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+ βu)|x=a = ψ(y). Then we can find u = u(1) + u(2) where u(1) solves problem with g = h = 0 and u(2) solves problem with ϕ = ψ = 0 (explain how it follows from the linearity). The second problem is also “our” problem with x and y permuted. (c) Assume that we have Neumann boundary conditions everywhere–as x = 0, x = a, ... |
′(2π) = Θ′(0). (6.4.5) We already know solution to (6.4.4)–(6.4.5): λ0 = 0, 1 2 Θ0 = λn = n2, n = 1, 2, . . . . Θn,1 = cos(nθ), Θn,2 = sin(nθ). (6.4.6) (6.4.7) Equation (6.4.3) is an Euler equation, we are looking for solutions R = rm. Then m(m − 1) + m − λ = 0 =⇒ m2 = λ. (6.4.8) Plugging λn = n2 into (6.4.8) we get m ... |
. (b) The same arguments work for Neumann and Robin boundary conditions and on each face of Γ could be its own boundary condition. (c) The same arguments work for many other equations, like heat equation, or Schr¨odinger equation. 6.6.2 Stationary equations Consider Laplace equation in the rectangular box Ω, with the D... |
lts hold in any dimension. Chapter 7. Laplace equation 238 7.2 Potential theory and around 7.2.1 Gauss formula and its applications Consider Gauss formula Ω ∇ · U dV = − Σ U · ν dS (7.2.1) where Ω is a bounded domain with the boundary Σ, dV is a volume element, dS is an area element, ν is a unit interior normal to Σ , ... |
l calculations are easy. Amazingly, the same is true for a shell between two proportional ellipsoids (Ivory theorem). Chapter 7. Laplace equation 247 7.3.2 Green’s function. I Recall (7.2.7)–(7.2.8) u(y) = Ω with G0(x, y)∆u(x) dV + Σ −u(x) ∂G0 ∂ν x (x, y) + G0(x, y) (x) dS ∂u ∂ν (7.2.7) |x − y|2−n n ̸= 2, G0(x, y) = − ... |
2z − y2z, 2z3 − 3x2z − 3y2z Table 8.1: Basis in the space of harmonic polynomials Then ΛY (ϕ, θ) = −l(l + 1)Y (ϕ, θ). (8.1.5) Definition 8.1.2. Solutions of Λv = 0 are called spherical harmonics. To find spherical harmonics we apply method of separation of variables again: Y (ϕ, θ) = Φ(ϕ)Θ(θ). Recalling Λ := ∂2 ϕ + cot... |
what values can take (l, m)? Problem 2. For equation to Φ find solutions in the form L(cos(ϕ), sin(ϕ)) where L is a polynomial of degree l with respect to cos(ϕ), sin(ϕ) in the following cases: (a) l = 0 and all possible values m; (b) l = 1 and all possible values m; (c) l = 2 and all possible values m; Chapter 8. Sep... |
1.25) This is spherical wave. Similar non-trivial solutions u = rαϕ(r − ct) − ϕ(−r − ct) do not exist for n = 2. 9.1.7 Remarks Remark 9.1.5. (a) Formula (9.1.12) could be generalized to the case of odd n ≥ 3: u(x, t) = c1−nκn ∂ ∂t 1 t ∂ ∂t +c1−nκn 1 t ∂ ∂t n−3 2 n−3 2 t−1 t−1 g(y) dσ S(x,c|t|) h(y) dσ (9.1.26) S(x,c|t|... |
h that φ(t0) = φ(t1) = 0 then f = 0 in I. I f (t)φ(t) dt = 0 Proof. Indeed, let us assume that f (¯t) > 0 at some point ¯t ∈ I (case f (¯t) < 0 is analyzed in the same way). Then f (t) > 0 in some vicinity V of ¯t. Consider function φ(x) which is 0 outside of V, φ ≥ 0 in V and φ(¯t) > 0. Then f (t)φ(t) has the same pro... |
undary condition ∂L ∂ ¨q + ∂M1 ∂ ˙q t=t1 = 0. (10.2.20) If there are no restriction, δq(t1) is also arbitrary and in addition to (10.2.20) we get − d dt ∂L ∂ ¨q + ∂L ∂ ˙q + ∂M1 ∂q t=t1 = 0. (10.2.21) We can also consider more general case of one restriction leading to αδq + βδ ˙q(t1) = 0 with (α, β) ̸= 0 and recover th... |
and we arrive to equation −∆u − λu − µ1u1 − . . . − µs−1us−1 = 0. (10.4.25) Taking an inner product with uk we arrive to −((∆ + λ)u, uk) − µk∥uk∥2 = 0 because we know that u1, . . . , us−1 are orthogonal. Further, ((∆ + λ)u, uk) = (u, (∆ + λ)uk) = (u, (−λk + λ)uk) = 0 and we conclude that µk = 0. Then (10.4.24) implies... |
rrive to −utt − K∆2u + f = 0 which is vibrating beam equation as n = 1 and vibrating plate equation as n = 2; further (10.5.36) H = 1 2 ρu2 t + i,j Ku2 xixj − 2f u dnx. (10.5.37) Example 10.5.7. Let ψ(x, t) be complex-valued function, ¯ψ(x, t) its complexconjugate and t1 S = t0 iℏ(ψ ¯ψt − ¯ψψt) − 2 ℏ2 2m 2m ∆ − V (x)) ... |
