S Chowla and SS Pillai
... ² The non-vanishing at s = 1 of L(s; Â) for non-trivial Dirichlet characters is the key fact used in the proof of Dirichlet's famous theorem on existence of in¯nitely many primes in any arithmetic progression an + b with (a; b) = 1. ² Kenkichi Iwasawa [12] showed in 1975 that the above result has co ...
... ² The non-vanishing at s = 1 of L(s; Â) for non-trivial Dirichlet characters is the key fact used in the proof of Dirichlet's famous theorem on existence of in¯nitely many primes in any arithmetic progression an + b with (a; b) = 1. ² Kenkichi Iwasawa [12] showed in 1975 that the above result has co ...
Full text
... Several conclusions are immediate from Theorem 2. By (a), 0 and g/2 are not contained in an F-cycle. Moreover, (g) is an F-cycle of length 1. This cycle is called the trivial cycle; all other cycles are proper cycles. Part (b) tells us that proper cycles exist if and only if g is not a power of 2. B ...
... Several conclusions are immediate from Theorem 2. By (a), 0 and g/2 are not contained in an F-cycle. Moreover, (g) is an F-cycle of length 1. This cycle is called the trivial cycle; all other cycles are proper cycles. Part (b) tells us that proper cycles exist if and only if g is not a power of 2. B ...
Lyashko–Looijenga morphisms and submaximal factorizations of a Coxeter element Vivien Ripoll
... The numbers Cat(p) (W ) are called Fuß–Catalan numbers of type W (and Catalan numbers for p = 1). When W is the symmetric (p+1)n group Sn , these are the classi1 . Those generalized Fuß–Catalan cal Catalan and Fuß–Catalan numbers pn+1 n numbers also appear in other combinatorial objects constructe ...
... The numbers Cat(p) (W ) are called Fuß–Catalan numbers of type W (and Catalan numbers for p = 1). When W is the symmetric (p+1)n group Sn , these are the classi1 . Those generalized Fuß–Catalan cal Catalan and Fuß–Catalan numbers pn+1 n numbers also appear in other combinatorial objects constructe ...
MA/CSSE 473 Day 9 Announcements and Summary
... Exam dates: Tuesday Sept 30, Tuesday November 4. In-class. Not in the schedule page yet. o If you are allowed extra time for the exam and plan to use that time, please talk with me soon about timing. I’ll be in my office hours 6-8 today. Tomorrow we will discuss the Donald Knuth interview. ...
... Exam dates: Tuesday Sept 30, Tuesday November 4. In-class. Not in the schedule page yet. o If you are allowed extra time for the exam and plan to use that time, please talk with me soon about timing. I’ll be in my office hours 6-8 today. Tomorrow we will discuss the Donald Knuth interview. ...
Pythagoras and the Pythagoreans
... and the proof is complete. (Try, p = 2, 3, 5, and 7 to get the numbers above.) There is just something about the word “perfect”. The search for perfect numbers continues to this day. By Euclid’s theorem, this means the search is for primes of the form (2p −1), where p is a prime. The story of and se ...
... and the proof is complete. (Try, p = 2, 3, 5, and 7 to get the numbers above.) There is just something about the word “perfect”. The search for perfect numbers continues to this day. By Euclid’s theorem, this means the search is for primes of the form (2p −1), where p is a prime. The story of and se ...
The Euclidean Algorithm and Its Consequences
... Because of the Key Fact, there is a procedure7 for finding the gcd through repeated division. The precise description of the procedure is somewhat cryptic, but an example will help to make the meaning clear. The Euclidean Algorithm Start with a nonnegative integer a and a positive integer b. Step 1: ...
... Because of the Key Fact, there is a procedure7 for finding the gcd through repeated division. The precise description of the procedure is somewhat cryptic, but an example will help to make the meaning clear. The Euclidean Algorithm Start with a nonnegative integer a and a positive integer b. Step 1: ...
