recursal/Fanatic-Fandom huggingface dataset:
Torricelli's Trumpet

## **Problem**

**Torricelli's trumpet** (also called **Gabriel's horn**) is a geometric figure that was first studied by Italian mathematician and physicist Evangelista Torricelli (1608 - 1647 AD). Torricelli called this figure the truncated acute hyperbolic solid in his 1643 paper *De solido hyperbolico acuto* (O*n the Acute Hyperbolic Solid*)*.*

Torricelli's trumpet is formed revolving the rectangular hyperbola

# y

1
x

{\displaystyle y=\frac{1}{x}}

around the x-axis with the domain

x
≥
1

{\displaystyle x \geq 1}

.

Use integration to prove that Torricelli's trumpet has finite volume but infinite surface area. Torricelli discovered this fact before the invention of calculus; he applied Cavalieri's principle to arrive at this result.

## **Solution**

**Volume**

The volume formula of a solid of revolution (revolved around the x-axis) is

# V
π

∫
a
b

y
2

d
x

{\displaystyle V = \pi\int\_a^b y^2 dx }

Using integration on the solid of revolution of

# y

1
x

{\displaystyle y=\frac{1}{x}}

revolved around the x-axis with the domain

x
≥
1

{\displaystyle x \geq 1}

, the volume of Torricelli's trumpet is:

V

=
π

∫
1
b

(

1
x

)

2

d
x

=
π

∫
1
b

1

x
2

d
x

=
π

(

1
−

1
b

)

{\displaystyle \begin{align}
V &= \pi\int\_1^b \left(\frac{1}{x}\right)^2 dx \\
&= \pi\int\_1^b \frac{1}{x^2} dx \\
&= \pi\left(1 - \frac{1}{b} \right) \\
\end{align}
}

Taking the limit as

b
→
∞

{\displaystyle b \to \infty}

 yields

lim

b
→
∞

V

=

lim

b
→
∞

π

(

1
−

1
b

)

=
π

lim

b
→
∞

(

1
−

1
b

)

=
π

{\displaystyle \begin{align}
\lim\_{b\to\infty} V &= \lim\_{b\to\infty} \pi\left(1 - \frac{1}{b}\right) \\
&= \pi \lim\_{b\to\infty}\left(1 - \frac{1}{b}\right) \\
&= \pi
\end{align}}

**Surface Area**

The surface area formula of a solid of revolution (revolved around the x-axis) is

# A
2
π

∫
a
b

y

1

(

d
y

d
x

)

2

d
x

{\displaystyle A = 2\pi\int\_a^b y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx }

so the surface area of Torricelli's trumpet is

A

=
2
π

∫
1
b

1
x

1

(

−

1

x
2

)

2

d
x

=
2
π

∫
1
b

1
x

1

1

x
4

d
x

{\displaystyle \begin{align}
A &= 2\pi\int\_1^b \frac{1}{x} \sqrt{1 + \left(-\frac{1}{x^2}\right)^2}dx \\
&= 2\pi\int\_1^b \frac{1}{x} \sqrt{1 + \frac{1}{x^4}}dx
\end{align}}

By comparing to the integral

2
π

∫
1
b

1
x

d
x

{\displaystyle 2\pi\int\_1^b \frac{1}{x} dx}

 we can deduce that

# A
2
π

∫
1
b

1
x

1

1

x
4

d
x
>
2
π

∫
1
b

1
x

d
# x
2
π
ln
⁡
a
.

{\displaystyle A = 2\pi\int\_1^b \frac{1}{x} \sqrt{1 + \frac{1}{x^4}} dx > 2\pi\int\_1^b \frac{1}{x} dx = 2\pi\ln a.}

Since

2
π

∫
1
b

1
x

d
# x
2
π
ln
⁡
(
b
)

{\displaystyle 2\pi\int\_1^b \frac{1}{x} dx = 2\pi\ln (b)}

and

lim

b
→
∞

2
π
ln
⁡
(
b
# )
∞

{\displaystyle \lim\_{b\to\infty} 2\pi\ln(b) = \infty}

we deduce that

# A
2
π

∫
1
∞

1
x

1

1

x
4

d
# x
∞
.

{\displaystyle A = 2\pi\int\_1^\infty \frac{1}{x} \sqrt{1 + \frac{1}{x^4}} dx = \infty.}

Therefore, Torricelli's horn has finite volume but infinite surface area!
Nicole Oresme

**By: Tao Steven Zheng (郑涛)**

**Born:** c. 1325 AD in Fleury-sur-Orne, Normandy

**Died:** 1382 AD in Lisieux, Normandy

**Nicole Oresme** or **Nicolas d'Oresme** was the most influential French philosopher, theologian, and mathematician of the late Middle Ages. He was one of the most original thinkers of the 14th century, developing concepts and ideas which anticipated the scientific breakthroughs of scientists centuries later. Oresme also served as the Bishop of Lisieux and the counselor of King Charles V of France. As an academic, he is known to have translated several works of Aristotle into French and wrote numerous works on mathematics, natural philosophy, economics, and theology.

## **Discoveries**

**Rates and Ratios**

Oresme developed a geometrical method of plotting speed or intensity over time or distance. He also further developed the theory of ratios by Thomas Bradwardine (c. 1295 - 1349 AD) to relate speed to force and resistance, and applied it to physics questions.

**Fractional Exponents**

In *Algorismus proportionum* and *De proportionibus proportionum*, Oresme developed the first use of fractional exponents, and calculations with irrational proportions*.* The basis of his method stems from the equalization of continuous magnitudes and discrete numbers, an idea that Oresme developed from his studies on music theory*.*

**Infinite Series**

In *Tractatus de configurationibus qualitatum et motuum,* Oresme investigated infinite series and the notion of the limit, which paved the way for the discovery of calculus. He proved the divergence of the harmonic series using the comparison test.

Harmonic Series
Weishu and Zhuishu 軎术和缀术

**By: Tao Steven Zheng (郑涛)**

**【历史纵横】**

《缀术》是南北朝时期祖冲之（公元429年 ~ 501年）、祖暅（约公元450年 ~ 520年）父子合撰的数学著作。此书汇集了他们的数学研究成果，比如计算圆周率与球的体积公式。《缀术》在唐代成为国子监算学课本之一。当时的学生需要四年时间学习，而且被认为非常深奥，以致“学官莫能究其深奥，故废而不理”（《隋书·律历志》）。因此，《缀术》终于北宋时期佚传。本节选来自沈括（公元1031 ~ 1095年）的《梦溪笔谈》。

**【Historical Note】**

The *Zhui Shu* (*Method of Interpolation*) is a mathematics text written during the Northern and Southern dynasties (386 – 589 AD) by Zu Chongzhi (429 – 501 AD) and his son Zu Geng (c. 450 – 520 AD). The book is a compilation of their mathematical findings, such as the calculation of *π* and the precise volume of the sphere. During the Tang dynasty, the *Zhui Shu* was included into the mathematics curriculum at the Guozijian (Imperial Academy). The book is said to require four years of study, and was considered to be so abstruse that “students could not study its profundity, so it is abandoned and ignored.” Unfortunately, the *Zhui Shu* was lost during the Northern Song Dynasty. This excerpt is a description of the *Zhui Shu* taken from the *Mengxi Bitan*, written by Shen Kuo (1031 – 1095 AD).

**《梦溪笔谈·卷18技艺3》**
审方面势，覆量高深远近，算家谓之“軎术”。軎文象形，如绳木[1]所用墨斗[2]也。求星辰之行，步气朔消长，谓之“缀术”。谓不可以形察，但以算数缀之而已。北齐祖亘[3] 有《缀术》二卷 [4] 。
***Mengxi Bitan* (Volume 18 Jiyi 3)**
Mathematicians call the method of surveying terrain, their heights and distances *weishu*. The character *wei* (軎) is a pictograph, which looks like a cord through the carpenter’s ink marker. The method of computing the movements of stars and the changes of the lunar calendar is called *zhuishu*. The name of this method suggests that the movement of celestial bodies cannot be known by their outward shapes and movements; rather, the movements can only be interpolated via a series of linked mathematical computations. During the Northern Qi dynasty, Zu Geng authored the *Zhui Shu* in two volumes.

