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If the pseudorandom number = occurring in the Pollard ρ algorithm were an actual random number, it would follow that success would be achieved half the time, by the birthday paradox in () (/) iterations. It is believed that the same analysis applies as well to the actual rho algorithm, but this is a heuristic claim, and rigorous analysis of ...
A method analogous to piece-wise linear approximation but using only arithmetic instead of algebraic equations, uses the multiplication tables in reverse: the square root of a number between 1 and 100 is between 1 and 10, so if we know 25 is a perfect square (5 × 5), and 36 is a perfect square (6 × 6), then the square root of a number greater than or equal to 25 but less than 36, begins with ...
A mathematical constant is a key number whose value is fixed by an unambiguous definition, often referred to by a symbol (e.g., an alphabet letter), or by mathematicians' names to facilitate using it across multiple mathematical problems. [1]
The square root of 2 is equal to the length of the hypotenuse of a right triangle with legs of length 1 and is therefore a constructible number. In geometry and algebra, a real number is constructible if and only if, given a line segment of unit length, a line segment of length | | can be constructed with compass and straightedge in a finite number of steps.
One can prove [citation needed] that = is the largest possible number for which the stopping criterion | + | < ensures ⌊ + ⌋ = ⌊ ⌋ in the algorithm above.. In implementations which use number formats that cannot represent all rational numbers exactly (for example, floating point), a stopping constant less than 1 should be used to protect against round-off errors.
x 1 = x; x 2 = x 2 for i = k - 2 to 0 do if n i = 0 then x 2 = x 1 * x 2; x 1 = x 1 2 else x 1 = x 1 * x 2; x 2 = x 2 2 return x 1 The algorithm performs a fixed sequence of operations ( up to log n ): a multiplication and squaring takes place for each bit in the exponent, regardless of the bit's specific value.
In words: the first two numbers in the sequence are both 2, and each successive number is formed by adding twice the previous Pell–Lucas number to the Pell–Lucas number before that, or equivalently, by adding the next Pell number to the previous Pell number: thus, 82 is the companion to 29, and 82 = 2 × 34 + 14 = 70 + 12.