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At the same time the student is generating a list of the multiples of the small number (i.e., partial quotients) that have so far been taken away, which when added up together would then become the whole number quotient itself. For example, to calculate 132 ÷ 8, one might successively subtract 80, 40 and 8 to leave 4:
Thought of quotitively, a division problem can be solved by repeatedly subtracting groups of the size of the divisor. [1] For instance, suppose each egg carton fits 12 eggs, and the problem is to find how many cartons are needed to fit 36 eggs in total. Groups of 12 eggs at a time can be separated from the main pile until none are left, 3 groups:
The process of adding one more partial quotient to a finite continued fraction is in many ways analogous to this process of "punching a hole" in an interval of real numbers. The size of the "hole" is inversely proportional to the next partial denominator chosen – if the next partial denominator is 1, the gap between successive convergents is ...
Instead, the division is reduced to small steps. Starting from the left, enough digits are selected to form a number (called the partial dividend) that is at least 4×1 but smaller than 4×10 (4 being the divisor in this problem). Here, the partial dividend is 9. The first number to be divided by the divisor (4) is the partial dividend (9).
Characteristics may fail to cover part of the domain of the PDE. This is called a rarefaction, and indicates the solution typically exists only in a weak, i.e. integral equation, sense. The direction of the characteristic lines indicates the flow of values through the solution, as the example above demonstrates.
In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. It is frequently used to transform the antiderivative of a product of functions into an ...
For example, can be expanded to the periodic continued fraction [;,,,...]. This article considers only the case of periodic regular continued fractions . In other words, the remainder of this article assumes that all the partial denominators a i ( i ≥ 1) are positive integers.
Of the cleanly formulated Hilbert problems, numbers 3, 7, 10, 14, 17, 18, 19, and 20 have resolutions that are accepted by consensus of the mathematical community. Problems 1, 2, 5, 6, [g] 9, 11, 12, 15, 21, and 22 have solutions that have partial acceptance, but there exists some controversy as to whether they resolve the problems.
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