orem 11.1.2. D′ ⊃ S′ ⊃ E′ 1 ⊃ K′ (11.1.4) 2 means not only that all elements of K2 are also elements of K′ K′ 2→ f . Also in (11.1.4) each smaller 2→ f implies that fn 1: for each f ∈ K′ 2 is dense in the larger one K′ 1 there exists a 2 converging to f in K′ 1. where K′ K1 but also that fn space K′ sequence fn ∈ K′ So... |
L is an operator of convolution: Lf = g ∗ f ; Chapter 11. Distributions and weak solutions 326 (b) One can extend convolution if none of f, g has a compact support but some other assumption is fulfilled. For example, in one–dimensional case we can assume that either supp(f ) ⊂ [a, ∞), supp(g) ⊂ [a, ∞) or that supp(f ) ... |
uations. Example 11.4.1. (a) Consider equation (ajku)xj xk + j,k j (bju)xj + cu = f (11.4.1) in domain Ω. In the smooth case this equation is equivalent to j,k u Ω ajkφxj xk − j bjφxj + cφ dx = Ω f φ dx (11.4.2) Chapter 11. Distributions and weak solutions 333 for all φ ∈ C 2 0 (Ω) and we call u ∈ C(Ω) (or even worse) ... |
ϵ such that ∥v∥L2(Σ) ≤ ϵ∥∇v∥ + Cϵ∥v∥. (13.1.6) Remark 13.1.1. Actually one need only assume that Σ+ is bounded and smooth. 13.1.2 Main variational principles Theorem 13.1.4. Let Ω be a bounded domain with the smooth boundary. Then (i) There exists a sequence of eigenvalues λ1 ≤ λ2 ≤ λ3 ≤ . . ., λk → ∞ and a sequence of... |
−1)/2 (13.2.7) and more precisely N (λ) = (2π)−dmesd(Ω)λd/2 ± 1 4 (2π)1−dmesd−1(∂Ω)λ(d−1)/2 + oλ(d−1)/2 (13.2.8) where “+” corresponds to Neumann and “−” to Dirichlet boundary conditions and mesk means k-dimensional volume. Chapter 13. Eigenvalues and eigenfunctions 349 Figure 13.1: For Dirichlet boundary condition we ... |
x = y). Consider now linear combinations of u31 and u13; Comparing two last pictures we see that crossing opens under small perturbations. (d) Further, λ7 = λ8 = 13, First figure shows nodal lines for u32 (and nodal lines for u23 are exactly like this but flipped over x = y). Consider now linear combinations of u32 an... |
well. However one needs a Spectral Theorem to deal with these issues properly. 13.4.2 Spectrum: examples Example 13.4.1. Schr¨odinger operator L = − 1 2 ∆ + V (x) (13.4.5) with potential V (x) → +∞ as |x| → ∞ has a discrete spectrum: its eignevalues En → +∞ have finite multiplicities. In dimension d = 1 all these eigen... |
ering Consider −∆ as unperturbed operator and −∆ + V (x) as perturbed where V (x) is smooth fast decaying at infinity potential. We ignore possible point Chapter 13. Eigenvalues and eigenfunctions 373 spectrum (which in this case will be finite and discrete). Let us consider perturbed wave equation utt − ∆u + V (x)u = ... |
L A · n ds = D (∇ · A) dS (A.1.1) where the left-hand side expression is a linear integral, the right-hand side expression is an area integral and n is a unit inner normal to L. This is Green formula. Let V be a bounded domain in R3 and Σ = ∂V be its boundary. Then − Σ A · n dS = (∇ · A) dV D (A.1.2) where the left-han... |
edges of B(n), there must be an open left-right crossing of B(n). To connect up the two infinite closed paths starting from the top and bottom of C(n), there must be a closed top-bottom crossing. But these cannot both happen, since this would require an open primal edge crossing a closed dual edge, which is impossible.... |
t 2 [|C(0)|α3] < ∞ for some α3, we observe that this expecta- n P 1 2 (|C(0)|α3 ≥ n). (iv) To prove the last part, note that θ(p) = Pp(|C(0)| = ∞) ≤ Pp(0 ↔ ∂Bn) 29 1 Percolation III Percolation and Random Walks on Graphs for all n. By the corollary of Russo’s formula, and since {0 ↔ ∂Bn} only depends on the edges in Bn... |
resistance Reff (a, z) of an electric network is defined to be the ratio Reff (a, z) = W (a) − W (z) I for any voltage W with associated current I. The effective conductance is Ceff (a, z) = Reff (a, z)−1. Proposition. Take a weighted random walk on G. Then Pa(τz < τ + a ) = 1 c(a)Reff (a, z) , where τ + a = min{t ≥ 1 : Xt =... |
)(v(x) − v(y)) = lim n→∞ c(x, y)(vn(x) − vn(y)) = lim n→∞ in(x, y). Then by dominated convergence, we know E(i) ≤ M , and also i is a flow from 0 to ∞. Note that this connection with electrical networks only works for reversible Markov chains. Corollary. Let G ⊆ G be connected graphs. (i) If a random walk on G is recurr... |
