Search results
Results from the WOW.Com Content Network
An n-variable instance of 3-SAT can be realized as a positivity problem on a polynomial with n variables and d=4. This proves that positivity testing is NP-Hard . More precisely, assuming the exponential time hypothesis to be true, v ( n , d ) = 2 Ω ( n ) {\displaystyle v(n,d)=2^{\Omega (n)}} .
Problems 1, 2, 5, 6, [a] 9, 11, 12, 15, and 22 have solutions that have partial acceptance, but there exists some controversy as to whether they resolve the problems. That leaves 8 (the Riemann hypothesis), 13 and 16 [b] unresolved. Problems 4 and 23 are considered as too vague to ever be described as solved; the withdrawn 24 would also be in ...
17 indivisible camels. The 17-animal inheritance puzzle is a mathematical puzzle involving unequal but fair allocation of indivisible goods, usually stated in terms of inheritance of a number of large animals (17 camels, 17 horses, 17 elephants, etc.) which must be divided in some stated proportion among a number of beneficiaries.
Semi-log plot of solutions of + + = for integer , , and , and .Green bands denote values of proven not to have a solution.. In the mathematics of sums of powers, it is an open problem to characterize the numbers that can be expressed as a sum of three cubes of integers, allowing both positive and negative cubes in the sum.
Many mathematical problems have been stated but not yet solved. These problems come from many areas of mathematics, such as theoretical physics, computer science, algebra, analysis, combinatorics, algebraic, differential, discrete and Euclidean geometries, graph theory, group theory, model theory, number theory, set theory, Ramsey theory, dynamical systems, and partial differential equations.
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 ...
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 ]
[4] [5] The Babylonians could have used the tables to solve cubic equations, but no evidence exists to confirm that they did. [6] The problem of doubling the cube involves the simplest and oldest studied cubic equation, and one for which the ancient Egyptians did not believe a solution existed. [7]