Homotopy idempotents on manifolds and Bass` conjectures
... This note has its origins in talks discussing Bass’ trace conjecture. After one such lecture (by IC), R Geoghegan kindly mentioned his geometric perspective on the matter. Then, when another of us (AJB) spoke about the conjecture at the Kinosaki conference, he thought that a topological audience mig ...
... This note has its origins in talks discussing Bass’ trace conjecture. After one such lecture (by IC), R Geoghegan kindly mentioned his geometric perspective on the matter. Then, when another of us (AJB) spoke about the conjecture at the Kinosaki conference, he thought that a topological audience mig ...
The Natural Number System: Induction and Counting
... initial step. It is also possible to greatly increase the power of proof by induction by refining the inductive step. To see how we might improve our situation with respect to the inductive step, consider a property q(i) that is in general stronger than p(i), so q(i) ⇒ p(i) for all i ∈ N. We could o ...
... initial step. It is also possible to greatly increase the power of proof by induction by refining the inductive step. To see how we might improve our situation with respect to the inductive step, consider a property q(i) that is in general stronger than p(i), so q(i) ⇒ p(i) for all i ∈ N. We could o ...
MATH 103A Homework 1 Solutions Due January 11, 2013
... c 12. Then a divides c (12 6 2), b divides c (12 4 3), but ab does not divide c (ab 24 12 c and a larger number can’t divide a smaller one). (3) (Gallian Chapter 0 # 11) Let n and a be positive integers and let d that the equation ax mod n 1 has a solution if and only if d ...
... c 12. Then a divides c (12 6 2), b divides c (12 4 3), but ab does not divide c (ab 24 12 c and a larger number can’t divide a smaller one). (3) (Gallian Chapter 0 # 11) Let n and a be positive integers and let d that the equation ax mod n 1 has a solution if and only if d ...
Chapter 9: Transcendental Functions
... techniques can often be employed to obtain approximations of the inverse function. Thus, theorem 9.1.8 and proposition 9.1.12 provide useful criteria for deciding whether a function is invertible. We now turn to the calculus of inverse functions. THEOREM 9.1.16 Let A be an open interval and let f : ...
... techniques can often be employed to obtain approximations of the inverse function. Thus, theorem 9.1.8 and proposition 9.1.12 provide useful criteria for deciding whether a function is invertible. We now turn to the calculus of inverse functions. THEOREM 9.1.16 Let A be an open interval and let f : ...
PPT
... interested in the number of whorls in each ring of the spiral. The ratio of consecutive ring lengths approaches the Golden Ratio. ...
... interested in the number of whorls in each ring of the spiral. The ratio of consecutive ring lengths approaches the Golden Ratio. ...
2013 - Fermat - CEMC - University of Waterloo
... A positive integer larger than one is a perfect square if and only if each of its prime factors occurs an even number of times. Since the integers in the list above contain in total only one factor of 5 and one factor of 7, then neither 5 nor 7 can be chosen to form a product that is a perfect squar ...
... A positive integer larger than one is a perfect square if and only if each of its prime factors occurs an even number of times. Since the integers in the list above contain in total only one factor of 5 and one factor of 7, then neither 5 nor 7 can be chosen to form a product that is a perfect squar ...
to the PDF file
... { ab }-gon, that is, a connected sequence of edges that visits every ath vertex of a regular convex b-gon (see Figure 1). We may then regard the regular N -gon as a regular star { N1 }-gon. It turns out that we can only fold a regular star { ab }-gon by a period-2 procedure if we can fold a regular ...
... { ab }-gon, that is, a connected sequence of edges that visits every ath vertex of a regular convex b-gon (see Figure 1). We may then regard the regular N -gon as a regular star { N1 }-gon. It turns out that we can only fold a regular star { ab }-gon by a period-2 procedure if we can fold a regular ...
On the divisor class group of 3
... by Reid in [8]. For results about canonical singularities and their minimal models, see also [9] and [10]. In the case of a Gorenstein variety, canonical is equivalent to rational. The aim of this note is to calculate ρ(X) in the case of an isolated canonical hypersurface singularity, which is assum ...