**【注释】**

[1] 绳木：绳子。

Cord, rope, string.

[2] 墨斗：中国木匠用的工具，由墨仓、线轮、墨线、墨签构成。

Ink marker, a tool used by Chinese carpenters for marking lines.

[3] 祖亘：祖暅 [gèng] （约公元 456年—536 年），一作祖暅之，是中国南北朝时期数学家、天文学家。

Zu Geng (c. 456-536), or Zu Gengzhi, was a Chinese mathematician and astronomer during the Northern and Southern Dynasties.

[4] 据《隋书·经籍志》，《缀术》有六卷。

According to *Sui Shu* (*Book of Sui*), the *Zhui Shu* has six volumes.

**【今译】**

推求方位和地势，测量高低和远近，算学家称它为“軎术”。是个象形字，像在木头上划直线用的墨斗。求天体的运行，推算节气朔望的盈缩变化，叫作“缀术”。这是指不可以用形体来考究，只能用算学的方法把它们连缀起来计算而已。北齐祖亘（祖暅）著有《缀术》二卷。
Logarithmic Differentiation

# y

x
x

{\displaystyle y = x^x }

, calculate its derivative.

## **Solution**

Take the logarithm on both sides of the equation

ln
⁡
(
y
# )
x
ln
⁡
(
x
)

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

.
Take the derivative on both sides with respect to

x

{\displaystyle x}

:

d

d
x

ln
⁡
(
y
# )

d

d
x

x
ln
⁡
(
x
)

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

1
y

d
y

d
x

=
ln
⁡
(
x
)
x

(

1
x

)

{\displaystyle \frac{1}{y} \frac{dy}{dx} = \ln(x) + x\left(\frac{1}{x}\right) }

d
y

d
x

=
y
[
ln
⁡
(
x
)
1
]

{\displaystyle \frac{dy}{dx} = y[\ln(x) + 1] }

.
Since

# y

x
x

{\displaystyle y = x^x }

,

d
y

d
x

=

x
x

[
ln
⁡
(
x
)
1
]

{\displaystyle \frac{dy}{dx} = x^x [\ln(x) + 1] }

d

d
x

x
x

=

x
x

[
ln
⁡
(
x
)
1
]

{\displaystyle \frac{d}{dx} x^x = x^x [\ln(x) + 1] }

.

## **Additional Remark**

For a general composite exponent function

f
(
x
# )
g
(
x

)

h
(
x
)

{\displaystyle f(x)=g(x)^{h(x)} }

, the logarithm transforms the exponentiation into a product:

ln
⁡
(
f
(
x
)
# )
ln
⁡

(

g
(
x

)

h
(
x
)

)

{\displaystyle \ln(f(x)) = \ln\left(g(x)^{h(x)}\right) }

ln
⁡
(
f
(
x
)
# )
h
(
x
)
l
n

(

g
(
x
)

)

{\displaystyle \ln(f(x)) = h(x) ln\left(g(x)\right) }

.
Differentiating by applying the chain rule and the sum rule yields

f
′

(
x
)

f
(
x
)

=

h
′

(
x
)
ln
⁡
(
g
(
x
)
)
h
(
x
)

g
′

(
x
)

g
(
x
)

{\displaystyle \frac{f'(x)}{f(x)} = h'(x) \ln(g(x)) + h(x)\frac{g'(x)}{g(x)} }

which becomes

f
′

(
x
# )
f
(
x
)

[

h
′

(
x
)
ln
⁡
(
g
(
x
)
)
h
(
x
)

g
′

(
x
)

g
(
x
)

]

{\displaystyle f'(x) = f(x)\left[ h'(x) \ln(g(x)) + h(x)\frac{g'(x)}{g(x)}\right] }

f
′

(
x
# )

g
(
x
)

h
(
x
)

[

h
′

(
x
)
ln
⁡
(
g
(
x
)
)
h
(
x
)

g
′

(
x
)

g
(
x
)

]

{\displaystyle f'(x) = {g(x)}^{h(x)}\left[ h'(x) \ln(g(x)) + h(x)\frac{g'(x)}{g(x)}\right] }

.
If we apply this function with

g
(
x
# )
x

{\displaystyle g(x) = x}

,

h
(
x
# )
x

{\displaystyle h(x) = x }

,

g
′

(
x
# )
1

{\displaystyle g'(x) = 1 }

, and

h
′

(
x
# )
1

{\displaystyle h'(x) = 1 }

, we will uncover the same result

d

d
x

x
x

=

x
x

[
ln
⁡
(
x
)
1
]

{\displaystyle \frac{d}{dx} x^x = x^x [\ln(x) + 1] }

.
Silver Items

## **Vintage Oneida WM. A. Rogers Triple Spoon**

**Materials:** Silver-plated

**Length:** 7.3 in / 18.5 cm

**Mass:** 1.5 oz / 44 g

**Year:** 1930s

## **Vintage Oneida Silversmiths Plate**

**Materials:** Silver-plated

**Diameter:** 6.1 in / 15.5 cm

**Height:** 0.92 in / 2.34 cm

**Mass:** 4.3 oz / 123 g

**Year:** 1940s

## **Vintage Oneida Silversmiths Paul Revere Reproduction Bowl**

**Materials:** Silver-plated

**Diameter:** 4.0 in / 10.16 cm

**Height:** 2.0 in / 5.08 cm

**Mass:** 3.6 oz / 104 g

**Year:** 1940s
Painted Cubes

## **Problem**

Suppose we have four cubes that are painted on every face:

2
×
2
×
2

{\displaystyle 2 \times 2 \times 2 }

,

3
×
3
×
3

{\displaystyle 3 \times 3 \times 3 }

,

4
×
4
×
4

{\displaystyle 4 \times 4 \times 4 }

 , and

5
×
5
×
5

{\displaystyle 5 \times 5 \times 5 }

.

**Part 1:** For each cube, determine the number of

1
×
1
×
1

{\displaystyle 1 \times 1 \times 1 }

 cubes that have three painted faces, two painted faces, only one painted face, and have no paint? Construct a chart.

**Part 2:** Determine the general formula for an

n
×
n
×
n

{\displaystyle n \times n \times n }

 cube.

**Part 3:** Check that the sum of all the general formulas add up to

n
3

{\displaystyle n^3 }

.

## **Solution**

**Part 1**

**Part 2**

**Part 3**

8
12
(
n
−
2
)
6
(
n
−
2

)
2

(
n
−
2

)
3

=
8
(
n
−
2
)

[

12
6
(
n
−
2
)
(
n
−
2

)
2

]

[
5
p
t
]

=
8
(
n
−
2
)

[

12
6
n
−
12

n
2

−
4
n
4

]

[
5
p
t
]

=
8
(
n
−
2
)

[

n
2

2
n
4

]

[
5
p
t
]

=
8

n
3

2

n
2

4
n
−
2

n
2

−
4
n
−
8

[
5
p
t
]

=

n
3

{\displaystyle \begin{align}
8 + 12(n-2) + 6 (n-2)^2 + (n-2)^3 &= 8 + (n-2) \left[12 + 6 (n-2) + (n-2)^2 \right] \\ [5 pt]
&= 8 + (n-2) \left[12 + 6n - 12 + n^2 - 4n + 4 \right] \\ [5 pt]
&= 8 + (n-2) \left[n^2 + 2n + 4 \right] \\ [5 pt]
&= 8 + n^3 + 2n^2 + 4n - 2n^2 - 4n - 8 \\ [5 pt]
&= n^3
\end{align}}
Trapezoidal Force Distribution

## **Problem**

This problem is from *Engineering Mechanics* (Hibbeler and Yap).

If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities

w
1

{\displaystyle w\_1 }

 and

w
2

{\displaystyle w\_2 }

 of this distribution needed to support the column loads. Answer in

k
N

/

m

{\displaystyle kN/m }

.

## **Hint**

(1) Split the trapezoidal distribution of load into a rectangle and a right triangle.

