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From this equation one gets the following properties of the evolute: At points with ′ = the evolute is not regular. That means: at points with maximal or minimal curvature (vertices of the given curve) the evolute has cusps. (See the diagrams of the evolutes of the parabola, the ellipse, the cycloid and the nephroid.)
The pedal equation can be found by eliminating x and y from these equations and the equation of the curve. The expression for p may be simplified if the equation of the curve is written in homogeneous coordinates by introducing a variable z, so that the equation of the curve is g(x, y, z) = 0. The value of p is then given by [2]
Finding a given Latin square's isomorphism class can be a difficult computational problem for squares of large order. To reduce the problem somewhat, a Latin square can always be put into a standard form known as a reduced square. A reduced square has its top row elements written in some natural order for the symbol set (for example, integers ...
If a tangent contains the point (x 0, y 0), off the parabola, then the equation = + = holds, which has two solutions m 1 and m 2 corresponding to the two tangents passing (x 0, y 0). The free term of a reduced quadratic equation is always the product of its solutions.
Abramowitz, Milton; Stegun, Irene A., eds. (1972). "Chapter 3". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables.
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
The square of an integer may also be called a square number or a perfect square. In algebra, the operation of squaring is often generalized to polynomials, other expressions, or values in systems of mathematical values other than the numbers. For instance, the square of the linear polynomial x + 1 is the quadratic polynomial (x + 1) 2 = x 2 ...
The variants described in this section are based on the formula = {() () /, /,. If one applies recursively this formula, by starting with y = 1, one gets eventually an exponent equal to 0, and the desired result is then the left factor.