Distribution Functions

 n= m= x= x-n x1=(x-n)/m x2=(x+n)/m f1(x)=(1/x2)*e-1/x f2(x)=xe-x / (1+x)2 (Gama function)f3(x)=xn-1e-x (Riemann-Zeta function)f4(x)=xn-1/(ex-1) f5a(x1)=1/(ex1  +1) f5b(x2)=1/(ex2  +1) f5(x)=f5a(x1) - f5b(x2) Gradshteyn and Ryzhik function  :f6(x)=ln(1+2cosm*e-x +e-2x) f7(x)=lnx/(xn-1) f8(x)=xn/(ex+1) f9(x)=xnex / (ex-1)2 f10(x)=xne-mx f1(x) max at x=1/2........f1(x) max=4/e2 f1(x) min at x=infinity........f1(x) min=0 area=(the function has singularity at x=0)......x ≠0∫∞f(x)dx=1-e-x z2=(√2-1) z2a=(√2+1) f2(x) max at x=-1+√2........f2(x) max=(1/2)*z2/ez2 f2(x) min at x=-1-√2........f2(x) min=(-1/2)*z2a*ez2a point of inflexion x=0 and at x=infinity,  f2(x)=0 ∫f2(x)dx = ∫e-x(1/[1+x] -1/[1+x]2)dx  =∫f2a(x)dx + ∫f'2a(x)dx =f2a(x)|=e-x(1/[1+x])| (infinity 2 zero) +e-x(1/[1+x])| (-infinity 2 zero) =1-1 (each area is 1 in positive and negative segment) Area on positive segment from 0 to x is e-x(1/[1+x]) - 1= f3(x) max at x=(n-1),f3(x) max=(n-1)n-1 e-(n-1) f3(x) min at x=0 and x=infinity 0∫∞f3(x,n)dx =Γ(n)=(n-1)! f4a=xn-1 f4c=xex = f4b=ex-1= n-1= xex /(ex-1) = f4(x) max at xex / (ex-1)=(n-1) f4(x) min at x=0 and x=infinity 0∫∞f4(x,n)dx =Γ(n)*ζ(n) 0∫∞f8(x,n)dx =n!*ζ(n+1)(1-2-n) for Re(n) > 0 0∫∞f9(x,n)dx =n!*ζ(n) 0∫∞f10(x,n,m)dx =ζ(n+1) / mn+1 zeta distribution is a discrete probability distribution fn (x). The probability that x  takes an  integer value is given by the probability mass function (pmf). n is to be a positive integer. Riemann -zeta function is undefined for n=1. Here n=1 to infinity and x=1,2,3..........Mean of the distribution = ζ(n-1) /  ζ(n) for n >2 fn (x) = x-n/ζ(n) ζ(0)=-1/2 ζ(1/2)= ζ(2)=π2/6 (Basel Problem) ζ(3)= ζ(4)=π4/90= 2π4/5!+ζ(2)π2/3! ζ(6)=π6/945=3π6/7!+ζ(2)π2/5!+ζ(4)π2/3! ζ(8)=π8/9450 gammatable Γ(1)=0! Γ(2)=1! Γ(3)=2! Γ(4)=3! Γ(1/2)=√π Γ(-1/2)=-2√π Γ(3/2)=√π/2 Γ(5/2)=(3/4)√π Γ(7/2)=(15/8)√π

* Standard Normal Distribution:

0exp(-x2/2)dx=Γ(1/2). Take z=(x2/2);  dz=xdx  or dx=(1/x) dz =(1/√2)z-1/2dz

0exp(-x2/2)dx=(1/√2) z-1/2*e-zdz=(1/√2)Γ(1/2)=√(π/2)

Characteristic of f4(x) function :

* The extremum value of f4 is when f '4  =0 , The condition is  xex /(ex-1) = n-1 =k(say) or ex(k-x)=k .