## **Solution**

The total length of the footing is 8 meters long:

1

m
2.5

m
3.5

m
1

# m
8

m

{\displaystyle 1 \; m + 2.5 \; m + 3.5 \; m + 1 \; m = 8 \; m}

. Let the pivot be the left side of the footing.

Separate the trapezoidal force distribution beneath the platform into a rectangle and a right triangle. The base of the rectangle is

8

m

{\displaystyle 8 \; m}

 long, and the height is equal to

w
2

k
N

/

m

{\displaystyle w\_2 \; kN/m }

. Beneath the rectangle is the right triangle. The base of the rectangle is

8

m

{\displaystyle 8 \; m}

 long, and the height is equal to

w
1

−

w
2

k
N

/

m

{\displaystyle w\_1 - w\_2 \; kN/m }

.

For the rectangular section, the equivalent force

F
1

{\displaystyle F\_1}

 is located at the center at

4

m

{\displaystyle 4 \; m }

 from the left. The magnitude of this equivalen force is calculated using the area of the rectangle

F
1

=
8
×

w
2

=
8

w
2

.

{\displaystyle \begin{align}
F\_1 &= 8 \times w\_2 \\[5 pt]
&= 8w\_2.
\end{align} }

For the triangular section, the equivalent force

F
2

{\displaystyle F\_2}

 is located at the one-third mark of the right triangle's base, which is

8
3

m

{\displaystyle \frac{8}{3} \; m }

 from the left. The magnitude of this equivalent force is calculated using the area of the triangle

F
2

=

1
2

×
8
×
(

w
1

−

w
2

)

=
4
(

w
1

−

w
2

)
.

{\displaystyle \begin{align}
F\_2 &= \frac{1}{2} \times 8 \times (w\_1 - w\_2) \\[5 pt]
&= 4(w\_1 - w\_2).
\end{align} }

The forces and moments (torque) must be balanced to maintain equilibrium.

**Forces**

F
1

F
2

=
60

k
N
80

k
N
50

k
N

F
1

F
2

=
190

k
N

{\displaystyle \begin{align}
F\_1 + F\_2 &= 60\; kN + 80 \; kN + 50 \; kN \\[5 pt]
F\_1 + F\_2 &= 190 \; kN
\end{align} }

**Moments**

4

F
1

8
3

F
2

=
(
60

k
N
⋅
1

m
)
(
80

k
N
⋅
3.5

m
)
(
50

k
N
⋅
7

m
)

4

F
1

8
3

F
2

=
490

k
N
⋅
m

{\displaystyle \begin{align}
4 F\_1 + \frac{8}{3} F\_2 &= (60\; kN \cdot 1 \;m ) + (80 \; kN \cdot 3.5 \;m ) + (50 \; kN \cdot 7 \;m ) \\[5 pt]
4F\_1 + \frac{8}{3} F\_2 &= 490 \; kN \cdot m
\end{align} }

Replacing each force with the distributions

w
1

{\displaystyle w\_1 }

 and

w
2

{\displaystyle w\_2 }

, we get the system of linear equations:

{

4

w
1

4

w
2

=
190

32
3

w
1

64
3

w
2

=

{\displaystyle \begin{cases}
4w\_1 + 4w\_2 &= 190 \\[5 pt]
\frac{32}{3}w\_1 + \frac{64}{3}w\_2 &= 690.
\end{cases} }

Solving this system of linear equations yields

w
1

=
30.3125

k
N

{\displaystyle w\_1 = 30.3125 \; kN }

 and

w
2

=
17.1875

k
N

{\displaystyle w\_2 = 17.1875 \; kN }

.
Springs in Series and Parallel

## **Problem**

Consider two springs with different spring constants

k
1

{\displaystyle k\_1 }

 and

k
2

{\displaystyle k\_2 }

.

**Part 1:** Determine the equivalent spring constant when the two springs are connected in series.

**Part 2:** Determine the equivalent spring constant when the two springs are connected in parallel.

## **Solution**

**Part 1: Springs connected in series (same forces, different length)**

When two springs are connected in series, the result is essentially a longer and flimsier spring. When a force is applied to the combined spring, the same force is applied to each individual spring. Since the springs have different spring constants, the displacements are different. Thus we get three equations:

1. F

a
p
p

=

k

e
q

(

x
1

x
2

)

{\displaystyle {F}\_{app} = {k}\_{eq} (x\_1 + x\_2) }
2. F

a
p
p

=

k
1

x
1

{\displaystyle {F}\_{app} = k\_1 x\_1 }
3. F

a
p
p

=

k
2

x
2

{\displaystyle {F}\_{app} = k\_2 x\_2 }

First equate equations 2 and 3 and isolate for

x
2

{\displaystyle x\_2}

k
1

x
1

=

k
2

x
2

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

x
2

=

k
1

x
1

k
2

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

.
Substitute

x
2

{\displaystyle x\_2}

 into equation 1.

F

a
p
p

=

k

e
q

(

x
1

k
1

x
1

k
2

)

{\displaystyle {F}\_{app} = {k}\_{eq} \left( x\_1 + \frac{k\_1 x\_1}{k\_2}\right) }

F

a
p
p

=

k

e
q

(

1

k
1

k
2

)

x
1

{\displaystyle {F}\_{app} = {k}\_{eq} \left( 1 + \frac{k\_1}{k\_2}\right)x\_1 }

To solve for the equivalent spring constant, set the above equation with equation 2. The strategy is to divide away

x
1

{\displaystyle x\_1}

.

k

e
q

(

1

k
1

k
2

)

x
1

=

k
1

x
1

{\displaystyle {k}\_{eq} \left( 1 + \frac{k\_1}{k\_2}\right)x\_1 = k\_1 x\_1 }

k

e
q

=

k
1

k
2

k
1

k
2

{\displaystyle {k}\_{eq} = \frac{k\_1 k\_2}{k\_1 + k\_2} }

In general, given

n

{\displaystyle n}

 springs added in series

1

k

e
q

=

1

k
1

1

k
2

1

k
3

.
.
.

1

k
n

{\displaystyle \frac{1}{{k}\_{eq}} = \frac{1}{k\_1} + \frac{1}{k\_2} + \frac{1}{k\_3} + ... + \frac{1}{k\_n} }

.
**Part 2: Springs connected in parallel (same length, different forces)**

When two springs are connected in parallel, the result is essentially two springs working together. When a force is applied to the combined spring, the force applied on each individual spring is different. Thus we get three equations:

1. F

1

=

k
1

x

{\displaystyle {F}\_{1} = k\_1 x }
2. F

2

=

k
2

x

{\displaystyle {F}\_{2} = k\_2 x }
3. F

a
p
p

=

k

e
q

# x

F
1

F
2

{\displaystyle {F}\_{app} = {k}\_{eq} x = F\_1 + F\_2 }

Thus

k

e
q

# x

k
1

x

k
2

x

{\displaystyle {k}\_{eq} x = k\_1 x + k\_2 x }

k

e
q

# x
(

k
1

k
2

)
x

{\displaystyle {k}\_{eq} x = (k\_1 + k\_2) x }

.
Therefore,

k

e
q

=

k
1

k
2

{\displaystyle {k}\_{eq} = k\_1 + k\_2 }

.
In general, given

n

{\displaystyle n}

 springs added in parallel

k

e
q

=

k

1

k

2

k

3

.
.
.

k

n

{\displaystyle {k}\_{eq} = {k}\_{1} + {k}\_{2} + {k}\_{3} +... + {k}\_{n}}

.
Variance of Sums

## **Problem**

The covariance of two random variables can be calculated using the expectation of each variable

c
o
v
(
X
,
Y
# )
E
[
X
Y
]
−
E
[
X
]
E
[
Y
]

{\displaystyle cov(X,Y) = E[XY] - E[X]E[Y] }

.

For a proof, check out this page.

Prove that

V
a
r
(
X
Y
# )
V
a
r
(
X
)
V
a
r
(
Y
)
2
c
o
v
(
X
,
Y
)

{\displaystyle Var(X+Y)=Var(X)+Var(Y)+2cov(X,Y)}

.

## **Solution**

V
a
r
(
X
Y
# )
E
[

(
X
Y
)

2

]
−

(

E
[
X
Y
]

)

2

{\displaystyle Var(X+Y)=E[{(X+Y)}^{2}]-{\left(E[X+Y]\right)}^{2}}

The expectation of sums is the sum of expectations.

