# Bailey-Borwein-Plouffe Formular for Pi

With the BBP formular, we can find the 10th billion digit of Pi without calculting all the digits before it.

# e^pi vs. pi^e

Prove:

$\boxed{e^{\pi} > \pi^{e}}$

# Pi hiding in prime regularities

Three mysterious Math Objects:

• Pi,
• Complex Numbers,
• Prime Numbers

are hiding in circle.

5岁儿童背 Pi 600小数位:

# Pi in the Bible — 1 Kings 7:23

This is the first appearance of
$\boxed {\pi = 3 }$
in any human’s text in history.

45 feet / 15 feet = 3

# Transcendental Numbers: e, pi

The French Mathematician and Physicist Joseph Fourier proved
e is irrational,

Another French mathematician Charles Hermite went further: e belongs to another mathematical world:
e is transcendental.

Hermite’s German student Lindermann followed the same method, proved:
pi is also transcendental.

# Machin’s Pi

1706 John Machin developed a formula for calculating Pi to an arbitrary number of decimal places:

$\boxed{\displaystyle \pi = \text {4 [4.arccot(5) - arccot(239)]} }$

$\boxed {\displaystyle \text{arccot (x)} = \frac{1}{x} - \frac {1}{3{x^{3}}} + \frac {1}{5{x^{5}}}-\frac {1}{7{x^{7}}} + \cdots }$

# 100-digit Pi

One fine day when we reach above 80 years old, if the doctor accuses us of having dementia, then prove the doctor wrong by shocking him with 100-digit Pi memory 🙂

With Chinese single-syllable sound for numbers, better still if can sing it as a song, memorizing 100-digit pi is easy!

# Transcendental Number

Transcendental numbers: e, Π, L…

What about $e^{e}, \pi^{\pi} ,\pi^{e}$ ?

Aleksander (Alexis) Osipovich Gelfond (1906-68):

Gelfond-Schneider Theorem

$a^ b$ transcendental if
a is algebraic, not 0 or 1
b irrational algebraic number

Examples:
$\sqrt{6}^{\sqrt{5}}, 3^{\sqrt{7}}$
Hilbert Number: $2^{\sqrt {2}}$ (Hilbert Problem proven by Gelfond}

Is log 2 transcendental ?
[log = logarithm Base 10]

Proof:
$10^{log 2} = 2$

1) Sufficient to prove log 2 irrational
Assume log 2 rational
log 2= p/q, p and q integers
$10^ {log 2} = 2 = 10^ {p/q}$
raise power q
$2^{q} = 10^{p} = (2.5)^{p}$
$2^{q} = 2^{p}.5^{p}$

Case 1: p>q
$1 = 2^{p-q}.5^{p}$
=> False

Case 2: q>p
$2^{q-p}= 5^{p}$
Left is even : $2^{m} \text { = even}$
Right is odd: $5^{n} \text {= ....5}$
=> False

Therefore p,q do not exist,
=> log 2 irrational

Reference: Top 15 Transcendental Numbers: http://sprott.physics.wisc.edu/pickover/trans.html

# π in 1King 7:23

King Solomon – The Temple’s Furnishing:

He made the Sea of cast metal, circular in shape,
measuring 10 cubits from rim to rim and 5 cubits
high. It took a line of 30 cubits to measure around
it.” – 1King 7: 23

[10 cubits = 15 feet,  30 cubits = 45 feet]

So mathematically by today’s primary school geometry:

Circumference = 30 = 2 *π *r
Diameter = 10 = 2 * r

π = 30/10 = 3

Note: Solomon’s period is equivalent to China’s Zhou 周 dynasty of the famous Prince “Zhou Gong” 周公