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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.
The problem is uninteresting for K of characteristic 2, since over such fields every sum of squares is a square, and we exclude this case. It is believed that otherwise admissibility is independent of the field of definition. [1]: 137
Hurwitz's theorem implies that multiplicative formulas for sums of squares can only occur in 1, 2, 4 and 8 dimensions, a result originally proved by Hurwitz in 1898. It is a special case of the Hurwitz problem, solved also in Radon (1922).
The number of ways to write a natural number as sum of two squares is given by r 2 (n).It is given explicitly by = (() ())where d 1 (n) is the number of divisors of n which are congruent to 1 modulo 4 and d 3 (n) is the number of divisors of n which are congruent to 3 modulo 4.
Legendre's three-square theorem states which numbers can be expressed as the sum of three squares; Jacobi's four-square theorem gives the number of ways that a number can be represented as the sum of four squares. For the number of representations of a positive integer as a sum of squares of k integers, see Sum of squares function.
If the divisor has a fractional part, one can restate the problem by moving the decimal to the right in both numbers until the divisor has no fraction, which can make the problem easier to solve (e.g., 10/2.5 = 100/25 = 4). Division can be calculated with an abacus. [14]
A result of Albrecht Pfister [8] shows that a positive semidefinite form in n variables can be expressed as a sum of 2 n squares. [9] Dubois showed in 1967 that the answer is negative in general for ordered fields. [10] In this case one can say that a positive polynomial is a sum of weighted squares of rational functions with positive ...
Therefore, the theorem states that it is expressible as the sum of two squares. Indeed, 2450 = 7 2 + 49 2. The prime decomposition of the number 3430 is 2 · 5 · 7 3. This time, the exponent of 7 in the decomposition is 3, an odd number. So 3430 cannot be written as the sum of two squares.
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