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Although the inner shaded region has the same area as the outer shaded annulus, it appears to be larger. Since the rings are equally spaced, A_(inner) = pi·3^2=9pi (1) ...

A circle bundle pi:E->M is a fiber bundle whose fibers pi^(-1)(x) are circles. It may also have the structure of a principal bundle if there is an action of SO(2) that ...

Laplace's integral is one of the following integral representations of the Legendre polynomial P_n(x), P_n(x) = 1/piint_0^pi(du)/((x+sqrt(x^2-1)cosu)^(n+1))du (1) = ...

Niven's theorem states that if x/pi and sinx are both rational, then the sine takes values 0, +/-1/2, and +/-1. Particular cases include sin(pi) = 0 (1) sin(pi/2) = 1 (2) ...

A constant appearing in formulas for the efficiency of the Euclidean algorithm, B = (12ln2)/(pi^2)[-1/2+6/(pi^2)zeta^'(2)]+C-1/2 (1) = 0.06535142... (2) (OEIS A143304), where ...

The second de Villiers point is the perspector of the reference triangle and the excenter analog of the BCI triangle, which is Kimberling center X_(1128) has triangle center ...

An n-gonal cupola Q_n is a polyhedron having n obliquely oriented triangular and n rectangular faces separating an {n} and a {2n} regular polygon, each oriented horizontally. ...

A test for the convergence of Fourier series. Let phi_x(t)=f(x+t)+f(x-t)-2f(x), then if int_0^pi(|phi_x(t)|dt)/t is finite, the Fourier series converges to f(x) at x.

int_0^pi(sin[(n+1/2)x])/(2sin(1/2x))dx=1/2pi, where the integral kernel is the Dirichlet kernel.

The equations are x = 2/(sqrt(pi(4+pi)))(lambda-lambda_0)(1+costheta) (1) y = 2sqrt(pi/(4+pi))sintheta, (2) where theta is the solution to ...