Problem Set 1 - University of Oxford
... Solution: From the theory of cyclic groups, a homomorphism from the abelian group Z to any group is completely determined by the image of 1 ∈ Z: if f : Z → (R, +) is a homomorphism of abelian groups then f (n) = f (1)+f (1)+. . . f (1) (n terms on the righthand side) when n ≥ 0 and f (n) = −f (−n) ...
... Solution: From the theory of cyclic groups, a homomorphism from the abelian group Z to any group is completely determined by the image of 1 ∈ Z: if f : Z → (R, +) is a homomorphism of abelian groups then f (n) = f (1)+f (1)+. . . f (1) (n terms on the righthand side) when n ≥ 0 and f (n) = −f (−n) ...
18.781 Problem Set 3
... Due Monday, September 26 in class. In each of these problems, the last part is meant to be somewhat harder than the others. 1(a). Give a rule analogous to “casting out nines” to find the remainder when the decimal numeral ak ak−1 · · · a0 a1 is divided by seven. (Hint: for three-digit numbers, the r ...
... Due Monday, September 26 in class. In each of these problems, the last part is meant to be somewhat harder than the others. 1(a). Give a rule analogous to “casting out nines” to find the remainder when the decimal numeral ak ak−1 · · · a0 a1 is divided by seven. (Hint: for three-digit numbers, the r ...
3.3 Polynomial Division: Factors and Zeros
... Be sure to write for missing coefficients . Beginning from the left, add each column and then write the product of the result and above the line in the next column. The final number written is the remainder. The other numbers are the coefficients of the quotient. ...
... Be sure to write for missing coefficients . Beginning from the left, add each column and then write the product of the result and above the line in the next column. The final number written is the remainder. The other numbers are the coefficients of the quotient. ...
Typed - CEMC
... Instructor’s Comments: This is the 25 minute mark Theorem: (Remainder Theorem (RT)) Suppose that f (x) ∈ F[x] and that c ∈ F. Then, the remainder when f (x) is divided by x − c is f (c). Proof: By the Division Algorithm for Polynomials, there exists unique q(x) and r(x) in F[x] such that f (x) = (x ...
... Instructor’s Comments: This is the 25 minute mark Theorem: (Remainder Theorem (RT)) Suppose that f (x) ∈ F[x] and that c ∈ F. Then, the remainder when f (x) is divided by x − c is f (c). Proof: By the Division Algorithm for Polynomials, there exists unique q(x) and r(x) in F[x] such that f (x) = (x ...
INTRODUCTION TO ALGEBRA II MIDTERM 1 SOLUTIONS Do as
... b.– How many generators are there in this group ? (it is not necessary to compute them all, though that’s certainly a method to answer the question). Solution: We have seen in class, and actually proved again in the last ...
... b.– How many generators are there in this group ? (it is not necessary to compute them all, though that’s certainly a method to answer the question). Solution: We have seen in class, and actually proved again in the last ...
Sol 1 - D-MATH
... and Nick will all have pizza on the 333rd day of 2014, Saturday, November 29. Note that exercise 4 holds more generally for a finite family of ideals I1 , . . . , In with the property that Ik +Il = R for any two distinct ideals (i.e. k 6= l). Then (a) translates to I1 · · · · · In = I1 ∩ · · · ∩ In ...
... and Nick will all have pizza on the 333rd day of 2014, Saturday, November 29. Note that exercise 4 holds more generally for a finite family of ideals I1 , . . . , In with the property that Ik +Il = R for any two distinct ideals (i.e. k 6= l). Then (a) translates to I1 · · · · · In = I1 ∩ · · · ∩ In ...
Full text
... [1] the generalization of the first kind, hence the title of this report. Later, Andre-Jeannin [2] also generalized the recurrence relation (1); thus, from the line-sequential point of view, generalized the M-space itself We call this latter way of generalization the generalization of the second kin ...
... [1] the generalization of the first kind, hence the title of this report. Later, Andre-Jeannin [2] also generalized the recurrence relation (1); thus, from the line-sequential point of view, generalized the M-space itself We call this latter way of generalization the generalization of the second kin ...
Another form of the reciprocity law of Dedekind sum
... We prove Theorem 1.1 by two mutually distinct ways. In the first proof in §2, we use the reciprocity law successively and also use the Matsumoto-Montesinos formula [7], which is proved by purely elementary number theoretic method (see also Remark 2.2). The proof of Proposition 1.2 is elementary, whic ...
... We prove Theorem 1.1 by two mutually distinct ways. In the first proof in §2, we use the reciprocity law successively and also use the Matsumoto-Montesinos formula [7], which is proved by purely elementary number theoretic method (see also Remark 2.2). The proof of Proposition 1.2 is elementary, whic ...
[Part 3]
... to M = 200. Also, these numbers have other unusual characteristics. Add any two and the sum will be some one or the other of the numbers or, if the sum is greater than 200, subtract M = 200 and the remainder will be found somewhere in the list Subtract any two numbers with the same result. Of course ...
