods)

## **Wang Yangming (1472 – 1529)**

### 持志如心痛。一心在痛，人安有工夫說閒語，管閒事。

保持自己的志向，就如同心臟在疼痛，注意力都被這種疼痛吸引了，哪裡有空閒說無意義的話和管無意義的事。《傳習錄·卷上·門人薛侃錄》

Keeping one's own ambition is like a pain in the heart, and attention is attracted by this pain, where there is no time to say meaningless words and deal with meaningless things.

注：世間本無事，庸人自擾之。與其為周遭大小事操碎心，倒不如專注於眼前所熱愛之事。

Comment: There is nothing in the world, and no one bothers about it. Instead of worrying about the big and small things around you, focus on what you love in front of you.

### 知者行之始。行者知之成。

以知為指導的行才能行之有效，脫離知的行則是盲動。同樣，以行驗證的知才是真知灼見，脫離行的知則是空知。《傳習錄·卷上·門人陸澄錄》

注：有了想要努力變更好的想法，一定不要再光說空話，而要盡快付諸行動，做實事才好。

Comment: When you have the idea that you want to work hard to improve, you must stop talking empty words, but put it into action as soon as possible, and do practical things.

### 授書不在徒多，但貴精熟。

傳授知識不追求數量與速度，只以使人精通與熟練掌握為貴。《傳習錄·卷下·右南大吉錄》

注：古語云，“欲速則不達”，學習知識也一樣，不能急於求成，要有耐心去與打磨。

The imparting of knowledge does not pursue quantity and speed, but only to make people patient and proficient.

Comment: As the old saying goes, "haste is not enough", the same is true for learning knowledge, you can't be in a hurry, you have to be patient and polish.
Markov Market

## **Problem**

Consider the stock market model above.

**Part 1:** Construct the transition matrix for this market.

**Part 2:** Determine the long-term trajectory as a ordered triple

(
x
,
y
,
z
)

{\displaystyle (x, y, z)}

.

## **Solution**

**Part 1**

Construct a table by fallowing the arrows. Make sure each column adds up to 1.

 **Bear** | 0.150
 **Stag** | 0.250

Let's call the matrix **M**.

M

=

[

0.90

0.075

0.025

0.15

0.80

0.05

0.25

0.25

0.50

]

{\displaystyle \mathbf{M} =
\begin{bmatrix}
0.90 & 0.075 & 0.025 \\
0.15 & 0.80 & 0.05 \\
0.25 & 0.25 & 0.50
\end{bmatrix} }

**Part 2**

To solve for the long-term trajectory, row reduce the following system.

[

x
∞

y
∞

z
∞

]

[

0.90

0.075

0.025

0.15

0.80

0.05

0.25

0.25

0.50

]

=

[

x
∞

y
∞

z
∞

]

{\displaystyle \begin{bmatrix} x\_\infty & y\_\infty & z\_\infty \end{bmatrix}
\begin{bmatrix}
0.90 & 0.075 & 0.025 \\
0.15 & 0.80 & 0.05 \\
0.25 & 0.25 & 0.50
\end{bmatrix} = \begin{bmatrix} x\_\infty & y\_\infty & z\_\infty \end{bmatrix} }

{

0.90

x
∞

0.15

y
∞

0.25

z
∞

=

x
∞

0.075

x
∞

0.80

y
∞

0.25

z
∞

=

y
∞

0.025

x
∞

0.05

y
∞

0.50

z
∞

=

z
∞

x
∞

y
∞

z
∞

=
1

{\displaystyle
\begin{cases} 0.90x\_\infty + 0.15y\_\infty + 0.25z\_\infty &= x\_\infty \\
0.075x\_\infty + 0.80y\_\infty + 0.25z\_\infty &= y\_\infty \\
0.025x\_\infty + 0.05y\_\infty + 0.50z\_\infty &= z\_\infty \\
x\_\infty + y\_\infty + z\_\infty &= 1 \ \end{cases}
}

or equivalently

M

=

[

−
0.10

0.075

0.025

0

0.15

−
0.20

0.05

0

0.25

0.25

−
0.50

0

1

1

1

1

]

{\displaystyle \mathbf{M} =
\begin{bmatrix}
-0.10 & 0.075 & 0.025 & 0 \\
0.15 & -0.20 & 0.05 & 0 \\
0.25 & 0.25 & -0.50 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix} }

Use the MATLAB/OCTAVE command **rref(M)** to row reduce the matrix. For the full solution all written out, use this site: Linear Algebra Toolkit.

r
r
e
f

(

M

)

=

[

1

0

0

0.6250

0

1

0

0.3125

0

0

1

0.0625

0

0

0

0

]

{\displaystyle rref\left(\mathbf{M}\right) =
\begin{bmatrix}
1 & 0 & 0 & 0.6250 \\
0 & 1 & 0 & 0.3125 \\
0 & 0 & 1 & 0.0625 \\
0 & 0 & 0 & 0
\end{bmatrix} }

This means

(
x
,
y
,
z
# )
(
0.625
,
0.3125
,
0.0625
)

{\displaystyle (x,y,z) = (0.625,0.3125,0.0625) }

, or in percentage terms

(
x
,
y
,
z
# )
(
62.5
%
,
31.25
%
,
6.25
%
)

{\displaystyle (x,y,z)=(62.5\%,31.25\%,6.25\%)}

.
Jiuzhang Suanshu Fangcheng 13

## **Problem**

This classic problem from the *Jiuzhang Suanshu* (九章算術) is given the name "well shared by five families" (五家共井 wǔ jiā gòng jǐng).

答曰：井深七丈二尺一寸。甲綆長二丈六尺五寸，乙綆長一丈九尺一寸，丙綆長一丈四尺八寸，丁綆長一丈二尺九寸，戊綆長七尺六寸。

術曰：如方程，以正負術入之。

Answer: The depth of the well is 7 *zhang* 2 *chi* 1 *cun*. A’s rope is 2 *zhang* 6 *chi* 5 *cun* long; B’s rope is 1 *zhang* 9 *chi* 1 *cun* long; C’s rope is 1 *zhang* 4 *chi* 8 *cun* long; D’s rope is 1 *zhang* 2 *chi* 9 *cun* long; E’s rope is 7 *chi* 6 *cun* long.

