Search Results for ""

The crossed trough is the surface z=x^2y^2. (1) The coefficients of its first fundamental form are E = 1+4x^2y^4 (2) F = 4x^3y^3 (3) G = 1+4x^4y^2 (4) and of the second ...

sum_(n=0)^(N-1)e^(inx) = (1-e^(iNx))/(1-e^(ix)) (1) = (-e^(iNx/2)(e^(-iNx/2)-e^(iNx/2)))/(-e^(ix/2)(e^(-ix/2)-e^(ix/2))) (2) = (sin(1/2Nx))/(sin(1/2x))e^(ix(N-1)/2), (3) ...

Given two circles, draw the tangents from the center of each circle to the sides of the other. Then the line segments AB and CD are of equal length. The theorem can be proved ...

Gieseking's constant is defined by G = int_0^(2pi/3)ln(2cos(1/2x))dx (1) = Cl_2(1/3pi) (2) = (3sqrt(3))/4[1-sum_(k=0)^(infty)1/((3k+2)^2)+sum_(k=1)^(infty)1/((3k+1)^2)] (3) = ...

A group representation of a group G on a vector space V can be restricted to a subgroup H. For example, the symmetric group on three letters has a representation phi on R^2 ...

A 4-hyperboloid has negative curvature, with R^2=x^2+y^2+z^2-w^2 (1) 2x(dx)/(dw)+2y(dy)/(dw)+2z(dz)/(dw)-2w=0. (2) Since r=xx^^+yy^^+zz^^, (3) it follows that ...

Given the binary quadratic form ax^2+2bxy+cy^2 (1) with polynomial discriminant b^2-ac, let x = pX+qY (2) y = rX+sY. (3) Then a(pX+qY)^2+2b(pX+qY)(rX+sY)+c(rX+sY)^2 ...

To inscribe an equilateral triangle in an ellipse, place the top polygon vertex at (0,b), then solve to find the (x,y) coordinate of the other two vertices. ...

Let a triangle have angles A, B, and C, then inequalities include sinA+sinB+sinC<=3/2sqrt(3) (1) 1<=cosA+cosB+cosC<=3/2 (2) sin(1/2A)sin(1/2B)sin(1/2C)<=1/8 (3) ...

Values of the trigonometric functions can be expressed exactly for integer multiples of pi/20. For cosx, cos(pi/(20)) = 1/4sqrt(8+2sqrt(10+2sqrt(5))) (1) cos((3pi)/(20)) = ...