Search Results for ""
4041 - 4050 of 5010 for Circle Passing Through Two PointsSearch Results
de Rham cohomology is a formal set-up for the analytic problem: If you have a differential k-form omega on a manifold M, is it the exterior derivative of another differential ...
A k-subset is a subset of a set on n elements containing exactly k elements. The number of k-subsets on n elements is therefore given by the binomial coefficient (n; k). For ...
Given a number n, Fermat's factorization methods look for integers x and y such that n=x^2-y^2. Then n=(x-y)(x+y) (1) and n is factored. A modified form of this observation ...
According to Pólya, the Cartesian pattern is the resolution method for arithmetical or geometrical problems based on equations. The first step is to translate the question ...
Every "large" even number may be written as 2n=p+m where p is a prime and m in P union P_2 is the set of primes P and semiprimes P_2.
The connected sum M_1#M_2 of n-manifolds M_1 and M_2 is formed by deleting the interiors of n-balls B_i^n in M_i^n and attaching the resulting punctured manifolds M_i-B^._i ...
The Fermat quotient for a number a and a prime base p is defined as q_p(a)=(a^(p-1)-1)/p. (1) If pab, then q_p(ab) = q_p(a)+q_p(b) (2) q_p(p+/-1) = ∓1 (3) (mod p), where the ...
Extend Hilbert's inequality by letting p,q>1 and 1/p+1/q>=1, (1) so that 0<lambda=2-1/p-1/q<=1. (2) Levin (1937) and Stečkin (1949) showed that (3) and ...
The roots of a semisimple Lie algebra g are the Lie algebra weights occurring in its adjoint representation. The set of roots form the root system, and are completely ...
A variable x is memoryless with respect to t if, for all s with t!=0, P(x>s+t|x>t)=P(x>s). (1) Equivalently, (P(x>s+t,x>t))/(P(x>t)) = P(x>s) (2) P(x>s+t) = P(x>s)P(x>t). (3) ...
...
View search results from all Wolfram sites (69290 matches)