... by Reid in [8]. For results about canonical singularities and their minimal models, see also [9] and [10]. In the case of a Gorenstein variety, canonical is equivalent to rational. The aim of this note is to calculate ρ(X) in the case of an isolated canonical hypersurface singularity, which is assum ...
A Musician`s Guide to Prime Numbers
... tones corresponding to prime intervals will change. Given a starting pitch S, can we come up with a method of showing precisely which tones (with respect to S) might correspond to a prime interval and precisely which tones will never correspond to a prime interval? Let us formulate some conjectures ...
... tones corresponding to prime intervals will change. Given a starting pitch S, can we come up with a method of showing precisely which tones (with respect to S) might correspond to a prime interval and precisely which tones will never correspond to a prime interval? Let us formulate some conjectures ...
CHAP02 Linear Congruences
... from his celebrated “Last Theorem”. Fermat's Last Theorem states that for all integers n ≥ 3 there are no solutions to the equation xn + yn = zn for non-zero integers x, y and z. We all know that 32 + 42 = 52 and 52 + 122 = 132. There infinitely many such integer solutions to the equation x2 + y2 = ...
... from his celebrated “Last Theorem”. Fermat's Last Theorem states that for all integers n ≥ 3 there are no solutions to the equation xn + yn = zn for non-zero integers x, y and z. We all know that 32 + 42 = 52 and 52 + 122 = 132. There infinitely many such integer solutions to the equation x2 + y2 = ...
16.4 Reasoning and Proof
... Must any two planes intersect? Why or why not? Name planes in the classroom that support your answer. No, if the planes are parallel they will never intersect. Possible example: opposite walls in the classroom ...
... Must any two planes intersect? Why or why not? Name planes in the classroom that support your answer. No, if the planes are parallel they will never intersect. Possible example: opposite walls in the classroom ...
Lecture notes, sections 2.1 to 2.3
... not require heavy background in pure mathematics. But elementary does not mean easy!! There are many extremely hard problems in elementary number theory, for example the following: Problem 1. Find all solutions to the following equation: ...
... not require heavy background in pure mathematics. But elementary does not mean easy!! There are many extremely hard problems in elementary number theory, for example the following: Problem 1. Find all solutions to the following equation: ...
An explicit version of Birch`s Theorem
... M and R, is established in §4. The proof of Theorem 1 is then completed routinely in §5 by making use of the main conclusions of §§3 and 4. Finally, in §6, we provide an appendix in which we discuss Schmidt’s method and its consequences for Birch’s Theorem. Throughout, implicit constants in Vinograd ...
... M and R, is established in §4. The proof of Theorem 1 is then completed routinely in §5 by making use of the main conclusions of §§3 and 4. Finally, in §6, we provide an appendix in which we discuss Schmidt’s method and its consequences for Birch’s Theorem. Throughout, implicit constants in Vinograd ...
Combinatorial Geometry with Algorithmic Applications János Pach
... It is easy to verify that C4 , a cycle of length 4, cannot be drawn as a thrackle, but any other cycle can [?]. If a graph cannot be drawn as a thrackle, then the same is true for all graphs that contain it as a subgraph. Thus, a thrackle does not contain a cycle of length 4, and, according to an ol ...
... It is easy to verify that C4 , a cycle of length 4, cannot be drawn as a thrackle, but any other cycle can [?]. If a graph cannot be drawn as a thrackle, then the same is true for all graphs that contain it as a subgraph. Thus, a thrackle does not contain a cycle of length 4, and, according to an ol ...
Linear independence of the digamma function and a variant of a conjecture of Rohrlich
... Motivated by the above theorem, the authors, in the same paper, conjectured the following: Conjecture. Let K be any number field over which the qth cyclotomic polynomial is irreducible. Then the ϕ (q) numbers ψ(a/q) with 1 a q and (a, q) = 1 are linearly independent over K . In this context, they ...
... Motivated by the above theorem, the authors, in the same paper, conjectured the following: Conjecture. Let K be any number field over which the qth cyclotomic polynomial is irreducible. Then the ϕ (q) numbers ψ(a/q) with 1 a q and (a, q) = 1 are linearly independent over K . In this context, they ...