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If x takes only nonnegative values, then P(x>=a)<=(<x>)/a. (1) To prove the theorem, write <x> = int_0^inftyxP(x)dx (2) = int_0^axP(x)dx+int_a^inftyxP(x)dx. (3) Since P(x) is ...

f(x) approx t_n(x)=sum_(k=0)^(2n)f_kzeta_k(x), where t_n(x) is a trigonometric polynomial of degree n such that t_n(x_k)=f_k for k=0, ..., 2n, and ...

The zeros of the derivative P^'(z) of a polynomial P(z) that are not multiple zeros of P(z) are the positions of equilibrium in the field of force due to unit particles ...

Let V(r) be the volume of a ball of radius r in a complete n-dimensional Riemannian manifold with Ricci curvature tensor >=(n-1)kappa. Then V(r)<=V_kappa(r), where V_kappa is ...

Let {a_n} be a nonnegative sequence and f(x) a nonnegative integrable function. Define A_n = sum_(k=1)^(n)a_k (1) B_n = sum_(k=n)^(infty)a_k (2) and F(x) = int_0^xf(t)dt (3) ...

This is sometimes knows as the "bars and stars" method. Suppose a recipe called for 5 pinches of spice, out of 9 spices. Each possibility is an arrangement of 5 spices ...

For a positive integer n, (2pi)^((n-1)/2)n^(1/2-nz)Gamma(nz)=product_(k=0)^(n-1)Gamma(z+k/n),

At rational arguments p/q, the digamma function psi_0(p/q) is given by psi_0(p/q)=-gamma-ln(2q)-1/2picot(p/qpi) +2sum_(k=1)^([q/2]-1)cos((2pipk)/q)ln[sin((pik)/q)] (1) for ...

An affine isoperimetric inequality.

If 0<=a,b,c,d<=1, then (1-a)(1-b)(1-c)(1-d)+a+b+c+d>=1. This is a special case of the general inequality product_(i=1)^n(1-a_i)+sum_(i=1)^na_i>=1 for 0<=a_1,a_2,...,a_n<=1. ...