Method: Use the *fang cheng* method and solve by the *zheng fu* method (正負術 *zheng fu shu*).

## **Solution**

This problem is an indeterminate system of linear equations of 5 equations in 6 unknowns:

{

2
a
# b
x

3
b
# c
x

4
c
# d
x

5
d
# e
x

6
e
# a
x

{\displaystyle \begin{cases}
2a + b = x \\
3b + c = x \\
4c + d = x \\
5d + e = x \\
6e + a = x \\
\end{cases} }

Set up the initial matrix.

[

1

2

3

1

4

1

5

1

6

1

1

1

1

1

1

]

{\displaystyle
\begin{bmatrix}
1 & & & & 2 \\
 & & & 3 & 1 \\
 & & 4 & 1 & \\
 & 5 & 1 & & \\
6 & 1 & & & \\
1 & 1 & 1 & 1 & 1 \\
\end{bmatrix} }

[

2

1

0

0

0

1

0

3

1

0

0

1

0

0

4

1

0

1

0

0

0

5

1

1

1

0

0

0

6

1

]

{\displaystyle
\begin{bmatrix}
2 & 1 & 0 & 0 & 0 & 1 \\
0 & 3 & 1 & 0 & 0 & 1 \\
0 & 0 & 4 & 1 & 0 & 1 \\
0 & 0 & 0 & 5 & 1 & 1 \\
1 & 0 & 0 & 0 & 6 & 1 \\
\end{bmatrix} }

The step-by-step solution for this problem is lengthy.

[

721

721

721

721

721

76

129

148

191

256

]

{\displaystyle
\begin{bmatrix}
 & & & & 721 \\
 & & & 721 & \\
 & & 721 & & \\
 & 721 & & & \\
721 & & & & \\
76 & 129 & 148 & 191 & 256 \\
\end{bmatrix} }

[

1

0

0

0

0

256
721

0

1

0

0

0

191
721

0

0

1

0

0

148
721

0

0

0

1

0

129
721

0

0

0

0

1

76
721

]

{\displaystyle
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & \frac{256}{721} \\
0 & 1 & 0 & 0 & 0 & \frac{191}{721} \\
0 & 0 & 1 & 0 & 0 & \frac{148}{721} \\
0 & 0 & 0 & 1 & 0 & \frac{129}{721} \\
0 & 0 & 0 & 0 & 1 & \frac{76}{721} \\
\end{bmatrix} }

Although there are infinitely many solutions, the *Jiuzhang Suanshu* gives the solution with smallest integers: the well being 721 *cun*, A’s rope is 265 *cun*, B’s rope is 191 *cun*, C’s rope is 148 *cun*, D’s rope is 129 *cun*, E’s rope is 76 *cun*.
Higher Mathematics Curriculum

## **Level 1**

1. Complex number system
2. Vectors
3. Matrix algebra
4. Linear systems
5. Conics
6. Sequences and series

## **Level 2**

1. Limits
2. Differentiation
3. Applications of differentiation
4. Integration
5. Applications of integration
6. Taylor series

## **Level 3**

1. Multivariable functions
2. Partial differentiation
3. Multiple integration
4. Linear differential equations
5. System of linear differential equations
6. Fourier series
Chinese Science and Technology: Four Great Inventions

**By: Tao Steven Zheng 郑涛**

**Francis Bacon** (1561 - 1626 AD), the leading English philosopher of the 17th century, once wrote: "It is well to observe the force and virtue and consequence of discoveries. These are to be seen nowhere more clearly than those three which were unknown to the ancients [the Greeks], and of which the origin, though recent, is obscure and inglorious; namely printing, gunpowder, and the magnet. For these three have changed the whole face and stage of things throughout the world, the first in literature, the second in warfare, the third in navigation; whence have followed innumerable changes; insomuch that no empire, no sect, no star, seems to have exerted greater power and influence in human affairs than these three mechanical discoveries."

Bacon had no idea where these things had come from, but historians now know that all three were invented in China.

The magnetic compass, papermaking, gunpowder, and printing, are inventions from ancient China that are celebrated by the Chinese as symbols of their scientific and technological legacy. Between the 8th to 15th centuries, these four inventions have spread from China to other parts of the world. Due to their considerable impact on the progress of human civilization, they are collectively known as the **Four Great Inventions** (四大發明 *si da fa ming*).
Fibonacci-Pythagorean Triples

Given four consecutive Fibonacci numbers

F

n

{\displaystyle {F}\_{n}}

,

F

n
1

{\displaystyle {F}\_{n+1}}

,

F

n
2

{\displaystyle {F}\_{n+2}}

,

F

n
3

{\displaystyle {F}\_{n+3}}

, show that

(

F

n

F

n
3

,
2

F

n
1

F

n
2

,

F

n
1

2

F

n
2

2

)

{\displaystyle ({F}\_{n}{F}\_{n+3}, 2{F}\_{n+1}{F}\_{n+2}, {{F}\_{n+1}^2}+{{F}\_{n+2}^2})}

 is a Pythagorean triple.

## **Solution**

The relation to be proven is the Pythagorean theorem

(

F

n

F

n
3

)

2

(

2

F

n
1

F

n
2

)

2

=

(

F

n
1

2

F

n
2

2

)

2

.

{\displaystyle
\left({F}\_{n}{F}\_{n+3}\right)^2 + \left(2{F}\_{n+1}{F}\_{n+2}\right)^2 = \left({{F}\_{n+1}}^{2}+{{F}\_{n+2}}^2 \right)^2.
}

Let

F

n
1

=
a

{\displaystyle {F}\_{n+1} = a}

 and

F

n
2

=
b

{\displaystyle {F}\_{n+2} = b}

, then

F

n

=
b
−
a

{\displaystyle {F}\_{n} = b -
 a}

 and

F

n
3

=
b
a

{\displaystyle {F}\_{n+3} = b + a}

. Substituting these values into the Pythagorean theorem yields

(

(
b
−
a
)
(
b
a
)

)

2

(

2
a
b

)

2

=

(

a
2

b
2

)

2

.