V
a
r
(
X
Y
# )
E
[

(
X
Y
)

2

]
−

(

E
[
X
]
E
[
Y
]

)

2

{\displaystyle Var(X+Y)=E[{(X+Y)}^{2}]-{\left(E[X]+E[Y]\right)}^{2}}

V
a
r
(
X
Y
# )
E
[

X

2

2
X
Y

Y

2

]
−

(

E
[
X
]

)

2

−
2
E
[
X
]
E
[
Y
]
−

(

E
[
Y
]

)

2

{\displaystyle Var(X+Y)=E[X^{2}+2XY+Y^{2}]-{\left(E[X]\right)}^{2}-2E[X]E[Y]-{\left(E[Y]\right)}^{2}}

V
a
r
(
X
Y
# )
E
[

X

2

]
2
E
[
X
Y
]
E
[

Y

2

]
−

(

E
[
X
]

)

2

−
2
E
[
X
]
E
[
Y
]
−

(

E
[
Y
]

)

2

{\displaystyle Var(X+Y)=E[X^{2}]+2E[XY]+E[Y^{2}]-{\left(E[X]\right)}^{2}-2E[X]E[Y]-{\left(E[Y]\right)}^{2}}

V
a
r
(
X
Y
# )

(

E
[

X

2

]
−

(

E
[
X
]

)

2

)

(

E
[

Y

2

]
−

(

E
[
Y
]

)

2

)

2

(

E
[
X
Y
]
−
E
[
X
]
E
[
Y
]

)

{\displaystyle Var(X+Y)=\left(E[X^{2}]-{\left(E[X]\right)}^{2}\right)+\left(E[Y^{2}]-{\left(E[Y]\right)}^{2}\right)+2\left(E[XY]-E[X]E[Y]\right)}

Since

V
a
r
(
X
# )
E
[

X

2

]
−

(

E
[
X
]

)

2

{\displaystyle Var(X)=E[X^{2}]-{\left(E[X]\right)}^{2}}

,

V
a
r
(
X
Y
# )
V
a
r
(
X
)
V
a
r
(
Y
)
2
c
o
v
(
X
,
Y
)

{\displaystyle Var(X+Y)=Var(X)+Var(Y)+2cov(X,Y)}

.
Yi Jing

## **Description**

The ***Yi Jing*** or ***I Ching*** (易經 yì jīng ), often translated into English as the *Book of Changes*, is one of the earliest Chinese classics. Dating to the Western Zhou dynasty (1046 – 771 BCE), the *Yi Jing* was an important text in several Chinese philosophical schools, being part of the **Daoist Canon** (道藏 dào zàng), the original **Thirteen Confucian Classics** (十三經 shí sān jīng), and later the **Five Classics** (五經 wǔ jīng) memorized by erudite scholars and imperial examinees. There exists different editions for each philosophical tradition.

## **Contents**

The basic text, called the *Zhou Yi* (周易 zhōu yì), was originally a manual for divination using hexagrams and numerology. Each hexagram, called *gua* (卦 guà), is an abstract figure consisting of six stacked lines, with each line being broken or solid. Broken lines represented the abstract principle *yin* (陰 yīn), whereas solid lines represented *yang* (陽 yáng). According to Chinese mytho-history, the great Sage King Fu Xi (伏羲 fú xī) invented the eight trigrams (八卦 bā guà). It is said that, according to tradition, King Wen of Zhou (周文王, c. 1112 – c. 1050 BCE) derived the 64 hexagrams by stacking combinations of two *ba gua* trigrams. Thus, the hexagrams of the *Zhou Yi* are collectively called the **King Wen’s Sequence** (文王卦序 wén wáng guà xù).

Each hexagram is given a name and is followed by a textual explanation on the significance of both of the hexagram as a whole and its individual lines. The text is referred to as the **hexagram statement** (卦辭 guà cí). The interpretation of the hexagrams is then used as a prognostication of future events or the appropriate decisions to be made.  However, during the Spring-Autumn and Warring States periods (770 – 221 BCE), the great thinkers of China began to develop a philosophical understanding of the divination system. Many commentaries written during this time period, commonly known as the **Ten Wings** (十翼 shí yì), gave rise to a more coherent cosmology imbued with moralistic and metaphysical constructs.
Squaring a Number Ending in 5

## **Problem**

There is a well-known trick for squaring a number ending in 5:

1. Take the digits to the left of the 5
2. Treat those digits as

n

{\displaystyle n}
3. Multiply

n

{\displaystyle n}

 by

n
1

{\displaystyle n+1}
4. Append the digits 25 after the result

n
(
n
1
)

{\displaystyle n(n+1)}

Prove why this trick works.

## **Solution**

This arithmetic trick can be easily shown with a little algebra. Express the number as

10
n
5

{\displaystyle 10n + 5}

. Squaring this binomial gives

(
10
n
5

)
2

=
100

n
2

100
n

{\displaystyle (10n + 5)^2 = 100n^2 + 100n + 25.}

Notice that the first two terms can be factored as

100
n
(
n
1
)

{\displaystyle 100n(n+1)}

, giving the expression

(
10
n
5

)
2

=
100
n
(
n
1
)

{\displaystyle (10n + 5)^2 = 100n(n+1) + 25.}

This means the last two digits is always 25, and what is left of the tens digit is the product

n
(
n
1
)

{\displaystyle n(n+1)}

.
Derivative of Tangent Functions

## **Problem**

Let the rational function

f
(
x
)

g
(
x
)

{\displaystyle \frac{f(x)}{g(x)} }

 be composed of two differentiable functions

f
(
x
)

{\displaystyle f(x)}

and

g
(
x
)

{\displaystyle g(x)}

, where

g
(
x
)
≠
0

{\displaystyle g(x) \ne 0}

. The **quotient rule** states that the derivative of this rational function is

d

d
x

[

f
(
x
)

g
(
x
)

]

=

g
(
x
)

d

d
x

[
f
(
x
)
]
−
f
(
x
)

d

d
x

[
g
(
x
)
]

[
g
(
x
)

]
2

{\displaystyle \frac{d}{dx} \left [ \frac{f(x)}{g(x)} \right ] = \frac{g(x)\frac{d}{dx}[f(x)] - f(x) \frac{d}{dx}[g(x)]}{[g(x)]^2} }

. More compactly, the quotient rule can be expressed as

(

f
g

)

′

=

g

f
′

−
f

g
′

g
2

{\displaystyle \left( \frac{f}{g} \right)' = \frac{gf'- fg'}{g^2} }

.

Use the quotient rule to show that

d

d
x

tan
⁡
# x

sec
2

⁡
x

{\displaystyle \frac{d}{dx} \tan x = \sec^2 x}

.

## **Hints**

(1)

tan
⁡
# x

sin
⁡
x

cos
⁡
x

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

(2)

cos
2

⁡
x

sin
2

⁡
# x
1

{\displaystyle \cos^2 x +\sin^2 x = 1 }

(3)

sec
⁡
# x

1

cos
⁡
x

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

## **Solution**

Begin with the quotient identity of the tangent function (Hint 1), then apply the quotient rule of derivatives.

d

d
x

tan
⁡
# x

d

d
x

[

sin
⁡
x

cos
⁡
x

]

{\displaystyle \frac{d}{dx} \tan x = \frac{d}{dx} \left[\frac{\sin x }{\cos x}\right] }

d

d
x

[

sin
⁡
x

cos
⁡
x

]

=

cos
⁡
x

⋅

d

d
x

sin
⁡
x
−
sin
⁡
x
⋅

d

d
x

cos
⁡
x

cos
2

⁡
x

=

cos
⁡
x
⋅
cos
⁡
x
−
sin
⁡
x
⋅
(
−
sin
⁡
x
)

cos
2

⁡
x

=

cos
2

⁡
x

sin
2

⁡
x

cos
2

⁡
x

.

