Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
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Theorem 39. Let \( G \) be a Lie group. A function \( f \) defined on \( G \) is of class \( {C}^{p} \) if and only if, for all \( X \in {G}^{\prime }\left( e\right) \), the derivative \[ R\left( X\right) f\left( g\right) = {D}_{0}f\left\lbrack {g \cdot \exp \left( {tX}\right) }\right\rbrack \] exists and is a function... | Let us first show that the map \( X \mapsto R\left( X\right) f \) is linear. Indeed, suppose that, for given \( X \), the derivative \( R\left( X\right) f \) exists and is a continuous function of \( g \) . Replacing \( g \) by \( g.\exp \left( {tX}\right) \), the function \( t \mapsto f\left\lbrack {g.\exp \left( {tX}... | Yes |
Theorem 40. Let \( \mathcal{H} \) be a Banach space, \( G \) a \( {C}^{\infty } \) manifold and p a map from \( G \) to the dual \( {\mathcal{H}}^{\prime } \) of \( \mathcal{H} \) . \( p \) is \( {C}^{\infty } \) if and only if the function \( \langle \mathbf{x} \mid p\left( g\right) \rangle \) is \( {C}^{\infty } \) f... | The question being local, \( G \) may be assumed to be open in \( {\mathbb{R}}^{n} \) . We first suppose that \( n = 1 \) and set \( D = d/{dt} \) . By its very definition, the derivative \( D\langle \mathbf{x} \mid p\left( t\right) \rangle \) is the limit of a sequence of continuous linear functionals in \( \mathbf{x}... | Yes |
Corollary 1. Let \( p \) be a map from a manifold \( G \) to a Banach space \( \mathcal{H} \) . \( p \) is \( {C}^{\infty } \) if and only if \( \langle p\left( g\right) \mid \mathbf{x}\rangle \) is \( {C}^{\infty } \) for all \( \mathbf{x} \in {\mathcal{H}}^{\prime } \) . | The proof follows by regarding \( p \) as a map with values in the dual \( {}^{109} \) of \( {\mathcal{H}}^{\prime } \) . | No |
Corollary 2. Let \( \left( {\mathcal{H},\pi }\right) \) be a continuous linear representation of a Lie group \( G \) on a Banach space \( \mathcal{H} \) . Then \( {\mathcal{H}}^{\infty } \) is the set of \( \mathbf{a} \in \mathcal{H} \) such that \( \langle \pi \left( g\right) \mathbf{a} \mid \mathbf{x}\rangle \) is \(... | Apply corollary 1 to \( p\left( g\right) = \pi \left( g\right) \mathbf{a} \) . | No |
Theorem 41. For any representation \( \left( {\pi ,\mathcal{H}}\right) \) of a Lie group \( G \) on a Banach space, the Gårding subspace is identical to the subspace \( {\mathcal{H}}^{\infty } \) : every \( \mathbf{a} \in {\mathcal{H}}^{\infty } \) is of the form\n\n\[ \mathbf{a} = \pi \left( {\varphi }_{1}\right) {\ma... | For all \( i \), the supports of \( {\varphi }_{i} \) can also be required to be in a given arbitrary neighbourhood of \( e \) . Inspired by methods from PDE theory, Dixmier’s and Malli-avin's very ingenious ten page proof requires little knowledge and even covers the case of representations on Fréchet spaces. The firs... | No |
Theorem 45. Let \( \left( {\mathcal{H},\pi }\right) \) be an irreducible unitary representation of \( G \) and \( \mu \) a distribution with compact support. If \( \mu \) is central, the operator \( {\pi }^{\infty }\left( \mu \right) \) is a scalar. | For all scalars \( \lambda \), the operator\n\n\[ S = \left\lbrack {{\pi }^{\infty }\left( \widetilde{\mu }\right) + \bar{\lambda }1}\right\rbrack \left\lbrack {{\pi }^{\infty }\left( \mu \right) + ╏}\right\rbrack : {\mathcal{H}}^{\infty } \rightarrow {\mathcal{H}}^{\infty } \]\n\nis positive symmetric by (16) and comm... | Yes |
