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Sum of Natural Numbers (second proof and extra footage) includes demonstration of Euler's method. What do we get if we sum all the natural numbers? response to comments about video by Tony Padilla; Related article from New York Times; Why –1/12 is a gold nugget follow-up Numberphile video with Edward Frenkel
The digital root (also repeated digital sum) of a natural number in a given radix is the (single digit) value obtained by an iterative process of summing digits, on each iteration using the result from the previous iteration to compute a digit sum. The process continues until a single-digit number is reached.
For example, the sum of the first n natural numbers can be denoted as ∑ i = 1 n i {\displaystyle \sum _{i=1}^{n}i} For long summations, and summations of variable length (defined with ellipses or Σ notation), it is a common problem to find closed-form expressions for the result.
For example, if the summands are uncorrelated random numbers with zero mean, the sum is a random walk, and the condition number will grow proportional to . On the other hand, for random inputs with nonzero mean the condition number asymptotes to a finite constant as n → ∞ {\displaystyle n\to \infty } .
In number theory, a polite number is a positive integer that can be written as the sum of two or more consecutive positive integers. A positive integer which is not polite is called impolite. [1] [2] The impolite numbers are exactly the powers of two, and the polite numbers are the natural numbers that are not powers of two.
The natural numbers 0 and 1 are trivial sum-product numbers for all , and all other sum-product numbers are nontrivial sum-product numbers. For example, the number 144 in base 10 is a sum-product number, because 1 + 4 + 4 = 9 {\displaystyle 1+4+4=9} , 1 × 4 × 4 = 16 {\displaystyle 1\times 4\times 4=16} , and 9 × 16 = 144 {\displaystyle 9 ...
For example, one can add N numbers either by a simple loop that adds each datum to a single variable, or by a D&C algorithm called pairwise summation that breaks the data set into two halves, recursively computes the sum of each half, and then adds the two sums. While the second method performs the same number of additions as the first and pays ...
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)} .