{\displaystyle \begin{align}
\frac{d}{dx} \left[\frac{\sin x}{\cos x}\right] &= \frac{{\cos x} \cdot \frac{d}{dx}\sin x-\sin x \cdot \frac{d}{dx}\cos x}{\cos^2 x} \\[5 pt]
&= \frac{\cos x \cdot \cos x -\sin x \cdot (-\sin x)}{\cos^2 x} \\[5 pt]
&= \frac{\cos^2 x +\sin^2 x }{\cos^2 x}.
\end{align} }

Since

cos
2

⁡
x

sin
2

⁡
# x
1

{\displaystyle \cos^2 x +\sin^2 x = 1 }

, we get

d

d
x

[

sin
⁡
x

cos
⁡
x

]

=

1

cos
2

⁡
x

{\displaystyle \frac{d}{dx} \left[\frac{\sin x}{\cos x}\right] = \frac{1}{\cos^2 x} }

and because

sec
⁡
# x

1

cos
⁡
x

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

, we get

d

d
x

tan
⁡
# x

sec
2

⁡
x
.

{\displaystyle \frac{d}{dx} \tan x = \sec^2 x. }
Temperature Scales

# C

5
9

(
F
−
32
)

{\displaystyle C = \frac{5}{9} (F-32) }

 converts temperatures measured in degrees Fahrenheit to degrees Celsius. If the normal human body temperature is around 37 degrees Celsius, what is its reading in degrees Fahrenheit?

## **Solution**

Isolate for

F

{\displaystyle F}

:

# C

5
9

(
F
−
32
)

{\displaystyle C = \frac{5}{9} (F-32) }

# F

9
5

C
32

{\displaystyle F = \frac{9}{5}C + 32}

.
Substitute 37 into

C

{\displaystyle C}

:

# F

9
5

(
37
)
# 32
98.6

{\displaystyle F = \frac{9}{5} (37) + 32=98.6 }

.
Zhang Qiujian Suanjing Upper Scroll 10

## **Problem**

This classic problem from the *Zhang Qiujian Suanjing* (張邱建算經) is given the name "path around a hill" (封山周棧 fēng shān zhōu zhàn).

答曰：十日、六分日之五。

術曰：置甲、乙、丙行里數，求等數為法。以周棧里數為實。實如法而得一。

草曰：置甲、乙、丙行里數，甲行一百五十，乙行一百二十，丙行九十。各求等數，得三十，為法。除周棧數得十日，法三十，餘二十五，各以五除之，法得六，餘得五。各以三十約之甲、乙、丙行數，乃甲得五周，乙得四周，丙得三周。合前問。

Answer:

10

5
6

{\displaystyle 10 \frac{5}{6} }

 days.

Method: Set the number of *li* travelled [as a rate] for A, B, and C. Determine their greatest common divisor (等數 *deng shu*) and let that be the divisor (法 *fa*). Let the length of the path be the dividend (實 *shi*). Divide the dividend by the divisor.

Explanation: Set the number of *li* travelled [as a rate] for A, B, and C: A travels 150, B travels 120, and C travels 90. Determine their greatest common divisor (等數 *deng shu*), which is 30, and let that be the divisor (法 *fa*). Divide the length of the path [by the divisor] to get 10 days with a remainder of 25 over 30. Divide each [numerator and denominator of the remainder] by 5 to get a remainder of 5 over 6. Reduce each of the three travel numbers A, B, C by 30; therefore, we determine that A runs 5 laps, B runs 4 laps, and C runs 3 laps. This solves the problem.

**Note:** Assume that all three runners start at the same location.

## **Solution**

First calculate the greatest common divisor (GCD) of the given rates. Note that the GCD is also termed as the highest common factor (HCF) or the greatest common factor (GCF). Here I present the method of multiplying the lowest powers of all prime factors that appear in all factorizations.

150

=
2
×
3
×

5
2

[
5
p
t
]
120

=

2
3

×
3
×
5

[
5
p
t
]
90

=
2
×

3
2

×
5

[
10
p
t
]
gcd
(
150
,
120
,
90
)

=
2
×
3
×
# 5
30

{\displaystyle \begin{align}
150 &= 2 \times 3 \times 5^2 \\ [5 pt]
120 &= 2^3 \times 3 \times 5 \\ [5 pt]
90 &= 2 \times 3^2 \times 5 \\ [10 pt]
\gcd(150, 120, 90) &= 2 \times 3 \times 5 = 30
\end{align} }

Calculate the time for all three to meet again (in days) by dividing the length of the path by the GCD.

325
30

=
10

5
6

{\displaystyle \frac{325}{30} = 10 \frac{5}{6} }

Therefore, the three individuals will meet again in

10

5
6

{\displaystyle 10 \frac{5}{6} }

 days.

The explanation section also gives an explanation for calculating the number of laps each person runs. This is calculated by dividing each travelling rate by the GCD

150
÷
30

=
5

[
5
p
t
]
120
÷
30

=
4

[
5
p
t
]
90
÷
30

=
3

{\displaystyle \begin{align}
150 \div 30 &= 5 \\ [5 pt]
120 \div 30 &= 4 \\ [5 pt]
90 \div 30 &= 3
\end{align} }
Sum of Consecutive Odd Numbers

# 1
1

1
# 3
4

1
3
# 5
9

1
3
5
# 7
16

1
3
5
6
7
# 9

{\displaystyle \begin{align}
&1 = 1 \\
&1 + 3 = 4 \\
&1 + 3 + 5 = 9 \\
&1 + 3 + 5 + 7 = 16 \\
&1 + 3 + 5 + 6 + 7 + 9 = 25.
\end{align}}

In each equation, the left hand side is the sum of consecutive odd numbers; the right hand side are perfect squares. Prove that the sum of consecutive odd numbers (beginning with 1) is a perfect square.

## **Solution**

The sum of consecutive odd numbers is an arithmetic series. There are two important equations to use:

(
1
)

t
n

=

t
1

(
n
−
1
)
d

(
2
)

S
n

=

n
2

(

t
1

t
n

)

{\displaystyle
\begin{align}
&(1) \; t\_n = t\_1 + (n-1) d \\[5 pt]
&(2) \; S\_n = \frac{n}{2}(t\_1 + t\_n)
\end{align} }

The first term is

t
1

=
1

{\displaystyle t\_1 = 1 }

 and the common difference is

# d
2

{\displaystyle d=2}

. This means the final term is

t
n

=
1
(
n
−
1
)
# 2
2
n
−
1

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

. Therefore,

1
3
5
⋯
2
n
−
1

=

n
2

(

1
2
n
−
1

)

1
3
5
⋯
2
n
−
1

=

n
2

×
2
n

1
3
5
.
.
.
2
n
−
1

=

n
2

.

{\displaystyle \begin{align}
1 + 3 + 5 + \cdots + 2n - 1 &= \frac{n}{2}\left(1 + 2n-1 \right) \\[5 pt]
1 + 3 + 5 + \cdots + 2n - 1 &= \frac{n}{2}\times 2n \\[5 pt]
1 + 3 + 5 + ... + 2n - 1 &= n^2.
\end{align}}
Product of Two Negative Numbers

## **Problem**

The product of two negative real numbers is often counter-intuitive when first introduced. Prove why the product of two negative real numbers is a positive real number.

## **Solution**

Let *A* and *B* be two positive real numbers. Subsequently, *–A* and *–B* are the
respective additive inverses. We begin with two true statements:

−
A
⋅
−
B
−
A
⋅
# B
−
A
⋅
(
−
B
B
)

{\displaystyle -A \cdot -B + -A \cdot B = -A \cdot (-B + B) }

and

−
A
⋅
(
−
B
B
# )
−
A
⋅
# 0
0

{\displaystyle -A \cdot (-B + B) = -A \cdot 0 = 0 }

.
Thus,

−
A
⋅
−
B
−
A
⋅
# B
0

{\displaystyle -A \cdot -B + -A \cdot B = 0 }

.
Add

A
⋅
B

{\displaystyle A \cdot B}

 to both sides of the equation:

−
A
⋅
−
B
−
A
⋅
B
A
⋅
# B
0
A
⋅
B

{\displaystyle -A \cdot -B + -A \cdot B + A \cdot B = 0 + A \cdot B }

−
A
⋅
−
B
(
−
A
A
)
⋅
# B
0
A
⋅
B

{\displaystyle -A \cdot -B + (-A + A) \cdot B = 0 + A \cdot B }

−
A
⋅
−
B
0
⋅
# B
0
A
⋅
B

{\displaystyle -A \cdot -B + 0 \cdot B = 0 + A \cdot B }

which simplifies to

−
A
⋅
−
# B
A
⋅
B

{\displaystyle -A \cdot -B = A \cdot B }

.

