recursal/Fanatic-Fandom huggingface dataset:
Euler's Factorial Integral

## **Problem**

Show that the integral

∫
0
1

(

−
ln
⁡
s

)

n

d
# s
n
!

{\displaystyle \int\_0^1 {\left(-\ln s\right)}^{n} ds = n!}

 for non-negative integers

n

{\displaystyle n}

.

# t
−
ln
⁡
s

{\displaystyle t = -\ln s }

, then

# s

e

−
t

{\displaystyle s = {e}^{-t} }

 and subsequently

d
# s
−

e

−
t

d
t

{\displaystyle ds = -{e}^{-t}dt }

.

When

# s
0
,

# t
∞

{\displaystyle s = 0, \; t = \infty}

, and when

# s
1
,

# t
0

{\displaystyle s = 1, \; t = 0}

; thus, the original integral transforms to

−

∫
∞
0

t

n

e

−
t

d
t

{\displaystyle -\int\_\infty^0 {t}^{n} {e}^{-t} dt }

 or

∫
0
∞

t

n

e

−
t

d
t

{\displaystyle \int\_0^\infty {t}^{n} {e}^{-t} dt }

. This is the Gamma function, and it is defined as

Γ
(
n
1
# )

∫
0
∞

t

n

e

−
t

d
t

{\displaystyle \Gamma (n+1) = \int\_0^\infty {t}^{n} {e}^{-t} dt }

If

n

{\displaystyle n}

 is a non-negative integer,

Γ
(
n
1
# )

∫
0
∞

t

n

e

−
t

d
# t
n
!

{\displaystyle \Gamma (n+1) = \int\_0^\infty {t}^{n} {e}^{-t} dt = n! }

To prove why the Gamma function is an integral that evaluates the factorial for non-negative integers, we need to show two facts:

(
1
)

Γ
(
n
1
# )
n
⋅
Γ
(
n
)

(
2
)

Γ
(
1
# !
1

{\displaystyle \begin{align}
&(1) \quad \Gamma (n+1) = n \cdot \Gamma (n) \\[5 pt]
&(2) \quad \Gamma (1) = 0! = 1
\end{align}
}

which demonstrate the definition of the factorial

n
# !
n
×
(
n
−
1
)
×
(
n
−
2
)
×
⋯
×
3
×
2
×
1

{\displaystyle n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1}

and

0
# !
1

{\displaystyle 0! = 1 }

.

The second fact is easy to show.

Γ
(
1
)

=

∫
0
∞

e

−
t

d
t

=
−

e

−
t

0
∞

=
−
(
0
−
1
)

=
1

{\displaystyle \begin{align}
\Gamma (1) &= \int\_0^\infty {e}^{-t} dt \\[5 pt]
&= -{e}^{-t} \Big|\_0^\infty \\[5 pt]
&= -(0 - 1) \\[5 pt]
&= 1
\end{align}
}

The first fact requires more work. Using integration by parts

Γ
(
n
1
# )

∫
0
∞

t

n

e

−
t

d
# t
n
!

{\displaystyle \Gamma (n+1) = \int\_0^\infty {t}^{n} {e}^{-t} dt = n! }

becomes

Γ
(
n
1
# )
−

t

n

e

−
t

0
∞

∫
0
∞

n

t

n
−
1

e

−
t

d
t

{\displaystyle \Gamma (n+1) = -{t}^{n}{e}^{-t} \Big|\_0^\infty + \int\_0^\infty n\; {t}^{n-1} {e}^{-t} dt }

or

Γ
(
n
1
# )
−
(
0
−
0
)
n
⋅

∫
0
∞

t

n
−
1

e

−
t

d
t

{\displaystyle \Gamma (n+1) = -(0-0) + n \cdot \int\_0^\infty {t}^{n-1} {e}^{-t} dt }

By the definition of the Gamma function, we get

Γ
(
n
# )

∫
0
∞

t

n
−
1

e

−
t

d
t

{\displaystyle \Gamma (n) = \int\_0^\infty {t}^{n-1} {e}^{-t} dt }

. Consequently,

Γ
(
n
1
# )
n
⋅
Γ
(
n
)

{\displaystyle \Gamma (n+1) = n \cdot \Gamma (n) }

 is a fundamental property of the Gamma function.

Therefore,

∫
0
1

(

−
ln
⁡
s

)

n

d
# s
n
!

{\displaystyle \int\_0^1 {\left(-\ln s\right)}^{n} ds = n!}

## **Historical Note**

In 1730, Leonhard Euler (1707 - 1783 AD) wrote a letter to Christian Goldbach (1690 - 1764 AD) with the integral

∫
0
1

(

ln
⁡

(

1
t

)

)

n
−
1

d
t

{\displaystyle \int\_0^1 {\left(\ln \left(\frac{1}{t}\right)\right)}^{n-1} dt }

. It was Adrien-Marie Legendre (1752 - 1833 AD) who named it the Gamma function.
Indian Combinatorics

## **Problem**

**Problem 1: Lord Shiva (From Bhaskara's *Lilavati*, c. 1150 AD)**

The God Shiva had ten items (one item in each hand): the rope, the hook, the snake, the drum, the skull, the trident, the bedstead, the dagger, the arrow, and the bow. How many different ways can these items be exchanged?

**Problem 2: Flavour Combinations (From Mahavira's *Ganita-Sara-Sangraha*, c. 850 AD)**

There are six flavours: astringent, bitter, sour, spicy, salty, and sweet. In how many ways can different flavours be used in combination?

## **Solution**

**Problem 1: Lord Shiva (From Bhaskara's *Lilivati*, c. 1150 AD)**

This is a permutation problem. The number of possible arrangements of Shiva’s

items is

10
×
9
×
8
×
7
×
6
×
5
×
4
×
3
×
2
×
# 1

{\displaystyle 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1 = 3628800. }

This calculation can be written more compactly using the factorial symbol:

10
# !

{\displaystyle 10! = 3628800. }

**Problem 2: Flavour Combinations (From Mahavira's *Ganita-Sara-Sangraha*, c. 850 AD)**

This problem is the sum of combinations

C

0

6

C

1

6

C

2

6

C

3

6

C

4

6

C

5

6

C

6

6

{\displaystyle {C}^{6}\_{0} + {C}^{6}\_{1} + {C}^{6}\_{2} + {C}^{6}\_{3} + {C}^{6}\_{4} + {C}^{6}\_{5} + {C}^{6}\_{6}}

where

C

r

n

=

n
!

r
!
(
n
−
r
)
!

{\displaystyle {C}^{n}\_{r} = \frac{n!}{r!(n-r)!} }

C

0

6

C

1

6

C

2

6

C

3

6

C

4

6

C

5

6

C

6

6

=
1
6
15
20
15
6
# 1

{\displaystyle {C}^{6}\_{0} + {C}^{6}\_{1} + {C}^{6}\_{2} + {C}^{6}\_{3} + {C}^{6}\_{4} + {C}^{6}\_{5} + {C}^{6}\_{6} = 1 + 6 + 15 + 20 + 15 + 6 + 1 = 64. }

**Shortcut**

There is a known property of combinations

∑

# r
0

n

C

r

n

=

2
n

.

{\displaystyle \sum\_{r=0}^{n} {C}\_{r}^{n} = 2^n. }

Using this property gives

∑

# r
0

6

C

r

6

=

2
6

=

{\displaystyle \sum\_{r=0}^{6} {C}\_{r}^{6} = 2^6 = 64. }
Improved Euler's Method with MATLAB

## **Task**

Write a program that plots the exact solution and approximation by the improved Euler’s method of the equation differential equation

y
′

=
y
cos
⁡
t
,

y
(
0
# )
1

{\displaystyle y' = y\cos t, \quad y(0) = 1 }

over the interval

0
≤
t
≤
6

{\displaystyle 0 \le t \le 6}

 with step size

# h

2

−
3

.

{\displaystyle h = 2^{-3}.}

 The program simultaneously plots the exact solution and computes the error of the approximation at the final time.

