Search results
Results from the WOW.Com Content Network
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.
It is unknown whether these constants are transcendental in general, but Γ( 1 / 3 ) and Γ( 1 / 4 ) were shown to be transcendental by G. V. Chudnovsky. Γ( 1 / 4 ) / 4 √ π has also long been known to be transcendental, and Yuri Nesterenko proved in 1996 that Γ( 1 / 4 ), π, and e π are algebraically independent.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
As another example, in radix 5, a string of digits such as 132 denotes the (decimal) number 1 × 5 2 + 3 × 5 1 + 2 × 5 0 = 42. This representation is unique. Let b be a positive integer greater than 1. Then every positive integer a can be expressed uniquely in the form
The real numbers can be constructed as a completion of the rational numbers, in such a way that a sequence defined by a decimal or binary expansion like (3; 3.1; 3.14; 3.141; 3.1415; ...) converges to a unique real number—in this case π. For details and other constructions of real numbers, see Construction of the real numbers.
Trigonometric identities may help simplify the answer. [1] [2] Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
Today, a more standard phrasing of Archimedes' proposition is that the partial sums of the series 1 + 1 / 4 + 1 / 16 + ⋯ are: + + + + = +. This form can be proved by multiplying both sides by 1 − 1 / 4 and observing that all but the first and the last of the terms on the left-hand side of the equation cancel in pairs.