Search results
Results from the WOW.Com Content Network
and −2 is the least absolute remainder. In the division of 42 by 5, we have: 42 = 8 × 5 + 2, and since 2 < 5/2, 2 is both the least positive remainder and the least absolute remainder. In these examples, the (negative) least absolute remainder is obtained from the least positive remainder by subtracting 5, which is d. This holds in general.
Modular exponentiation is the remainder when an integer b (the base) is raised to the power e (the exponent), and divided by a positive integer m (the modulus); that is, c = b e mod m. From the definition of division, it follows that 0 ≤ c < m. For example, given b = 5, e = 3 and m = 13, dividing 5 3 = 125 by 13 leaves a remainder of c = 8.
In matrix inversion however, instead of vector b, we have matrix B, where B is an n-by-p matrix, so that we are trying to find a matrix X (also a n-by-p matrix): = =. We can use the same algorithm presented earlier to solve for each column of matrix X. Now suppose that B is the identity matrix of size n.
More generally, we can factor a complex m×n matrix A, with m ≥ n, as the product of an m×m unitary matrix Q and an m×n upper triangular matrix R.As the bottom (m−n) rows of an m×n upper triangular matrix consist entirely of zeroes, it is often useful to partition R, or both R and Q:
The remainder of the divide step is to solve for the eigenvalues (and if desired the eigenvectors) of ^ and ^, that is to find the diagonalizations ^ = and ^ =. This can be accomplished with recursive calls to the divide-and-conquer algorithm, although practical implementations often switch to the QR algorithm for small enough submatrices.
If A is an m × n matrix and B is a p × q matrix, then the Kronecker product A ⊗ B is the pm × qn block matrix: = [], more explicitly: = []. Using / / and % to denote truncating integer division and remainder, respectively, and numbering the matrix elements starting from 0, one obtains
The matrix X on the left is a Vandermonde matrix, whose determinant is known to be () = < (), which is non-zero since the nodes are all distinct. This ensures that the matrix is invertible and the equation has the unique solution A = X − 1 ⋅ Y {\displaystyle A=X^{-1}\cdot Y} ; that is, p ( x ) {\displaystyle p(x)} exists and is unique.
Divide it by the orthogonal polynomial p n to get = () + (). where q(x) is the quotient, of degree n − 1 or less (because the sum of its degree and that of the divisor p n must equal that of the dividend), and r(x) is the remainder, also of degree n − 1 or less (because the degree of the remainder is always less than that of the divisor).