## **Advanced Remark**

Real numbers are an example of an algebraic structure called a **field**. Fields satisfy the following axioms for the operations **addition** and **multiplication**.

**1. Associative property**

Addition:

(
a
b
)
# c
a
(
b
c
)

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

Multiplication:

(
a
b
)
# c
a
(
b
c
)

{\displaystyle (ab)c = a(bc) }

**2. Commutative property**

Addition:

a
# b
b
a

{\displaystyle a + b = b + a}

Multiplication:

a
# b
b
a

{\displaystyle ab = ba}

**3. Existence of identity element**

The additive element for real numbers is 0. The multiplicative element is 1.

Addition:

a
# 0
a

{\displaystyle a + 0 = a}

Multiplication:

a
(
1
# )
a

{\displaystyle a(1) = a }

**4. Existence of inverses**

The additive inverses are negative numbers

a
(
−
a
# )
0

{\displaystyle a + (-a) = 0 }

.

The multiplicative inverses are reciprocals

a
(

a

−
1

# )
1

{\displaystyle a + ({a}^{-1}) = 1 }

.

**5. Distributive property**

a
(
b
c
# )
a
b
a
c

{\displaystyle a(b+c)=ab+ac}
PV Diagrams and Thermodynamic Processes

## **PV Diagrams**

The work done by a gas undergoing expansion and compression in a closed system can be determined through use of a pressure-volume diagram. The work done during an incremental volume change equals the area under the PV curve. The following curves in the diagram below illustrate four thermodynamic processes: adiabatic, isochoric, isobaric, and isothermal.

### **Adiabatic Process**

In an adiabatic process, no heat flows into or out of the system. An adiabatic process usually occurs when a gas is compressed or expands so rapidly that there is no time for the heat to flow in or out of the system.

Δ
# U
−
W

# Q
0

{\displaystyle \begin{align}
&\Delta U = -W \\
& Q = 0 \\
\end{align}}

### **Isochoric Process**

In an isochoric process, the volume of the gas remains constant.

Δ
# V
0

Δ

# P
k
T

{\displaystyle \begin{align}
&\Delta V = 0 \\
&\Delta U = Q \\
& W = 0 \\
& P = kT
\end{align}}

### **Isobaric Process**

In an isobaric process, the pressure of the gas remains constant.

Δ

# W
P
Δ
V

# V
k
T

{\displaystyle \begin{align}
&\Delta P = 0 \\
& W = P\Delta V \\
& V = kT
\end{align}}

### **Isothermal Process**

In an isothermal process, the temperature of the gas remains constant.

Δ
# T
0

Δ

# Q
W

P
# V
k

{\displaystyle \begin{align}
&\Delta T = 0 \\
&\Delta U = 0 \\
& Q = W \\
& PV = k
\end{align}}
Seki Kowa (関 孝和)

**By: Tao Steven Zheng (郑涛)**

**Born:** 1642 AD

**Died:** 1708 AD

**Seki Kowa** was a Japanese mathematician who contributed greatly to the study of algebra in East Asia. As Chinese mathematical research declined during the Ming dynasty (1368 - 1644 AD), Seki Kowa and his contemporaries continued where Chinese mathematicians left off, igniting Japan's first wave of mathematical research called *wasan* (和算 lit. Japanese Computation). Several of Seki Kowa's discoveries predated or were contemporary to the discoveries of Western mathematicians. These discoveries include the determinant, Bernoulli numbers, infinite series, and the Aitken extrapolation method.

## **Discoveries**

**Algebraic Works**

Following the vein of Chinese mathematical tradition, much of Seki Kowa's research is algebraic in nature. Seki Kowa studied systems of linear equations by the elimination method. In 1683, he completed the *Kai-fukudai-no-ho* (解伏題之法), which detailed the calculation of resultants and determinants.

**Bernoulli Numbers**

Seki Kowa discovered the Bernoulli numbers around the same time of Jacob Bernoulli's (1654 - 1705 AD) discovery. Both mathematicians recorded their findings in books published posthumously. Seki Kowa's work *Katsuyo sanpo* (括要算法) was published in 1712 AD, while Jacob Bernoulli's work *Ars Conjectandi* was published in 1713 AD. The picture below is actually a tabulation of the binomial coefficients in the *Katsuyo sanpo*, not the Bernoulli numbers.

**Calculation of pi**

Seki Kowa developed an algorithm that was similar to the Aitken extrapolation method to arrive at a value of pi accurate to 10 decimal places (3.1415926536). Aitken extrapolation is a series transformation that is used for accelerating the rate of convergence of a series. The general Aitken extrapolation was developed by the New Zealand mathematician Alexander Aitken (1895 - 1967 AD).

**Napkin-Ring Problem**

Seki Kowa investigated the famous napkin-ring problem, which states that the volume of a cored out sphere depends only on its height

# V

π

h
3

6

{\displaystyle V = \frac{\pi h^3}{6} }

.
Napkin Ring Problem
Telescoping Series 2

## **Problem**

   Calculate the infinite series

∑

# n
1

∞

18
n
16

n

3

3

n

2

2
n

.

{\displaystyle \sum \_{n=1}^{\infty }{\frac {18n+16}{n^{3}+3n^{2}+2n}}.}

## **Hint**

(1) Use partial fractions; this is an essential technique for evaluating telescoping series.

## **Solution**

Write the infinite series using summation notation:

∑

# n
1

∞

18
n
16

n

3

3

n

3

2
n

.

{\displaystyle \sum \_{n=1}^{\infty }{\frac {18n+16}{n^{3}+3n^{3}+2n}}.}

Decompose the pattern by partial fractions

18
n
16

n
(
n
1
)
(
n
2
)

=

A
n

B

n
1

C

n
2

{\displaystyle {\frac {18n+16}{n(n+1)(n+2)}}={\frac {A}{n}}+{\frac {B}{n+1}}+{\frac {C}{n+2}}}

where the coefficients

A
,
B
,
C

{\displaystyle A, B, C}

are real numbers.
Thus,

18
n
16

n
(
n
1
)
(
n
2
)

=

A
(
n
1
)
(
n
2
)
B
(
n
)
(
n
2
)
C
(
n
)
(
n
1
)

n
(
n
1
)
(
n
2
)

{\displaystyle {\frac {18n+16}{n(n+1)(n+2)}}={\frac {A(n+1)(n+2)+B(n)(n+2)+C(n)(n+1)}{n(n+1)(n+2)}}}

By inspecting the numerator, we get

18
n
# 16
A
(
n
1
)
(
n
2
)
B
(
n
)
(
n
2
)
C
(
n
)
(
n
1
)
.

{\displaystyle 18n+16=A(n+1)(n+2)+B(n)(n+2)+C(n)(n+1).}

Setting

# n
0

{\displaystyle n=0}

, we get

# 16
2
A

{\displaystyle 16=2A}

.
Setting

# n
−
1

{\displaystyle n = -1}

, we get

−
# 2
−
B

{\displaystyle -2=-B}

.

Setting

# n
−
2

{\displaystyle n=-2 }

, we get

−
# 20
2
C

{\displaystyle -20=2C}

.

Solving for each coefficient yields:

{

A

=
8

B

=
2

C

=
−

{\displaystyle {\begin{cases}A&=8\\B&=2\\C&=-10.\end{cases}}}

Substituting the solved coefficients into the original infinite series, we get

∑

# n
1

∞

18
n
16

n

3

3

n

3

2
n

=

∑

# n
1

∞

8
n

∑

# n
1

∞

2

n
1

∑

# n
1

∞

−
10

n
2

{\displaystyle \sum \_{n=1}^{\infty }{\frac {18n+16}{n^{3}+3n^{3}+2n}}=\sum \_{n=1}^{\infty }{\frac {8}{n}}+\sum \_{n=1}^{\infty }{\frac {2}{n+1}}+\sum \_{n=1}^{\infty }{\frac {-10}{n+2}}}

or

∑

# n
1

∞

18
n
16

n

3

3

n

3

2
n

=
8

∑

# n
1

∞

1
n

2

∑

# n
1

∞

1

n
1

−
10

∑

# n
1

∞

1

n
2

.