{\displaystyle
\left((b-a)(b+a) \right)^2 + \left(2ab \right)^2 = \left(a^2+b^2 \right)^2.
}

(

b
2

−

a
2

)

2

(

2
a
b

)

2

=

(

a
2

b
2

)

2

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

The LHS and the RHS of the equation must match to prove the above relation.

b
4

−
2

a
2

b
2

a
4

4

a
2

b
2

=

a
4

2

a
2

b
2

b
4

{\displaystyle
b^4 - 2a^2b^2 + a^4 + 4a^2b^2 = a^4 + 2a^2b^2 +b^4
}

b
4

2

a
2

b
2

a
4

=

a
4

2

a
2

b
2

b
4

{\displaystyle
b^4 + 2a^2b^2 + a^4 = a^4 + 2a^2b^2 +b^4
}

Both sides of the equation indeed match.
Coated Glass Lens

## **Problem**

A glass lens is coated on one side with a thin film of magnesium fluoride (

MgF

2

{\displaystyle \ce{MgF2}}

) to reduce reflection from the lens surface. The index of refraction fo air is taken to be 1.00, the index of refraction of

MgF

2

{\displaystyle \ce{MgF2}}

 is 1.38, and the index of refraction of the glass is 1.50. What is the least coating thickness that eliminates (via interference) the reflections at the middle of the visible spectrum (

# λ
550

n
m

{\displaystyle \lambda = 550 \; nm}

). Assume the light is approximately perpendicular to the lens surface.

## **Solution**

In this scenario, the reflection is eliminated if the film thickness

L

{\displaystyle L}

 enables the light waves reflected from the two thin film interfaces to be out of phase.

At the first interface, the incident light is in air, which has a smaller index of refraction than the

MgF

2

{\displaystyle \ce{MgF2}}

 coating. Thus, the reflection phase shift is half wavelength

0.5
λ

{\displaystyle 0.5 \lambda}

 at the first interface.

At the second interface, the incident light is in the

MgF

2

{\displaystyle \ce{MgF2}}

coating, which has a smaller index of refraction than the glass. Thus, the reflection phase shift is half wavelength

0.5
λ

{\displaystyle 0.5 \lambda}

 at the second interface.

Because both reflections cause the same phase shift, they tend to put the two waves in phase. Since we want those waves to be out of phase, their path difference

2
L

{\displaystyle 2L}

 must be an odd number of half-wavelengths:

2
# L

m
λ

2

n
2

,

# m
1
,
3
,
5
,
.
.
.

{\displaystyle 2L = \frac{m \lambda}{2n\_2}, \quad m = 1,3,5, ... }

To calculate the least thickness for the coating, we set

# m
1

{\displaystyle m = 1}

 and solve for

L

{\displaystyle L}

:

# L

1
×
550

n
m

4
×
1.38

=
99.6

n
m
.

{\displaystyle L = \frac{1 \times 550 \; nm}{4 \times 1.38} = 99.6 \; nm. }
Gauss' Mean Value Theorem

## **Problem**

Let

f
(
z
)

{\displaystyle f(z)}

 be a complex-valued function that is analytic inside and on a circle

C

{\displaystyle C}

 centered at

z
0

.

{\displaystyle z\_0 . }

 Use Cauchy's integral formula to prove Gauss' mean value theorem

f
(

z
0

# )

1

2
π

∫

0

2
π

f
(

z
0

r

e

i
θ

)

d
θ
.

{\displaystyle f(z\_0) = \frac{1}{2\pi} \int\_{0}^{2\pi} f(z\_0 + r{e}^{i\theta}) \,d\theta. }

**Additional Remark:** This formula is a mean value theorem because it has the form

f
(
c
# )

1

b
−
a

∫

a

b

f
(
x
)

d
x
.

{\displaystyle f(c) = \frac{1}{b-a} \int\_{a}^{b} f(x)\, dx. }

## **Solution**

The equation for the circle

C

{\displaystyle C}

 is

z
−

z
0

=
r

{\displaystyle |z - z\_0| = r }

 or

# z

z
0

r

e

i
θ

{\displaystyle z = z\_0 + r{e}^{i\theta} }

. It follows that

d
# z
i
r

e

i
θ

d
θ

{\displaystyle dz = ir{e}^{i\theta} \,d\theta }

. Thus, we can transform the Cauchy's integral formula in terms of

θ
:

{\displaystyle \theta: }

f
(

z
0

)

=

1

2
π
i

∮
C

f
(
z
)

z
−

z
0

d
z

=

1

2
π
i

∫

0

2
π

f
(

z
0

r

e

i
θ

)
i
r

e

i
θ

r

e

i
θ

d
θ

=

1

2
π

∫

0

2
π

f
(

z
0

r

e

i
θ

)

d
θ
.

{\displaystyle \begin{align}
f(z\_0) &= \frac{1}{2\pi i}\oint\_C \frac{f(z)}{z-z\_0}dz \\[5 pt]
&= \frac{1}{2\pi i} \int\_{0}^{2\pi} \frac{f(z\_0 + r{e}^{i\theta})ir{e}^{i\theta}}{r{e}^{i\theta}} d\theta \\[5 pt]
&= \frac{1}{2\pi} \int\_{0}^{2\pi} f(z\_0 + r{e}^{i\theta})\, d\theta.
\end{align} }
Cauchy Integral Formula

## **Problem**

If

f
(
z
)

{\displaystyle f(z)}

 is analytic inside and on the boundary

C

{\displaystyle C}

 of a simply-connected region, prove the Cauchy integral formula

f
(

z
0

# )

1

2
π
i

∮
C

f
(
z
)

z
−

z
0

d
z

{\displaystyle f(z\_0) = \frac{1}{2\pi i}\oint\_C \frac{f(z)}{z-z\_0}dz }

.

