Week 7: Non-Linear Equations and review of topic.
... and vice versa Plot complex numbers on an Argand diagram De Moivres theorem for powers Finding equations to describe loci where the points satisfy conditions linked to modulus, argument, real or imaginary parts of z=x+iy. ...
... and vice versa Plot complex numbers on an Argand diagram De Moivres theorem for powers Finding equations to describe loci where the points satisfy conditions linked to modulus, argument, real or imaginary parts of z=x+iy. ...
Full text
... Readers of The Fibonacci Quarterly will be pleased to know that many of its problems can now be searched electronically (at no charge) on the World Wide Web at ...
... Readers of The Fibonacci Quarterly will be pleased to know that many of its problems can now be searched electronically (at no charge) on the World Wide Web at ...
THE CHARNEY-DAVIS QUANTITY FOR CERTAIN GRADED POSETS
... Proof. Given w ∈ S(12 22 · · · N 2 ), append a 0 to its right, creating a multiset permutation ŵ ∈ S(01 12 22 · · · N 2 ) that ends with 0. As in the proof of Proposition 2.1, encode ŵ ∈ S(012 22 · · · N 2 ) as a decreasing binary tree on vertices labelled 0, 1, 1, 2, 2, . . . , N, N , with root l ...
... Proof. Given w ∈ S(12 22 · · · N 2 ), append a 0 to its right, creating a multiset permutation ŵ ∈ S(01 12 22 · · · N 2 ) that ends with 0. As in the proof of Proposition 2.1, encode ŵ ∈ S(012 22 · · · N 2 ) as a decreasing binary tree on vertices labelled 0, 1, 1, 2, 2, . . . , N, N , with root l ...
Irish Intervarsity Mathematics Competition 2002 University College Dublin Time allowed: Three hours
... 10. A sector of a place circular disc of radius r and central angle θ is folded to give an open right circular cone. What value of θ, to the nearest degree, gives a cone of maximum volume? ...
... 10. A sector of a place circular disc of radius r and central angle θ is folded to give an open right circular cone. What value of θ, to the nearest degree, gives a cone of maximum volume? ...
PART I. THE REAL NUMBERS
... but T does not have a rational least upper bound. The Archimedean Property THEOREM 4. (The Archimedean Property) The set N of natural numbers is unbounded above. Proof: Suppose N is bounded above. Let m = sup N. By Theorem 3 there exists a positive integer k such that m − 1 < k ≤ m. But then k + 1 i ...
... but T does not have a rational least upper bound. The Archimedean Property THEOREM 4. (The Archimedean Property) The set N of natural numbers is unbounded above. Proof: Suppose N is bounded above. Let m = sup N. By Theorem 3 there exists a positive integer k such that m − 1 < k ≤ m. But then k + 1 i ...