at Wilson’s method gives the correct distribution. In Wilson’s algorithm, we pop cycles by erasing loops in the order of cycle creation. This procedure will stop will probability 1. With this procedure, we will reveal a finite set of coloured cycles O lying over a spanning tree. Let X be the set of all (O, T ), where O ... |
’s Elements has stood throughout the ages as the model of organized, rational thought carried to its ultimate perfection. Mathematicians and philosophers in every generation have tried to imitate its lucid perfection and flawless simplicity. Descartes and Leibniz dreamed of organizing all human knowledge into an axioma... |
On a different level, the positive integers form the basis of “recursion theory,” which singles out the particular way positive integers may be constructed, beginning with 1 and adding 1 each time. It therefore happens that the traditional subdivision of mathematics into subject matters has been radically altered. No ... |
a neutral element for addition, and 1 is a neutral element for multiplication. If a is any element of A, and x is an element of A such that a * x = e and x * a = e (4) then x is called an inverse of a. Roughly speaking, Equation (4) tells us that when an element is combined with its inverse it produces the neutral elem... |
des with 0]. This group, the group of integers modulo n, is represented by the symbol n. Often when working with finite groups, it is useful to draw up an “operation table.” For example, the operation table of 6 is The basic format of this table is as follows: with one row for each element of the group and one column f... |
rors in a single word (a very unlikely occurrence) might go undetected. In practice, a code is constructed as follows: in every codeword, certain positions are information positions, and the remaining positions are redundancy positions. For instance, in our code C1, the first three positions of every codeword are the i... |
ement of G. 1 Prove that if ab = e, then ba = e. (HINT: See Theorem 2.) 2 Prove that if abc = e, then cab = e and bca = e. 3 State a generalization of parts 1 and 2 Prove the following: 4 If xay = a−1, then yax = a−1. 5 Let a, b, and c each be equal to its own inverse. If ab = c, then bc = a and ca = b. 6 If abc is its... |
p of continuous function, and the negative –f of any continuous function f is a continuous function. Because any ( ), with the operation +, is a group. It is subgroup of a group is itself a group, we may conclude that denoted by 〈 ( ), +〉, or simply ( ). ( ) represents the set of all the differentiable functions from (... |
gram: Moving in the forward direction of the arrow → means multiplying by b, whereas moving in the backward direction of the arrow means multiplying by b–1: (Note that “multiplying x by b” is understood to mean multiplying on the right by b: it means xb, not bx.) It is also a convention that if a2 = e (hence a = a–1), ... |
element of A. This is the same as saying that B is the range of f. Now, suppose that f is both injective and surjective. By Definitions 1 and 2, each element of B is the image of at least one element of A, and no more than one element of A. So each element of B is the image of exactly one element of A. In this case, f... |
hematical notion, the notion of finite automata, also known as finitestate machines. A finite automaton receives information which consists of sequences of symbols from some alphabet A. A typical input sequence is a word x = x1x2 … xn, where xl x2,. . . are symbols in the alphabet A. The machine has a set of internal c... |
n A, and it is represented by the symbol SA. For any positive integer n, the symmetric group on the set {1, 2, 3,. . ., n} is called the symmetric group on η elements, and is denoted by Sn. Let us take a look at S3. First, we list all the permutations of the set {1,2,3}: This notation for functions was explained on pag... |
