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Since x 2 represents the area of a square with side of length x, and bx represents the area of a rectangle with sides b and x, the process of completing the square can be viewed as visual manipulation of rectangles. Simple attempts to combine the x 2 and the bx rectangles into a larger square
If two primes which end in 3 or 7 and surpass by 3 a multiple of 4 are multiplied, then their product will be composed of a square and the quintuple of another square. In other words, if p, q are of the form 20k + 3 or 20k + 7, then pq = x 2 + 5y 2. Euler later extended this to the conjecture that
This corresponds to a set of y values whose product is a square number, i.e. one whose factorization has only even exponents. The products of x and y values together form a congruence of squares. This is a classic system of linear equations problem, and can be efficiently solved using Gaussian elimination as soon as the number of rows exceeds ...
As the integrand is the third-degree polynomial y(x) = 7x 3 – 8x 2 – 3x + 3, the 2-point Gaussian quadrature rule even returns an exact result. In numerical analysis , an n -point Gaussian quadrature rule , named after Carl Friedrich Gauss , [ 1 ] is a quadrature rule constructed to yield an exact result for polynomials of degree 2 n − 1 ...
The inscribed square problem, also known as the square peg problem or the Toeplitz' conjecture, is an unsolved question in geometry: Does every plane simple closed curve contain all four vertices of some square? This is true if the curve is convex or piecewise smooth and in other special cases. The problem was proposed by Otto Toeplitz in 1911. [1]
An equivalent definition is to say that the square of the function itself (rather than of its absolute value) is Lebesgue integrable.For this to be true, the integrals of the positive and negative portions of the real part must both be finite, as well as those for the imaginary part.
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent.
Denoting the two roots by r 1 and r 2 we distinguish three cases. If the discriminant is zero the fraction converges to the single root of multiplicity two. If the discriminant is not zero, and |r 1 | ≠ |r 2 |, the continued fraction converges to the root of maximum modulus (i.e., to the root with the greater absolute value).