... to M = 200. Also, these numbers have other unusual characteristics. Add any two and the sum will be some one or the other of the numbers or, if the sum is greater than 200, subtract M = 200 and the remainder will be found somewhere in the list Subtract any two numbers with the same result. Of course ...
Characterizing integers among rational numbers
... It is not known whether there exists an algorithm for the analogous problem with Z replaced by the field Q of rational numbers. But Robinson showed that the full first-order theory of Q is undecidable: she reduced the problem to the corresponding known result for Z by showing that Z could be defined ...
... It is not known whether there exists an algorithm for the analogous problem with Z replaced by the field Q of rational numbers. But Robinson showed that the full first-order theory of Q is undecidable: she reduced the problem to the corresponding known result for Z by showing that Z could be defined ...
Special Products – Blue Level Problems In
... In Exercises 11 - 13, Rick Claims that if you multiply four consecutive integers together and add 1, you always get a perfect square. 11. Show that Rick’s statement is true if the smallest of the integers is 2. 12. Show that Rick’s statement is true if the smallest of the integers is 3. 13. Suppose ...
... In Exercises 11 - 13, Rick Claims that if you multiply four consecutive integers together and add 1, you always get a perfect square. 11. Show that Rick’s statement is true if the smallest of the integers is 2. 12. Show that Rick’s statement is true if the smallest of the integers is 3. 13. Suppose ...
Number Theory Week 9
... in the example given in Lecture 16. For example, the element x ∈ Z105 such that x ≡ 2 (mod 3), x ≡ 1 (mod 5) and x ≡ 5 (mod 7) is 26. The other two solutions of x3 = 41 are 94 (corresponding to (2,1,3)) and 41 (corresponding to (2,1,6)). Let f (x) = xk +a1 xk−1 +· · ·+ak−1 x+ak be a polynomial over ...
... in the example given in Lecture 16. For example, the element x ∈ Z105 such that x ≡ 2 (mod 3), x ≡ 1 (mod 5) and x ≡ 5 (mod 7) is 26. The other two solutions of x3 = 41 are 94 (corresponding to (2,1,3)) and 41 (corresponding to (2,1,6)). Let f (x) = xk +a1 xk−1 +· · ·+ak−1 x+ak be a polynomial over ...
MS Word
... @- First, I would like to thank William for his talk last week in which he explained some of the many ways in which people have misinterpreted Godels first incompleteness theorem -In this talk I hope, among other things, to give a proof of this theorem -I will then leave you to make your own misinte ...
... @- First, I would like to thank William for his talk last week in which he explained some of the many ways in which people have misinterpreted Godels first incompleteness theorem -In this talk I hope, among other things, to give a proof of this theorem -I will then leave you to make your own misinte ...
Full text
... in the theory of numbers. Such representations are useful in lattice point problems, crystallography, and certain problems in mechanics [6, pp. 1-4]. If rk{ri) denotes the number of representations of an integer n as a sum of k squares, Jacobi's two- and four-square theorems [9] are: ...
... in the theory of numbers. Such representations are useful in lattice point problems, crystallography, and certain problems in mechanics [6, pp. 1-4]. If rk{ri) denotes the number of representations of an integer n as a sum of k squares, Jacobi's two- and four-square theorems [9] are: ...
Full text
... where / ( / ) is a differentiate function defined on the real number interval [a, b), and x may be a real or complex number with x*0 and x*l. Obviously, the case for x = l of (1.1) could be generally treated by means of the well-known Euler-Maclaurin summation formula. The object of this paper is to ...
... where / ( / ) is a differentiate function defined on the real number interval [a, b), and x may be a real or complex number with x*0 and x*l. Obviously, the case for x = l of (1.1) could be generally treated by means of the well-known Euler-Maclaurin summation formula. The object of this paper is to ...
4.1-4.3 Review
... 4.1 Polynomial Functions Complex numbers o Notation o Addition, subtraction, multiplication, solving with i Fundamental Theorem of Algebra o Use to find roots o Use to write polynomials for given roots 4.2 Quadratic Equations Quadratic formula Completing the square Use the discriminant to ...
... 4.1 Polynomial Functions Complex numbers o Notation o Addition, subtraction, multiplication, solving with i Fundamental Theorem of Algebra o Use to find roots o Use to write polynomials for given roots 4.2 Quadratic Equations Quadratic formula Completing the square Use the discriminant to ...
Full text
... f-k^x) ~ xl t n e Euler totient function
... f-k^x) ~ xl t n e Euler totient function
t(n) [7] for k = 1 and f\(x) = x(x + 1 ) ... (x + t - 1 ) , t > 1; also the totients investigated by Nagell [5], Alder [1], and others (cf. [8]). The aim of this paper is to establish an asymptotic formula for the summato ...
Full text
... Ewell [3] gave two recurrence formulas for q(2l) and q(2i + 1) for nonnegative integers £ in a slightly different, but equivalent, form to that in formula (2). This paper presents a recursion-type formula for p*(n), the number of partitions of n into parts not divisible by k9 where k is some given i ...