son belongs to clan ki, that person’s father belongs to clan c−1(ki). If a woman belongs to clan kj, her husband belongs to clan w−1(kj). 5 If any man is in the same clan as his son, then c = ε. If any woman is in the same clan as her son, then c = w. 6 If a person belongs to clan ki, the son of his mother’s sister bel... |
en tk −1 tk = (ca)(xa) (ca)(xa) = (xc)(ca) But We replace tk −1 tk by (xa)(bc) in Equation (1), as in Case II. Case IV tk −1 = (bc), where b ≠ x, a and c ≠ x, a Then tk −tk = (bc)(xa) (bc)(xa) = (xa)(bc) But We replace tk −1 tk by (xa)(bc) in Equation (1), as in Cases II and III. In Case I, we are done. In Cases II, II... |
en the two palindromes of the preceding example, for it transforms the first palindrome into the second. Our next and final example is from algebra. Consider the two groups G1 and G2 described below: Table of G1 Table of G2 G1 and G2 are different, but isomorphic. Indeed, if in G1 we replace 0 by e, 1 by a, and 2 by b,... |
c to G*. To do this, we must find an isomorphism f : G → G*. Let f be the function f(a) = πa In other words, f matches each element a in G with the permutation πa in G*. We can quickly show that f is an isomorphism: f is injective: Indeed, if f(a) = f(b) then πa = πb. Thus, πa(e) = πb(e), that is, ae = be, so, finally,... |
compute their tables. (If the group is abelian, find its regular representation.) 1 P2, the group of subsets of a two-element set. (See Chapter 3, Exercise C.) 2 4. 3 The group G of matrices described on page 28 of the text. CHAPTER TEN ORDER OF GROUP ELEMENTS Let G be an arbitrary group, with its operation denoted mul... |
rime, then ord(ab) = mn. (HINT: Use part 2.) 5 Let a and b commute. There is an element c in G whose order is lcm(m, n). (HINT: Use part 4, above, together with Exercise D3. Let c = aib where ai is a certain power of a.) 6 Give an example to show that part 1 is not true if a and b do not commute. Thus, there is no simp... |
er k which divides n. 5 Let G be an abelian group of order mn, where m and n are relatively prime. If G has an element of order m and an element of order n, G is cyclic. (See Chapter 10, Exercise E4.) 6 Let 〈a〉 be a cyclic group of order n. If n and m are relatively prime, then the function f(x) = xm is an automorphism... |
lent to y. For example, let us return to the jarful of coins we discussed earlier. If A is the set of coins in the jar, call any two coins “equivalent” if they have the same value: thus, pennies are equivalent to pennies, nickels are equivalent to nickels, and so on. If x is a particular nickel in A, then [x], the equi... |
, when we need to prove that two cosets Ha and Hb are equal, we must show that they are equal sets. What this means, of course, is that every element x ∈ Ha is in Hb, and conversely, every element y ∈ Hb is in Ha. For example, let us prove the following elementary fact: If a ∈ Hb, then Ha = Hb (1) We are given that a ∈... |
: The equation ba = ab2 tells us that we may move a factor a from the right to the left of a factor b, but in so doing, we must square b. To prove an equation such as the preceding one, move all factors a to the left of all factors b.) 5 If ba = ab3, prove that ba2 = a2b9 = a2b4, and conclude that b = b4. This is impos... |
t their parity ; their parity alone (with its arithmetic) is retained, and faithfully preserved. Another example will make this point clearer. Remember that D4 is the group of the symmetries of the square. Now, every symmetry of the square either interchanges the two diagonals here labeled 1 and 2, or leaves them as th... |
Homomorphism and the Order of Elements If f : G → H is a homomorphism, prove each of the following: 1 For each element a ∈G, the order of f(a) is a divisor of the order of a. 2 The order of any element b ≠ e in the range of f is a common divisor of |G| and |H|. (Use part 1.) 3 If the range of f has n elements, then xn... |
) Ha = H iff ab−1 ∈ H and iff a ∈ H PROOF: If Ha = Hb, then a ∈ Hb, so a = hb for some h ∈ H. Thus, ab−1 = h ∈ H If ab−1 ∈ H, then ab−1 = h for h ∈ H, and therefore a = hb ∈ Hb. It follows by Property (1) of Chapter 13 that Ha = Hb. This proves (i). It follows that Ha = He iff ae−1 = a ∈ H, which proves (ii). ■ For our... |