(a) when x=k, RHS =k=0 and x=0]

(b) when x=∞, k=∞ and n=∞

(c) when x, k ≠ 0 , the value of x to be found out by trial and error method.

(d) Except the extremum value of f4(x) , when 0<x < k , ex(k-x) >0. When x > K, ex(k-x) < 0

(e) If f4(x) has extremum value at x=1, then e(k-1)=k or n=(2e-1)/(e-1) and n is not an integer.

(f) We can also prefix the value of k as integer, and by trial and error determine value of x satisfying the equation xex /(ex-1) = n-1 =k
 value of k value of x at which f4(x) is extremum value of f4(x) 1 0.000000001399 1.000000013063 2 1.59362427 0.64761024 3 2.82143938 1.42143547

Poisson Distribution : It is a discrete distribution and is a limiting case of binomial distribution when number of trials are indefinitely large. Here, time interval is fixed. Mean is fixed and is equal to variance. If λ=mean=variance, x is the no. of occurrences , P(x) = f(x) =     (λx / x!)e
Example: In a cafe, the customer arrives at a mean rate of 2/min. Find the probability of arrival of 5 customers/minute  using Poisson distribution. λ=2, x=5 . If mean is large, Poisson distribution approaches Normal distribution,
Zipf's Law: Y =KXα where X is independent variable , K is a constant and α is also a constant known as law's exponent/ shape parameter and is > 1, =0 or >0. The law states that given a list of most frequent words in an arbitrary book, the most frequent word will appear twice as of the second most frequent word and the second most frequent word will appear twice as the 3rd most frequent word. This is same as the Power Law. If we take K=(i=1 to n)Σ(1/i)n , and Y=f(x)= 1/KXα  ,where  α  > 1, =0 or >0 and this is zeta distribution.
In ζ(n) , normally n is a complex number and n= σ+ it. fσ(t)= ζ(n) / ζ(σ) and σ >1 .

which is absolutely convergent for all complex s with real part greater than one. One of the first properties of this is that, as shown by Riemann, it extends to an analytic function on the entire complex plane, other than a simple pole at x=1. By the theory of analytic continuation this extension is necessarily unique, so the importance of the result lies in showing that an extension exists. One way of doing this is to find an alternative expression for the zeta function which is well defined everywhere. For example, it can be expressed as an absolutely convergent integral, as performed by Riemann himself in his original 1859 paper on the subject. This leads to an explicit expression for the zeta function, scaled by an analytic prefactor, as the integral of x n multiplied by a function of x over the range x>0. In fact, this can be done in a way such that the function of x is a probability density function, and hence expresses the Riemann zeta function over the entire complex plane in terms of the generating function ${{\mathbb E}[X^s]}$ of a positive random variable X. The probability distributions involved here are not the standard ones taught to students of probability theory, so may be new to many people. Although these distributions are intimately related to the Riemann zeta function they also, intriguingly, turn up in seemingly unrelated contexts involving Brownian motion.

We define xi function as ξ(n) =(1/2)n*n-1*π-n/2 *Γ(n/2)*ζ(n)
0f4(x)dx=0 xn-1/(Aex-1)dx where A=1. In fact, 1/(ex-1)dx is the Bose-Einstein distribution function where A=1 refers to spin 1 particles, photons. n=4 gives rise to Stefan - Boltzman distribution. u=total energy radiated by a black
0xn-1e-ax*x =(1/2)a-n/2  * Γ(n/2)  where a >0
0 xn/(ex+1) dx  gives rise to Fermi distribution in the early history of the universe when n=3 and x is replaced by x/kBT . There is a multiplicative constant 4π/h3 before the integral.

Hence   0 [x3/(ex/kB+1)] dx=(7π4/ 120)*(kBT)4

Spectral energy density  =(8πν2 /c3)*hν /(ehν/kT  - 1) When << kT , then =1+hν /kT +... - 1=hν /kT and spectral energy density=(8πν2 /c3)*kT