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The exercise of working through this problem may be used to explain and demonstrate exponents and the quick growth of exponential and geometric sequences. It can also be used to illustrate sigma notation. When expressed as exponents, the geometric series is: 2 0 + 2 1 + 2 2 + 2 3 + ... and so forth, up to 2 63. The base of each exponentiation ...
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
For example, the sequence 2, 6, 18, 54, ... is a geometric progression with a common ratio of 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with a common ratio of 1/2. Examples of a geometric sequence are powers r k of a fixed non-zero number r, such as 2 k and 3 k. The general form of a geometric sequence is
Bellman's lost-in-a-forest problem is an unsolved minimization problem in geometry, originating in 1955 by the American applied mathematician Richard E. Bellman. [1] The problem is often stated as follows: "A hiker is lost in a forest whose shape and dimensions are precisely known to him.
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.
The solution to this problem for a given set of coin ... the Frobenius number of a set in a geometric sequence. ... 31, 34, 37, and 43 (sequence A065003 ...
There is a unique sequence of Heronian triangles that are "almost equilateral" because the three sides are of the form n − 1, n, n + 1. A method for generating all solutions to this problem based on continued fractions was described in 1864 by Edward Sang, [29] and in 1880 Reinhold Hoppe gave a closed-form expression for the solutions. [30]
As explained above in Applications, the polynomial interpolation problem for () = + + + + satisfying () =, …, = is equivalent to the matrix equation =, which has the unique solution =. There are other known formulas which solve the interpolation problem, which must be equivalent to the unique a = V − 1 y {\displaystyle a=V^{-1}y} , so they ...
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