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It may be used to prove Nicomachus's theorem that the sum of the first cubes equals the square of the sum of the first positive integers. [2] Summation by parts is frequently used to prove Abel's theorem and Dirichlet's test.
For example, the overall sum of a roll-up is just the sum of the sub-sums in each cell. Functions that can be decomposed in this way are called decomposable aggregation functions, and include COUNT, MAX, MIN, and SUM, which can be computed for each cell and then directly aggregated; these are known as self-decomposable aggregation functions. [13]
Fix a complex number .If = for and () =, then () = ⌊ ⌋ and the formula becomes = ⌊ ⌋ = ⌊ ⌋ + ⌊ ⌋ +. If () >, then the limit as exists and yields the ...
The algorithm performs summation with two accumulators: sum holds the sum, and c accumulates the parts not assimilated into sum, to nudge the low-order part of sum the next time around. Thus the summation proceeds with "guard digits" in c , which is better than not having any, but is not as good as performing the calculations with double the ...
In the mathematics of convergent and divergent series, Euler summation is a summation method. That is, it is a method for assigning a value to a series, different from the conventional method of taking limits of partial sums. Given a series Σa n, if its Euler transform converges to a sum, then that sum is called the Euler sum of the original ...
This algorithm can easily be adapted to compute the variance of a finite population: simply divide by n instead of n − 1 on the last line.. Because SumSq and (Sum×Sum)/n can be very similar numbers, cancellation can lead to the precision of the result to be much less than the inherent precision of the floating-point arithmetic used to perform the computation.
There are (at least) three slightly different methods called Borel summation. They differ in which series they can sum, but are consistent, meaning that if two of the methods sum the same series they give the same answer. Throughout let A(z) denote a formal power series = =,
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)} .