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In this vein, the discriminant is a symmetric function in the roots that reflects properties of the roots – it is zero if and only if the polynomial has a multiple root, and for quadratic and cubic polynomials it is positive if and only if all roots are real and distinct, and negative if and only if there is a pair of distinct complex ...
SRS can be solved in polynomial time in the Real RAM model. [3] However, its run-time complexity in the Turing machine model is open, as of 1997. [1] The main difficulty is that, in order to solve the problem, the square-roots should be computed to a high accuracy, which may require a large number of bits.
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 ...
We are not taking the square root of any negative values here, since both and are necessarily positive. But we have lost the solution x = − 2. {\displaystyle x=-2.} The reason is that x {\displaystyle x} is actually not in general the positive square root of x 2 . {\displaystyle x^{2}.}
The 1/e-law of best choice is due to F. Thomas Bruss. [ 26 ] Ferguson has an extensive bibliography and points out that a similar (but different) problem had been considered by Arthur Cayley in 1875 and even by Johannes Kepler long before that, who spent 2 years investigating 11 candidates for marriage during 1611 -- 1613 after the death of his ...
In cases where the function in question has multiple roots, it can be difficult to control, via choice of initialization, which root (if any) is identified by Newton's method. For example, the function f ( x ) = x ( x 2 − 1)( x − 3)e −( x − 1) 2 /2 has roots at −1, 0, 1, and 3. [ 18 ]
In the case of two nested square roots, the following theorem completely solves the problem of denesting. [2]If a and c are rational numbers and c is not the square of a rational number, there are two rational numbers x and y such that + = if and only if is the square of a rational number d.
The square root of a positive integer is the product of the roots of its prime factors, because the square root of a product is the product of the square roots of the factors. Since p 2 k = p k , {\textstyle {\sqrt {p^{2k}}}=p^{k},} only roots of those primes having an odd power in the factorization are necessary.