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... Our procedure in demonstrating the truth of the Theorem is to adapt and extend the technique used in [2] for positively subscripted Pell numbers. Two important differences between the criteria (1.2) in our Theorem for P_n («>0) and those in [2] for Pn (n > 0) must be noted: (i) In [2], sf=2=> £,._! ...
... Our procedure in demonstrating the truth of the Theorem is to adapt and extend the technique used in [2] for positively subscripted Pell numbers. Two important differences between the criteria (1.2) in our Theorem for P_n («>0) and those in [2] for Pn (n > 0) must be noted: (i) In [2], sf=2=> £,._! ...
The Sum of Two Squares
... Fermat was among the first to state a solution to this problem. Although he did not give a proof (perhaps it could not fit in the margin), the theorem is generally named after him. The first to give a proof was Euler, but his proof is very complicated. The following proof is in the style of Dedekind ...
... Fermat was among the first to state a solution to this problem. Although he did not give a proof (perhaps it could not fit in the margin), the theorem is generally named after him. The first to give a proof was Euler, but his proof is very complicated. The following proof is in the style of Dedekind ...
INTEGER PARTITIONS: EXERCISE SHEET 6 (APRIL 27 AND MAY 4)
... Exercise 2: Refinement of Schur’s theorem The goal of this exercise is to prove the following refinement of Schur’s theorem due to Gleissberg. Theorem 1 (Gleissberg). Let C(m, n) denote the number of partitions of n into m distinct parts congruent to 1 or 2 modulo 3. Let D(m, n) denote the number of ...
... Exercise 2: Refinement of Schur’s theorem The goal of this exercise is to prove the following refinement of Schur’s theorem due to Gleissberg. Theorem 1 (Gleissberg). Let C(m, n) denote the number of partitions of n into m distinct parts congruent to 1 or 2 modulo 3. Let D(m, n) denote the number of ...
Diophantine Olympics and World Champions: Polynomials and
... have sequences generatedby polynomialfunctionstake the field. Theorem 3. Let f(x) be a nonconstant polynomial with integer coefficients whose leading coefficient is positive. Then there exists a sequence of integers nl, n2, n3, ... such that f(n1), f(n2), f(n3), ... is a complete world champion sequ ...
... have sequences generatedby polynomialfunctionstake the field. Theorem 3. Let f(x) be a nonconstant polynomial with integer coefficients whose leading coefficient is positive. Then there exists a sequence of integers nl, n2, n3, ... such that f(n1), f(n2), f(n3), ... is a complete world champion sequ ...
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... to have been reached or even to be imminent. By way of contrast, even if it is theoretically correct to have done so, one might query whether such a problem would ever have been seriously raised in practice If it had not been for the nonexistence of certain desired natural densities. For, suppose th ...
... to have been reached or even to be imminent. By way of contrast, even if it is theoretically correct to have done so, one might query whether such a problem would ever have been seriously raised in practice If it had not been for the nonexistence of certain desired natural densities. For, suppose th ...
WaiCheungChingHo
... If the testing number satisfies either cases, it will be said as “inconclusive”. That means it could be a prime number. From Fermat’s Theorem, it concludes 341 is a prime but it is 11 * 31! Now try to use Rabin-Miller’s Algorithm. ...
... If the testing number satisfies either cases, it will be said as “inconclusive”. That means it could be a prime number. From Fermat’s Theorem, it concludes 341 is a prime but it is 11 * 31! Now try to use Rabin-Miller’s Algorithm. ...
DIFFERENTIABILITY OF A PATHOLOGICAL FUNCTION
... If x = p/q ∈ Q, let us take a sequence {xn } of irrational numbers such that xn → x when n → ∞; then f (xn ) = 0 for every n and the sequence {f (xn )} does not converge to f (x) = 1/q, so f is not continuous at x. On the other hand, for x ∈ R \ Q, let us see that f is continuous at x by checking th ...
... If x = p/q ∈ Q, let us take a sequence {xn } of irrational numbers such that xn → x when n → ∞; then f (xn ) = 0 for every n and the sequence {f (xn )} does not converge to f (x) = 1/q, so f is not continuous at x. On the other hand, for x ∈ R \ Q, let us see that f is continuous at x by checking th ...
Asymptotic formulas for coefficients of inverse theta functions
... conformal field theory [12, 24], black hole physics [10, 13], Hilbert schemes of points [18], Donaldson invariants [19], and many other topics. This paper focuses on a class of negative index Jacobi forms with a single order pole in the elliptic variable w. The analysis of the coefficients of such f ...
... conformal field theory [12, 24], black hole physics [10, 13], Hilbert schemes of points [18], Donaldson invariants [19], and many other topics. This paper focuses on a class of negative index Jacobi forms with a single order pole in the elliptic variable w. The analysis of the coefficients of such f ...
Algebraic Fractions
... We can only add/subtract fractions when they have the same denominator, so change every term so that it has the same lowest common denominator Where there are fractions in the numerator/denominator multiply throughout by the lowest common denominator to eliminate these ‘fractions within fractions’. ...
... We can only add/subtract fractions when they have the same denominator, so change every term so that it has the same lowest common denominator Where there are fractions in the numerator/denominator multiply throughout by the lowest common denominator to eliminate these ‘fractions within fractions’. ...
REU 2006 · Discrete Math · Lecture 2
... n = a1 + · · · + ak . To produce a different partition we can modify each ai by reducing 0 or 1, and fill the partition with 1’s to sum up to the same value. For example, 44 = 3 + 7 + 13 + 21 might change to 2 + 6 + 13 + 20 + 1 + 1 + 1. By this procedure we can create 2k distinct partitions (2 choic ...