## **Hint**

(1) The improved Euler's method is

k
1

=
f
(

t
n

,

y
n

)

k
2

=
f
(

t
n

h
,

y
n

h

k
1

)

y

n
1

=

y
n

h
(

k
1

k
2

)

/

{\displaystyle \begin{align}
k\_1 &= f(t\_n,y\_n) \\[5 pt]
k\_2 &= f(t\_n + h,y\_n + hk\_1) \\[5 pt]
y\_{n+1} &= y\_n + h(k\_1 + k\_2)/2.
\end{align}}

## **Code**

```
% Define the differential equation: y' = f(t,y)
f = @(t,y) y\*cos(t);

% Choose the step size h and create the vector
% of t values from t0 to tf incremented by h
t0 = 0;
tf = 6;
h = 2^-3;
t = t0:h:tf;

% Initialize a vector of y values as a zero vector
% and set the initial value: y(t0) = y0
y = zeros(size(t));
y0 = 1;
y(1) = y0;

% Implement Euler's method to create the vector of t values
for n = 1:(length(t)-1)
 k1 = f(t(n),y(n));
 k2 = f(t(n)+h,y(n)+h\*k1);
 y(n+1) = y(n) + h\*(k1+k2)/2;
end

% Plot the approximation with the exact solution
t\_exact = t0:0.02:tf;
y\_exact = exp(sin(t\_exact));
plot(t,y,'b.-',t\_exact,y\_exact)
grid on
ylim([0,3])
savefig('improved\_euler.fig')

% Compare to exact answer and compute error at tf
error = abs(y\_exact(end) - y(end))

```
Dicey Primes

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

## **Problem**

A gambling game asks the player to select either a six-sided die (D6) or a twenty-sided die (D20). Is there a higher probability getting three prime numbers from seven rolls of a six-sided die, or getting three prime numbers from 6 rolls of a twenty-sided die? Assume the dice are fair.

## **Solution**

The probability of dice rolling follows the binomial distribution.

P
(
n
,
k
,
p
# )

(

n
k

)

(
p

)
k

(
1
−
p

)

n
−
k

{\displaystyle P(n,k,p) = \binom{n}{k}(p)^k(1-p)^{n-k} }

where

(

n
k

)

=

n
!

k
!
(
n
−
k
)
!

.

{\displaystyle \binom{n}{k} = \frac{n!}{k! (n-k)!}.}

But first, we must determine the probability of getting prime numbers for each die.

**D6**

There are three primesnumbers on a D6 die: 2, 3, 5.

# p

3
6

=
0.5

{\displaystyle p = \frac{3}{6} = 0.5 }

Thus the probability of getting three prime numbers from seven rolls of a six-sided die is

P
(
# n
7
,
# k
3
,
# )

(

7
3

)

(
0.5

)
3

(
1
−
0.5

)

7
−
3

{\displaystyle P(n=7, k=3, p=0.5) = \binom{7}{3}(0.5)^3(1-0.5)^{7-3} }

P
(
# n
7
,
# k
3
,
# )

(

7
3

)

(
0.5

)
3

(
0.5

)

4

=
0.273475

{\displaystyle P(n=7, k=3, p=0.5) = \binom{7}{3}(0.5)^3(0.5)^{4} = 0.273475 }

Probability is

27.35
%

{\displaystyle 27.35\%}

.

**D20**

There are eight primes on a D20 die: 2, 3, 5, 7, 11, 13, 17, 19.

# p

8
20

=
0.4

{\displaystyle p = \frac{8}{20} = 0.4 }

Thus the probability of getting three prime numbers from six rolls of a twenty-sided die is

P
(
# n
6
,
# k
3
,
# )

(

6
3

)

(
0.4

)
3

(
1
−
0.4

)

6
−
3

{\displaystyle P(n=6, k=3, p=0.4) = \binom{6}{3}(0.4)^3(1-0.4)^{6-3} }

P
(
# n
6
,
# k
3
,
# )

(

6
3

)

(
0.4

)
3

(
0.6

)

3

=
0.27648

{\displaystyle P(n=6, k=3, p=0.4) = \binom{6}{3}(0.4)^3(0.6)^{3} = 0.27648 }

Probability is

27.65
%

{\displaystyle 27.65\%}

.

The odds are close, but it seems option 2, getting three primes from six rolls of a twenty-sided die is slightly higher.
Converting Non-linear ODE into Matrix Form

## **Problem**

Convert the following second-order differential equation

t
2

x
″

3
t

x
′

−
(

t
2

−
8
)
# x
2
sin
⁡
(
3
t
)

{\displaystyle t^2 x'' + 3t x' - (t^2 - 8)x = 2\sin(3t)}

into an equivalent system of first-order differential equations.

# v

d
x

d
t

=

x
′

{\displaystyle v = \frac{dx}{dt} = x'}

 and

v
′

=

d
2

x

d

t
2

=

x
″

{\displaystyle v' = \frac{d^2 x}{dt^2} = x''}

; then, the differential equation

t
2

x
″

3
t

x
′

−
(

t
2

−
8
)
# x
2
sin
⁡
(
3
t
)

{\displaystyle t^2 x'' + 3t x' - (t^2 - 8)x = 2\sin(3t)}

becomes

t
2

v
′

3
t
v
−
(

t
2

−
8
)
# x
2
sin
⁡
(
3
t
)

{\displaystyle t^2 v' + 3t v - (t^2 - 8)x = 2\sin(3t)}

and

v
′

=
−

3
v

t

(

t
2

−
8
)
x

t
2

2
sin
⁡
(
3
t
)

t
2

.

{\displaystyle v' = -\frac{3 v}{t} + \frac{(t^2 - 8)x}{t^2} + \frac{2\sin(3t)}{t^2}.}

Thus the equivalent matrix is

(

x
′

v
′

)

=

(

0

1

t
2

−
8

t
2

−

3
t

)

(

x

v

)

(

0

2
sin
⁡
(
3
t
)

t
2

)

.

{\displaystyle \begin{align}
\begin{pmatrix} x' \\ v' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ \frac{t^2 - 8}{t^2} & -\frac{3}{t} \end{pmatrix} \begin{pmatrix} x \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{2\sin(3t)}{t^2} \end{pmatrix}.
\end{align}}
Jiuzhang Suanshu Shanggong 15

## **Problem**

答曰：九十三尺、少半尺。

術曰：廣袤相乘，以高乘之，三而一。

Answer:

93

1
3

{\displaystyle 93 \frac{1}{3}}

 [cubic] *chi*.

Method: Multiply the width by the length, then multiply by the height, divide by 3.

1 Literally translates to “male horse”, which in actuality is an architectural feature.

## **Derivation**

The *yang ma* (陽馬) is a rectangular pyramid (Figure 1). A rectangular prism can be dissected into three rectangular pyramids of equal volume (Figure 2). The volume of the rectangular prism is

# V
a
b
c

{\displaystyle V=abc }

, where

a
,
b
,
c

{\displaystyle a,b,c}

 denote the width, length, and height respectively. Therefore, ithe volume of the right triangular prism is

V

y
a
n
g
m
a

=

1
3

a
b
c
.

{\displaystyle V\_{yangma} = \frac{1}{3} abc. }

## **Calculation**

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

V

=

1
3

a
b
c

=

1
3

×
5
×
7
×
8

=

280
3

=
93

1
3

{\displaystyle \begin{align}
V &= \frac{1}{3}abc \\[5 pt]
& = \frac{1}{3} \times 5 \times 7 \times 8 \\[5 pt]
&= \frac{280}{3} \\[5 pt]
&= 93 \frac{1}{3}
\end{align}}

The volume of the rectangular prism (陽馬 *yang ma*) is

93

1
3

{\displaystyle 93 \frac{1}{3}}

 cubic *chi*.
Green's Theorem and Areas

## **Problem**

Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and D be the region bounded by C. If P and Q are functions of (*x*, *y*) defined on an open region containing D and have continuous partial derivatives, **Green's theorem** states that

∮
C

P
d
x
Q
d
# y

∬
D

(

∂
Q

∂
x

−

∂
P

∂
y

)

d
x

d
y
.

{\displaystyle \oint\_C P dx + Q dy = \iint\_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \,dx\,dy. }

Use Green's theorem to prove that the area bounded by a simple closed curve can be calculated with the integral

1
2

∮
C

x
d
y
−
y
d
x
.