Theorem 46 (M. H. Stone). Let \( t \mapsto U\left( t\right) \) be a unitary representation of \( \mathbb{R} \) on a Hilbert space \( \mathcal{H} \). (i) There is a spectral measure \( M \) on \( \mathbb{R} \) such that \( \left( {26.17}\right) \)\[ U\left( t\right) = \int \exp \left( {i\lambda t}\right) {dM}\left( \lam... | Proposition (i) holds for every commutative lcg. It can be obtained directly from the spectral theory of \( \mathrm{u} \) Chap. XI, \( {\mathrm{n}}^{ \circ }{22} \) applied to the GN algebra generated by operators \( U\left( f\right) = \int f\left( t\right) {dt} \) where \( f \in {L}^{1} \) . Like F. Riesz for \( G = \... | No |
Theorem 47. Let \( \Gamma \) be a discrete subgroup of \( G \) and \( f \) a generalized automorphic form of weight \( r \) for \( \Gamma \) . Then the functions \[ {Z}_{r}f\left( z\right) = {rf}\left( z\right) {y}^{-1} + {Df}\left( z\right) , \] \[ {\bar{Z}}_{r}f\left( z\right) = {y}^{2}\bar{D}f\left( z\right) \] \[ {... | Starting from the equality \( f\left( {\gamma z}\right) = {\left( cz + d\right) }^{r}f\left( z\right) \), we get \[ \bar{D}\left\lbrack {f\left( {\gamma z}\right) }\right\rbrack = {\left( cz + d\right) }^{r}\bar{D}f\left( z\right) \] since \( {\left( cz + d\right) }^{r} \) is holomorphic. As \( \bar{D} \) is a differen... | No |
Theorem 48. Let \( \Gamma \) be a discrete subgroup of \( G, f\left( z\right) \) a holomorphic automorphic form of weight \( r > 0 \) for \( \Gamma \) and\n\n\[ \varphi \left( g\right) = {f}_{r}\left( g\right) = {\left( ci + d\right) }^{-r}f\left( {gi}\right) \]\n\nthe associated function on \( G \) . Let \( {HC}\left(... | For parabolic forms, i.e. such that \( \varphi \in {L}^{2}\left( {\Gamma \smallsetminus G}\right) \), this result is the infinitesimal analogue of theorem 27 of \( {\mathrm{n}}^{ \circ }{20} \) which says that the representation of \( G \) on the closed invariant subspace \( \mathcal{H}\left( \varphi \right) \) of \( {... | Yes |
For any irreducible unitary representation \( \left( {\mathcal{H},\pi }\right) \) of \( G \) and any character \( \chi \) of \( K \), the subspace \( \mathcal{H}\left( \chi \right) \) of solutions of\n\n\[ \pi \left( k\right) \mathbf{a} = \chi \left( k\right) \mathbf{a} \]\n\nhas dimension \( \leq 1 \) . | Let us suppose that \( \mathcal{H}\left( \chi \right) \) is non-trivial and let \( \mathbf{a} \neq 0 \) and \( \mathbf{b} \) be two elements of this space. The representation being irreducible, \( \mathbf{b} \) is the limit of vectors \( \pi \left( f\right) \mathbf{a} \) where \( f \in L\left( G\right) \) . If \( E\lef... | Yes |
To show that the equation has no other real root, we use Rolle's Theorem and argue that, no matter how much we enlarge by contradiction. Suppose that it had two roots \( a \) and \( b \) . Then \( f\left( a\right) = 0 = f\left( b\right) \) and, the viewing rectangle, we can never since \( f \) is a polynomial, it is di... | But\n\n\n\nFIGURE 2\n\n\[ \n{f}^{\prime }\left( x\right) = 3{x}^{2} + 1 \geq 1\;\text{ for all }x \]\n\n(since \( {x}^{2} \geq 0 \) ) so \( {f}^{\prime }\left( x\right) \) can never be 0 . This gives a contradiction.... | Yes |
EXAMPLE 3 Find the local maximum and minimum values of the function\n\n\\[ g\\left( x\\right) = x + 2\\sin x\\;0 \\leq x \\leq {2\\pi } \\] | SOLUTION As in Example 1, we start by finding the critical numbers. The derivative is\n\n\\[ {g}^{\\prime }\\left( x\\right) = 1 + 2\\cos x \\]\n\nso \\( {g}^{\\prime }\\left( x\\right) = 0 \\) when \\( \\cos x = - \\frac{1}{2} \\) . The solutions of this equation are \\( {2\\pi }/3 \\) and \\( {4\\pi }/3 \\) . Because... | Yes |