## **Solution**

The function

f
(

z
0

)

z
−

z
0

{\displaystyle \frac{f(z\_0)}{z-z\_0} }

 is analytic inside and on the boundary

C

{\displaystyle C}

 except at the point

# z

z
0

{\displaystyle z = z\_0}

. Consider a circle

C
0

{\displaystyle C\_0}

 of radius

r

0

{\displaystyle r\_{0}}

 that is centered at

z
0

{\displaystyle z\_0}

. The equation of the circle

C
0

{\displaystyle C\_0}

 is

z
−

z
0

=

r
0

{\displaystyle |z - z\_0| = r\_0 }

 or in polar form

z
−

z
0

=

r
0

e

i
θ

{\displaystyle z - z\_0 = r\_0 e^{i\theta} }

 where

0
≤
θ
<
2
π

{\displaystyle 0 \leq \theta<2\pi}

. Thus

# z

z
0

e

i
θ

{\displaystyle z =
 z\_0 + e^{i\theta} }

 and

d
# z
i

r
0

e

i
θ

{\displaystyle dz = i r\_0 e^{i\theta} }

. Consequently,

∮
C

f
(
z
)

z
−

z
0

d
# z

∮

C
0

f
(
z
)

z
−

z
0

d
z

{\displaystyle \oint\_C \frac{f(z)}{z-z\_0}dz = \oint\_{C\_0} \frac{f(z)}{z-z\_0}dz }

∮
C

f
(
z
)

z
−

z
0

d
# z

∫

0

2
π

f
(

z
0

r
0

e

i
θ

)
i

r
0

e

i
θ

r
0

e

i
θ

d
θ

{\displaystyle \oint\_C \frac{f(z)}{z-z\_0}dz = \int\_{0}^{2\pi} \frac{f(z\_0 + r\_0 e^{i\theta})i r\_0 e^{i\theta}}{r\_0 e^{i\theta}} d\theta }

∮
C

f
(
z
)

z
−

z
0

d
# z
i

∫

0

2
π

f
(

z
0

r
0

e

i
θ

)
d
θ
.

{\displaystyle \oint\_C \frac{f(z)}{z-z\_0}dz = i\int\_{0}^{2\pi} f(z\_0 + r\_0 e^{i\theta}) d\theta. }

Taking the limit shows that

lim

r
0

→
0

i

∫

0

2
π

f
(

z
0

r
0

e

i
θ

)
d
# θ
i
(
2
π
)
f
(

z
0

)
.

{\displaystyle \lim\_{r\_0 \to 0} i\int\_{0}^{2\pi} f(z\_0 + r\_0 e^{i\theta}) d\theta = i(2 \pi) f(z\_0). }

Therefore,

∮
C

f
(
z
)

z
−

z
0

d
# z
2
π
i
f
(

z
0

)

{\displaystyle \oint\_C \frac{f(z)}{z-z\_0}dz = 2\pi i f(z\_0) }

and

f
(

z
0

# )

1

2
π
i

∮
C

f
(
z
)

z
−

z
0

d
z
.

{\displaystyle f(z\_0) = \frac{1}{2\pi i}\oint\_C \frac{f(z)}{z-z\_0}dz. }
Laplacian of a Scalar Field

## **Problem**

Calculate the Laplacian of the scalar function

ϕ

(

x
,
y
,
z

)

=

e

x

[

x
cos
⁡

(
y
)

−
y
sin
⁡

(
y
)

]

{\displaystyle \phi \left(x,y,z\right)={e}^{x}\left[x\cos {\left(y\right)}-y\sin {\left(y\right)}\right]}

.

Is this scalar field harmonic?

## **Solution**

\left( \right)
\left[ \right]

Use the Laplace operator (a.k.a Laplacian) in rectangular coordinates.

∇

2

# ϕ

∂

2

ϕ

∂

x

2

∂

2

ϕ

∂

y

2

∂

2

ϕ

∂

z

2

{\displaystyle {\nabla }^{2}\phi ={\frac {{\partial }^{2}\phi }{\partial {x}^{2}}}+{\frac {{\partial }^{2}\phi }{\partial {y}^{2}}}+{\frac {{\partial }^{2}\phi }{\partial {z}^{2}}}}

Calculate each second-order partial derivative.

∂

2

ϕ

∂

x

2

=

e

x

[

(

x
2

)

cos
⁡

(
y
)

−
y
sin
⁡

(
y
)

]

{\displaystyle {\frac {{\partial }^{2}\phi }{\partial {x}^{2}}}={e}^{x}\left[\left(x+2\right)\cos {\left(y\right)}-y\sin {\left(y\right)}\right]}

∂

2

ϕ

∂

y

2

=

e

x

[

−

(

x
2

)

cos
⁡

(
y
)

y
sin
⁡

(
y
)

]

{\displaystyle {\frac {{\partial }^{2}\phi }{\partial {y}^{2}}}={e}^{x}\left[-\left(x+2\right)\cos {\left(y\right)}+y\sin {\left(y\right)}\right]}

∂

2

ϕ

∂

z

2

=
0

{\displaystyle {\frac {{\partial }^{2}\phi }{\partial {z}^{2}}}=0}

Therefore,

∇

2

=
0

{\displaystyle {\nabla }^{2}=0}

Since the Laplacian equals 0, the final result shows that this scalar field is harmonic.
Casus grammaticus

Casus est forma mutabilis, quam vocabulum in se recipit ad indicandum munus suum grammaticum erga alia vocabula eiusdem periodi.

**English translation:** The case is a changeable form, which accepts a word in itself to indicate its grammatical function with respect to other words of the same period.