{\displaystyle \frac{1}{2} \oint\_C xdy - ydx. }

## **Solution**

Let

D

{\displaystyle D}

 be the simply-connected domain bounded by a simple closed curve

C

{\displaystyle C}

 that is oriented in the positive direction. Green's theorem states that

∮
C

P
d
x
Q
d
# y

∬
D

(

∂
Q

∂
x

−

∂
P

∂
y

)

d
x

d
y

{\displaystyle \oint\_C P dx + Q dy = \iint\_D \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \,dx\,dy }

for two continuous functions

P
(
x
,
y
)

{\displaystyle P(x, y)}

 and

Q
(
x
,
y
)

{\displaystyle Q(x,y) }

 that have continuous first partial derivatives on

D

{\displaystyle D}

.

If

# P
−
y

{\displaystyle P = -y }

 and

# Q
x

{\displaystyle Q = x }

, then

∮
C

−
y
d
x
x
d
# y

∬
D

(

∂

∂
x

(
x
)
−

∂

∂
y

(
−
y
)

)

d
x

d
y

{\displaystyle \oint\_C -y dx + x dy = \iint\_D \left(\frac{\partial}{\partial x}(x) - \frac{\partial }{\partial y}(-y) \right) \,dx\,dy }

∮
C

x
d
y
−
y
d
# x

∬
D

2

d
x

d
y

{\displaystyle \oint\_C xdy - ydx = \iint\_D 2 \,dx\,dy }

∮
C

x
d
y
−
y
d
# x
2
A

{\displaystyle \oint\_C xdy - ydx = 2A }

.
Therefore,

# A

1
2

∮
C

x
d
y
−
y
d
x
.

{\displaystyle A = \frac{1}{2} \oint\_C xdy - ydx. }
Heating Baking Soda

## **Problem**

When sodium bicarbonate

NaHCO

3

{\displaystyle \ce{NaHCO3}}

, commonly known as baking soda, is heated to temperatures above 80 degrees Celsius, it starts to break down into three compounds: sodium carbonate

Na

2

CO

3

{\displaystyle \ce{Na2CO3}}

, water

H

2

O

{\displaystyle \ce{H2O}}

, and carbon dioxide

CO

2

{\displaystyle \ce{CO2}}

.

**Part 1:** Write a balanced chemical equation for the decomposition of sodium bicarbonate.

**Part 2:** Calculate the mass of sodium bicarbonate required to produce 20.5 g of carbon dioxide.

## **Solution**

**Part 1**

Start by writing the decomposition reaction:

NaHCO

3

⟶

Na

2

CO

3

H

2

O

CO

2

{\displaystyle \ce{NaHCO3 -> Na2CO3 + H2O + CO2}}

The balanced chemical equation is

2

NaHCO

3

⟶

Na

2

CO

3

H

2

O

CO

2

{\displaystyle \ce{2NaHCO3 -> Na2CO3 + H2O + CO2}}

**Part 2**

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

 Oxygen

Add the net molar mass of each element to obtain the net molar mass of carbon dioxide.

12.01

g

m
o
l

32.00

g

m
o
l

=
44.01

g

m
o
l

{\displaystyle 12.01 \; \frac{g}{mol} + 32.00 \; \frac{g}{mol} = 44.01 \; \frac{g}{mol} }

Determine the number of moles of 20.5 g carbon dioxide.

20.5

g
×

1

m
o
l

44.01

g

=
0.4658

m
o
l

{\displaystyle 20.5 \;g \times \frac{1 \, mol}{44.01 \, g} = 0.4658 \, mol}

According to the chemical equation we require twice the number of moles of sodium bicarbonate to carbon dioxide.

2
×
0.4658

m
o
# l
0.9316

m
o
l

{\displaystyle 2 \times 0.4658 \, mol = 0.9316 \, mol}

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

 Hydrogen
 Carbon
 Oxygen

Add the net molar mass of each element to obtain the net molar mass of carbon dioxide.

22.99

g

m
o
l

1.01

g

m
o
l

12.01

g

m
o
l

48.00

g

m
o
l

=
84.01

g

m
o
l

{\displaystyle 22.99 \; \frac{g}{mol} + 1.01 \; \frac{g}{mol} + 12.01 \; \frac{g}{mol} + 48.00 \; \frac{g}{mol} = 84.01 \; \frac{g}{mol} }

Lastly, calculate the mass of sodium bicarbonate to three significant figures.

0.9316

m
o
l
×

84.01

g

m
o
l

=
78.3

g

{\displaystyle 0.9316 \, mol \times \frac{84.01 \, g}{mol} = 78.3 \, g}

Therefore, 78.3 g of sodium bicarbonate is required to produce 20.5 g carbon dioxide.
Maximizing Cobb-Douglas Production

## **Problem**

The Cobb-Douglas production function models total production as a function of labour

L

{\displaystyle L}

 and capital investment

K

{\displaystyle K}

:

P
(
L
,
K
# )
c

L

α

K

β

{\displaystyle P(L,K) = c {L}^ {\alpha} {K}^{\beta} }

where

c
>
0
,
α
>
0
,
β
>
0

{\displaystyle c > 0, \alpha > 0, \beta > 0 }

.

If the budget constraint is

m
L
n
# K
p

{\displaystyle mL + nK = p }

, where

m
,
n

{\displaystyle m,n}

 are the unit cost of labour and capital. Determine the labour and capital values that maximizes production in terms of

α
,
β
,
m
,
n
,
p

{\displaystyle \alpha, \beta, m, n, p }

.

## **Solution**

This problem requires the use of Lagrange multipliers. Consider the following functions:

P
(
L
,
K
# )
c

L

α

K

β

{\displaystyle P(L,K) = c {L}^ {\alpha} {K}^{\beta} }

g
(
L
,
K
# )
m
L
n
K
−
p
.

{\displaystyle g(L,K) = mL + nK - p. }

Take the following partial derivatives in order to generate the equations for

∇
# P
λ
∇
g

{\displaystyle \nabla P = \lambda \nabla g }

:

∂
P

∂
L

=
c
α

L

α
−
1

K

β

{\displaystyle \frac{\partial P}{\partial L} = c \alpha {L}^{\alpha - 1} {K}^{\beta} }

∂
P

∂
K

=
c
β

L

α

K

β
−
1

{\displaystyle \frac{\partial P}{\partial K} = c \beta {L}^{\alpha} {K}^{\beta - 1} }

∂
g

∂
L

=
m

{\displaystyle \frac{\partial g}{\partial L} = m }

∂
g

∂
K

=
n
.

{\displaystyle \frac{\partial g}{\partial K} = n. }

Therefore, we get the equations

c
α

L

α
−
1

K

β

=
λ
m

{\displaystyle c \alpha {L}^{\alpha - 1} {K}^{\beta} = \lambda m }

c
β

L

α

K

β
−
1

=
λ
n
.

{\displaystyle c \beta {L}^{\alpha} {K}^{\beta - 1} = \lambda n. }

Isolate for

λ

{\displaystyle \lambda}

 for both of the above equations and isolate for either

K

{\displaystyle K}

 or

L

{\displaystyle L}

 (here I chose

K

{\displaystyle K}

).

# λ

c
α

m

L

α
−
1

K

β

=

c
β

n

L

α

K

β
−
1

{\displaystyle \lambda = \frac{c \alpha}{m} {L}^{\alpha - 1} {K}^{\beta} = \frac{c \beta}{n} {L}^{\alpha} {K}^{\beta - 1} }

Subsequently, we get

# K

β
m

α
n

L
.

{\displaystyle K = \frac{\beta m}{\alpha n} L. }

Substitute this value of

K

{\displaystyle K}

 into the budget constraint to solve for

L

{\displaystyle L}

. Then use the newfound value of

L

{\displaystyle L}

 to solve for

K

{\displaystyle K}

.

m
L
n

(

β
m

α
n

L

)

=
p

{\displaystyle mL + n\left(\frac{\beta m}{\alpha n} L\right) = p }

m
L

(

1

β
α

)

=
p

{\displaystyle mL \left(1 + \frac{\beta}{\alpha} \right) = p }

# L

α

α
β

p
m

{\displaystyle L = \frac{\alpha}{\alpha + \beta}\frac{p}{m} }

# K

β
m

α
n

⋅

α

α
β

⋅

p
m

{\displaystyle K = \frac{\beta m}{\alpha n} \cdot \frac{\alpha}{\alpha + \beta} \cdot \frac{p}{m} }

# K

β

α
β

p
n

{\displaystyle K = \frac{\beta}{\alpha + \beta} \frac{p}{n} }

Therefore, the solution set that maximizes production is

(

L
,
K

)

=

(

α

α
β

p
m

,

β

α
β

p
n

)

.

