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An inverse problem in science is the process of calculating from a set of observations the causal factors that produced them: for example, calculating an image in X-ray computed tomography, source reconstruction in acoustics, or calculating the density of the Earth from measurements of its gravity field.
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.
It is possible, using classical results, to construct explicitly a polynomial whose Galois group over is the cyclic group Z/nZ for any positive integer n.To do this, choose a prime p such that p ≡ 1 (mod n); this is possible by Dirichlet's theorem.
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In mathematics and physics, the inverse scattering problem is the problem of determining characteristics of an object, based on data of how it scatters incoming radiation or particles. [1] It is the inverse problem to the direct scattering problem , which is to determine how radiation or particles are scattered based on the properties of the ...
A problem with a low condition number is said to be well-conditioned, while a problem with a high condition number is said to be ill-conditioned. In non-mathematical terms, an ill-conditioned problem is one where, for a small change in the inputs (the independent variables) there is a large change in the answer or dependent variable. This means ...
The problem has a solution; The solution is unique; The solution's behavior changes continuously with the initial conditions; Examples of archetypal well-posed problems include the Dirichlet problem for Laplace's equation, and the heat equation with specified initial conditions. These might be regarded as 'natural' problems in that there are ...
The answer to the first question is 2 / 3 , as is shown correctly by the "simple" solutions. But the answer to the second question is now different: the conditional probability the car is behind door 1 or door 2 given the host has opened door 3 (the door on the right) is 1 / 2 .
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