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G(3) is at least 4 (since cubes are congruent to 0, 1 or −1 mod 9); for numbers less than 1.3 × 10 9, 1 290 740 is the last to require 6 cubes, and the number of numbers between N and 2N requiring 5 cubes drops off with increasing N at sufficient speed to have people believe that G(3) = 4; [17] the largest number now known not to be a sum of ...
Consider the sum = = = (). The two sequences are non-increasing, therefore a j − a k and b j − b k have the same sign for any j, k.Hence S ≥ 0.. Opening the brackets, we deduce:
Bennett's inequality, an upper bound on the probability that the sum of independent random variables deviates from its expected value by more than any specified amount Bhatia–Davis inequality , an upper bound on the variance of any bounded probability distribution
Occasionally, chained notation is used with inequalities in different directions, in which case the meaning is the logical conjunction of the inequalities between adjacent terms. For example, the defining condition of a zigzag poset is written as a 1 < a 2 > a 3 < a 4 > a 5 < a 6 > ... . Mixed chained notation is used more often with compatible ...
Hoeffding's inequality was proven by Wassily Hoeffding in 1963. [1] Hoeffding's inequality is a special case of the Azuma–Hoeffding inequality and McDiarmid's inequality. It is similar to the Chernoff bound, but tends to be less sharp, in particular when the variance of the random variables is small. [2]
Euler was aware of the equality 59 4 + 158 4 = 133 4 + 134 4 involving sums of four fourth powers; this, however, is not a counterexample because no term is isolated on one side of the equation. He also provided a complete solution to the four cubes problem as in Plato's number 3 3 + 4 3 + 5 3 = 6 3 or the taxicab number 1729.
In Cleveland, the income inequality ratio is 5.49. The 80th percentile earns $115,915 and the 20th percentile makes $21,102. ... That's at least 4.06 times as much as the 20th percentile, which is ...
We can use Boole's Inequality to solve this problem. By finding the complement of event "all five are good", we can change this question into another condition: P( at least one estimation is bad) = 0.05 ≤ P( A 1 is bad) + P( A 2 is bad) + P( A 3 is bad) + P( A 4 is bad) + P( A 5 is bad) One way is to make each of them equal to 0.05/5 = 0.01 ...