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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 ...
Here the function is and therefore the three real roots are 2, −1 and −4. In algebra, a cubic equation in one variable is an equation of the form in which a is not zero. The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients a, b, c, and d of the cubic ...
Fermat–Catalan conjecture. In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c satisfy the equation an + bn = cn for any integer value of n greater than 2. The cases n = 1 and n = 2 have been known since antiquity to have infinitely many ...
t. e. The Millennium Prize Problems are seven well-known complex mathematical problems selected by the Clay Mathematics Institute in 2000. The Clay Institute has pledged a US $1 million prize for the first correct solution to each problem. The Clay Mathematics Institute officially designated the title Millennium Problem for the seven unsolved ...
The lower bound of multiplications needed is 2mn+2n−m−2 (multiplication of n×m-matrices with m×n-matrices using the substitution method, m⩾n⩾3), which means n=3 case requires at least 19 multiplications and n=4 at least 34. [38] For n=2 optimal 7 multiplications 15 additions are minimal, compared to only 4 additions for 8 multiplications.
The P versus NP problem is a major unsolved problem in theoretical computer science. Informally, it asks whether every problem whose solution can be quickly verified can also be quickly solved. Here, quickly means an algorithm that solves the task and runs in polynomial time exists, meaning the task completion time is bounded above by a ...
Solving an equation symbolically means that expressions can be used for representing the solutions. For example, the equation x + y = 2x – 1 is solved for the unknown x by the expression x = y + 1, because substituting y + 1 for x in the equation results in (y + 1) + y = 2 (y + 1) – 1, a true statement. It is also possible to take the ...
(¬x n−3 ∨ l n−2 ∨ x n−2) ∧ (¬x n−2 ∨ l n−1 ∨ l n) where x 2, ⋯ , x n−2 are fresh variables not occurring elsewhere. Although the two formulas are not logically equivalent, they are equisatisfiable. The formula resulting from transforming all clauses is at most 3 times as long as its original; that is, the length growth ...