{\displaystyle \left(L, K \right) = \left(\frac{\alpha}{\alpha + \beta}\frac{p}{m}, \frac{\beta}{\alpha + \beta} \frac{p}{n} \right). }
Enthalpy of Formation for Nitromethane

## **Problem**

Consider the following combustion equation:

2

CH

3

NO

2

(
l
)

3

2

O

2

(
g
)

⟶

2

CO

2

(
g
)

3

H

2

O

(
g
)

N

2

(
g
)

{\displaystyle \ce{ {{2CH3NO2}\_{(l)}} + {\frac{3}{2} {O2}\_{(g)}}
 -> {{2CO2}\_{(g)}} + {{3H2O}\_{(g)}} + {{N2}\_{(g)}} }}

The enthalpy of reaction for the above combustion process is

−
709.2
k
J

/

m
o
l

{\displaystyle -709.2 kJ/mol }

. Calculate the standard enthalpy of formation for nitromethane

CH

3

NO

2

{\displaystyle \ce{ CH3NO2 }}

.

## **Solution**

Make a table like this one.

 Nitrogen
 Carbon dioxide
 Water
 Nitromethane

Use the net enthalpy equation to solve for the enthalpy of formation for nitromethane.

Δ

H

r
x
n

∘

=
∑
Δ

H

f
,
P

∘

−
∑
Δ

H

f
,
R

∘

{\displaystyle \Delta {H}\_{rxn}^{\circ} = \sum \Delta {H}\_{f,P}^{\circ} - \sum \Delta {H}\_{f,R}^{\circ} }

Let

x

{\displaystyle x}

 be the enthalpy of formation for nitromethane

CH

3

NO

2

{\displaystyle \ce{ CH3NO2 }}

. Set up the equation and solve for the unknown.

2
(
−
709.2
# )
[
2
(
−
393.5
)
3
(
−
285.8
)
1
(
0
)
]
−
[
2
(
x
)
1.5
(
0
)
]

{\displaystyle 2(-709.2) = [2(-393.5) + 3(-285.8)+ 1(0)] - [2(x) + 1.5(0)] }

−
# 1418.4
−
1644.4
−
2
x

{\displaystyle -1418.4 = -1644.4 - 2x }

# x
−
113.2

{\displaystyle x = -113.2 }

Therefore, the enthalpy of formation for nitromethane is

Δ

H

f

∘

=
−
113.2
k
J

/

m
o
l

{\displaystyle \Delta {H}\_{f}^{\circ} = -113.2 kJ/mol }

.
Integral Test

## **Problem**

Use the integral test to determine whether the series

∑

# n
1

∞

ln
3

⁡
(
n
)

n
6

{\displaystyle \sum\_{n=1}^{\infty} \frac{\ln^3(n)}{n^6} }

converges of diverges.

## **Hint**

(1) Use the Gamma function:

∫

0

∞

t
n

e

−
a
t

d
# t

Γ
(
n
1
)

a

n
1

=

n
!

a

n
1

.

{\displaystyle \int\_{0}^{\infty} t^n {e}^{-at}dt = \frac{\Gamma (n+1)}{{a}^{n+1}} = \frac{n!}{{a}^{n+1}}. }

## **Solution**

"Convert" the series in an integral. If the integral converges, then the series converges. If the integral diverges, then the series diverges.

Let

# y
ln
⁡
(
x
)

{\displaystyle y = \ln(x) }

, then

# x

e
y

{\displaystyle x = e^y }

 and

d
# y

1
x

d
x

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

. By the method of substitution, the integral becomes

∫

0

∞

y
3

e

−
5
y

d
y
.

{\displaystyle \int\_{0}^{\infty} y^3 {e}^{-5y} dy. }

One can use integration by parts several times or use the Gamma function to evaluate this integral.

∫

0

∞

t
n

e

−
a
t

d
# t

Γ
(
n
1
)

a

n
1

=

n
!

a

n
1

{\displaystyle \int\_{0}^{\infty} t^n {e}^{-at}dt = \frac{\Gamma (n+1)}{{a}^{n+1}} = \frac{n!}{{a}^{n+1}} }

This means

∫

0

∞

y
3

e

−
5
y

d
y

=

3
!

5
4

=

6
625

.

{\displaystyle \begin{align}
\int\_{0}^{\infty} y^3 {e}^{-5y} dy &= \frac{3!}{5^4} \\[5 pt]
&= \frac{6}{625}.
\end{align} }

Since the integral converges, the above infinite series is convergent.
Double-Slit Interference

## **Problem**

What is the distance on screen between adjacent maxima near the center of the interference pattern? The wavelength

λ

{\displaystyle \lambda}

 of the light is 546 nm, the slit separation *d* is 0.12 mm, and the slit-screen separation *D* is 55 cm. Assume that

θ

{\displaystyle \theta}

 is small enough to permit use of the approximations

sin
⁡
θ
≈
tan
⁡
θ
≈
θ

{\displaystyle \sin \theta \approx \tan \theta \approx \theta }

, in which

θ

{\displaystyle \theta}

 is expressed in radian measure.

## **Hints**

(1) Where the bright fringes (maxima) are located is given by the formula

d
sin
⁡
# θ
n
λ
,

# n
0
,
1
,
2
,
.
.
.

{\displaystyle d\sin \theta =n\lambda ,\quad n=0,1,2,...}

and the dark fringes (minima) is

d
sin
⁡
# θ

(

n

1
2

)

λ
,

# n
0
,
1
,
2
,
.
.
.

{\displaystyle d\sin \theta =\left(n+{\frac {1}{2}}\right)\lambda ,\quad n=0,1,2,...}

## **Solution**

The maximum's vertical distance

y
n

{\displaystyle y\_n}

 from the center of the interference pattern on the screen is related to its angle from the central axis:

tan
⁡
# θ

y

n

D

.

{\displaystyle \tan \theta ={\frac {y\_{n}}{D}}.}

The *n*th-order maximum is given by

sin
⁡
# θ

n
λ

d

.

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

Using the small-angle approximation

sin
⁡
θ
≈
tan
⁡
θ
≈
θ

{\displaystyle \sin \theta \approx \tan \theta \approx \theta }

, we can equate the right-hand side of each equation:

y

n

D

=

n
λ

d

.

{\displaystyle {\frac {y\_{n}}{D}}={\frac {n\lambda }{d}}.}

Solving for

y
n

{\displaystyle y\_n}

 yields

y

n

=

n
λ
D

d

.

{\displaystyle y\_{n}={\frac {n\lambda D}{d}}.}

The next maximum from the interference pattern's center is

y

n
1

=

(
n
1
)
λ
D

d

{\displaystyle y\_{n+1}={\frac {(n+1)\lambda D}{d}}}

thus, the distance between two adjacent maxima is

Δ
# y

y

n
1

−

y

n

=

λ
D

d

.

{\displaystyle \Delta y=y\_{n+1}-y\_{n}={\frac {\lambda D}{d}}.}

Calculate the separation for this problem:

Δ
y

=

(
546
×

10

−
9

m
)
(
55
×

10

−
2

m
)

0.12
×

10

−
3

m

=
2.50
×

10

−
3

m
.

{\displaystyle {\begin{aligned}\Delta y&={\frac {(546\times 10^{-9}\;m)(55\times 10^{-2}\;m)}{0.12\times 10^{-3}\;m}}\\[5pt]&=2.50\times 10^{-3}\;m.\end{aligned}}}

Therefore, the distance between two adjacent maxima is 2.5 mm.
Cayley-Hamilton Theorem

## **Problem**

State the Cayley-Hamilton theorem. Verify the Cayley-Hamilton for the matrix

A

=

[

3

2

−
1

4

]

{\displaystyle \mathbf{A} =
 \begin{bmatrix}
 3 & 2 \\
 -1 & 4 \\
 \end{bmatrix}
}

.

