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For example, the largest amount that cannot be obtained using only coins of 3 and 5 units is 7 units. The solution to this problem for a given set of coin denominations is called the Frobenius number of the set. The Frobenius number exists as long as the set of coin denominations is setwise coprime.
The 1-form dz − y dx. on R 3 maximally violates the assumption of Frobenius' theorem. These planes appear to twist along the y-axis.It is not integrable, as can be verified by drawing an infinitesimal square in the x-y plane, and follow the path along the one-forms.
The theorem is also known as straightening out of a vector field. The Frobenius theorem in differential geometry can be considered as a higher-dimensional generalization of this theorem. Proof
Some solutions of a differential equation having a regular singular point with indicial roots = and .. In mathematics, the method of Frobenius, named after Ferdinand Georg Frobenius, is a way to find an infinite series solution for a linear second-order ordinary differential equation of the form ″ + ′ + = with ′ and ″.
In mathematics, more specifically in abstract algebra, the Frobenius theorem, proved by Ferdinand Georg Frobenius in 1877, characterizes the finite-dimensional associative division algebras over the real numbers. According to the theorem, every such algebra is isomorphic to one of the following: R (the real numbers) C (the complex numbers) H ...
Frobenius reciprocity theorem in group representation theory describing the reciprocity relation between restricted and induced representations on a subgroup Perron–Frobenius theorem in matrix theory concerning the eigenvalues and eigenvectors of a matrix with positive real coefficients
The postage stamp problem (also called the Frobenius Coin Problem and the Chicken McNugget Theorem [1]) is a mathematical riddle that asks what is the smallest postage value which cannot be placed on an envelope, if the latter can hold only a limited number of stamps, and these may only have certain specified face values.
Since z = 1 − x, the solution of the hypergeometric equation at x = 1 is the same as the solution for this equation at z = 0. But the solution at z = 0 is identical to the solution we obtained for the point x = 0, if we replace each γ by α + β − γ + 1. Hence, to get the solutions, we just make this substitution in the previous results.