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The conjecture is that there is a simple way to tell whether such equations have a finite or infinite number of rational solutions. More specifically, the Millennium Prize version of the conjecture is that, if the elliptic curve E has rank r, then the L-function L(E, s) associated with it vanishes to order r at s = 1.
The 3x + 1 semigroup has been used to prove a weaker form of the Collatz conjecture. In fact, it was in such context the concept of the 3 x + 1 semigroup was introduced by H. Farkas in 2005. [ 2 ] Various generalizations of the 3 x + 1 semigroup have been constructed and their properties have been investigated.
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 ...
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 variable y to be the unknown, and then the equation is solved by y = x – 1.
[4] [5] [6] Cramer's rule, implemented in a naive way, is computationally inefficient for systems of more than two or three equations. [7] In the case of n equations in n unknowns, it requires computation of n + 1 determinants, while Gaussian elimination produces the result with the same computational complexity as the computation of a single ...
has no real number solution since no real number squared equals −1. Sometimes a quadratic equation has a root of multiplicity 2, such as: (+) = For this equation, −1 is a root of multiplicity 2. This means −1 appears twice, since the equation can be rewritten in factored form as
They are the solutions of a system of 4 equations of degree 5 in 3 variables. Such an overdetermined system has no solution in general (that is if the coefficients are not specific). If it has a finite number of solutions, this number is at most 5 3 = 125, by Bézout's theorem. However, it has been shown that, for the case of the singular ...
For very simple problems, say a function of two variables subject to a single equality constraint, it is most practical to apply the method of substitution. [4] The idea is to substitute the constraint into the objective function to create a composite function that incorporates the effect of the constraint.