## **Descriptio (Description)**

### **Numerus casuum (Number of Cases)**

Lingua Gallica antiqua, exempli gratia, duos tantum casus habebat; lingua Hungarica plus quam viginti. Latina autem sex casibus de more utitur. Lingua Protoindoeuropaea usa est octo; ad illos Latinos instrumentalem addiderunt, qui in lingua nostra in ablativo mersus est. Omnes linguae Indoeuropaeae aliquam subclassem ipsarum habent. In lingua Vasconica etiam ergativus stat.

**English translation:** The ancient French language, for example, had only two cases; Hungarian language is more than twenty. However, Latin typically uses six cases. The Proto-Indo-Europaean language used eight; to those Latins they added an instrumental, which in our language has been drowned in the ablative. All Indo-European languages have some subclass of them. In the Basque language, it also stands.

### **Casus linguae Latinae (The cases of the Latin language )**

Casus Latini, uti supra demonstratum est, sunt sex, quibus additur locativus paucis vocabulis relictus:

1. Nominativus
2. Genetivus (s. Genitivus)
3. Dativus
4. Accusativus
5. Vocativus
6. Ablativus
7. Locativus

**English translation:** The Latin cases, as was mentioned above, are six, to which a locative is added, left to a few words:

1. Nominative
2. Genitive
3. Dative
4. Accusative
5. Vocative
6. Ablative
7. Locative
Double-angled Sine

## **Problem**

**Part 1:** Prove that

sin
⁡
(
2
x
# )
2
sin
⁡
(
x
)
cos
⁡
(
x
)

{\displaystyle \sin(2x)=2\sin(x)\cos(x)}

.

**Part 2:** Express the result in part 1 with only the tangent ratio.

## **Solution**

**Part 1**

Use the angle sum formula

sin
⁡
(
x
y
# )
sin
⁡
(
x
)
cos
⁡
(
y
)
cos
⁡
(
x
)
sin
⁡
(
y
)

{\displaystyle \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) }

 and make the substitution

# y
x

{\displaystyle y = x}

.

sin
⁡
(
x
x
# )
sin
⁡
(
x
)
cos
⁡
(
x
)
cos
⁡
(
x
)
sin
⁡
(
x
)

{\displaystyle \sin(x+x) = \sin(x)\cos(x) + \cos(x)\sin(x) }

Subsequently,

sin
⁡
(
2
x
# )
2
sin
⁡
(
x
)
cos
⁡
(
x
)
.

{\displaystyle \sin(2x) = 2\sin(x)\cos(x). }

**Part 2**

Since

tan
⁡
(
x
# )

sin
⁡
(
x
)

cos
⁡
(
x
)

{\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}}

 and

cos
⁡
(
x
# )

1

sec
⁡
(
x
)

{\displaystyle \cos(x) = \frac{1}{\sec(x)}}

, consider the product

2
sin
⁡
(
x
)
cos
⁡
(
x
# )
2

(

sin
⁡
(
x
)

cos
⁡
(
x
)

)

cos
2

⁡
(
x
)
.

{\displaystyle 2\sin(x)\cos(x) = 2\left(\frac{\sin(x)}{\cos(x)}\right)\cos^2(x). }

Then

sin
⁡
(
2
x
)

=
2
tan
⁡
(
x
)

cos
2

⁡
(
x
)

sin
⁡
(
2
x
)

=

2
tan
⁡
(
x
)

sec
2

⁡
(
x
)

.

{\displaystyle \begin{align}
\sin(2x) &= 2\tan(x)\cos^2(x) \\[5 pt]
\sin(2x) &= \frac{2\tan(x)}{\sec^2(x)}.
\end{align} }

Since

1

tan
2

⁡
(
x
# )

sec
2

⁡
(
x
)

{\displaystyle 1 + \tan^2(x) = \sec^2(x) }

, we can express

sin
⁡
(
2
x
)

{\displaystyle \sin(2x) }

 in terms of the tangent ratio:

sin
⁡
(
2
x
# )

2
tan
⁡
(
x
)

1

tan
2

⁡
(
x
)

.

{\displaystyle \sin(2x) = \frac{2\tan(x)}{1+ \tan^2(x)}. }
From Imaginary to Real

## **Problem**

What is the principle value of

i
i

{\displaystyle i^i}

?

Hint: first read about this problem. Euler's Gem

## **Solution**

From Euler's Gem, we have the result

e

i
x

=
cos
⁡
(
x
)
i
sin
⁡
(
x
)

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

.

Since

i

{\displaystyle i}

 has a positive imaginary component and no real component, it can be expressed as

# i
cos
⁡

(

π
2

)

i
sin
⁡

(

π
2

)

{\displaystyle i = \cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)}

.

This means

# i

e

i
π

2

{\displaystyle i = {e}^{\frac{i\pi}{2}} }

and

i
i

=

(

e

i
π

2

)

i

=

(

e

i
⋅
i
π

2

)

{\displaystyle i^i = \left({e}^{\frac{i\pi}{2}}\right)^i = \left({e}^{\frac{i\cdot i\pi}{2}}\right) }

.
Since

i
⋅
# i

i
2

=
−
1

{\displaystyle i \cdot i = i^2 = -1 }

, we discover that

i
i

=

e

−

π
2

{\displaystyle i^i = {e}^{-\frac{\pi}{2}} }

.
Bhaskara's Proof of the Pythagorean Theorem

## **Problem**

The diagram below refers to a right triangle

A
B
C

{\displaystyle ABC}

, where the apex is a right angle. A perpendicular line is drawn from the apex to the base at point

H

{\displaystyle H}

. The lines

B
C
,
A
C
,
A
B

{\displaystyle BC, AC, AB}

 are given side lengths

a
,
b
,
c

{\displaystyle a,b,c}

 respectively. Use proportion to prove

a
2

b
2

=

c
2

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

.

## **Solution**

There are three similar triangles

A
B
C
,
A
H
C
,
B
H
C

{\displaystyle ABC,AHC,BHC }

.