... n = a1 + · · · + ak . To produce a different partition we can modify each ai by reducing 0 or 1, and fill the partition with 1’s to sum up to the same value. For example, 44 = 3 + 7 + 13 + 21 might change to 2 + 6 + 13 + 20 + 1 + 1 + 1. By this procedure we can create 2k distinct partitions (2 choic ...
MAS110 Problems for Chapter 2: Summation and Induction
... 14. The tomatina, sort of. A number of people take part in a tomato throwing contest. They stand in a field in such a way that each person has a unique nearest neighbour, and each person throws a tomato to their nearest neighbour. (This is completely reasonable: distances between contestants will al ...
... 14. The tomatina, sort of. A number of people take part in a tomato throwing contest. They stand in a field in such a way that each person has a unique nearest neighbour, and each person throws a tomato to their nearest neighbour. (This is completely reasonable: distances between contestants will al ...
Transcendental values of certain Eichler integrals,
... S2k (K) ∩ S2k (K) = {0}. This along with a result of Manin [24] implies that, for any normalized Hecke eigen cuspform f , L(j + 1, f )/π j+1 is either zero or transcendental for all 0 j 2k − 2. Applying Kohnen’s conjecture to the classical Eisenstein series, we have the transcendence of ζ(2k + 1 ...
... S2k (K) ∩ S2k (K) = {0}. This along with a result of Manin [24] implies that, for any normalized Hecke eigen cuspform f , L(j + 1, f )/π j+1 is either zero or transcendental for all 0 j 2k − 2. Applying Kohnen’s conjecture to the classical Eisenstein series, we have the transcendence of ζ(2k + 1 ...
Sketch of Lecture 18
... complicated than trying to nd these factors, but the situation is the opposite when the numbers get large. Also note. If 234 had worked out to be congruent to 1 modulo 35, then we wouldn't have learned anything for certain: 35 might be a prime, or it might not. However, we would have gathered some ...
... complicated than trying to nd these factors, but the situation is the opposite when the numbers get large. Also note. If 234 had worked out to be congruent to 1 modulo 35, then we wouldn't have learned anything for certain: 35 might be a prime, or it might not. However, we would have gathered some ...
3 - Sophia Smith
... Theorem 1 Let a, b, and c are integers. Then 1. if a | b and a | c, then a | b + c. 2. if a | b, then a | nb for all integers n. 3. if a | b and b | c, then a | c. Corollary If a, b, and c are integers such that a | b and a | c, then a | mb + nc. Theorem 2 (Division Algorithm) Let a be an integer an ...
... Theorem 1 Let a, b, and c are integers. Then 1. if a | b and a | c, then a | b + c. 2. if a | b, then a | nb for all integers n. 3. if a | b and b | c, then a | c. Corollary If a, b, and c are integers such that a | b and a | c, then a | mb + nc. Theorem 2 (Division Algorithm) Let a be an integer an ...
Methods of Proof Ch 11
... colour any map so that no countries with a common border had the same colour. This conjecture was finally proven in 1976, having first been put forward in 1852. Fermat’s Last Theorem: This stated for n, an integer greater than two, there are no positive integer values x, y, and z such that xn + yn = ...
... colour any map so that no countries with a common border had the same colour. This conjecture was finally proven in 1976, having first been put forward in 1852. Fermat’s Last Theorem: This stated for n, an integer greater than two, there are no positive integer values x, y, and z such that xn + yn = ...
Solutions - Full
... Now we see that xy is the product of 4 and an integer (namely, the integer k1 k2 ). So xy is divisible by 4. 6(a) Let a and b be real numbers. Claim 5 |ab| = |a| |b| . Proof. If a = 0 or b = 0, then |ab| = 0 = |a| |b| . Otherwise, there are four cases. 1. If a > 0 and b > 0, then |a| = a and |b| = b ...
... Now we see that xy is the product of 4 and an integer (namely, the integer k1 k2 ). So xy is divisible by 4. 6(a) Let a and b be real numbers. Claim 5 |ab| = |a| |b| . Proof. If a = 0 or b = 0, then |ab| = 0 = |a| |b| . Otherwise, there are four cases. 1. If a > 0 and b > 0, then |a| = a and |b| = b ...
Slides for Rosen, 5th edition
... What you might write • Theorem: x,y: Rational(x) Irrational(y) → Irrational(x+y) • Proof: Let x, y be any rational and irrational numbers, respectively. … (universal generalization) • Now, just from this, what do we know about x and y? You should think back to the definition of rational: • … Sinc ...
... What you might write • Theorem: x,y: Rational(x) Irrational(y) → Irrational(x+y) • Proof: Let x, y be any rational and irrational numbers, respectively. … (universal generalization) • Now, just from this, what do we know about x and y? You should think back to the definition of rational: • … Sinc ...
Solutions 1
... The trick here is to follow what we did in class when we showed that the open interval (0, 1) is uncountable, i.e., we assume it is countable and prove (by contradiction) that if all of its elements were to be enumerated, and therefore in 1-1 correspondence with N, that one could construct a sequenc ...
... The trick here is to follow what we did in class when we showed that the open interval (0, 1) is uncountable, i.e., we assume it is countable and prove (by contradiction) that if all of its elements were to be enumerated, and therefore in 1-1 correspondence with N, that one could construct a sequenc ...
(1) E x\ = n
... order, or by the sign of any xt. The following results are classical: Necessary and sufficient conditions that (1) should have solutions are: for 5 = 1, that n should be a square; for 5 = 2, that the highest power at which any prime p = 3 (mod 4) divides n should be even (possibly zero); for s = 3, ...
... order, or by the sign of any xt. The following results are classical: Necessary and sufficient conditions that (1) should have solutions are: for 5 = 1, that n should be a square; for 5 = 2, that the highest power at which any prime p = 3 (mod 4) divides n should be even (possibly zero); for s = 3, ...