Discrete Mathematics: Solutions to Homework 3 1. (10%) Devise an
... (b) The (logx)4 is insignificant compared to the x5 term, so the answer is n = 5. Formally we an take C = 4 and k = 1 as witnesses. (c) For large x, this fraction is fairly close to 1. (This can be seen by dividing numerator and denominator by x4 .) Therefore we can take n = 0. In other words, this ...
... (b) The (logx)4 is insignificant compared to the x5 term, so the answer is n = 5. Formally we an take C = 4 and k = 1 as witnesses. (c) For large x, this fraction is fairly close to 1. (This can be seen by dividing numerator and denominator by x4 .) Therefore we can take n = 0. In other words, this ...
Applications of eigenvalues
... has stationary points exactly when v is an eigenvector of H. Optimizing the Rayleigh quotient is therefore example of a non-convex global optimization problem that I know how to solve in polynomial time. Such examples are rare, and so it is tempting to try to massage other nonconvex optimization pro ...
... has stationary points exactly when v is an eigenvector of H. Optimizing the Rayleigh quotient is therefore example of a non-convex global optimization problem that I know how to solve in polynomial time. Such examples are rare, and so it is tempting to try to massage other nonconvex optimization pro ...
on dominant dimension of noetherian rings
... that if R is left noetherian and left QF-3 then it is also right QF-3. Thus, if R is left and right noetherian, R is left QF-3 if and only if it is right QF-3. Generalizing this, we will prove the following Theorem. Let R be left and right noetherian. pR^nif and only if dom dim RR^n. ...
... that if R is left noetherian and left QF-3 then it is also right QF-3. Thus, if R is left and right noetherian, R is left QF-3 if and only if it is right QF-3. Generalizing this, we will prove the following Theorem. Let R be left and right noetherian. pR^nif and only if dom dim RR^n. ...
The Frobenius Coin Problem Upper Bounds on The
... Indeed, if n is expressed as above, consider n0 = (p1 − 1)(p2 − 1) − 1 − n = p1 p2 − p1 − p2 − xp1 − yp2 = (p2 − 1 − x)p1 + (−1 − y)p2 . Since 0 ≤ x ≤ p2 − 1, we have 0 ≤ p2 − 1 − x ≤ p2 − 1. It follows that if y ≥ 0, then n is representable and n0 is not, while if y < 0, then n is not representable ...
... Indeed, if n is expressed as above, consider n0 = (p1 − 1)(p2 − 1) − 1 − n = p1 p2 − p1 − p2 − xp1 − yp2 = (p2 − 1 − x)p1 + (−1 − y)p2 . Since 0 ≤ x ≤ p2 − 1, we have 0 ≤ p2 − 1 − x ≤ p2 − 1. It follows that if y ≥ 0, then n is representable and n0 is not, while if y < 0, then n is not representable ...
ON THE FIELDS GENERATED BY THE LENGTHS OF CLOSED
... To formalize the idea of “polynomial relations” between the weak length spectra of Riemannian manifolds, we need to introduce some additional notations and definitions. For a Riemannian manifold M, we let F (M ) denote the subfield of R generated by the set L(M ). Given two Riemannian manifolds M1 a ...
... To formalize the idea of “polynomial relations” between the weak length spectra of Riemannian manifolds, we need to introduce some additional notations and definitions. For a Riemannian manifold M, we let F (M ) denote the subfield of R generated by the set L(M ). Given two Riemannian manifolds M1 a ...