## **Solution**

Cayley-Hamilton theorem states: **A matrix satisfies its characteristic equation.**

Calculate the characteristic equation

det

(
A
−
λ
I
)

=
0

{\displaystyle \det{(A - \lambda I)} = 0 }

. For this problem

I

{\displaystyle I}

 is the 2x2 identity matrix

[

1

0

0

1

]

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

.

3
−
λ

2

−
1

4
−
λ

=
0

{\displaystyle \begin{vmatrix}
 3 - \lambda & 2 \\
 -1 & 4 - \lambda \\
 \end{vmatrix} =0
}

(
3
−
λ
)
(
4
−
λ
)
−
(
2
)
(
−
1
# )
0

{\displaystyle (3-\lambda)(4-\lambda)-(2)(-1)=0 }

λ
2

−
7
λ
# 14
0

{\displaystyle \lambda^2 - 7\lambda + 14 = 0 }

.
Verify that

A
2

−
7
A
14
# I
0

{\displaystyle A^2 - 7A + 14I = 0 }

.

A
2

−
7
A
14
# I

[

3

2

−
1

4

]

[

3

2

−
1

4

]

−
7

[

3

2

−
1

4

]

14

[

1

0

0

1

]

{\displaystyle
A^2-7A+14I= \begin{bmatrix}
 3 & 2 \\
 -1 & 4 \\
 \end{bmatrix}
\begin{bmatrix}
 3 & 2 \\
 -1 & 4 \\
 \end{bmatrix} -7 \begin{bmatrix}
 3 & 2 \\
 -1 & 4 \\
 \end{bmatrix} +14 \begin{bmatrix}
 1 & 0 \\
 0 & 1 \\
 \end{bmatrix}
}

A
2

−
7
A
14
# I

[

7

14

−
7

14

]

−

[

21

14

−
7

28

]

[

14

0

0

14

]

{\displaystyle
A^2-7A+14I= \begin{bmatrix}
 7 & 14 \\
 -7 & 14 \\
 \end{bmatrix}
 - \begin{bmatrix}
 21 & 14 \\
 -7 & 28 \\
 \end{bmatrix} + \begin{bmatrix}
 14 & 0 \\
 0 & 14 \\
 \end{bmatrix}
}

A
2

−
7
A
14
# I

[

0

0

0

0

]

{\displaystyle A^2-7A+14I = \begin{bmatrix}
 0 & 0 \\
 0 & 0 \\
 \end{bmatrix} }

Thus the Cayley-Hamilton theorem is verified.
Fossilized Bamboo Shoots in Yanzhou 延州石笋

**《梦溪笔谈·卷21异事17》**
近岁延州永宁关大河岸崩，入地数十尺，土下得竹笋一林，凡数百茎，根干相连，悉化为石。适有中人过，亦取数茎去，云欲进呈。延郡素无竹，此入在数十尺土下，不知其何代物。无乃旷古以前，地卑气湿而宜竹邪？婺（wù）州金华山有松石，又如 核桃、芦根、鱼、蟹之类，皆有成石者。然皆其地本有之物，不足深怪。此深地中所无，又非本土所有之物，特可异耳。

***Mengxi Bitan* Chapter 21 Yishi 17**
In recent years there was a landslide on the bank of the Yellow River in Yongning Fortress near Yanzhou. The bank collapsed, opening a space of several dozens of feet, and under the ground a forest of bamboo shoots was revealed. It contained several hundred bamboos with their roots and trunks closely connected and all turned to stone. A high official from the royal court chanced to pass by He took several of them away, saying that he would present them to the emperor. As people in Yanzhou never planted bamboo, no one knew the origin of these bamboo shoots which had been buried deep underground. Probably in the distant past this place was a low-lying area with a humid climate fit for the growth of bamboo. In Mount Jinhua in Wuzhou there are stone pine cones, stone peach pits, stone reed rhizomes, stone fishes, stone crabs, and so on. As they are all native to this place, their existence does not cause a great stir. But these stone bamboo shoots usually do not appear under the ground. What is more, they are not the traditional local products in Yanzhou. So their appearance is really strange.

**【今译】**

近年，延州永宁关黄河岸崩塌，裂缝有几十尺深，土下面发现了一片竹笋林，总共有好几百根，竹笋的根和干连在一块，全变为石头了。当时正好有宦官经过这里，也拿走了几根竹茎，说要献给皇帝。延州从来没有过“竹子”，这些竹子深入地下几十尺深，不知是什么时代的东西。是不是在很古很古以前，这里地势低洼气候湿润而适合生长“竹子”呢？婺州的金华山有松石，还有核桃、芦根、 鱼、螃蟹等等，都有变成石头的。但是所有这些，都是当地就有的东西，没有什么奇怪。而竹笋在这 样深的地中是不可能存在的，而且本地也没有生这些东西，这才真是奇怪呢。
Area of a Circle

## **Problem**

**Part 1:** Come up with a convincing method to transform a circle into a triangle. Use this transformation to derive the formula for the area of a circle.

**Part 2**: Express the area of a circle in terms of its diameter.

## **Solution**

   **Part 1**

Take each ring of the circle straighten it out. Stack the straightened rings such that the base is the circumference of the circle, while the center of the circle becomes the apex of the triangle. The end result is a triangle with a height equal to the radius of the circle, and a base equal to the circumference of the circle.

A

c
i
r
c
l
e

=

1
2

×
2
π
r
×
r

[
5
p
t
]

=
π

r
2

.

{\displaystyle \begin{align}
{A}\_{circle} &= \frac{1}{2} \times 2\pi r \times r \\ [5 pt]
&= \pi r^2. \\
\end{align} }

**Part 2**

The diameter of a circle is twice its radius

# d
2
r

{\displaystyle d = 2r}

, subsequently,

A

c
i
r
c
l
e

=
π

(

d
2

)

2

=

π

d
2

4

.

{\displaystyle
\begin{align}
{A}\_{circle} &= \pi \left(\frac{d}{2}\right)^2 \\[5 pt]
&= \frac{\pi d^2}{4}.
\end{align}
}
Lune of Hippocrates

## **Problem**

Prove that the lune (crescent shape) in the diagram below is equal in area to the isosceles right triangle inside the circular quadrant.

## **Solution**

Let

O
A

¯

=
r

{\displaystyle \overline{OA} = r }

O
B

¯

=
r

{\displaystyle \overline{OB} = r }

.
Since the right triangle is isosceles

A
B

¯

=

2

r

{\displaystyle \overline{AB} = \sqrt{2} r }

.
and the area of the isosceles right triangle is therefore

r
2

2

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

.
Let

A

{\displaystyle A}

 be the area of the lune, and

B

{\displaystyle B}

 be the area of the unshaded region between the lune and the right triangle. The sum of

A

{\displaystyle A}

 and

B

{\displaystyle B}

 forms a semicircle

A
B

{\displaystyle AB}

.
The semicircle

A
B

{\displaystyle AB}

 has the area,

1
2

π

(

2

r

2

)

2

=

π

r
2

4

{\displaystyle \frac{1}{2}\pi {\left(\frac{\sqrt{2}r}{2}\right)}^{2} = \frac{\pi r^2}{4} }

.
The area of

B

{\displaystyle B}

 is the area of the quadrant minus the area of the isosceles right triangle:

# B

π

r
2

4

−

r
2

2

{\displaystyle B = \frac{\pi r^2}{4} - \frac{r^2}{2} }

To determine the area of the lune

A

{\displaystyle A}

, subtract the area of

B

{\displaystyle B}

 from the semicircle

A
B

{\displaystyle AB}

.

# A

π

r
2

4

−

(

π

r
2

4

−

r
2

2

)

{\displaystyle A = \frac{\pi r^2}{4} - \left(\frac{\pi r^2}{4} - \frac{r^2}{2} \right) }

# A

r
2

2

{\displaystyle A = \frac{r^2}{2} }

Therefore the area of the lune is equal to the area of the isosceles right triangle.
Gougu Theorem

What was the Pythagorean theorem called in ancient China?

