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Since |(a+ib)(c+id)| = |a+ib||c+di| (1) |(ac-bd)+i(bc+ad)| = sqrt(a^2+b^2)sqrt(c^2+d^2), (2) it follows that (a^2+b^2)(c^2+d^2) = (ac-bd)^2+(bc+ad)^2 (3) = e^2+f^2. (4) This ...

Let a, b, and k be integers with k>=1. For j=0, 1, 2, let S_j=sum_(i=j (mod 3))(-1)^i(k; i)a^(k-i)b^i. Then 2(a^2+ab+b^2)^(2k)=(S_0-S_1)^4+(S_1-S_2)^4+(S_2-S_0)^4.

For even h, (1) (Nagell 1951, p. 176). Writing out symbolically, sum_(n=0)^h((-1)^nproduct_(k=0)^(n-1)(1-x^(h-k)))/(product_(k=1)^(n)(1-x^k))=product_(k=0)^(h/2-1)1-x^(2k+1), ...

A solution of a linear homogeneous ordinary differential equation with polynomial coefficients.

The identity (xy)x^2=x(yx^2) satisfied by elements x and y in a Jordan algebra.

The Lebesgue identity is the algebraic identity (Nagell 1951, pp. 194-195).

This is proven in Rademacher and Toeplitz (1957).

(b-c)/a = (sin[1/2(B-C)])/(cos(1/2A)) (1) (c-a)/b = (sin[1/2(C-A)])/(cos(1/2B)) (2) (a-b)/c = (sin[1/2(A-B)])/(cos(1/2C)). (3)

Let a triangle have side lengths a, b, and c with opposite angles A, B, and C. Then (b+c)/a = (cos[1/2(B-C)])/(sin(1/2A)) (1) (c+a)/b = (cos[1/2(C-A)])/(sin(1/2B)) (2) ...

A perfect cubic polynomial can be factored into a linear and a quadratic term, x^3+y^3 = (x+y)(x^2-xy+y^2) (1) x^3-y^3 = (x-y)(x^2+xy+y^2). (2)