# Maths is a potential life changer — from wages to dating

https://www.ft.com/content/f039c3ca-d762-11e8-aa22-36538487e3d0

When you are stuck with too many choices in life, whether buying houses recommended by agents, finding schools for children, hiring staff among the hundreds of CVs…apply ‘e’ the number from nature (logarithm) to make smart and efficient decision under the constraints of time and resources.

Theory of Optimal Stopping” = 1/e ~ 37%

where

e = 2.7 18281828 45 90 45…

If you do a house search with 20 properties, then by the Theory of Optimal Stopping, pick the first property which is better than the first 7 properties you see (7. 4 = 20 x 37%).

[Note] Ponder over the hidden Philosophy behind – “Brilliant Limit” : $\displaystyle \boxed{ \lim_{n \to \infty}\left({\frac{n!}{n^n}}\right)^{\frac{1} {n}} = \frac{1} {e}}$

# e^pi vs. pi^e

Prove: $\boxed{e^{\pi} > \pi^{e}}$ # 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.

# Draw e^ax

My nephew gets confused in drawing exponential curves with many variations: $\boxed {y = e^{ax} }$ If $-e^{ax}$, then reverse the y values, that is flip the curve over x-axis.

Note: $\frac {dy}{dx} e^{x} = e^{x}$ $\int e^{x} dx = e^{x} + c$

# 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

# Irrational ‘e’

In secondary school, we know how to prove √2 is irrational, how about e ?

e= 1 + 1/1!  +1/2!  + 1/3!  + 1/4!  +…

Assume e= p/q as rational
multiply both sides by q!
LHS: e. q!= (p/q) .q! = p.(q-1)!  => integer

RHS:  q!+q! + (3.4…q)+ (4.5…q) +…1 + 1/(q+1) +…. => fraction