Since \( {f}^{\prime }\left( 2\right) = {10} \), the tangent line to \( y = {x}^{3} - {2x} - 5 \) at \( \left( {2, - 1}\right) \) has equation \( y = {10x} - {21} \) | so its \( x \) -intercept is \( {x}_{2} = {2.1} \) . \n\[ \n= 2 - \frac{{2}^{3} - 2\left( 2\right) - 5}{3{\left( 2\right) }^{2} - 2} = {2.1} \n\] | No |
Example 4 (the current) approaches\n\[ - \frac{1}{3}\ln \left| {{15} - {3I}}\right| = t + C \] | its limiting value. Comparison with\n\nFigure 9.2.10 shows that we were able\n\[ \left| {{15} - {3I}}\right| = {e}^{-3\left( {t + C}\right) } \]\n\nto draw a fairly accurate solution curve\n\nfrom the direction field.\n\[ {15} - {3I} = \pm {e}^{-{3C}}{e}^{-{3t}} = A{e}^{-{3t}} \]\n\n\[ I = 5 - \frac{1}{3}A{e}^{-{3t}} \... | Yes |
We already know how to find the length \( L \) of a curve \( C \) given in the form \( y = F\left( x\right) \) , \( a \leq x \leq b \) . Formula 8.1.3 says that if \( {F}^{\prime } \) is continuous, then\n\n\[ L = {\int }_{a}^{b}\sqrt{1 + {\left( \frac{dy}{dx}\right) }^{2}}{dx} \] | Suppose that \( C \) can also be described by the parametric equations \( x = f\left( t\right) \) and \( y = g\left( t\right) \) , \( \alpha \leq t \leq \beta \), where \( {dx}/{dt} = {f}^{\prime }\left( t\right) > 0 \) . This means that \( C \) is traversed once, from left to right, as \( t \) increases from \( \alpha... | Yes |
\[ y\left( 1\right) = {c}_{1}{e}^{-1} + {c}_{2}{e}^{-1} = 3 \] | The first condition gives \( {c}_{1} = 1 \), so the second condition becomes\n\n\[ {e}^{-1} + {c}_{2}{e}^{-1} = 3 \]\n\nSolving this equation for \( {c}_{2} \) by first multiplying through by \( e \), we get\n\n\[ 1 + {c}_{2} = {3e}\;\text{ so }\;{c}_{2} = {3e} - 1 \]\n\nThus the solution of the boundary-value problem ... | Yes |
Theorem 1.6.1. The exponential generating function of the Bell numbers is \( {e}^{{e}^{x} - 1} \), i.e., the coefficient of \( {x}^{n}/n \) ! in the power series expansion of \( {e}^{{e}^{x} - 1} \) is the number of partitions of a set of \( n \) elements. | First, the theorem tells us that\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}\frac{b\left( n\right) }{n!}{x}^{n} = {e}^{{e}^{x} - 1} \]\n\n(1.6.12)\n\nWe are going to carry out a very standard operation on this equation, but the first time this operation appears it seems to be anything but standard.\n\n## \(\text{The}x\lef... | Yes |
Find a closed formula for the sum of the series \( \mathop{\sum }\limits_{{n > 0}}\left( {{n}^{2} + {4n} + 5}\right) /n! \) . | According to the rule, the answer is the value at \( x = 1 \) of the series\n\n\[ \left\{ {{\left( xD\right) }^{2} + 4\left( {xD}\right) + 5}\right\} {e}^{x} = \left\{ {{x}^{2} + x}\right\} {e}^{x} + {4x}{e}^{x} + 5{e}^{x} \]\n\n\[ = \left( {{x}^{2} + {5x} + 5}\right) {e}^{x}. \]\n\nTherefore the answer to the question... | Yes |
Let \( f\left( {n, k}\right) \) denote the number of ways that the nonnegative integer \( n \) can be written as an ordered sum of \( k \) nonnegative integers. Find \( f\left( {n, k}\right) \) . For instance, \( f\left( {4,2}\right) = 5 \) because \( 4 = 4 + 0 = 3 + 1 = 2 + 2 = 1 + 3 = 0 + 4 \) . | To find \( f \), consider the power series \( 1/{\left( 1 - x\right) }^{k} \) . Since \( 1/\left( {1 - x}\right) \overset{ops}{ \rightarrow }\{ 1\} \) , by (2.2.5) we have\n\n\[ 1/{\left( 1 - x\right) }^{k}\overset{\text{ ops }}{ \leftrightarrow }\{ f\left( {n, k}\right) {\} }_{n = 0}^{\infty }.\]\n\nBy (1.5.5), \( f\l... | Yes |