(for some positive integer t), what element of G has order p? 2 Suppose ord(a) is not equal to a multiple of p. Then G/〈a〉 is a group having fewer than k elements. (Explain why.) The order of G/〈a〉 is a multiple of p. (Explain why.) 3 Why must G/(a) have an element of order p? 4 Conclude that G has an element of order... |
S}. Prove: 1 S* is a subgroup of G. 2 K S*. 3 Let g be the restriction of f to S.*[That is, g(x) = f(x) for every x ∈ S*, and S* is the domain of g.] Then g is a homomorphism from S* onto S, and K = ker g. 4 S ≅ S*/K. † K. Cauchy’s Theorem Prerequisites: Chapter 13, Exercise I, and Chapter 15, Exercises G and H. If G ... |
ons of a ring. By a ring we mean a set A with operations called addition and multiplication which satisfy the following axioms: (i) A with addition alone is an abelian group. (ii) Multiplication is associative. (iii) Multiplication is distributive over addition. That is, for all a, b, and c in A, and a(b + c) = ab + ac... |
e that the ring in part 2 is a field, and indicate the multiplicative inverse of an arbitrary nonzero element. 7 Do the same for the ring in part 3. B. Ring of Real Functions 1 Verify that ( ) satisfies all the axioms for being a commutative ring with unity. Indicate the zero and unity, and describe the negative of any... |
b) ∈ B. That is, 0 is in B. Thus, B satisfies all the requirements for being a ring. For example, is a subring of because the sum of two rational numbers is rational, the product of two rational numbers is rational, and the negative of every rational number is rational. By the way, if B is a nonempty subset of A, there... |
mutative ring is a commutative ring. Any homomorphic image of a field is a field. 7 If the domain A of the homomorphism f is a field, and if the range of f has more than one element, then f is injective. (HINT: Use Exercise D6.) G. Examples of Isomorphisms 1 Let A be the ring of Exercise A2 in Chapter 17. Show that the... |
there are many practical instances in which it is possible to select an ideal J of A so as to “factor out” unwanted traits of A, and obtain a quotient ring A/J with “desirable” features. As a simple example, let A be a ring, not necessarily commutative, and let J be an ideal of A which contains all the differences ab –... |
lation property, that is, if a ≠ 0 and ab = ac then b = c (1) At the end of Chapter 17 we saw that an integral domain may also be defined as a commutative ring with unity having no divisors of zero, which is to say that if ab = 0 then a = 0 or b = 0 (2) for as we saw, (1) and (2) are equivalent properties in any commut... |
ement of A is either a divisor of zero or invertible. (HINT: Use an argument analogous to the proof of Theorem 4.) 2 Prove: If a ≠ 0 is not a divisor of zero, then some positive power of a is equal to 1. (HINT: Consider a, a2, a3,.... Since A is finite, there must be positive integers n < m such that an = am.) 3 Use pa... |
principle of mathematical induetion is applied, let Sn be the statement that that is, the sum of the first n positive integers is equal to n(n + 1)/2. Then S1 is simply which is clearly true. Suppose, next, that k is any positive integer and that Sk is true. In other words, Then, by adding k + 1 to both sides of this ... |
re very few invertible elements. As a matter of fact, r|s Theorem 2 The only invertible elements of are 1 and −1. PROOF: If s is invertible, this means there is an integer r such that rs = 1 Clearly r ≠ 0 and s ≠ 0 (otherwise their product would be 0). Furthermore, r and s are either both positive or both negative (oth... |
then gcd(a, bn) = 1. (Prove by induction.) 6 Suppose gcd(a, b) = 1 and c|ab. Then there exist integers r and s such that c = rs, r|a, s|b, and gcd(r, s) = 1. E. A Property of the gcd Let a and b be integers. Prove parts 1 and 2: # 1 Suppose a is odd and b is even, or vice versa. Then gcd(a, b) = gcd(a + b, a − b). 2 Su... |