In the modern day, the Chinese call the Pythagorean theorem 勾股定理 gōu gǔ dìng lǐ (gougu theorem). However, the word "theorem" did not exist in the Chinese language before Jesuit missionaries introduced European mathematics and astronomy to the Chinese Imperial Court during the late Ming-early Qing dynasties. Traditionally, the Pythagorean theorem was called 勾股求弦 gōu gǔ qiú xián, which literally means "determining the chord [hypotenuse] of a right triangle".

Here is an animation of a traditional Chinese proof of the Pythagorean theorem. This dissection proof uses a traditional Chinese technique called 出入相補 chū rù xiàng bǔ (out-in complementary principle). The original proof comes from the 周髀算經 "Zhoubi Suanjing" (Mathematical Canon of the Zhou Gnomon), an ancient Chinese text on mathematical astronomy compiled by generations of scholars between the 11th century to 2nd centuries BC.

zh:勾股定理
Chord Ratio

## **Problem**

Although **Hipparchus of Nicaea** (c. 190 – c. 120 BC) invented trigonometry, he did not use the three trigonometric ratios sine, cosine, and tangent common used today [1]. Hipparchus used the chord ratio instead (Figure 1). Use the half-angle identity to show that

c
r
d

(
θ
)

=
2
sin
⁡

(

θ
2

)

.

{\displaystyle \mathrm{crd} \left(\theta \right) = 2 \sin{\left(\frac{\theta}{2} \right)}. }

## **Solution**

A right triangle is formed with the sine and the versine. The versine is another outdated trigonometric ratio that can be visually explained be equivalent to

1
−
cos
⁡

θ

{\displaystyle 1 - \cos{\theta}}

.

By the Pythagorean theorem,

c
r
d

(
θ
)

=

(

sin
⁡

θ

)

2

(

1
−
cos
⁡

θ

)

2

=

sin
2

⁡

θ

1
−
2
cos
⁡

θ

cos
2

⁡

θ

.

{\displaystyle \begin{align}
\mathrm{crd} \left(\theta \right) &= \sqrt{{\left(\sin{\theta} \right)}^{2} + {\left(1 - \cos{\theta}\right)}^{2}} \\[5 pt]
&= \sqrt{\sin^2{\theta} + 1 - 2\cos{\theta} + \cos^2{\theta}}.
\end{align} }

Since

sin
2

⁡

θ

cos
2

⁡

θ

=
1

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

,

c
r
d

(
θ
)

=

2
−
2
cos
⁡

θ

=

2

1
−
cos
⁡

θ

.

{\displaystyle \begin{align}
\mathrm{crd} \left(\theta \right) &= \sqrt{2 - 2\cos{\theta}} \\[5 pt]
&= \sqrt{2} \sqrt{1 - \cos{\theta}}.
\end{align} }

Now deliberately write the equation as such

c
r
d

(
θ
)

=

2

2

(

1
−
cos
⁡

θ

)

2

.

{\displaystyle \mathrm{crd} \left(\theta \right) = \sqrt{2} \sqrt{\frac{2\left(1 - \cos{\theta} \right)}{2}}. }

Thus,

c
r
d

(
θ
)

=
2

1
−
cos
⁡

θ

2

=
2
sin
⁡

(

θ
2

)

.

{\displaystyle \begin{align}
\mathrm{crd} \left(\theta \right) &= 2 \sqrt{\frac{1 - \cos{\theta} }{2}} \\[5 pt]
&= 2 \sin{\left(\frac{\theta}{2} \right)}.
\end{align} }
Gradient of a Scalar Field

## **Problem**

Calculate the gradient of the scalar field

ϕ

(

x
,
y
,
z

)

=

(

y
2

)

cos
⁡

(
x
)

x
cos
⁡

(

y
−
2

)

{\displaystyle \phi \left(x,y,z \right) = \left( y+2 \right) \cos{\left(x \right)} + x \cos{\left(y-2\right)} }

.

## **Solution**

Use the gradient formula in rectangular coordinates:

∇
# ϕ

∂
ϕ

∂
x

i

∂
ϕ

∂
y

j

∂
ϕ

∂
z

k
.

{\displaystyle \nabla \phi = \frac{\partial \phi}{\partial x} \boldsymbol{i} + \frac{\partial \phi}{\partial y} \boldsymbol{j} + \frac{\partial \phi}{\partial z} \boldsymbol{k} .}

Calculate each partial derivative.

∂
ϕ

∂
x

=
−

(

y
2

)

sin
⁡

(
x
)

cos
⁡

(

y
−
2

)

∂
ϕ

∂
y

=
cos
⁡

(
x
)

−
x
sin
⁡

(

y
−
2

)

∂
ϕ

∂
z

=
0

{\displaystyle \begin{align}
\frac{\partial \phi}{\partial x} &= - \left(y+2 \right)\sin{\left(x \right)} + \cos{\left(y-2 \right)} \\[5 pt]
\frac{\partial \phi}{\partial y} &= \cos{\left(x \right)} - x \sin{\left(y-2 \right)} \\[5 pt]
\frac{\partial \phi}{\partial z} &= 0
\end{align}}

Therefore,

∇
# ϕ

[

−

(

y
2

)

sin
⁡

(
x
)

cos
⁡

(

y
−
2

)

]

i

[

cos
⁡

(
x
)

−
x
sin
⁡

(

y
−
2

)

]

j
.

{\displaystyle \nabla \phi = \left[- \left(y+2 \right)\sin{\left(x \right)} + \cos{\left(y-2 \right)}\right] \boldsymbol{i} + \left[\cos{\left(x \right)} - x \sin{\left(y-2 \right)} \right] \boldsymbol{j} .}
Graphing Tangent Lines with Python

**By: Tao Steven Zheng (鄭濤)**

## **Code**

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

'''Part 1: Calculating Derivatives on Python'''

def deriv(f,x):
     h = 0.000000001                 #step-size
     return (f(x+h) - f(x))/h        #definition of derivative

'''Part 2: Plot function with tangent'''

def tangent\_line(f,x\_0,a,b):
     x = np.linspace(a,b,200)
     y = f(x)
     y\_0 = f(x\_0)
     y\_tan = deriv(f,x\_0) \* (x - x\_0) + y\_0

   #plotting
     plt.plot(x,y,'r-')
     plt.plot(x,y\_tan,'b-')
     plt.axis([a,b,a,b])
     plt.xlabel('x')
     plt.ylabel('y')
     plt.title('Plot of a function with tangent line')
     plt.show()

'''Part 3: Tests and results'''
#Test number 1

def f1(x):
     return x\*\*2

 tangent\_line(f1,1,-2,2)

#Test number 2

 def f2(x):
     return np.exp(-x\*\*2)