Prove that the Fibonacci numbers satisfy\n\n\[ \n{F}_{0} + {F}_{1} + {F}_{2} + \cdots + {F}_{n} = {F}_{n + 2} - 1\;\left( {n \geq 0}\right) .\n\] | By Rule 5, the opsgf of the sequence on the left side is \( F/\left( {1 - x}\right) \), where \( F \) is the opsgf of the Fibonacci numbers, which we found in section 1.3 to be \( x/\left( {1 - x - {x}^{2}}\right) \) . By Rule 1, the opsgf of the sequence on the right hand side\n\nis\n\[ \n\frac{F - x}{{x}^{2}} - \frac... | No |
Lemma 2.3.1. If \( k \geq 1 \) and \( g\left( k\right) \) is the number of primitive legal strings, and \( f\left( k\right) \) is the number of all legal strings of \( {2k} \) parentheses, then\n\n\[ g\left( k\right) = f\left( {k - 1}\right) . \] | Proof. Given any legal string of \( k - 1 \) pairs of parentheses, make a primitive one of length \( {2k} \) by adding an initial left parenthesis and a terminal right parenthesis to it. Conversely, given a primitive string of length \( {2k} \), if its initial left and terminal right parentheses are deleted, what remai... | Yes |
There exists a number \( R,0 \leq R \leq + \infty \), called the radius of convergence of the series \( f \), such that the series converges for all values of \( z \) with \( \left| z\right| < R \) and diverges for all \( z \) such that \( \left| z\right| > R \) . The number \( R \) is expressed in terms of the sequenc... | Proof of theorem 2.4.1. Let \( R \) be the number shown in (2.4.1), and suppose first that \( 0 < R < \infty \) . Choose \( z \) such that \( \left| z\right| < R \) . We will show that the series converges at \( z \) .\n\nFor the given \( z \), we can find \( \epsilon > 0 \) such that\n\n\[ \left| z\right| < \frac{R}{1... | Yes |
Theorem 2.4.2. Suppose the power series \( \sum {a}_{n}{z}^{n} \) converges for all \( z \) in \( \left| z\right| < R \), and let \( f\left( z\right) \) denote its sum. Then \( f\left( z\right) \) is an analytic function in \( \left| z\right| < R \) . If furthermore the series diverges for \( \left| z\right| > R \), th... | Proof. If \( f \) has no singularity on its circle of convergence \( \left| z\right| = R \), then about each point of that circle we can draw an open disk in which \( f \) remains analytic. By the Heine-Borel theorem, a finite number of these disks cover the circle \( \left| z\right| = R \), and therefore \( f \) must ... | Yes |
For fixed \( n \), find\n\n\[ \n{\lambda }_{n} = \mathop{\sum }\limits_{k}{\left( -1\right) }^{k}\left( \begin{matrix} n \\ {3k} \end{matrix}\right) \n\] | We could do this one if we knew the function\n\n\[ \nf\left( x\right) = \mathop{\sum }\limits_{k}\left( \begin{matrix} n \\ {3k} \end{matrix}\right) {x}^{3k} \n\]\n\nbecause \( {\lambda }_{n} = f\left( {-1}\right) \) . But \( f\left( x\right) \) picks out every third term from the series \( F\left( x\right) = {\left( 1... | Yes |
Theorem 3.4.1 (The exponential formula). Let \( \\mathcal{F} \) be an exponential family whose deck and hand enumerators are \( \\mathcal{D}\\left( x\\right) \) and \( \\mathcal{H}\\left( {x, y}\\right) \), respectively. Then\n\n\[ \n\\mathcal{H}\\left( {x, y}\\right) = {e}^{y\\mathcal{D}\\left( x\\right) }\n\] | Proof. In \( {3.4.3}) \) we have proved this result in the special case where there is only one nonempty deck. But a general exponential family with a full sequence of nonempty decks \( {\\mathcal{D}}_{1},{\\mathcal{D}}_{2},\\ldots \) is the merger of the special families \( {\\mathcal{F}}_{r}\\left( {r = 1,2,\\ldots }... | Yes |
