Full text
... The purpose of this article is to extend these results to higher-order sequences, their zero-one number systems, and associated partitions. To this end, and for the remainder of the article, let m be an arbitrary fixed integer greater than 2. Define a sequence {s^} inductively as follows: Si = 1 ...
... The purpose of this article is to extend these results to higher-order sequences, their zero-one number systems, and associated partitions. To this end, and for the remainder of the article, let m be an arbitrary fixed integer greater than 2. Define a sequence {s^} inductively as follows: Si = 1 ...
1 Basic Combinatorics
... is the Taylor series for ex . Replacing x with −1, we obtain that the number of derangements of [n] is close to n!/e. Using the remainder formula for Taylor’s Theorem, the number of derangements is in fact exactly bn!/e + 1/2c for n ≥ 1, where bxc is the greatest integer less than or equal to x. For ...
... is the Taylor series for ex . Replacing x with −1, we obtain that the number of derangements of [n] is close to n!/e. Using the remainder formula for Taylor’s Theorem, the number of derangements is in fact exactly bn!/e + 1/2c for n ≥ 1, where bxc is the greatest integer less than or equal to x. For ...
Proof of the Fundamental Theorem of Algebra
... We now complete the proof of the Fundamental Theorem of Algebra. In the following proof, we use circle to denote a path of the form (r, t), t ∈ [0, 2π ], where r > 0 is constant. Fundamental Theorem of Algebra. Let f be a nonconstant polynomial. Then f has at least one complex root. Proof of Fundam ...
... We now complete the proof of the Fundamental Theorem of Algebra. In the following proof, we use circle to denote a path of the form (r, t), t ∈ [0, 2π ], where r > 0 is constant. Fundamental Theorem of Algebra. Let f be a nonconstant polynomial. Then f has at least one complex root. Proof of Fundam ...
1.3 Using Reasoning to Find a Counterexample
... Foundations of Mathematics 11 Chapter 1- Inductive and Deductive Reasoning 1.3 Using Reasoning to Find a Counterexample Today’s Goal: To find examples that contradict given conjectures. Let’s define Counterexample: In order to disprove or make a conjecture false we must find one example of the state ...
... Foundations of Mathematics 11 Chapter 1- Inductive and Deductive Reasoning 1.3 Using Reasoning to Find a Counterexample Today’s Goal: To find examples that contradict given conjectures. Let’s define Counterexample: In order to disprove or make a conjecture false we must find one example of the state ...
The application of a new mean value theorem to the fractional parts
... Us (P, H, P η ) H 2s−1+ε P λs +ε . We may now apply Lemma 3.6 with t = 2, and make use of Lemma 3.2. Thus, by the argument of the proof of Theorem 2.1 we may finally conclude with the following estimate. Lemma 3.7. Let λs be defined as in the statement of the corollary to Theorem 2.1. Then under t ...
... Us (P, H, P η ) H 2s−1+ε P λs +ε . We may now apply Lemma 3.6 with t = 2, and make use of Lemma 3.2. Thus, by the argument of the proof of Theorem 2.1 we may finally conclude with the following estimate. Lemma 3.7. Let λs be defined as in the statement of the corollary to Theorem 2.1. Then under t ...
Warm ups
... enrollment of incoming freshmen at a high school over the last four years. The school wants to predict the number of freshmen for next year. Make a conjecture about the enrollment for next year. A. ...
... enrollment of incoming freshmen at a high school over the last four years. The school wants to predict the number of freshmen for next year. Make a conjecture about the enrollment for next year. A. ...
The Factor Theorem and a corollary of the
... Copyright © 2006 by Murray Eisenberg. All rights reserved. For your convenience, the input cells in this notebook are already evaluated, so that you may read it with MathReader. When you have access to Mathematica itself, you should carry out the evaluations one-by-one. As you work through this note ...
... Copyright © 2006 by Murray Eisenberg. All rights reserved. For your convenience, the input cells in this notebook are already evaluated, so that you may read it with MathReader. When you have access to Mathematica itself, you should carry out the evaluations one-by-one. As you work through this note ...
New Point Addition Formulae for ECC Applications
... This addition involves 5M and 2S. As they require special conditions, our formulae are logically more efficient than any general or mixed addition formulae. What is more striking is the fact that they are more efficient than any doubling formulae (the best doubling is obtained using modified Jacobian coo ...
... This addition involves 5M and 2S. As they require special conditions, our formulae are logically more efficient than any general or mixed addition formulae. What is more striking is the fact that they are more efficient than any doubling formulae (the best doubling is obtained using modified Jacobian coo ...
Course Description
... BSCS-411 — Discrete Mathematics (Mathematical Induction) — Week 6 Mathematical Induction • A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large. • Essentially a “domino effect” principle. • Based on a predicate-logic inference rule: ...
... BSCS-411 — Discrete Mathematics (Mathematical Induction) — Week 6 Mathematical Induction • A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large. • Essentially a “domino effect” principle. • Based on a predicate-logic inference rule: ...
REVISED 3/23/14 Ms C. Draper lesson elements for Week of ___3
... expressions and equations. We will move forward with Pythagorean theorem and continue to review & practice basic skills. Students are now moving in and out of their “power groups” and returning to home teams to share and/or peer teach concepts at 4-teired readiness levels. This week tiered groups wi ...
... expressions and equations. We will move forward with Pythagorean theorem and continue to review & practice basic skills. Students are now moving in and out of their “power groups” and returning to home teams to share and/or peer teach concepts at 4-teired readiness levels. This week tiered groups wi ...
Full text
... modulo m. Many moduli 777, characterized in [1], have the property that every residue modulo 777 occurs in each period. (Indeed, 8 and 11 are the smallest moduli which do not have this property.) However, moduli m with the property that all m residues modulo m appear in one period the some nwnhev of ...
... modulo m. Many moduli 777, characterized in [1], have the property that every residue modulo 777 occurs in each period. (Indeed, 8 and 11 are the smallest moduli which do not have this property.) However, moduli m with the property that all m residues modulo m appear in one period the some nwnhev of ...
Full text
... In fact, if we restrict a to be an irrational which is not equivalent to T = (1 + A/5)/2 = {1, 1, 1, . ..} (the Golden Mean), then we are able to find 0 < 3 < 1 for which there are an infinite number of solutions. For example, if a is equivalent to A/2, then from Le Veque [3, p. 252] we have 3 = VTO ...
... In fact, if we restrict a to be an irrational which is not equivalent to T = (1 + A/5)/2 = {1, 1, 1, . ..} (the Golden Mean), then we are able to find 0 < 3 < 1 for which there are an infinite number of solutions. For example, if a is equivalent to A/2, then from Le Veque [3, p. 252] we have 3 = VTO ...
Full text
... This is our main result and is presented in Theorem 3. We also show in Theorem 4 that the structure of the permutation function Qn (x) is determined by the parities of the residues r0 (x), . . . , rn−1 (x). 2. The 3x + 1 Congruence Class Triangle The following useful lemma follows immediately from t ...
... This is our main result and is presented in Theorem 3. We also show in Theorem 4 that the structure of the permutation function Qn (x) is determined by the parities of the residues r0 (x), . . . , rn−1 (x). 2. The 3x + 1 Congruence Class Triangle The following useful lemma follows immediately from t ...