 tangent\_line(f2,1,-2,2)

```

## **Results**

**Test 1**

**Test 2**
Answers to Number Puzzles by Steven Zheng

**By: Tao Steven Zheng (郑涛)**

## **Puzzle 1 (February 24, 2018)**

**About this problem:** Created this problem while tinkering with the Chinese abacus.

If 1719, 2954961, 36, 6

Then 391, ?, ?, ?

**Answer:** 152881, 25, 5

Square the first number to get the second. Add the digits of the second number to get the third. Take the square root of the third to get the fourth.
Addition of Unit Fractions

## **Problem**

Determine the sum

1
12

1
18

1
21

1
25

1
35

.

{\displaystyle \frac{1}{12} + \frac{1}{18} + \frac{1}{21} + \frac{1}{25} + \frac{1}{35}.}

## **Solution**

When adding fractions, the first step is to find the lowest common multiple (LCM) of the denominators. Begin by prime factorizing each denominator.

12

=

2
2

×
3

18

=
2
×

3
2

21

=
3
×
7

25

=

5
2

35

=
5
×
7

{\displaystyle \begin{align}
12 &= 2^2 \times 3 \\[5 pt]
18 &= 2 \times 3^2 \\[5 pt]
21 &= 3 \times 7 \\[5 pt]
25 &= 5^2 \\[5 pt]
35 &= 5 \times 7
\end{align}}

To calculate the LCM, take the highest power of each prime factor and multiply them together. Thus,

L
C
M

(
12
,
18
,
21
,
25
,
35
)

=

2
2

×

3
2

×

5
2

×
7

=

{\displaystyle \begin{align}
\mathrm{LCM}(12, 18, 21, 25, 35) &= 2^2 \times 3^2 \times 5^2 \times 7 \\[5 pt]
&= 6300.
\end{align}}

Next, divide the LCM by each denominator to get the scale factor for their respective numerators.

6300
÷
12

=
525

6300
÷
18

=
350

6300
÷
21

=
300

6300
÷
25

=
252

6300
÷
35

=
180

{\displaystyle \begin{align}
6300 \div 12 &= 525 \\[5 pt]
6300 \div 18 &= 350 \\[5 pt]
6300 \div 21 &= 300 \\[5 pt]
6300 \div 25 &= 252 \\[5 pt]
6300 \div 35 &= 180
\end{align}}

Therefore,

1
12

1
18

1
21

1
25

1
35

=

525
6300

350
6300

300
6300

252
6300

180
6300

=

1607
6300

.

{\displaystyle \begin{align}
\frac{1}{12} + \frac{1}{18} + \frac{1}{21} + \frac{1}{25} + \frac{1}{35} &= \frac{525}{6300} + \frac{350}{6300} + \frac{300}{6300} + \frac{252}{6300} + \frac{180}{6300} \\[5 pt]
&= \frac{1607}{6300} .
\end{align}}
Archimedes' Triumph

## **Problem**

In Volume I of On the *Sphere and the Cylinder*, Archimedes determined the volumetric ratio of a sphere to a circumscribed cylinder. The height and width of the cylinder is equal to the diameter of the sphere. What is this ratio?

## **Solution**

Consider a sphere of radius

r

{\displaystyle r}

. The circumscribed cylinder must share the same height and width as the sphere. Let

r

{\displaystyle r}

 the radius of the cylinder's base, and

# h
2
r

{\displaystyle h = 2r }

 be the height of the cylinder.

The volume of a sphere is

V

s
p
h
e
r
e

=

4
π

r
3

3

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

.

The volume of a cylinder is

V

c
y
l
i
n
d
e
r

=
π

r
2

h

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

.

To determine the volume circumscribed cylinder is, substitute

# h
2
r

{\displaystyle h = 2r }

:

V

c
y
l
i
n
d
e
r

=
π

r
2

(
2
r
)

[
5
p
t
]

=
2
π

r
3

{\displaystyle \begin{align}
V\_{cylinder} &= \pi r^2 (2r) \\ [5 pt]
& = 2\pi r^3
\end{align} }

Therefore the volumetric ratio of a sphere to its circumscribed cylinder must be

V

s
p
h
e
r
e

:

V

c
y
l
i
n
d
e
r

=

4
π

r
3

3

:
2
π

r
3

{\displaystyle V\_{sphere}:V\_{cylinder} = \frac{4\pi r^3}{3}:2\pi r^3 }

which simplifies to

V

s
p
h
e
r
e

:

V

c
y
l
i
n
d
e
r

=
2
:

{\displaystyle V\_{sphere}:V\_{cylinder} = 2:3. }
Chemistry Learning Resources

Chemical Equations Online - Great encyclopedia of chemical reactions with detailed descriptions on the reaction type.

Old Chemistry 12 Provincial Exams - Collection of past chemistry 12 provincial exams for British Columbia from 1994 - 2004

Oxidation Numbers Calculator - Instant oxidation numbers calculator!
Electrostatic Oscillator with Python

**By: Tao Steven Zheng (郑涛)**

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

#Define constants
k = 9 \* (10\*\*9) #Coulomb constant
Q = 0.000001 #charge
m = 0.1 #mass
R = 1 #radius
N = 10
dt = 1
x\_0 = 10

#Set up arrays for position and time
time = [0]
acc = [(-1\*k\*Q\*Q / m) \*x\_0 / (x\_0\*\*2 + R\*\*2)\*\*(3/2)]
vel = [0]
pos = [x\_0]

#Define a function that returns the force or acceleration
# a = (-1\*k\*Q\*Q / m)
#Set initial position and velocity
t = 0
x = x\_0
a = (-1\*k\*Q\*\*2 / m) \*x / (x\*\*2 + R\*\*2)\*\*(3/2)
v = 0

#Integrate equation of motion
for i in range(0,1000):
# v[i+1] = v[i] + a[i]\*dt
    t = t + dt
    v1 = v + a \* dt /2
    x = x + v1 \* dt
    a = (-1\*k\*Q\*Q / m) \*x / (x\*\*2 + R\*\*2)\*\*(3/2)
    v = v1 + a \* dt /2

    time.append(t)
    acc.append(a)
    vel.append(v)
    pos.append(x)

#Plot trajectory
plt.plot(time, pos ,'b-')
plt.xlabel('Time (s)') #x-axis label
plt.ylabel('Position (m)') #y-axis label
plt.title('Trajectory of Charged Particle Due to a Charged Ring')
plt.show()

```
Exponential Distribution

## **Problem**

Calculate the following attributes for the probability distribution function (PDF)

p
(
x
# )
λ

e

−
λ
x

{\displaystyle p(x) = \lambda {e}^{-\lambda x} }

defined on the interval

x
∈
[
0
,
∞
)

{\displaystyle x\in[0,\infty)}

.

**Part 1:** Mean

**Part 2:** Median

**Part 3:** Variance and standard deviation

## **Solution**

**Part 1**

The mean is the first moment of the PDF.

# μ

∫

−
∞

∞

x
⋅
p
(
x
)
d
x

{\displaystyle \mu = \int\_{-\infty}^{\infty} x \cdot p(x) dx }

# μ

∫

0

∞

λ
x

e

−
λ
x

d
x

{\displaystyle \mu = \int\_{0}^{\infty} \lambda x{e}^{-\lambda x} dx }

Using integration by parts yields

# μ

lim

b
→
∞

[

−

e

−
λ
x

(

x

1
λ

)

0
b

]

{\displaystyle \mu = \lim\_{b \to \infty } \left[-{e}^{-\lambda x} \left(x+\frac{1}{\lambda}\right) \Biggr|\_0^b\right] }

# μ

1
λ

{\displaystyle \mu = \frac{1}{\lambda}}

.
**Part 2**

To calculate the median, one must obtain the cumulative distribution function (CDF).

P
(
x
# )

∫

0

x

λ

e

−
λ
t

d
t

{\displaystyle P(x) = \int\_{0}^{x} \lambda e^{-\lambda t}dt }

P
(
x
# )
−

e

−
λ
t

0
x

{\displaystyle P(x) = -{e}^{-\lambda t} \Big|\_0^x }

P
(
x
# )
1
−

e

−
λ
x

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

.
The median is located where the probability is 50%, so

P
(
x
# )
1
−

e

−
λ
x

=
0.5

{\displaystyle P(x) = 1 - e^{-\lambda x} = 0.5 }

e

−
λ
x

=
0.5

{\displaystyle e^{-\lambda x} = 0.5 }

# x

ln
⁡
(
0.5
)

−
λ

{\displaystyle x = \frac{\ln(0.5)}{-\lambda} }

# x

ln
⁡
(
2
)

λ

{\displaystyle x = \frac{\ln(2)}{\lambda} }

.
**Part 3**

The variance is calculated by subtracting the square of the mean from the second moment.

V
a
r
(
X
# )

∫

−
∞

∞

x
2

⋅
p
(
x
)
d
x
−

μ

2

{\displaystyle Var(X) = \int\_{-\infty}^{\infty} x^2 \cdot p(x) dx - {\mu}^2 }

V
a
r
(
X
# )

∫

0

∞

λ

x
2

e

−
λ
x

d
x
−

μ

2

{\displaystyle Var(X) = \int\_{0}^{\infty} \lambda x^2 {e}^{-\lambda x} dx - {\mu}^2 }

Using integration by parts yields (twice), and the mean calculated in **part 1** yields

V
a
r
(
X
# )

lim

b
→
∞

[

−

e

−
λ
x

(

x
2

2
x

λ

2

λ

2

)

0
b

]

−

1

λ

2

{\displaystyle Var(X) = \lim\_{b \to \infty } \left[-{e}^{-\lambda x} \left(x^2+\frac{2x}{\lambda} + \frac{2}{{\lambda}^2}\right) \Biggr|\_0^b\right] - \frac{1}{{\lambda}^2} }

V
a
r
(
X
# )

1

λ

2

{\displaystyle Var(X) = \frac{1}{{\lambda}^2}}

.
The standard deviation is the square root of the variance

# σ

V
a
r

{\displaystyle \sigma = \sqrt{Var} }

# σ

1
λ

{\displaystyle \sigma = \frac{1}{\lambda} }

.
Exponential and Logarithmic Equations

## **Problem**

Solve the following equations.
(1)

e

a
x

=
c

e

−
b
x

{\displaystyle {e}^{ax} = c{e}^{-bx} }

 where

a
,
b
,
c

{\displaystyle a,b,c}

 are positive constants.

