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... Remarks: (1), (2), (3) are well-known properties of Dedekind sums. (See [1], p. 62.) (4) through (8) are well-known properties of linear recurrences of order 2. (See [2], p. 193-194). (9) can be proved via (6) or by induction on n. THE MAIN RESULTS Theorem 1: Let {un } be a linear recurrence of orde ...
... Remarks: (1), (2), (3) are well-known properties of Dedekind sums. (See [1], p. 62.) (4) through (8) are well-known properties of linear recurrences of order 2. (See [2], p. 193-194). (9) can be proved via (6) or by induction on n. THE MAIN RESULTS Theorem 1: Let {un } be a linear recurrence of orde ...
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... Using DeMoivre's Theorem with the power being a rational exponent (and therefore meaning a root), we can develop a method for finding complex roots. This leads to the following formula: ...
... Using DeMoivre's Theorem with the power being a rational exponent (and therefore meaning a root), we can develop a method for finding complex roots. This leads to the following formula: ...
A,B
... Sweet Clyde’s Inversion Theorem. Any permutation on a set of n individuals created in the body switching machine can be restored by introducing at most two extra individuals and adhering to the cerebral immune response rule (i.e. no repeated transpositions). ...
... Sweet Clyde’s Inversion Theorem. Any permutation on a set of n individuals created in the body switching machine can be restored by introducing at most two extra individuals and adhering to the cerebral immune response rule (i.e. no repeated transpositions). ...
Soergel diagrammatics for dihedral groups
... m-th root of unity. The statement that q 2 = ζm is equivalent to the statement that [m] = 0 and [n] 6= 0 for n < m. So we can specialize Z[q + q −1 ] algebraically to the case where q 2 = ζm by setting the appropriate polynomial in [2] equal to zero. In the case m odd, q 2 = ζm allows q itself to be ...
... m-th root of unity. The statement that q 2 = ζm is equivalent to the statement that [m] = 0 and [n] 6= 0 for n < m. So we can specialize Z[q + q −1 ] algebraically to the case where q 2 = ζm by setting the appropriate polynomial in [2] equal to zero. In the case m odd, q 2 = ζm allows q itself to be ...
Test - Mu Alpha Theta
... 12. How many integers strictly between 100 and 10,000 have an odd number of positive integral factors? (A) 91 ...
... 12. How many integers strictly between 100 and 10,000 have an odd number of positive integral factors? (A) 91 ...