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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 ...
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.
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.
144 (one hundred [and] forty-four) is the natural number following 143 and preceding 145. It is coincidentally both the square of twelve (a dozen dozens , or one gross .) and the twelfth Fibonacci number , and the only nontrivial number in the sequence that is square.
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}.}
Tonelli–Shanks cannot be used for composite moduli: finding square roots modulo composite numbers is a computational problem equivalent to integer factorization. [ 1 ] An equivalent, but slightly more redundant version of this algorithm was developed by Alberto Tonelli [ 2 ] [ 3 ] in 1891.
A number where some but not all prime factors have multiplicity above 1 is neither square-free nor squareful. The Liouville function λ(n) is 1 if Ω(n) is even, and is -1 if Ω(n) is odd. The Möbius function μ(n) is 0 if n is not square-free. Otherwise μ(n) is 1 if Ω(n) is even, and is −1 if Ω(n) is odd.
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 ]