in J. Furthermore, every integer in J is a multiple of gcd(a, n). Thus, b is a linear combination of a and n iff b ∈ J iff b is a multiple of gcd(a, n). This completes the proof of our theorem. Now that we are able to recognize when a congruence has a solution, let us see what such a solution looks like. Consider the c... |
−1 (mod p) to the power (p − 1)/2, and use Fermat’s little theorem.] # 4 Let p and q be distinct primes. Then pq − 1 + qp − 1 ≡ 1(mod pq). 5 Let p be a prime. (a) If, (p − 1) | m, then am ≡ 1 (mod p) provided that p a. (b) If, (p − 1)| m, then am + 1 ≡ a(mod pq) for all integers a. # 6 Let p and q be distinct primes. ... |
by deg a(x) For example, 1 + 2x − 3x2 + x3 is a polynomial degree 3. The polynomial 0 + 0x + 0x2 + ⋯ all of whose coefficients are equal to zero is called the zero polynomial, and is symbolized by 0. It is the only polynomial whose degree is not defined (because it has no nonzero coefficient). If a nonzero polynomial a... |
C. Rings A[x] Where A Is Not an Integral Domain 1 Prove: If A is not an integral domain, neither is A[x]. 2 Give examples of divisors of zero, of degrees 0, 1, and 2, in 4[x]. 3 In 10[x], (2x + 2)(2x + 2) = (2x + 2)(5x3 + 2x + 2), yet (2x + 2) cannot be canceled in this equation. Explain why this is possible in 10[x],... |
cient. Any polynomial whose leading coefficient is equal to 1 is called monk. Thus, every nonzero polynomial a(x) has a unique monic associate. For example, the monic associate of 3 + 4x + 2x3 is . A polynomial d(x) is called a greatest common divisor of a(x) and b(x) if d(x) divides a(x) and b(x), and is a multiple of... |
. As such, it is called a polynomial function on F. The difference between a polynomial and a polynomial function is mainly a difference of viewpoint. Given a(x) with coefficients in F: if x is regarded merely as a placeholder, then a(x) is a polynomial; if x is allowed to assume values in F, then a(x) is a polynomial ... |
OMIALS OVER AND One of the most far-reaching theorems of classical mathematics concerns polynomials with complex coefficients. It is so important in the frame work of traditional algebra that it is called the fundamental theorem of algebra. It states the following: Every nonconstant polynomial with complex coefficients... |
onclusion that in p[x], xp − x can be factored as xp − x = x(x − 1)(x − 2) ⋯ [x − (p − 1)] 3 Prove that if a(x) and b(x) determine the same function in p[x], then (xp − x)|(a(x)−b(x)) In the next four parts, let F be any finite field. # 4 Let a(x) and b(x) be in F[x]. Prove that if a(x) and b(x) determine the same func... |
of σc. By the smallest field containing F and c we mean the field which contains F and c and is contained in any other field containing F and c. It is called the field generated by F and c, and is denoted by the important symbol Now, here is what we have, in a nutshell: σc is a homomorphism with domain F[x], range F(c... |
s a root of p(cx). ) ≅ F( ) . G. Questions Relating to Transcendental Elements Let F be a field, and let c be transcendental over F. Prove the following: 1 {a(c):a(x) ∈ F[x]} is an integral domain isomorphic to F[x]. # 2 F(c) is the field of quotients of {a(c): a(x) ∈ F[x]}, and is isomorphic to F(x), the field of quot... |
tions of a1,a2, …, an is a subspace of V. (This fact is exceedingly easy to verify.) If U is the subspace consisting of all the linear combinations of a1,a2, …, an, we call U the subspace spanned by a1,a2, …, an. An equivalent way of saying the same thing is as follows: a space (or subspace) U is spanned by a1,a2, …, a... |
pendent. # 6 If {a, b, c} is linearly independent, so is {a + b, b + c, a + c}. 7 If {a1, …, an} is a basis of V, so is {k1a1, …, k nan } for any nonzero scalars 8 The space spanned by {a1, …,an} is the same as the space spanned by {b1, …, bm} iff each ai is a linear combination of b1, …, bm, and each bj is a linear co... |
every element of K is algebraic over F PROOF: Indeed, suppose K is of degree n; over F, and let c be any element of K. Then the set {1,c,c2, …, cn) is linearly dependent, because it has n + 1 elements in a vector space K of dimension n. Consequently, there are scalars a0,…,an ∈ F, not all zero, such that a0 + a1c + ⋯ +... |
