Integrals Don`t Have Anything to Do with Discrete Math, Do They?
... two distinct flavours: continuous and discrete. The former is embodied by the calculus, into which many math majors delve extensively, while the latter has its own introductory course (often entitled Discrete Mathematics) whose overlap with calculus is slight. The distinction persists as we learn mo ...
... two distinct flavours: continuous and discrete. The former is embodied by the calculus, into which many math majors delve extensively, while the latter has its own introductory course (often entitled Discrete Mathematics) whose overlap with calculus is slight. The distinction persists as we learn mo ...
pythagorean theorem applications2 and outline - Mrs-Cook
... To determine if it is a right triangle, I used the Pythagorean theorem. If the theorem is proven, then the triangle is a right triangle. I plugged the numbers into the theorem (10 had to be c because it is the longest side). I squared the legs and added them together. I compared that the square of t ...
... To determine if it is a right triangle, I used the Pythagorean theorem. If the theorem is proven, then the triangle is a right triangle. I plugged the numbers into the theorem (10 had to be c because it is the longest side). I squared the legs and added them together. I compared that the square of t ...
Algebraic Numbers - Département de Mathématiques d`Orsay
... However, the same question with π and eπ has been answered: Theorem 2.2 (Nesterenko, 1996) The numbers π and eπ are algebraically independent. Concerning logarithms of algebraic numbers, log(2) and log(3) are conjectured to be algebraically independent. A more general conjecture is the following: if ...
... However, the same question with π and eπ has been answered: Theorem 2.2 (Nesterenko, 1996) The numbers π and eπ are algebraically independent. Concerning logarithms of algebraic numbers, log(2) and log(3) are conjectured to be algebraically independent. A more general conjecture is the following: if ...
Letter to the Editor
... I'm afraid there was an error in the February issue of The Fibonacci Quarterly. Mr. Shallit's proof that phi is irrational is correct up to the point where he claims that 1/0 can't be an integer. He has no basis for making that claim, as 0 was defined as a rational number, not an integer. The proof ...
... I'm afraid there was an error in the February issue of The Fibonacci Quarterly. Mr. Shallit's proof that phi is irrational is correct up to the point where he claims that 1/0 can't be an integer. He has no basis for making that claim, as 0 was defined as a rational number, not an integer. The proof ...
journal of number theory 13, 446
... perhaps to mere convergence of Y 1/jx„ . However, a weakening to x n = O(n), for instance, would probably require a completely new approach, if such a result were true at all . Of course the results of [4 ] or even [3 ] show that Theorems 2 .1-2 .3 are false for x„ = 0(n' - '), e > 0 . In the other ...
... perhaps to mere convergence of Y 1/jx„ . However, a weakening to x n = O(n), for instance, would probably require a completely new approach, if such a result were true at all . Of course the results of [4 ] or even [3 ] show that Theorems 2 .1-2 .3 are false for x„ = 0(n' - '), e > 0 . In the other ...
On integers with many small prime factors
... basic tools for nearly all proofs in this paper. The results used for the proofs of Theorems 1-5 have all been published. Since it is our main interest to indicate the possibility of the applications, we did not bother too much to get as good results as possible, but it should be noted that it is po ...
... basic tools for nearly all proofs in this paper. The results used for the proofs of Theorems 1-5 have all been published. Since it is our main interest to indicate the possibility of the applications, we did not bother too much to get as good results as possible, but it should be noted that it is po ...
The Fundamental Theorem of Arithmetic: any integer greater than 1
... 1. The Fundamental Theorem of Arithmetic: Any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers. eg. 13608 23 35 7 2. The remainder theorem: For any positive integers a b , we can find unique integers k and r such that a kb r , whe ...
... 1. The Fundamental Theorem of Arithmetic: Any integer greater than 1 can be written as a unique product (up to ordering of the factors) of prime numbers. eg. 13608 23 35 7 2. The remainder theorem: For any positive integers a b , we can find unique integers k and r such that a kb r , whe ...
ON CONGRUENCE PROPERTIES OF CONSECUTIVE VALUES OF
... The study of congruence properties of the general partition function p(n) has been a wellspring of mathematical research from the moment Ramanujan [9] first made his ground breaking studies nearly eighty years ago. Inspired by Ramanujan’s work, the results of Watson,[10] Atkin,[3] Dyson,[4] and Andr ...
... The study of congruence properties of the general partition function p(n) has been a wellspring of mathematical research from the moment Ramanujan [9] first made his ground breaking studies nearly eighty years ago. Inspired by Ramanujan’s work, the results of Watson,[10] Atkin,[3] Dyson,[4] and Andr ...
Full text
... In this paper we show that for every n there are infinitely many n~gonal numbers that can be written as the product of two other n-gonal numbers, and in fact show how to generate infinitely many such products. We suspect that our method does not generate all of the solutions for every n, but we have ...
... In this paper we show that for every n there are infinitely many n~gonal numbers that can be written as the product of two other n-gonal numbers, and in fact show how to generate infinitely many such products. We suspect that our method does not generate all of the solutions for every n, but we have ...
Solutions to selected homework problems
... Solution: To prove that any number can be represented this way just mimic the proof of Theorem 2.1. For the uniqueness suppose: ek 3k + ek−1 3k−1 + · · · + e1 3 + e0 = e0k 3k + e0k−1 3k−1 + · · · + e01 3 + e00 . Then (ek − e0k )3k + · · · + (e1 − e01 )3 + (e0 − e00 ) = 0. Note that for all j, −3 < | ...