(2)

ln
⁡

(

ln
⁡

x

)

=
1

{\displaystyle \ln{\left(\ln{x}\right)} = 1 }

.

## **Solution**

(1) Divide both sides by

e

−
b
x

{\displaystyle {e}^{-bx} }

e

a
x

e

−
b
x

=
c

{\displaystyle \frac{{e}^{ax}}{{e}^{-bx}} = c }

e

(
a
b
)
x

=
c

{\displaystyle {e}^{(a+b)x} = c }

Take the logarithm on both sides of the equation.

ln
⁡

e

(
a
b
)
x

=
ln
⁡

c

{\displaystyle \ln{{e}^{(a+b)x}} = \ln{c} }

(
a
b
)
x
ln
⁡

e

=
ln
⁡

c

{\displaystyle (a+b)x\ln{e} = \ln{c} }

Since

ln
⁡

e

=
1

{\displaystyle \ln{e} = 1 }

 it follows that

(
a
b
)
# x
ln
⁡

c

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

Thus

# x

ln
⁡

c

a
b

{\displaystyle x = \frac{\ln{c}}{a+b} }

.

(2)

e

ln
⁡

(

ln
⁡

x

)

=

e

1

{\displaystyle {e}^{\ln{\left(\ln{x}\right)}} = {e}^{1} }

ln
⁡

x

=
e

{\displaystyle \ln{x} = e }

e

ln
⁡

x

=

e

e

{\displaystyle {e}^{\ln{x}} = {e}^{e} }

# x

e
e

{\displaystyle x = e^e }
Five Yellow Cars

## **Problem**

Suppose that on average, in any given hour, there are 3 yellow cabs travelling on a given stretch of road. Calculate the probability of seeing exactly five yellow cabs to four decimal places.

## **Solution**

This problem is best modelled as a **Poisson distribution**. The Poisson distribution describes events occurring at random in a fixed interval of time or space, with the condition that these events occur with a known constant mean rate

λ

{\displaystyle \lambda}

 and independently of the time since the last event.

For the Poisson distribution, the probability of some number of events

x

{\displaystyle x}

 occurring over the set timeframe is calculated using the formula

P
(
x
,
λ
# )

λ
x

x
!

e

−
λ

.

{\displaystyle P(x,\lambda)= \frac{\lambda^x}{x!}{e}^{-\lambda}.}

For this problem, the rate is

# λ
3

{\displaystyle \lambda = 3}

, and

# x
5

{\displaystyle x=5}

; therefore, the probability of seeing exactly five yellow cabs is

P
(
5
,
3
)

=

3
5

5
!

e

−
3

=
0.1008

{\displaystyle \begin{align}
P(5, 3) &= \frac{3^5}{5!}{e}^{-3} \\[5 pt]
&= 0.1008
\end{align}}

or

10.08
%

{\displaystyle 10.08\%}

.
Interval of Convergence

## **Problem**

Consider the following infinite series:

∑

# n
1

∞

4
n

n

(
x
−
3

)
n

.

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}(x-3)^n. }

Determine the radius of convergence and the interval of convergence.

## **Solution**

In order for the series to converge, the limit must be less than 1.

lim

n
→
∞

a

n
1

a
n

<
1

{\displaystyle \lim\_{n\to\infty} \left| \frac{a\_{n+1}}{a\_n} \right| < 1 }

For the series

∑

# n
1

∞

4
n

n

(
x
−
3

)
n

,

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}(x-3)^n, }

lim

n
→
∞

4

n
1

n
1

(
x
−
3
)

n
1

⋅

n

4
n

(
x
−
3
)

n

<
1

{\displaystyle \lim\_{n\to\infty} \left| \frac{4^{n+1}}{n+1} {(x-3)}^{n+1} \cdot \frac{n}{4^n {(x-3)}^{n}} \right| < 1 }

lim

n
→
∞

4
n

n
1

(
x
−
3
)

<
1

{\displaystyle \lim\_{n\to\infty} \left| \frac{4n}{n+1}(x-3) \right| < 1 }

lim

n
→
∞

4
n

n
1

x
−
3

<
1

4

x
−
3

<
1

{\displaystyle 4|x-3| < 1 }

x
−
3

<

1
4

.

{\displaystyle |x-3| < \frac{1}{4}. }

At this stage, we find that the radius of convergence is

# R

1
4

{\displaystyle R = \frac{1}{4} }

.

To solve for the interval of convergence, just solve for

x

{\displaystyle x}

.

−
1

4

<
x
−
3
<

1
4

{\displaystyle \frac{-1}{4} < x-3 < \frac{1}{4} }

11
4

<
x
<

13
4

.

{\displaystyle \frac{11}{4} < x < \frac{13}{4}. }

This is as far as the ratio test allows us to conclude. To determine if each endpoint converges or not, we must apply other tests.

When

# x
2.75

{\displaystyle x = 2.75 }

, the series becomes

∑

# n
1

∞

4
n

n

(

−
1

4

)

n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}{\left(\frac{-1}{4}\right)}^{n} }

∑

# n
1

∞

(
−
1

)
n

n

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

By the alternating series test, this series converges.

When

# x
3.25

{\displaystyle x = 3.25 }

, the series becomes

∑

# n
1

∞

4
n

n

(

1
4

)

n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}{\left(\frac{1}{4}\right)}^{n} }

∑

# n
1

∞

1
n

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

This is the Harmonic series, and it was shown here that it diverges.

Therefore, the radius of convergence for

∑

# n
1

∞

4
n

n

(
x
−
3

)
n

{\displaystyle \sum\_{n=1}^\infty\frac{4^n}{n}(x-3)^n }

is actually

11
4

≤
x
<

13
4

{\displaystyle \frac{11}{4} \le x < \frac{13}{4} }

.
Gougu Diagram

## **Problem**

In the *Zhoubi Suanjing*, the Gougu diagram as shown below demonstrates the Pythagorean theorem.

Use this diagram to prove the Pythagorean theorem in two ways.

## **Solution**

The Gougu diagram provides two proofs. Let the length of the bold square be

c

{\displaystyle c}

, the base of the right triangle is

b

{\displaystyle b}

 and the height of the right triangle

a

{\displaystyle a}

.

**Proof 1**

The first proof uses the smallest square in the center and four right triangles that are contained in the bold square. The side length of the smallest square is

b
−
a

{\displaystyle b-a}

. The area of the smallest square and the four triangles is equal to the area of the bold square.

c
2

=
4

(

1
2

a
b

)

(

a
−
b

)

2

c
2

=
2
a
b

a
2

−
2
a
b

b
2

c
2

=

a
2

b
2

{\displaystyle \begin{align}
c^2 &= 4\left(\frac{1}{2} ab \right) + \left(a-b\right)^2 \\[5 pt]
c^2 &= 2ab + a^2-2ab+b^2 \\[5 pt]
c^2 &= a^2 + b^2
\end{align}
}

**Proof 2**

The second proof uses the bolder square and the four right triangles outside the bold square. The area of the bold square and the four triangles is equal to the area of the largest square.

c
2

4

(

1
2

a
b

)

=

(

a
b

)

2

c
2

2
a
# b

a
2

2
a
b

b
2

c
2

=

a
2

b
2

{\displaystyle \begin{align}
&c^2 + 4\left(\frac{1}{2} ab \right) = \left(a+b\right)^2 \\[5 pt]
&c^2 + 2ab = a^2 + 2ab + b^2 \\[5 pt]
&c^2 = a^2 + b^2
\end{align}
}