Search results
Results from the WOW.Com Content Network
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 ...
The solution set of the equation x 2 / 4 + y 2 = 1 forms an ellipse when interpreted as a set of Cartesian coordinate pairs. Main article: Solution set The solution set of a given set of equations or inequalities is the set of all its solutions, a solution being a tuple of values, one for each unknown , that satisfies all the equations ...
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.
For example, the logarithm of 1000 to base 10 is 3, because 1000 is 10 to the 3 rd power: 1000 = 10 3 = 10 × 10 × 10. More generally, if x = b y, then y is the logarithm of x to base b, written log b x, so log 10 1000 = 3. As a single-variable function, the logarithm to base b is the inverse of exponentiation with base b.
Microsoft Math 2.0: Part of Microsoft Student 2007 Microsoft Math 3.0 : Standalone commercial product that requires product activation ; includes calculus support, digital ink recognition features and a special display mode for video projectors
The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
If only one root, say r 1, is real, then r 2 and r 3 are complex conjugates, which implies that r 2 – r 3 is a purely imaginary number, and thus that (r 2 – r 3) 2 is real and negative. On the other hand, r 1 – r 2 and r 1 – r 3 are complex conjugates, and their product is real and positive. [23]
Lucas numbers have L 1 = 1, L 2 = 3, and L n = L n−1 + L n−2. Primefree sequences use the Fibonacci recursion with other starting points to generate sequences in which all numbers are composite. Letting a number be a linear function (other than the sum) of the 2 preceding numbers.