Let

A
# H
y

{\displaystyle AH=y }

. By proportion, we deduce that

y
b

=

b
c

{\displaystyle \frac{y}{b} = \frac{b}{c} }

and

c
−
y

a

=

a
c

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

,
which implies

# y

b
2

c

{\displaystyle y = \frac{b^2}{c} }

and

c
2

−
c
# y

a
2

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

With a little algebra

c
2

−

b
2

=

a
2

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

or

a
2

b
2

=

c
2

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

.
Heat Kernel

## **Problem**

Verify that the normal distribution

u
(
x
,
t
# )

1

4
π
k
t

exp
⁡

(

−

x
2

4
k
t

)

{\displaystyle u(x,t) = \frac{1}{\sqrt{4 \pi kt}} \exp{\left(-\frac{x^2}{4kt}\right)}}

satisfies the heat equation

u
t

=
k

u

x
x

{\displaystyle u\_t = ku\_{xx} }

 for

t
>
0

{\displaystyle t > 0}

, subject to the initial condition

u
(
x
,
0
# )
δ
(
x
)

{\displaystyle u(x,0)=\delta(x) }

.

## **Solution**

Take the first partial derivatives of

u
(
x
,
t
)

{\displaystyle u(x, t)}

 with respect to

t

{\displaystyle t}

.

u
t

=

−
1

2
t

4
π
k
t

exp
⁡

(

−

x
2

4
k
t

)

1

4
π
k
t

(

x
2

4
k

t
2

)

exp
⁡

(

−

x
2

4
k
t

)

{\displaystyle u\_t = \frac{-1}{2t\sqrt{4 \pi kt}} \exp{\left(-\frac{x^2}{4kt}\right)} + \frac{1}{\sqrt{4\pi kt}}\left(\frac{x^2}{4kt^2}\right)\exp{\left(-\frac{x^2}{4kt}\right)}}

u
t

=

1

4
π
k
t

(

−
1

2
t

x
2

4
k

t
2

)

exp
⁡

(

−

x
2

4
k
t

)

{\displaystyle u\_t = \frac{1}{\sqrt{4\pi kt}} \left(\frac{-1}{2t} + \frac{x^2}{4kt^2}\right) \exp{\left(-\frac{x^2}{4kt}\right)} }

Take the first and second partial derivatives of

u
(
x
,
t
)

{\displaystyle u(x, t)}

 with respect to

x

{\displaystyle x}

.

u
x

=

1

4
π
k
t

(

−
2
x

4
k
t

)

exp
⁡

(

x
2

4
k
t

)

{\displaystyle u\_x = \frac{1}{\sqrt{4\pi kt}} \left(\frac{-2x}{4kt}\right) \exp{\left(\frac{x^2}{4kt}\right)}}

u

x
x

=

1

4
π
k
t

(

−
1

2
k
t

x
2

4

k
2

t
2

)

exp
⁡

(

−

x
2

4
k
t

)

{\displaystyle u\_{xx} = \frac{1}{\sqrt{4\pi kt}} \left(\frac{-1}{2kt} + \frac{x^2}{4k^2t^2}\right) \exp{\left(-\frac{x^2}{4kt}\right)} }

k

u

x
x

=

1

4
π
k
t

(

−
1

2
t

x
2

4
k

t
2

)

exp
⁡

(

−

x
2

4
k
t

)

{\displaystyle ku\_{xx} = \frac{1}{\sqrt{4\pi kt}} \left(\frac{-1}{2t} + \frac{x^2}{4kt^2}\right) \exp{\left(-\frac{x^2}{4kt}\right)} }

Both sides of the equation match; therefore, the partial differential equation

u
t

=
k

u

x
x

{\displaystyle u\_t = ku\_{xx} }

 is satisfied.

Now for the initial condition. The Dirac delta function is actually a distribution, not a function. In fact the limit

lim

t
→
0

1

4
π
k
t

exp
⁡

(

−

x
2

4
k
t

)

=
δ
(
x
)

{\displaystyle \lim\_{t \to 0}\frac{1}{\sqrt{4 \pi kt}} \exp{\left(-\frac{x^2}{4kt}\right)}= \delta (x) }

automatically verifies the validity of this initial condition.
Walking Man

## **Problem**

Here is a classic related rates problem:

A man is walking away from a street light at the rate of 5 ft/sec. If the man is 6 ft tall and the light is 15 ft high, how fast is the man's shadow lengthening when he is 10 ft from the base of the street light?

## **Solution**

Consider the diagram on the right (Figure 1). Let

s

{\displaystyle s}

 be the length of the man's shadow at time

t

{\displaystyle t}

, and let

x

{\displaystyle x}

 be the distance the man is from the base of the street light at time

t

{\displaystyle t}

. We need to determine the related rate

d
s

d
t

{\displaystyle \frac{ds}{dt} }

 when

# x
10

f
t

{\displaystyle x = 10\;ft }

, given that

d
x

d
t

=
5

f
t

/

s
e
c

{\displaystyle \frac{dx}{dt} = 5 \; ft/sec }

. There are two right triangles in Figure 1, and they are similar triangles, so by proportion

x
s

15

=

s
6

.

{\displaystyle \frac{x+s}{15} = \frac{s}{6}. }

Hence, with a little algebra we get

6
x
6
s

=
15
s

6
x

=
9
s

s

=

2
3

x
.

{\displaystyle \begin{align}
6x + 6s &= 15s \\[5 pt]
6x &= 9s \\[5 pt]
s & = \frac{2}{3}x.
\end{align} }

Taking the derivative on both sides with respect to time gives

d
s

d
t

=

2
3

d
x

d
t

.

