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Simplifying this further gives us the solution x = −3. It is easily checked that none of the zeros of x ( x + 1)( x + 2) – namely x = 0 , x = −1 , and x = −2 – is a solution of the final equation, so no spurious solutions were introduced.
Two fractions a / b and c / d are equal or equivalent if and only if ad = bc.) For example, 1 / 4 , 5 / 6 , and −101 / 100 are all irreducible fractions. On the other hand, 2 / 4 is reducible since it is equal in value to 1 / 2 , and the numerator of 1 / 2 is less than the ...
In case 2, the rate of convergence depends on the absolute value of the ratio between the two roots: the farther that ratio is from unity, the more quickly the continued fraction converges. This general solution of monic quadratic equations with complex coefficients is usually not very useful for obtaining rational approximations to the roots ...
Continued fractions can also be applied to problems in number theory, and are especially useful in the study of Diophantine equations. In the late eighteenth century Lagrange used continued fractions to construct the general solution of Pell's equation, thus answering a question that had fascinated mathematicians for more than a thousand years. [9]
[12] [13] In general, a common fraction is said to be a proper fraction if the absolute value of the fraction is strictly less than one—that is, if the fraction is greater than −1 and less than 1. [14] [15] It is said to be an improper fraction, or sometimes top-heavy fraction, [16] if the absolute value of the fraction is greater than or ...
The first term, as we see, is the first fraction; the first and second together give the second fraction, 22 / 7 ; the first, the second and the third give the third fraction 333 / 106 , and so on with the rest; the result being that the series entire is equivalent to the original value.
An item whose delay is times the length of a message must occupy a fraction of at least / of the time slots on the channel it is assigned to, so a solution to the scheduling problem can only come from a solution to the unit fraction bin packing problem with the channels as bins and the fractions / as item sizes.
For example, a fraction is put in lowest terms by cancelling out the common factors of the numerator and the denominator. [2] As another example, if a × b = a × c , then the multiplicative term a can be canceled out if a ≠0, resulting in the equivalent expression b = c ; this is equivalent to dividing through by a .