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The article by Boas analyzes two-digit cases in bases other than base 10, e.g., 32 / 13 = 2 / 1 and its inverse are the only solutions in base 4 with two digits. [2]An example of anomalous cancellation with more than two digits is 165 / 462 = 15 / 42 , and an example with different numbers of digits is 98 / 392 = 8 / 32 .
The polynomial x 2 + cx + d, where a + b = c and ab = d, can be factorized into (x + a)(x + b).. In mathematics, factorization (or factorisation, see English spelling differences) or factoring consists of writing a number or another mathematical object as a product of several factors, usually smaller or simpler objects of the same kind.
For univariate polynomials over the rationals (or more generally over a field of characteristic zero), Yun's algorithm exploits this to efficiently factorize the polynomial into square-free factors, that is, factors that are not a multiple of a square, performing a sequence of GCD computations starting with gcd(f(x), f '(x)). To factorize the ...
The limit, should it exist, is a positive real solution of the equation y = x y. Thus, x = y 1/y. The limit defining the infinite exponential of x does not exist when x > e 1/e because the maximum of y 1/y is e 1/e. The limit also fails to exist when 0 < x < e −e. This may be extended to complex numbers z with the definition:
That is, the goal is to minimize the maximum value of () (), where P(x) is the approximating polynomial, f(x) is the actual function, and x varies over the chosen interval.
Computing the square root of 2 (which is roughly 1.41421) is a well-posed problem. Many algorithms solve this problem by starting with an initial approximation x 0 to , for instance x 0 = 1.4, and then computing improved guesses x 1, x 2, etc. One such method is the famous Babylonian method, which is given by x k+1 = (x k + 2/x k)/2.
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The first condition is the Fermat primality test using base 2. In general, if p ≡ a (mod x 2 +4), where a is a quadratic non-residue (mod x 2 +4) then p should be prime if the following conditions hold: 2 p−1 ≡ 1 (mod p), f(1) p+1 ≡ 0 (mod p), f(x) k is the k-th Fibonacci polynomial at x.