{\displaystyle \frac{ds}{dt} = \frac{2}{3} \frac{dx}{dt}. }

Substituting the rate

d
x

d
t

=
5

f
t

/

s
e
c

{\displaystyle \frac{dx}{dt} = 5 \; ft/sec }

 yields:

d
s

d
t

=

2
3

⋅
5

f
t

/

s
e
c

=

10
3

f
t

/

s
e
c

{\displaystyle \begin{align}
\frac{ds}{dt} &= \frac{2}{3} \cdot 5 \; ft/sec \\[5 pt]
&= \frac{10}{3} \, ft/sec
\end{align} }

From this calculation we find that the man's shadow is lengthening at the rate of

10
3

f
t

/

s
e
c

{\displaystyle \frac{10}{3} \; ft/sec }

 no matter how far he is from the base of the street light!
Four Triangles Problem

## **Problem**

Consider triangle ABC in Figure 1. There are three triangles coloured yellow, blue, and red that meet at one point P. The lines through P are parallel to the sides of triangle ABC; hence, the uncovered areas of triangle ABC are parallelograms with diagonals PA, PB, and PC respectively. Given that the area of the yellow triangle, blue triangle, and red triangle are 4, 9, and 49 respectively, determine the area of triangle ABC.

## **Solution**

The yellow, blue, red triangles and triangle ABC are all similar triangles. Each side of of triangle ABC is the sum of the corresponding sides of all three coloured triangles. The ratio of each similar side of the yellow, blue, and red triangles is

s
Y

:

s
B

:

s
R

:

s

△
A
B
C

=
2
:
3
:
7
:

{\displaystyle s\_Y : s\_B : s\_R : s\_{\bigtriangleup ABC} = 2:3:7:12.}

By the principle of areal proportion, we get

A
Y

:

A
B

:

A
R

:

A

△
A
B
C

=
4
:
9
:
49
:

{\displaystyle A\_Y : A\_B : A\_R : A\_{\bigtriangleup ABC} = 4:9:49:144.}

Thus, the area of triangle ABC is 144.
Hat Party!

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

## **Problem**

Suppose you and five of your friends each brought a hat to a party. The hats are then put into a large box for a random hat draw. What is the probability that nobody selects his or her own hat?

## **Solution**

There are six people in total (you + 5 friends), so the number of possibilities that each person can get their hat is:

P

6

=
6
# !

{\displaystyle P\_{6}=6!=720.}

The number of possibilities that nobody selects their own hat is called the derangement problem. For six people, the number of derangements is:

D

6

=
6
!

∑

# k
0

6

(

−
1

)

k

k
!

=
265

{\displaystyle D\_{6}=6!\sum \_{k=0}^{6}{\frac {{\left(-1\right)}^{k}}{k!}}=265}

so the probability of each person get their wrong hat is:

# p

265
720

{\displaystyle p={\frac {265}{720}}}

# p

53
144

≈
0.368056.

{\displaystyle p={\frac {53}{144}}\approx 0.368056.}

Therefore, the probability that nobody selects his or her own hat is 36.81%.

Note that a faster way of calculating the number of derangements is

D

n

=

[

n
!

e

]

.

{\displaystyle D\_{n}=\left[{\frac {n!}{e}}\right].}
Lie Algebra Axioms

## **Problem**

Prove the four axioms of Lie algebra for the Lie bracket

[
A
,
B
# ]
A
B
−
B
A

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

, where

A
,
B

{\displaystyle A, B}

 are non-commutative operators, and

α
,
β

{\displaystyle \alpha, \beta}

 are scalars.

**1. Bilinearity**

[
α
A
β
B
,
C
# ]
α
[
A
,
C
]
β
[
B
,
C
]

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

**2. Alternativity**

[
A
,
A
# ]
0

{\displaystyle [A, A] = 0 }

**3. Anticommutativity**

[
A
,
B
# ]
−
[
B
,
A
]

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

**4. Jacobi identity**

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

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

## **Solution**

**1. Bilinearity**

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

{\displaystyle [\alpha A + \beta B, C] = (\alpha A + \beta B)C - C(\alpha A + \beta B) }

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

{\displaystyle [\alpha A + \beta B, C] = \alpha AC + \beta BC - (\alpha CA + \beta CB) }

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

{\displaystyle [\alpha A + \beta B, C] = (\alpha AC - \alpha CA) + (\beta BC - \beta CB) }

[
α
A
β
B
,
C
# ]
α
[
A
,
C
]
β
[
B
,
C
]

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

**2. Alternativity**

[
A
,
A
# ]
A
A
−
A
A

{\displaystyle [A, A] = AA - AA }

[
A
,
A
# ]
0

{\displaystyle [A, A] = 0 }

**3. Anticommutativity**

[
A
,
B
# ]
A
B
−
B
A

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

[
A
,
B
# ]
−
(
B
A
−
A
B
)

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

[
A
,
B
# ]
−
[
B
,
A
]

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

**4. Jacobi identity**

[
A
,
[
B
,
C
]
# ]
[
A
,
B
C
−
C
B
]

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

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

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

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

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

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

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

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

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

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

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

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

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

.
Free Computation Tools

## **Calculators and Graphing**

Graphmatica - Free graphing software to download. Great for learning and teaching coordinate geometry and single-variable calculus.

Octave Online - Essentially free MATLAB that can execute uploaded files through an online terminal.

Symbolab - Want Wolfram Alpha with the free solutions back, and without the information overload? This is the simpler and free symbolic computation software.

Integral Calculator - A neat online tool that solves integrals with steps shown!

Linear Algebra Toolkit - Excellent calculator for reduced row echelon form with every step written out.

Matrix Calculator - Need to multiply two matrices instantaneously?

GeoGebra Math Apps - Great tool for teaching and learning algebra, geometry, and probability visually.

Geogebra 3D - An online three-dimensional graphing tool!

Simplex Method Calculator - Excellent and easy-to-use online calculator for linear programming.

Matrix Game Calculator - Easy to use matrix game solver. Unfortunately it doesn't show the steps...

Normal Distribution Calculator - Easy to use normal distibution calculator that even shades in the area under the bell curve!

## **Python Programming Sites**

Automate the Boring Stuff with Python - Intro to what you can do to automate tasks using Python.

Python for Data Science - Excellent reference for Python programming and data science.
Eigenvalues of Hermitian Matrices

# H

H
†

{\displaystyle H = H^\dagger }

where the superscript dagger means the transpose matrix with complex-conjugate elements.

Prove that the eigenvalues of a Hermitian matrix are real numbers.

