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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 ...

Consider the family of ellipses (x^2)/(c^2)+(y^2)/((1-c)^2)-1=0 (1) for c in [0,1]. The partial derivative with respect to c is -(2x^2)/(c^3)+(2y^2)/((1-c)^3)=0 (2) ...

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 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) ...

The pentagonal antiprism is the antiprism having a regular pentagon for the top and bottom bases. It is also the uniform polyhedron with Maeder index 77 (Maeder 1997) and ...