{\displaystyle \sum \_{n=1}^{\infty }{\frac {18n+16}{n^{3}+3n^{3}+2n}}=8\sum \_{n=1}^{\infty }{\frac {1}{n}}+2\sum \_{n=1}^{\infty }{\frac {1}{n+1}}-10\sum \_{n=1}^{\infty }{\frac {1}{n+2}}.}

For this problem, write out the first three terms of each infinite series and calculate.

∑

# n
1

∞

18
n
16

n

3

3

n

3

2
n

=
8

(

1

1
2

1
3

⋯

)

2

(

1
2

1
3

1
4

⋯

)

−
10

(

1
3

1
4

1
5

⋯

)

=
8

(

1

1
2

)

2

(

1
2

)

10

(

1
3

1
4

1
5

⋯

)

−
10

(

1
3

1
4

1
5

⋯

)

=
13

{\displaystyle {\begin{aligned}\sum \_{n=1}^{\infty }{\frac {18n+16}{n^{3}+3n^{3}+2n}}&=8\left(1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots \right)+2\left({\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+\cdots \right)-10\left({\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots \right)\\[5pt]&=8\left(1+{\frac {1}{2}}\right)+2\left({\frac {1}{2}}\right)+10\left({\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots \right)-10\left({\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+\cdots \right)\\[5pt]&=13\end{aligned}}}

Therefore,

∑

# n
1

∞

18
n
16

n

3

3

n

2

2
n

=

{\displaystyle \sum \_{n=1}^{\infty }{\frac {18n+16}{n^{3}+3n^{2}+2n}}=13.}
Vector Laplacian

# A
P
(
x
,
y
,
z
)
i
Q
(
x
,
y
,
z
)
j
R
(
x
,
y
,
z
)
k

{\displaystyle \boldsymbol{A} = P(x,y,z) \boldsymbol{i} + Q(x,y,z) \boldsymbol{j} + R(x,y,z) \boldsymbol{k} }

 be a vector field with components that are continuous second partial derivatives in three dimensional space.

The vector Laplacian operator is defined as

∇

2

# A
∇
(
∇
⋅
A
)
−
∇
×
(
∇
×
A
)

{\displaystyle {\nabla}^{2} \boldsymbol{A} = \nabla(\nabla \cdot \boldsymbol{A}) - \nabla \times (\nabla \times \boldsymbol{A}) }

. Express the vector Laplacian operator in terms of

P
,
Q
,
R

{\displaystyle P, Q, R}

.

## **Solution**

First calculate

∇
(
∇
⋅
A
)

{\displaystyle \nabla(\nabla \cdot \boldsymbol{A}) }

.

∇
⋅
# A

P
x

Q
y

R
z

{\displaystyle \nabla \cdot \boldsymbol{A} = P\_x + Q\_y + R\_z }

∇
(
∇
⋅
A
# )

(

P

x
x

Q

y
x

R

z
x

P

x
y

Q

y
y

R

z
y

P

x
z

Q

y
z

R

z
z

)

{\displaystyle \nabla(\nabla \cdot \boldsymbol{A}) = \begin{pmatrix} P\_{xx} + Q\_{yx} + R\_{zx} \\ P\_{xy} + Q\_{yy} + R\_{zy} \\ P\_{xz} + Q\_{yz} + R\_{zz} \end{pmatrix} }

Then calculate

∇
×
(
∇
×
A
)

{\displaystyle \nabla \times (\nabla \times \boldsymbol{A}) }

.

∇
×
# A

(

R
y

−

Q
z

P
z

−

R
x

Q
x

−

P
y

)

{\displaystyle \nabla \times \boldsymbol{A} = \begin{pmatrix} R\_y - Q\_z \\ P\_z - R\_x \\ Q\_x - P\_y \end{pmatrix}}

∇
×
(
∇
×
A
# )

(

Q

x
y

−

P

y
y

−

P

z
z

R

x
z

−

Q

x
x

P

y
x

R

y
z

−

Q

z
z

P

z
x

−

R

x
x

−

R

y
y

Q

z
y

)

{\displaystyle \nabla \times (\nabla \times \boldsymbol{A}) = \begin{pmatrix} Q\_{xy} -P\_{yy} - P\_{zz} + R\_{xz} \\ -Q\_{xx} + P\_{yx} + R\_{yz} - Q\_{zz} \\ P\_{zx} - R\_{xx} - R\_{yy} + Q\_{zy} \end{pmatrix} }

Consequently,

∇
(
∇
⋅
A
)
−
∇
×
(
∇
×
A
# )

(

P

x
x

P

y
y

P

z
z

(

R

z
x

−

R

x
z

)
(

Q

y
x

−

Q

x
y

)

Q

x
x

Q

y
y

Q

z
z

(

P

x
y

−

P

y
x

)
(

R

z
y

−

R

y
z

)

R

x
x

R

y
y

R

z
z

(

P

x
z

−

P

z
x

)
(

Q

y
z

−

Q

z
y

)

)

.

{\displaystyle \nabla(\nabla \cdot \boldsymbol{A}) - \nabla \times (\nabla \times \boldsymbol{A}) = \begin{pmatrix} P\_{xx} + P\_{yy} + P\_{zz} + (R\_{zx} - R\_{xz}) + (Q\_{yx} - Q\_{xy}) \\ Q\_{xx} + Q\_{yy} + Q\_{zz} + (P\_{xy} - P\_{yx}) + (R\_{zy} - R\_{yz}) \\ R\_{xx} + R\_{yy} + R\_{zz} + (P\_{xz} - P\_{zx}) + (Q\_{yz} - Q\_{zy}) \end{pmatrix}. }

By Clairaut's theorem of mixed partials, each bracket in the above vector equal zero; therefore,

∇
(
∇
⋅
A
)
−
∇
×
(
∇
×
A
# )

(

P

x
x

P

y
y

P

z
z

Q

x
x

Q

y
y

Q

z
z

R

x
x

R

y
y

R

z
z

)

.

{\displaystyle \nabla(\nabla \cdot \boldsymbol{A}) - \nabla \times (\nabla \times \boldsymbol{A}) =
\begin{pmatrix} P\_{xx} + P\_{yy} + P\_{zz} \\ Q\_{xx} + Q\_{yy} + Q\_{zz} \\ R\_{xx} + R\_{yy} + R\_{zz} \end{pmatrix}. }

This expression can be compactly expressed as

∇

2

# A

(

∇

2

P

∇

2

Q

∇

2

R

)

{\displaystyle {\nabla}^{2} \boldsymbol{A} = \begin{pmatrix} {\nabla}^{2} P \\ {\nabla}^{2} Q \\ {\nabla}^{2} R \end{pmatrix} }

where the left-hand side is the vector Laplacian of

A

{\displaystyle \boldsymbol{A}}

, and the right-hand side is a vector of the scalar Laplacian of the components of

A

{\displaystyle \boldsymbol{A}}

.
Weight to Mass Conversion with Python

**By: Tao Steven Zheng (郑涛)**

## **Task**

Write a short program that asks: "What is your weight in pounds?", where the user can enter a number (float). The program will take the input and convert it to kilograms and round it to two decimal places. Remember 1 kg = 2.2 pounds.

## **Code**

```
mass = float(input("What is your mass in pounds?"))
weight = mass \* 9.8
weight\_rounded = round(weight,2)
print("Your weight is", mass\_rounded, "Ibs.")