## **Solution**

Let

λ

{\displaystyle \lambda}

 be the eigenvalue of a Hermitian matrix

H

{\displaystyle H}

. For every non-zero vector

v

{\displaystyle v}

H
# v
λ
v

{\displaystyle Hv = \lambda v }

(
H
v

)
†

=
(
λ
v

)
†

{\displaystyle (Hv)^\dagger =(\lambda v)^\dagger }

v
†

H
†

=

λ
∗

v
†

{\displaystyle v^\dagger H^\dagger = \lambda^\* v^\dagger }

Then

v
†

H
†

# v

λ
∗

v
†

v

{\displaystyle v^\dagger H^\dagger v = \lambda^\* v^\dagger v }

Since

# H

H
†

{\displaystyle H = H^\dagger }

v
†

λ
# v

λ
∗

v
†

v

{\displaystyle v^\dagger \lambda v = \lambda^\* v^\dagger v }

λ

v
†

# v

λ
∗

v
†

v

{\displaystyle \lambda v^\dagger v = \lambda^\* v^\dagger v }

This implies

# λ

λ
∗

{\displaystyle \lambda = \lambda^\* }

. If the eigenvalues were imaginary or complex, then

λ
∗

≠
λ

{\displaystyle \lambda^\* \ne \lambda }

. Therefore the eigenvalues of a Hermitian matrix must be real.
Density of Air

## **Problem**

The composition of air is roughly 80% nitrogen gas and 20% oxygen gas. Calculate the density of air under **normal temperature and pressure conditions (NTP)**.

Normal temperature and pressure (NTP) is defined as the condition when the temperature is 20 degrees Celsius and the pressure is 1 atmosphere.

### **Hints**

(1) Nitrogen gas and oxygen gas are diatomic.

(2) Gases in the Earth's atmosphere is of low concentration; this means the atmosphere can be modelled as an ideal gas.

## **Solution**

**Mass**

Nitrogen and oxygen gas are diatomic, so the molar mass is

0.80

(

2
×
14.01

g

m
o
l

)

0.20

(

2
×
16.00

g

m
o
l

)

=
28.80

g

m
o
l

.

{\displaystyle 0.80 \left(2 \times 14.01 \; \frac{g}{mol} \right) + 0.20 \left(2 \times 16.00 \; \frac{g}{mol} \right) = 28.80 \; \frac{g}{mol}. }

.
Calculate the mass of 1 mole of air:

1

m
o
l
×
28.80

g

m
o
l

×

1

k
g

1000

g

=
0.02880

k
g
.

{\displaystyle 1 \; mol \times 28.80 \; \frac{g}{mol} \times
\frac{1 \; kg}{1000 \; g} = 0.02880 \; kg. }

.
Therefore, 1 mole of air weighs 0.02880 kg.

**Volume**

To calculate the volume of 1 mole of air at NTP, the pressure and temperature are:

# P
1

a
t
# m
101325

P
a

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

# T

20

∘

C
# 273.15
293.15

K

{\displaystyle T = {20} \;^{\circ}C + 273.15 = 293.15 \; K }

Substitute the known values into the **ideal gas law** formula and solve for the volume:

P
# V
n
R
T

{\displaystyle PV = nRT}

(
101325

P
a
)
# V
(
1

m
o
l
)

(

8.31

J

K
⋅
m
o
l

)

(
293.15

K
)

{\displaystyle (101325 \; Pa)V = (1 \; mol)\left(8.31 \; \frac{J}{K\cdot mol}\right)(293.15 \; K) }

# V
0.024

m
3

.

{\displaystyle V = 0.024 \; m^3 . }

**Density**

Simply divide the mass by the volume to get the density of air :

ρ

a
i
r

=

0.02880

k
g

0.024

m
3

{\displaystyle {\rho}\_{air} = \frac{0.02880 \, kg}{0.024 \, m^3} }

ρ

a
i
r

=
1.2

k
g

m
3

{\displaystyle {\rho}\_{air} = 1.2 \, \frac{kg}{m^3} }
Adding Vectors

## **Problem**

A man walks 53 meters 20 degrees northeast, and then 34 meters 63 degrees northeast. How far and in which direction did he travel from his starting point?

# A
⟨
53
cos
⁡

20

∘

,
53
sin
⁡

20

∘

⟩

{\displaystyle \boldsymbol{A} = \langle 53 \cos{20^{\circ}} , 53\sin{20^{\circ}} \rangle }

 and

# B
⟨
34
cos
⁡

63

∘

,
34
sin
⁡

63

∘

⟩

{\displaystyle \boldsymbol{B} = \langle 34 \cos{63^{\circ}} , 34\sin{63^{\circ}} \rangle }

. Add up each component to get the resultant vector.

R

=

A
B

=
⟨
53
cos
⁡

20

∘

34
cos
⁡

63

∘

,
53
sin
⁡

20

∘

34
sin
⁡

63

∘

⟩

=
⟨
65.2393
,
48.4213
⟩
.

{\displaystyle \begin{align}
\boldsymbol{R} &= \boldsymbol{A + B} \\[5 pt]
&= \langle 53 \cos{20^{\circ}} + 34 \cos{63^{\circ}} , 53\sin{20^{\circ}} + 34 \sin{63^{\circ}} \rangle \\[5 pt]
&= \langle 65.2393 , 48.4213 \rangle.
\end{align} }

Calculate the norm to obtain the magnitude of the resultant vector.

R

=

(
65.2393

)
2

(
48.4213

)
2

[
5
p
t
]

=
81.25

m

{\displaystyle \begin{align}
&= 81.25 \;m
\end{align}
 }

Calculate the direction of the resultant vector using inverse tangent.

θ
R

=

tan

−
1

(

48.4213
65.2393

)

=

36.58

∘

{\displaystyle \begin{align}
\theta\_R &= {\tan}^{-1}\left(\frac{48.4213}{65.2393}\right) \\[5 pt]
&= 36.58^{\circ}
\end{align}}