Pólya conjecture

10=2*5 There are 2 prime numbers which are called prime factorization.

This is n=50.

4=2^2 is 2 prime numbers.

12=2^2*3 is 3 prime numbers.

I put even numbers in 1. There are 21 in 50. According to Pólya conjecture, there are more odd numbers in prime factorization.

Prime numbers are chaotic, so this conjecture isn't fractal like piling squares.

906,150,257 ≤ n ≤ 906,488,079

There are more even numbers in n.

It is hard to put circles in the square. I think that you need to adjust the range of the square to extinguish the circle, such as 906000000≦n≦906500000.

For example, there are 7 even numbers in 4≦n≦16. This is contradictory.

Wolstenholme's theorem

p is a prime number (p≧5).

When p is 5, p^3=125.

2376 is one more than a multiple of 125. 2376=125*19+1

When p is 7, p^3=343.

1716 is one more than a multiple of 343.

This is chaotic, but you can find it by your spreadsheet.

Moreover, this is Binomial coefficient. Pascal's triangle is well known as fractal.

This is harmonic numbers.

You see that this is almost zeta function.

Riemann's zeta function is ζ(s)=0.

Random graph

You are a dot in social media.

You connect with blue and yellow, although blue and yellow may be also connected. You don't care.

You are isolated. P=0.

This is perfect. P=1.

You see binomial distribution.

G(n,p)

n is vertex.

This contains 31 vertices.

This has 32291925 vertices.

This is called The Hadwiger-Nelson Problem.

Poisson distribution

This is e^-x.

You may have 10 heads in each 10 times.

P(X=10)=9.765625E-4≒9/10000

This is about 0.09%, and it is almost Zero.

This is Poisson distribution.

p = λ / n

This is called Poisson limit theorem, and binomial distribution is Bernoulli trial.

x=1000 heads, n=1000 times, p=0.5(heads or tails), f=9.33E-302

∴

Hoeffding’s inequality

8/10 is 8 heads in each 10 times. This is qite rare.

80% of the heads are in 99.9%. The red is the area.

t≧0

In this case, t is 0.8, and -t is 0.2. They are symmetry. 1/100 is the possibilty to have 80% of the head. This is very difficlt like finding rich men.

Var(Z)＜∞

∴

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You may know Bernoulli trial.

n=10, and k=8.

P(X=8)=0.044≒4/100.

Therefore, you may have 1～4 of (8/10) heads in each 100 times.You can't predict anything because of Zero.

Central limit theorem

Whenever you toss your coins, you are close to normal distribution.

You can see Gaussian function.

μ is the average, and σ^2 is the distribution.

Y is E(X).

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I tossed coins 100 times.

NORMDIST(x,0.5,(x-0.5)^2,0)

You may have 80% of the head once (1.7%). When you are around Y which is the average, it is almost Zero. You need to go far beyond that. It is quite rare.0.5 is null, and 0.8 and 0.2 are symmetry.

NORMDIST(0.8,0.5,(0.8-0.5)^2,1)≒0.999

X=0.6 is 100%

The average is normal and Zero, but 9/10 and 10/10 is the difficulty. It is narrower and close to Zero. You may talk about singularity.

Markov’s Inequality

It is hard to see rich men.

This is called Markov’s Inequality. E[X] is the average, although I don't reach it. X is a discrete random variable.

∴

a＞0