The partition function or configuration integral, as used in probability theory, information theory and dynamical systems, is a generalization of the definition of a partition function in statistical mechanics. It is a special case of a normalizing constant in probability theory, for the Boltzmann distribution. The partition function occurs in many problems of probability theory because, in situations where there is a natural symmetry, its associated probability measure, the Gibbs measure, has the Markov property. This means that the partition function occurs not only in physical systems w...
@Shisui Well, "just saying it" is not enough. You need to show it. You have $$(1+x)^n(1+x)^m = \left(\sum_{k=0}^n \binom{n}{k}x^k\right)\left(\sum_{l=0}^m\binom{m}{l}x^l\right) = \sum_{k=0}^n\sum_{l=0}^m \binom{n}{k}\binom{m}{l}x^{k+l}.$$ Group according to constant $k+l$.
@salbeira Do you use a comma to separate the integer part from the fractional part or are you referring to a number like 1,999 where the comma is used?
All real numbers can be written in the form n + r where n is an integer (the integer part) and the remaining fractional part r is a nonnegative real number less than one. For a positive number written in decimal notation, the fractional part corresponds to the digits appearing after the decimal point.
The fractional part of a real number x is x-\lfloor x\rfloor, where \lfloor\;\rfloor is the floor function. It is sometimes denoted \{x\}, \langle x \rangle or x\,\bmod\,1.
If x is rational, then the fractional part of x can be expressed in the form p / q, where p and q are integers and 0 ...
I realize this is a novice question, but I'm stuck here. Suppose you have an ordered basis $B = (v_1, v_2, v_3, v_4)$ of $R^4$. My problem is how to determine a system of equations that defines, relatively to the basis $B$, the linear subspace of $R^4$ generated by the vectors $v_1-v_2+v_4$ and $...
In number theory, natural density (or asymptotic density or arithmetic density) is one of the possibilities to measure how large a subset of the set of natural numbers is.
Intuitively, it is thought that there are more positive integers than perfect squares, since every perfect square is already positive, and many other positive integers exist besides. However, the set of positive integers is not in fact larger than the set of perfect squares: both sets are infinite and countable and can therefore be put in one-to-one correspondence. Nevertheless if one goes through the natural numbers, t...
Let $S$ be the set of all rational numbers which are squares $x = p^2/q^2$ for some integers $p$ and $q$. How can I show that $S$ is dense in the set of non-negative real numbers?
@skullpatrol Yes, the word density means different things. Some of the questions tagged density were about probability density functions. And those about topological density did not have the general-topology tag. Mo' tags, mo' tagging problems.