Counting and Numbering - of the Irish Mathematical Olympiad
... algebra, and combinatorics. They require no knowledge of higher mathematics such as calculus and analysis, and solutions are often short and elementary. However, they are usually disguised so as to make the process of finding the solutions difficult. ...
... algebra, and combinatorics. They require no knowledge of higher mathematics such as calculus and analysis, and solutions are often short and elementary. However, they are usually disguised so as to make the process of finding the solutions difficult. ...
3x3+5x2+8x+7 by 3x+2
... Suppose f(x) is divided by x - k using synthetic division. 1. If k≥ 0 and every number in the last line is nonnegative (pos. or zero), then k is an upper bound for the real zeros of f. 2. If k ≤ 0 and the numbers in the last line alternate nonnegative and nonpositive, then k is a lower bound for the ...
... Suppose f(x) is divided by x - k using synthetic division. 1. If k≥ 0 and every number in the last line is nonnegative (pos. or zero), then k is an upper bound for the real zeros of f. 2. If k ≤ 0 and the numbers in the last line alternate nonnegative and nonpositive, then k is a lower bound for the ...
Transcendental values of class group L-functions,
... formula. Indeed, our proof of Theorem 6 leads to a simple proof of the Chowla– Selberg formula which we give in Sect. 7. Naturally, this leads one to enquire about the transcendence of special values of√the -function. Not much is known in this context. It is well-known that (1/2) = π is transcende ...
... formula. Indeed, our proof of Theorem 6 leads to a simple proof of the Chowla– Selberg formula which we give in Sect. 7. Naturally, this leads one to enquire about the transcendence of special values of√the -function. Not much is known in this context. It is well-known that (1/2) = π is transcende ...
THE E.IRREGULAR PRIMES
... primes is infinite. The proof is similar to the corresponding proof mentioned above. one might therefore expect that it would also be easy to obtain results on the distribution of .E-irregular primes, by modifying suitable methods used in connection with the ordinary irregular primes. It seems, howe ...
... primes is infinite. The proof is similar to the corresponding proof mentioned above. one might therefore expect that it would also be easy to obtain results on the distribution of .E-irregular primes, by modifying suitable methods used in connection with the ordinary irregular primes. It seems, howe ...
Full text
... then have H(a9 b) = ( a - !)(& - 1)* Using this together with Lemma 1.2S we have H(k9 ri) = H(ad9 bd) = adbd - a M + d(a - 1) (fc - 1) = a d M - aid7 + a&d - da - db + d = nk - k - ri + d» From Corollary 1-1, it follows that the function H is commutative (symmetric), E(k9 n) =ff(n,/c). The function ...
... then have H(a9 b) = ( a - !)(& - 1)* Using this together with Lemma 1.2S we have H(k9 ri) = H(ad9 bd) = adbd - a M + d(a - 1) (fc - 1) = a d M - aid7 + a&d - da - db + d = nk - k - ri + d» From Corollary 1-1, it follows that the function H is commutative (symmetric), E(k9 n) =ff(n,/c). The function ...
Full text
... We now offer an alternate proof of Theorem 3, based on combinatorics. Proof. (Second.) There is a bijection between necklaces of n beads of which k are black, the rest white, and cyclic compositions of n into k parts as follows: Given such a necklace, start at any black bead. Let n1 denote the numbe ...
... We now offer an alternate proof of Theorem 3, based on combinatorics. Proof. (Second.) There is a bijection between necklaces of n beads of which k are black, the rest white, and cyclic compositions of n into k parts as follows: Given such a necklace, start at any black bead. Let n1 denote the numbe ...
Math Glossary - Waioneke School
... The list below explains the meaning of some of the key terms used in the math: Meaning ...
... The list below explains the meaning of some of the key terms used in the math: Meaning ...
Full tex
... [8], McCranie, [1], and Nash [5]. Several examples have been found of hyperperfect numbers with two, three and four different prime factors and one such number with five different prime factors was discovered be te Riele [8]. In this paper we include some new hyperperfect numbers with five different ...
... [8], McCranie, [1], and Nash [5]. Several examples have been found of hyperperfect numbers with two, three and four different prime factors and one such number with five different prime factors was discovered be te Riele [8]. In this paper we include some new hyperperfect numbers with five different ...
Complex Numbers - Berkeley City College
... irrational numbers to solve polynomial equations where the coefficients are only rational numbers. x2 + 1 = 0 This is an equation where the coefficients are all real numbers, but the solutions, ±i, are not real numbers, they are complex numbers. We need to involve the complex numbers to solve polyno ...
... irrational numbers to solve polynomial equations where the coefficients are only rational numbers. x2 + 1 = 0 This is an equation where the coefficients are all real numbers, but the solutions, ±i, are not real numbers, they are complex numbers. We need to involve the complex numbers to solve polyno ...
Monday, August 8: Samples of Proofs
... Let a b = a + b + 4. Then a k = a + k + 4 and if a + k + 4 = a, then k = -4. Consider k = 4. a 4 = a + 4 + 4 = a for all real values of a, and 4 a = 4 + a + 4 = a for all real numbers a. Hence if a b = a + b + 4 , then a 4 = 4 a = a for all a Reals. Proof by Induction: (Set up ...
... Let a b = a + b + 4. Then a k = a + k + 4 and if a + k + 4 = a, then k = -4. Consider k = 4. a 4 = a + 4 + 4 = a for all real values of a, and 4 a = 4 + a + 4 = a for all real numbers a. Hence if a b = a + b + 4 , then a 4 = 4 a = a for all a Reals. Proof by Induction: (Set up ...
12.1
... Example 2 – Finding the nth Term of a Sequence Find the nth term of a sequence whose first several terms are given. (a) (b) –2, 4, –8, 16, –32, . . . Solution: (a) We notice that the numerators of these fractions are the odd numbers and the denominators are the even numbers. Even numbers are of the ...
... Example 2 – Finding the nth Term of a Sequence Find the nth term of a sequence whose first several terms are given. (a) (b) –2, 4, –8, 16, –32, . . . Solution: (a) We notice that the numerators of these fractions are the odd numbers and the denominators are the even numbers. Even numbers are of the ...