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Let A be the sum of the negative values and B the sum of the positive values; the number of different possible sums is at most B-A, so the total runtime is in (()). For example, if all input values are positive and bounded by some constant C , then B is at most N C , so the time required is O ( N 2 C ) {\displaystyle O(N^{2}C)} .
7] = + = + = + = + = (+) (+) + = = + = + ( + ()) ( ()) An infinite series of any rational function of can be reduced to a finite series of polygamma ...
If the inputs are all non-negative, then the condition number is 1. Note that the 1 − ε log 2 n {\displaystyle 1-\varepsilon \log _{2}n} denominator is effectively 1 in practice, since ε log 2 n {\displaystyle \varepsilon \log _{2}n} is much smaller than 1 until n becomes of order 2 1/ε , which is roughly 10 10 15 in double precision.
The sum of four cubes problem [1] asks whether every integer is the sum of four cubes of integers. It is conjectured the answer is affirmative, but this conjecture has been neither proven nor disproven. [2] Some of the cubes may be negative numbers, in contrast to Waring's problem on sums of cubes, where they are required to be positive.
The sum of the reciprocals of the powerful numbers is close to 1.9436 . [4] The reciprocals of the factorials sum to the transcendental number e (one of two constants called "Euler's number"). The sum of the reciprocals of the square numbers (the Basel problem) is the transcendental number π 2 / 6 , or ζ(2) where ζ is the Riemann zeta ...
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In number theory, Waring's problem asks whether each natural number k has an associated positive integer s such that every natural number is the sum of at most s natural numbers raised to the power k. For example, every natural number is the sum of at most 4 squares, 9 cubes, or 19 fourth powers.
The Fletcher checksum cannot distinguish between blocks of all 0 bits and blocks of all 1 bits. For example, if a 16-bit block in the data word changes from 0x0000 to 0xFFFF, the Fletcher-32 checksum remains the same. This also means a sequence of all 00 bytes has the same checksum as a sequence (of the same size) of all FF bytes.