Search results
Results from the WOW.Com Content Network
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.
For instance, the first counterexample must be odd because f(2n) = n, smaller than 2n; and it must be 3 mod 4 because f 2 (4n + 1) = 3n + 1, smaller than 4n + 1. For each starting value a which is not a counterexample to the Collatz conjecture, there is a k for which such an inequality holds, so checking the Collatz conjecture for one starting ...
Quadratic formula. The roots of the quadratic function y = 1 2 x2 − 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.
The first four partial sums of the series 1 + 2 + 3 + 4 + ⋯.The parabola is their smoothed asymptote; its y-intercept is −1/8, and the area of the parabola ...
Knuth's up-arrow notation. In mathematics, Knuth's up-arrow notation is a method of notation for very large integers, introduced by Donald Knuth in 1976. [1] In his 1947 paper, [2] R. L. Goodstein introduced the specific sequence of operations that are now called hyperoperations. Goodstein also suggested the Greek names tetration, pentation ...
Happily the context of 51 and 52, together with the base, mid-line, and smaller triangle area (which are given as 4 + 1/2, 2 + 1/4 and 7 + 1/2 + 1/4 + 1/8, respectively) make it possible to interpret the problem and its solution as has been done here. The given paraphrase therefore represents a consistent best guess as to the problem's intent ...
where a = 5(4ν + 3) / ν 2 + 1 . Using the negative case of the square root yields, after scaling variables, the first parametrization while the positive case gives the second. The substitution c = −m / l 5 , e = 1 / l in the Spearman-Williams parameterization allows one to not exclude the special case a = 0, giving the ...
Continued fraction. A finite regular continued fraction, where is a non-negative integer, is an integer, and is a positive integer, for . In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this ...