... Solution: To prove that any number can be represented this way just mimic the proof of Theorem 2.1. For the uniqueness suppose: ek 3k + ek−1 3k−1 + · · · + e1 3 + e0 = e0k 3k + e0k−1 3k−1 + · · · + e01 3 + e00 . Then (ek − e0k )3k + · · · + (e1 − e01 )3 + (e0 − e00 ) = 0. Note that for all j, −3 < | ...
Full text
... where p, q axe positive integers,p ^ q9 and which have initial terms W0, W-L. Many properties of {wn} have been studied by Horadam [2; 3; 4] (and elsewhere), to whom some of the notation is due. We look at conditions for fewer than two intersections, exactly two intersections, and more than two inte ...
... where p, q axe positive integers,p ^ q9 and which have initial terms W0, W-L. Many properties of {wn} have been studied by Horadam [2; 3; 4] (and elsewhere), to whom some of the notation is due. We look at conditions for fewer than two intersections, exactly two intersections, and more than two inte ...
MIDTERM REVIEW FOR MATH 500 1. The limit Define limn→∞ an
... But if you are asked to to evaluate the limit by using the ε—N language, you can not apply the limit theorems directly. 2. The completeness Axiom and the least upper bound Let us first recall the definition of the least upper bound for a set A: we say L is the least upper bound of A if • L is an upp ...
... But if you are asked to to evaluate the limit by using the ε—N language, you can not apply the limit theorems directly. 2. The completeness Axiom and the least upper bound Let us first recall the definition of the least upper bound for a set A: we say L is the least upper bound of A if • L is an upp ...
Continued fractions in p-adic numbers
... converge in p-adic numbers, where the general terms qn are allowed to be non-integral rational numbers. We shall consider the continued fractions CF∗p ({qn }) with integral general terms qn having slightly different shapes from (1.1). The simplest such continued fractions are of the following type, w ...
... converge in p-adic numbers, where the general terms qn are allowed to be non-integral rational numbers. We shall consider the continued fractions CF∗p ({qn }) with integral general terms qn having slightly different shapes from (1.1). The simplest such continued fractions are of the following type, w ...
topologically equivalent measures in the cantor space
... Proof. This result follows immediately from Theorem 3.1 and the fact that the correspondence r -» p(r) is one-to-one. Theorem 3.2 can be generalized to the family of all Borel measures ft in X. Let K(p) denote the class of Borel measures which are topologically equivalent to ft. A Borel measure is d ...
... Proof. This result follows immediately from Theorem 3.1 and the fact that the correspondence r -» p(r) is one-to-one. Theorem 3.2 can be generalized to the family of all Borel measures ft in X. Let K(p) denote the class of Borel measures which are topologically equivalent to ft. A Borel measure is d ...
OF DIOPHANTINE APPROXIMATIONS
... Now repeat the application of the box principle with another pair (p',q') satisfying (3.2), where q' is larger than the common denominator q of the first solution vector. This gives rise to a second solution vector which is distinct from the first. Repeating this process, which is possible since (3. ...
... Now repeat the application of the box principle with another pair (p',q') satisfying (3.2), where q' is larger than the common denominator q of the first solution vector. This gives rise to a second solution vector which is distinct from the first. Repeating this process, which is possible since (3. ...
L. Caporaso COUNTING RATIONAL POINTS ON ALGEBRAIC
... number of rational points that any curve of fìxed genus can have. Let us look at a family X —* B of curves of genus 2 or more (we might very well view this as a family containing ali isomorphism classes of such curves). Faltings Theorem tells us that each fiber has fìnitely many rational points; the ...
... number of rational points that any curve of fìxed genus can have. Let us look at a family X —* B of curves of genus 2 or more (we might very well view this as a family containing ali isomorphism classes of such curves). Faltings Theorem tells us that each fiber has fìnitely many rational points; the ...
Exercises for Lectures 19 and 20
... c) Find the multiplicative inverse of 41 in Z43 . d) Find the multiplicative inverse of 43 in Z41 . [Note 43 ≡ 2 mod 41.] 3. Find the multiplicative inverses of all non-zero elements of Z13 . You need not use the Euclidean Algorithm. 4. In Z24 find all elements which have a multiplicative inverse an ...
... c) Find the multiplicative inverse of 41 in Z43 . d) Find the multiplicative inverse of 43 in Z41 . [Note 43 ≡ 2 mod 41.] 3. Find the multiplicative inverses of all non-zero elements of Z13 . You need not use the Euclidean Algorithm. 4. In Z24 find all elements which have a multiplicative inverse an ...
A note on two linear forms
... In 1967 H. Davenport and W. Schmidt [2] (see also Ch. 8 from Schmidt’s book [11]) proved that for any two independent linear forms L, P there exist infinitely many integer points x such that |L(x)| 6 C|P (x)| |x|−3, with a positive constant C depending on the coefficients of forms L, P . From this resu ...
... In 1967 H. Davenport and W. Schmidt [2] (see also Ch. 8 from Schmidt’s book [11]) proved that for any two independent linear forms L, P there exist infinitely many integer points x such that |L(x)| 6 C|P (x)| |x|−3, with a positive constant C depending on the coefficients of forms L, P . From this resu ...
Two Irrational Numbers That Give the Last Non
... digit of nn ; that is, N = 0.14765636 . . .. In a recent paper [1], R. Euler and J. Sadek showed that this N is a rational number with a period of twenty digits: N = 0.14765636901636567490. This is a nice result, and we might well wonder if it can be extended. Indeed, Euler and Sadek in [1] recommen ...
... digit of nn ; that is, N = 0.14765636 . . .. In a recent paper [1], R. Euler and J. Sadek showed that this N is a rational number with a period of twenty digits: N = 0.14765636901636567490. This is a nice result, and we might well wonder if it can be extended. Indeed, Euler and Sadek in [1] recommen ...