a away. The final resolution of these problems, by proving that the required constructions are impossible, came from a most unlikely source: it was a by-product of the arcane study of field extensions, in the upper reaches of modern algebra. To understand how all this works, we will see how the process of ruler-and-com... |
are constructible numbers. 2 cos 3 If cos α, cos 4 cos 5 If α and β are constructible angles, so are # 6 The following angles are constructible: 7 The following angles are not constructible: 20°; 40°, 140°. (HINT: Use the proof of Theorem 3.) , and nα for any positive integer n. , then cos (α + ß), cos iff cos . . . . ... |
swer, quite clearly, is to let send the polynomial with coefficients a0, a1, …, an to the polynomial with coefficients h(a0), h(a1), …, h(an): (a0 + a1x + ⋯ + anxn) = h(a0) + h(a1)x + ⋯ + h(an)xn It is child’s play to verify formally that polynomial isomorphism, a(x) is irreducible iff ha(x) is irreducible. (a(x)), obt... |
olynomial p(x) in F[x] is said to be separable over F if it has no multiple roots in any extension of F. If p(x) does have a multiple root in some extension, it is inseparable over F. 1 Prove that if F has characteristic 0, every irreducible polynomial in F[x] is separable. Thus, for characteristic 0, there is no quest... |
hat we can now use all of our accumulated knowledge about groups to help us analyze the solutions of polynomial equations. And that is precisely what Galois theory is all about. If K is the root field of a polynomial a(x) in F[x], the group of all the automorphisms of K which fix F is called the Galois group of a(x). W... |
page 323 to explain why every element of Gal( permutation of the three cube roots of 2. , i 5 Use part 4 to prove that Gal( ] = 6. ) : ) : ) : , i , ( ( ( ( ) ≅ S3. ), then show that [ designates the real cube root of 2. ( , i ): ( )] = 2. Conclude that ) has six elements. Then use the discussion following ) may be id... |
eorem: Theorem 1 Any finite group of nonzero elements in a field is a cyclic group. (The operation in the group is the field’s multiplication.) PROOF: If F* denotes the set of nonzero elements of F, suppose that G ⊆ F*, and that G, with the field’s “multiply” operation, is a group of n elements. We will compare G with ... |
nct 5-tuples of this kind are there? Well, if we select a, b, c, and d at random, there is a unique k = d−1c−1b−1a−1 in G making abcdk = e. Thus, there are n4 such 5-tuples. Call two 5-tuples equivalent if one is merely a cyclic permutation of the other. Thus, (a, b, c, d, k) is equivalent to exactly five distinct 5-tu... |
than or equal to 2. We may also symbolize this set by {x: x ∈ and x ≤ 2} {x ∈ : x ≤ 2} to be read as “the set of all x in satisfying x ≤ 2.” If A and B are sets, there are several ways of forming new sets from A and B: 1. The union of A and B (denoted by A elements of B. Here, the word “or” is used in the “inclusive” s... |
integers is equal to n(n + 1)/2. Let A consist of all the positive integers n for which Equation (1) is true. Then 1 is in A because Next, suppose that k is any positive integer in A; we must show that, in that case, k + 1 also is in A. To say that k is in A means that By adding k + 1 to both sides of this equation, we... |
n even number of 1s). To and Te may be described as follows: To(s0), = s1, To(s1) = s0 Te(s1) = s1 Te(s0), = s0, Now, Te ∘ To = Toe by part 3. Since e is a sequence with an even number of 1s and o is a sequence with an odd number of 1s, oe is a sequence with an odd number of 1s; hence Toe = To. Thus, Te ∘ To = To. Simi... |
es. E6 Every element of G6 If G is cyclic, then necessarily G ≅ p2. (Why?) / is a coset + (m/n). If G is not cyclic, then every element x ≠ e in G has order p. (Why?) Take any two elements a ≠ e and b ≠ e in G where b is not a power of a. Complete the problem. CHAPTER 16 D1 Let f ∈ Aut(G); that is, let f be an isomorph... |
ete.) E5 If a(x) = a0 + a1x + ⋯ anxn ∈ J, then a0 + a1 + ⋯ + an = 0. If b(x) is any element in A[x], say b(x) = b0 + b1x + ⋯ + bmxm, let B = b0 + b1 + ⋯ + bm. Explain why the sum of all the coefficients in a(x)b(x) is equal to (a0 + a1 ⋯ + an)B. Supply all remaining details. G3 If h is surjective, then every element of... |
