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.
2024年3月29日金曜日
2024年2月1日木曜日
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.
n is vertex. This contains 31 vertices. This has 32291925 vertices.
This is called The Hadwiger-Nelson Problem.
G(n,p)
n is vertex. This contains 31 vertices. This has 32291925 vertices.
This is called The Hadwiger-Nelson Problem.
2024年1月27日土曜日
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.
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 ∴
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 ∴
2023年6月13日火曜日
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.
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.
∴
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.
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)<∞
∴
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.
2023年6月6日火曜日
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).
I tossed coins 100 times.
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.
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.
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.
2023年6月5日月曜日
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
2023年6月2日金曜日
Gaussian function
This is an even function which is symmetry by the Y-axis.
∴
This is the inflection point. You still see the square.
f(0)=1
The average is the top which is called normal distribution. μ is the average, and σ^2>0. R=(-∞,∞)
登録:
投稿 (Atom)