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where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
A necessary (but not sufficient) condition for solvability is that n is not divisible by 4 or by a prime of form 4k + 3. [note 3] Thus, for example, x 2 − 3 y 2 = −1 is never solvable, but x 2 − 5 y 2 = −1 may be. [27] The first few numbers n for which x 2 − n y 2 = −1 is solvable are with only one trivial solution: 1
5 is halved (2.5) and 6 is doubled (12). The fractional portion is discarded (2.5 becomes 2). The figure in the left column (2) is even, so the figure in the right column (12) is discarded. 2 is halved (1) and 12 is doubled (24). All not-scratched-out values are summed: 3 + 6 + 24 = 33. The method works because multiplication is distributive, so:
Simplification is the process of replacing a mathematical expression by an equivalent one that is simpler (usually shorter), according to a well-founded ordering. Examples include:
49 is the smallest discriminant of a totally real cubic field. [2] 49 and 94 are the only numbers below 100 whose all permutations are composites but they are not multiples of 3, repdigits or numbers which only have digits 0, 2, 4, 5, 6 and 8, even excluding the trivial one digit terms. 49 = 7^2 and 94 = 2 * 47
In the second step, they were divided by 3. The final result, 4 / 3 , is an irreducible fraction because 4 and 3 have no common factors other than 1. The original fraction could have also been reduced in a single step by using the greatest common divisor of 90 and 120, which is 30. As 120 ÷ 30 = 4, and 90 ÷ 30 = 3, one gets
12 (twelve) is the natural number following 11 and preceding 13.. Twelve is the 3rd superior highly composite number, [1] the 3rd colossally abundant number, [2] the 5th highly composite number, and is divisible by the numbers from 1 to 4, and 6, a large number of divisors comparatively.
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.