Distribution Functions
* Standard Normal Distribution: 0∫∞exp(-x2/2)dx=Γ(1/2). Take z=(x2/2); dz=xdx or dx=(1/x) dz =(1/√2)z-1/2dz 0∫∞exp(-x2/2)dx=(1/√2) z-1/2*e-zdz=(1/√2)Γ(1/2)=√(π/2) -∞∫∞exp(-ax2/2 + bx)dx=√(2π/a)*exp(b2/2a) ; -∞∫∞exp-(ax2/2)dx=√(2π/a) 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
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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-λ | ||||||||||||
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 |
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We define xi function as ξ(n) =(1/2)n*n-1*π-n/2 *Γ(n/2)*ζ(n) | ||||||||||||
0∫∞f4(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 | ||||||||||||
0∫∞xn-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/kBT +1)] dx=(7π4/ 120)*(kBT)4 |
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Spectral energy density =(8πν2 /c3)*hν /(ehν/kT - 1) When hν << kT , then =1+hν /kT +... - 1=hν /kT and spectral energy density=(8πν2 /c3)*kT | ||||||||||||