... Ewell [3] gave two recurrence formulas for q(2l) and q(2i + 1) for nonnegative integers £ in a slightly different, but equivalent, form to that in formula (2). This paper presents a recursion-type formula for p*(n), the number of partitions of n into parts not divisible by k9 where k is some given i ...
A2 – Section 5.5 Date
... Factor Theorem: x – h is a factor of f(x) _______________________ f(h) = 0. Remainder Theorem: If f(x) is divided by (x – h), then, the remainder equals f(h), ____________. Synthetic Division: Uses _________________ to divide ...
... Factor Theorem: x – h is a factor of f(x) _______________________ f(h) = 0. Remainder Theorem: If f(x) is divided by (x – h), then, the remainder equals f(h), ____________. Synthetic Division: Uses _________________ to divide ...
Parry A
... Pythagoras (585-500 B.C.) had a partial solution of x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 Plato's (388 B.C.) determined a partial solution of x = 2n, y = n2 - 1, and z = n2 + 1 Neither of the above accounts for all the triplets, and it was not until Euclid wrote in his Elements the follow ...
... Pythagoras (585-500 B.C.) had a partial solution of x = 2n + 1, y = 2n^2 + 2n, and z = 2n^2 + 2n + 1 Plato's (388 B.C.) determined a partial solution of x = 2n, y = n2 - 1, and z = n2 + 1 Neither of the above accounts for all the triplets, and it was not until Euclid wrote in his Elements the follow ...
a reciprocity theorem for certain hypergeometric series
... In Entry 2 of Chapter 14 in his second notebook [4], Ramanujan states a beautiful reciprocity theorem (with no hypotheses or proof) for certain hypergeometric series. In his notebooks [4], Ramanujan recorded many “reciprocity theorems” or “modular relations” for infinite series, but we are unaware o ...
... In Entry 2 of Chapter 14 in his second notebook [4], Ramanujan states a beautiful reciprocity theorem (with no hypotheses or proof) for certain hypergeometric series. In his notebooks [4], Ramanujan recorded many “reciprocity theorems” or “modular relations” for infinite series, but we are unaware o ...
Talent 01V
... in R + . Let us fix a positive real number α and define x y = α(x + y), so that is certainly a binary operation on R + . Furthermore, it is easy to check directly that the weak associative law (x y) x = x (y x) holds. In fact, since it is clear that the commutative law x y = y x is satisfied, we get ...
... in R + . Let us fix a positive real number α and define x y = α(x + y), so that is certainly a binary operation on R + . Furthermore, it is easy to check directly that the weak associative law (x y) x = x (y x) holds. In fact, since it is clear that the commutative law x y = y x is satisfied, we get ...
MATH 363 Discrete Mathematics SOLUTIONS: Assignment 7 1
... Using the notation of the master theorem we have a = 9, b = 3, c = 20, d = 2. The condition on the recursive formula is satisfied for all n multiple of b = 3, in particular it is satisfied for all n which is a power of b. Since a = 9 = bd , we have that f (n) is O(n2 log n). • g(n) = 16g(n/4) + n n ...
... Using the notation of the master theorem we have a = 9, b = 3, c = 20, d = 2. The condition on the recursive formula is satisfied for all n multiple of b = 3, in particular it is satisfied for all n which is a power of b. Since a = 9 = bd , we have that f (n) is O(n2 log n). • g(n) = 16g(n/4) + n n ...
CMPS 12A
... 6. Use limits to prove the following (these are some of the exercises at the end of the asymptotic growth rates handout): a. If P (n) is a polynomial of degree k 0 , then P(n) (n k ) . b. For any positive real numbers and : n o(n ) iff , n (n ) iff , and n (n ) ...
... 6. Use limits to prove the following (these are some of the exercises at the end of the asymptotic growth rates handout): a. If P (n) is a polynomial of degree k 0 , then P(n) (n k ) . b. For any positive real numbers and : n o(n ) iff , n (n ) iff , and n (n ) ...
2011 U OF I FRESHMAN MATH CONTEST Solutions
... the period of this expansion. Now consider a block B = 0 . . . 0 consisting of p consecutive digits 0. Since any integer of the form 10k with k ≥ p contains p consecutive 0’s, this block must occur infinitely often in the decimal expansion of x. By our assumption that this expansion is ultimately pe ...
... the period of this expansion. Now consider a block B = 0 . . . 0 consisting of p consecutive digits 0. Since any integer of the form 10k with k ≥ p contains p consecutive 0’s, this block must occur infinitely often in the decimal expansion of x. By our assumption that this expansion is ultimately pe ...
Chinese remainder theorem
The Chinese remainder theorem is a result about congruences in number theory and its generalizations in abstract algebra. It was first published in the 3rd to 5th centuries by the Chinese mathematician Sun Tzu.In its basic form, the Chinese remainder theorem will determine a number n that, when divided by some given divisors, leaves given remainders. For example, what is the lowest number n that when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 2?