```

## **Results**

What is your weight in pounds?50
Your mass is 68. kg.
Omar Khayyam's Cubic

## **Problem**

   **Omar Khayyam** (1048 – 1131 AD) solved cubic equations of the form

x
3

a
# x
b

{\displaystyle x^3 + ax = b }

 by transforming the problem into finding the intersection of a circle and a parabola:

{

x
2

y
2

=
q
x

[
5
p
t
]

x
2

=
p
y

{\displaystyle
\begin{cases}
x^2 + y^2 = qx \\ [5 pt]
x^2 = py
\end{cases}
}

where

a
,
b
,
p
,
q

{\displaystyle a,b,p,q }

 are positive numbers.

**Part 1:** Determine

p
,
q

{\displaystyle p,q }

 in terms of

a
,
b

{\displaystyle a, b}

.

**Part 2:** Use Omar Khayyam’s method to solve the cubic equation

x
3

4
# x
16

{\displaystyle x^3 + 4x = 16 }

 by sketching the two conic sections and locating the intersection points.

## **Solution**

**Solution for Part 1**

Write the first equation as

y
2

=
x
(
q
−
x
)

{\displaystyle y^2 = x(q-x) }

 and the second equation as

# y

x
2

p

{\displaystyle y=\frac{x^2}{p} }

. Substitute the second equation into the first equation to eliminate

y

{\displaystyle y}

, and obtain

x
4

p
2

=
x
(
q
−
x
)

{\displaystyle \frac{x^4}{p^2} = x(q-x) }

.

Consequently,

x
3

=

p
2

(
q
−
x
)

{\displaystyle x^3 = p^2 (q-x)}

or

x
3

p
2

# x

p
2

q
.

{\displaystyle x^3 + p^2 x = p^2 q.}

Matching terms with the cubic equation

x
3

a
# x
b

{\displaystyle x^3 + ax = b }

 gives

{

p
2

=
a

[
5
p
t
]
# q

b
a

{\displaystyle
\begin{cases}
p^2 = a \\ [5 pt]
q = \frac{b}{a}
\end{cases}
}

Since

a
,
b
,
p
,
q

{\displaystyle a,b,p,q }

 are positive numbers,

{

# p

a

[
5
p
t
]
# q

b
a

.

{\displaystyle
\begin{cases}
p = \sqrt{a} \\ [5 pt]
q = \frac{b}{a}.
\end{cases}
}

**Solution for Part 2**

For the equation

x
3

4
# x
16

{\displaystyle x^3 + 4x = 16 }

,

# a
4

{\displaystyle a = 4 }

 and

# b
16

{\displaystyle b = 16 }

. Thus,

{

# p

4

=
2

[
5
p
t
]
# q

16
4

=
4

{\displaystyle
\begin{cases}
p = \sqrt{4} = 2 \\ [5 pt]
q = \frac{16}{4} = 4
\end{cases}
}

and we need to graph

{

x
2

y
2

=
4
x

[
5
p
t
]

x
2

=
2
y
.

{\displaystyle
\begin{cases}
x^2 + y^2 = 4x \\ [5 pt]
x^2 = 2y.
\end{cases}
}

**Intersection points:**

(
0
,
0
)

{\displaystyle (0,0) }

;

(
2
,
2
)

{\displaystyle (2,2) }

. When the values of

x

{\displaystyle x}

 is substituted into

x
3

4
# x
16

{\displaystyle x^3 + 4x = 16 }

, we find one solution to this cubic

# x
2

{\displaystyle x = 2 }

.
Power-log Combo

## **Problem**

**Part 1:** Calculate the indefinite integral

∫

x
p

ln
⁡
(
x
)

d
x
.

{\displaystyle \int x^p \ln(x) \,dx. }

**Part 2:** For what values

p

{\displaystyle p}

 does the integral

∫
0
1

x
p

ln
⁡
(
x
)

d
x

{\displaystyle \int\_0^1 x^p \ln(x) \,dx }

 converge?

## **Solution**

**Part 1**

This family of functions is the product of a power function with the natural logarithm function. Therefore the method of integration by parts should be used.

According to the LIATE strategy, let

# u
ln
⁡
(
x
)

{\displaystyle u = \ln(x) }

 and

d
# v

x
p

d
x

{\displaystyle dv = x^p dx }

. Then

d
# u

1
x

d
x

{\displaystyle du = \frac{1}{x}dx }

 and

x

p
1

p
1

{\displaystyle \frac{{x}^{p+1}}{p+1} }

. Therefore,

∫

x
p

ln
⁡
(
x
)

d
x

=
ln
⁡
(
x
)

x

p
1

p
1

−
∫

x

p
1

p
1

1
x

d
x

=
ln
⁡
(
x
)

x

p
1

p
1

−
∫

x

p

p
1

d
x

=
ln
⁡
(
x
)

x

p
1

p
1

−

x

p
1

(
p
1

)
2

C
.

{\displaystyle \begin{align}
\int x^p \ln(x) \,dx &= \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p+1}}{p+1} \frac{1}{x} \,dx \\[5 pt]
&= \ln(x) \frac{{x}^{p+1}}{p+1} - \int \frac{{x}^{p}}{p+1} \,dx \\[5 pt]
&= \ln(x) \frac{{x}^{p+1}}{p+1} - \frac{{x}^{p+1}}{(p+1)^2} + C.
\end{align} }

**Part 2**

From the result of the indefinite integral, it is obvious that

p
≠
−
1

{\displaystyle p \ne -1}

. Just look at the denominators. Now this means

∫
0
1

x
p

ln
⁡
(
x
)

d
# x
ln
⁡
(
x
)

x

p
1

p
1

0
1

−

x

p
1

(
p
1

)
2

0
1

.

{\displaystyle \int\_0^1 x^p \ln(x) \,dx = \ln(x) \frac{{x}^{p+1}}{p+1} \Biggr|\_0^1 - \frac{{x}^{p+1}}{(p+1)^2} \Biggr|\_0^1. }

Notice that if

p
<
−
1

{\displaystyle p < -1 }

, the power functions become reciprocal functions, which means the integral diverges when evaluating the lower bound. However if

p
>
−
1

{\displaystyle p > -1 }

, the integral converges to the expression

∫
0
1

x
p

ln
⁡
(
x
)

d
# x

−
1

(
p
1

)
2

.

{\displaystyle \int\_0^1 x^p \ln(x) \,dx = \frac{-1}{(p+1)^2}. }

Therefore the integral

∫
0
1

x
p

ln
⁡
(
x
)

d
x

{\displaystyle \int\_0^1 x^p \ln(x) \,dx }

 converges if

p
>
−
1

{\displaystyle p > -1 }

.
Vesica Piscis

## **Problem**

The **vesica piscis** (literally "fish bladder" in Latin) is the intersection of two congruent disks, each centered on the perimeter of the other. Determine the area coloured in turquoise in Figure 1.

## **Solution**

Let

r

{\displaystyle r}

 be the radius of each circle. The area of the vesica piscis is formed by two equilateral triangles and four circular segments of equal area. One equilateral triangle and one segment form a sector of one sixth of the circle. The area of the sector is then:

A

s
e
c
t
o
r

=

1
6

π

r
2

{\displaystyle A\_{sector} = \frac{1}{6} \pi r^2}

. Since the side of the equilateral triangle has length

r

{\displaystyle r}

, its area is

A

t
r
i
a
n
g
l
e

=

3

4

r
2

{\displaystyle A\_{triangle} = \frac{\sqrt{3}}{4} r^2}

.

Thus, the area of the segment is the difference between those two areas:

A

s
e
g
m
e
n
t

=

1
6

π

r
2

−

3

4

r
2

{\displaystyle A\_{segment} = \frac{1}{6} \pi r^2 - \frac{\sqrt 3}{4} r^2}

Add the areas of two equilateral triangles and four circular segments to obtain the area of the vesica piscis:

A

V
P

=
2
×

3

4

r
2

4
×

(

1
6

π

r
2

−

3

4

r
2

)

=

1
6

(

4
π
−
3

3

)

r
2

.

{\displaystyle \begin{align}
A\_{VP} &= 2\times \frac{\sqrt{3}}{4}r^2 + 4\times \left(\frac{1}{6} \pi r^2 - \frac{\sqrt 3}{4} r^2 \right) \\
&= \frac{1}{6} \left( 4\pi - 3\sqrt 3 \right)r^2.
\end{align} }