nd d2 is a root of x2 − [(a/2) + d1]. Explain why (a) F(d1, d2) = F(c) and (b) F(d1, d2) contains all the roots of p(x). , ) = + + ). , , , ( ( ( ( D3 From page 313, ) = As in the illustration for Theorem 2, taking t = 1 gives c = ); hence ( D5 Use the result of part 3. The degree of polynomial p(x) of degree 8 having ... |
bgroup, 165 Sylow’s theorem, 165 Symmetric difference, 30 Symmetric group, 71 Transcendental elements, 273 Transposition, 83 Unique factorization, 222, 255 Unity, ring with, 172 Vector space, 282-291 basis of, 286 dimension of, 287 Weight, 55 Well-ordering property, 210 Wilson’s theorem, 237 |
e. If not, we iterate this procedure. The procedure stops, because 0 ≤ D2 1 = (π∗ 1D2 + m2E2) < D2. Exercise. We have the following universal property of blow ups: If Z → X is a birational map of surfaces, and suppose f −1 is not defined at a point x ∈ X. Then f factors uniquely through the blow up Blx(X). Theorem. Let ... |
rem. We do not have a similarly precise statement for arbitrary varieties, but an “asymptotic” version is good enough for what we want. Theorem (Asymptotic Riemann–Roch). Let X be a projective normal variety over K = ¯K. Let D be a Cartier divisor, and E a Weil divisor on X. Then χ(X, OX (mD + E)) is a numerical polyno... |
or all irreducible components Xi. (iii) If V ⊆ X is a proper subscheme, and D is nef, then D|V is nef. (iv) If f : X → Y is a finite morphism of proper schemes, and D is nef on Y , then f ∗D is nef on X. The converse is true if f is surjective. The last one follows from the analogue of Nakai’s criterion, called Kleinman... |
(1 − p)(n−2)−(r−2) = n(n − 1)p2. The sum goes to 1 since it is the sum of all probabilities of a binomial N (n − 2, p) So E[X 2] = n(n − 1)p2 + E[X] = n(n − 1)p2 + np. So var(X) = E[X 2] − (E[X])2 = np(1 − p) = npq. Example (Poisson distribution). If X ∼ P (λ), then E[X] = λ, and var(X) = λ, since P (λ) is B(n, p) with... |
(Covariance). Given two random variables X, Y , the covariance is cov(X, Y ) = E[(X − E[X])(Y − E[Y ])]. 34 3 Discrete random variables IA Probability Proposition. (i) cov(X, c) = 0 for constant c. (ii) cov(X + c, Y ) = cov(X, Y ). (iii) cov(X, Y ) = cov(Y, X). (iv) cov(X, Y ) = E[XY ] − E[X]E[Y ]. (v) cov(X, X) = var(... |
vidual. At each iteration, the individual produces a random number of offsprings. In the next iteration, each offspring will individually independently reproduce randomly according to the same distribution. We will ask questions such as the expected number of individuals in a particular generation and the probability o... |
variable that is neither discrete nor continuous might look like this: P Note that we always have P(a < x ≤ b) = F (b) − F (a). This will be equal to b Definition (Uniform distribution). The uniform distribution on [a, b] has pdf a f (x) dx in the case of continuous random variables. f (x) = 1 b − a . F (x) = x a f (z)... |
ould be explicit in what we mean by that. Example (Buffon’s needle). A needle of length ℓ is tossed at random onto a floor marked with parallel lines a distance L apart, where ℓ ≤ L. Let A be the event that the needle intersects a line. What is P(A)? X ℓ θ L Suppose that X ∼ U [0, L] and θ ∼ U [0, π]. Then f (x, θ) = 1... |
ng into the first and last bins are f (y1)δ and f (yn)δ. Then the probability of Y1 ∈ [y1, y1 + δ) and Yn ∈ (yn − δ, yn] is n(n − 1)(F (yn) − F (y1))n−2f (y1)f (yn)δ2, and the result follows. We can also find the joint distribution of the order statistics, say g, since it is just given by g(y1, · · · yn) = n!f (y1) · ·... |
orem). If the random variables X1, X2, · · · have mgf’s m1(θ), m2(θ), · · · and mn(θ) → m(θ) as n → ∞ for all θ, then Xn →D the random variable with mgf m(θ). We now provide a sketch-proof of the central limit theorem: Proof. wlog, assume µ = 0, σ2 = 1 (otherwise replace Xi with Xi−µ ). σ Then mXi(θ) = E[eθXi] = 1 + θE... |
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