Discoveries of Steven Zheng

**By: Tao Steven Zheng (郑涛)**

 This is a log of my mathematical discoveries during my last two years of high school. I was fascinated by infinite series and integral calculus at the time, which led to my obsession to play around with such concepts and computations.

Grade 11 (September 2010 - June 2011) was my most productive year for mathematical research. I focused more on studying university level math and physics in grade 12.

## **December 27, 2010**

From my investigations of the Collatz Conjecture (December 23, 2010 - April 5, 2012), I defined two concepts called the **Attractor Numbers** and **Continuous Rays**. From these concepts I derived the modular congruence

2
i

≡
4

(
mod

6
)
.

{\displaystyle 2^i \equiv 4\pmod 6. }

I solved this modular congruence and found that the solution is

2

2
n
2

=
6

(

∑

# k
0

n
−
1

2

2
k
1

)

4

{\displaystyle {2}^{2n+2} = 6\left(\sum\_{k=0}^{n-1} {2}^{2k+1} \right) +4 }

for integers

n
≥
1

{\displaystyle n \geq 1}

.

## **January 25, 2011**

I found a simple approximation for calculating the zeta function. By observation, I found

ζ
(
n
# )

2
n

2
n

−
1

(

∑

# k
0

∞

1

(
2
k
1
)

n

)

.

{\displaystyle \zeta(n) = \frac{2^n}{2^n - 1}\left(\sum\_{k=0}^{\infty} \frac{1}{{(2k+1)}^n} \right). }

This led to the approximation

ζ
(
n
)
≈

2
n

2
n

−
1

.

{\displaystyle \zeta(n) \approx \frac{2^n}{2^n - 1}. }

## **February 17, 2011**

I studied more combinatorics around this time, and I came across the notion of counting multisets. From my investigations, I found this finite series

∑

# k
1

n

(

k
p
−
1

p

)

=

(

n
p

p
1

)

.

{\displaystyle \sum \_{ k=1 }^{ n }{ \left( \begin{matrix} k+p-1 \\ p \end{matrix} \right) } =\left( \begin{matrix} n+p \\ p+1 \end{matrix} \right). }

## **March 19, 2011 - May 25, 2011**

On March 19, 2011 I made an accidental discovery from fiddling around with the gamma function on Graphmatica, I noticed that

e
a

∫

a

∞

e

−
t

t
n

d
# t
i
n
t
e
g
e
r

{\displaystyle e^a \int\_{a}^{\infty} e^{-t} t^n dt = integer }

for integers

a

{\displaystyle a}

 and non-negative integers

n

{\displaystyle n}

. I later noticed that when

# n
1

{\displaystyle n=1}

, the integral

e

∫

1

∞

e

−
t

t

n
−
2

d
t

{\displaystyle e \int\_{1}^{\infty}e^{-t}t^{n-2}dt }

generates the solutions the the combinatorial problem on the number of paths between two points in a polygonal network. I posed this problem in 2010 but solved it in a different way.

This led me to define something called **permutation numbers**, where

P
a

(
n
# )

e
a

∫

a

∞

e

−
t

t
n

d
t
.

{\displaystyle P\_a(n) = e^a \int\_{a}^{\infty} e^{-t} t^n dt. }

Over the course of one month I found three representations of my **permutation numbers**. I proved that the integral equivalent to the finite series

P
a

(
n
# )
n
!

(

∑

# k
0

n

a
k

k
!

)

.

{\displaystyle P\_a (n) = n!\left(\sum\_{k=0}^{n} \frac{a^k}{k!} \right).}

This led to the approximation

P
a

(
n
)
≈
n
!

e
a

{\displaystyle P\_a (n) \approx n!e^a }

, which is good for

# a
−
1
,
0
,
1

{\displaystyle a = -1, 0, 1 }

. It is exact for

# a
0

{\displaystyle a = 0}

.

The last representation I found is a recurrence-relation:

P
a

(
n
# )
n
⋅

P
a

(
n
−
1
)

a
n

{\displaystyle P\_a (n) = n\cdot P\_a (n-1) + a^n }

where

P
a

(
0
# )
1

{\displaystyle P\_a (0) = 1 }

.

## **June 1, 2011 - June 17, 2011**

Starting with the bound

2
π
n

(

n
e

)

n

(

1

1

12
n

)

<
n
!
<

2
π
n

(

n
e

)

n

e

1

12
n

{\displaystyle \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^{n} \left(1+\frac{1}{12n}\right) < n! < \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^{n} e^{\frac{1}{12n}} }

, I made a series of observations and found that the bound can be refined to

2
π
n

(

n
e

)

n

(

1

1

12
n

)

<
n
!
<

2
π
n

(

e

−
1

(

n

1

12
n

)

)

n

.

{\displaystyle \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^{n} \left(1+\frac{1}{12n}\right) < n! < \sqrt{2\pi n}{\left(e^{-1}\left( n+\frac{1}{12n}\right) \right)}^{n}. }

## **December 19, 2011**

I drew this diagram using equations on Graphmatica in early December and painted the circles with various colours.

The radius of the blue circular quadrants is one. If the above pattern continues indefinitely, find the formula for calculating the radii of the nested circles (the nested circles do not include the two blue quadrants).

The total area of the nested circles is

# A

8

π

5

30

π

3

−
1035
π

360

≈
0.35221495

{\displaystyle A = \frac{8{\pi}^5 + 30{\pi}^3 -1035\pi}{360} \approx 0.35221495 }

.

Since the area between the quadrants is

2
−

π
2

≈
0.429203673

{\displaystyle 2 - \frac{\pi}{2} \approx 0.429203673 }

, the nested circles occupy 82.06% of this space.