Corollary 3.4.1. Let \( \mathcal{F} \) be an exponential family, let \( \mathcal{D}\left( x\right) \) be the egf of the sequence \( {\left\{ {d}_{n}\right\} }_{1}^{\infty } \) of sizes of the decks, and let \( \mathcal{H}\left( x\right) \overset{\text{ egf }}{ \leftrightarrow }{\left\{ {h}_{n}\right\} }_{0}^{\infty } \... | \[ \mathcal{H}\left( x\right) = {e}^{\mathcal{D}\left( x\right) } \] | Yes |
Theorem 3.8.1. Fix \( m > 0 \) . The numbers of permutations of \( n \) letters whose mth power is the identity permutation have the generating function\n\n\[ \exp \left( {\mathop{\sum }\limits_{{d \smallsetminus m}}\left( {{x}^{d}/d}\right) }\right) \] | Let’s try a special case of this theorem. Take \( m = 2 \) . Then we are talking about permutations whose square is 1 . These are called involutions. Involutions can have cycles of lengths 1 or 2 only, by the lemma above. If \( {t}_{n} \) is the number of involutions of \( n \) letters, then by (3.8.2) we have\n\n\[ \m... | Yes |
Theorem 3.10.1. The counting sequences \( \left\{ {d}_{n}\right\} \) and \( \left\{ {h}_{n}\right\} \), of decks and hands in an exponential family satisfy the recurrence\n\n\[ n{h}_{n} = \mathop{\sum }\limits_{k}\left( \begin{array}{l} n \\ k \end{array}\right) k{d}_{k}{h}_{n - k}\;\left( {n \geq 1;{h}_{0} = 1}\right)... | Proof. Apply the ’ \( {xD}\log \) ’ method of section 1.6 to the exponential formula \( \left| \left( {3.4.6}\right) \right| \) | No |
Theorem 3.11.1. Let \( \beta \left( n\right) \) denote the number of vertex labeled bipartite graphs of \( n \) vertices. Then\n\n\[ \mathop{\sum }\limits_{{n \geq 0}}\frac{\beta \left( n\right) }{n!}{x}^{n} = \sqrt{\mathop{\sum }\limits_{{n \geq 0}}\frac{{\gamma }_{n}}{n!}{x}^{n}} \]\n\n(3.11.5)\n\nwhere the \( {\gamm... | So all of the complications about multiple counting were resolved by taking the square root of the generating function that we started with! | No |
Theorem 3.12.1. For each \( n \geq 1 \) there are exactly \( {n}^{n - 2} \) labeled trees of \( n \) vertices. | Although many proofs are known, the one by generating functions, which uses the exponential formula, is particularly enchanting, and here it is:\n\nA rooted tree is a tree that has a distinguished vertex called the root. There are obviously \( n \) times as many labeled rooted trees of \( n \) vertices as there are tre... | Yes |
Theorem 3.14.1. In a prefab \( \mathcal{P} \) whose hand enumerator is \( \mathcal{H}\left( {x, y}\right) \) we have\n\n\[ \mathcal{H}\left( {x, y}\right) = \mathop{\prod }\limits_{{n = 1}}^{\infty }\frac{1}{{\left( 1 - y{x}^{n}\right) }^{{d}_{n}}} \]\n\nwhere \( {d}_{n} \) is the number of cards in the \( n \) th deck... | If we take the logarithm of both sides of (3.14.4)\n\n\[ \log \mathcal{H}\left( {x, y}\right) = \mathop{\sum }\limits_{{s = 1}}^{\infty }\log \frac{1}{{\left( 1 - y{x}^{s}\right) }^{{d}_{s}}} \]\n\n\[ = \mathop{\sum }\limits_{{s \geq 1}}{d}_{s}\log \frac{1}{\left( 1 - y{x}^{s}\right) } \]\n\n\[ = \mathop{\sum }\limits_... | Yes |
Theorem 3.14.2. Let the prefab \( \mathcal{P} \) contain decks of sizes \( {d}_{1},{d}_{2},\ldots \), and let \( W \) be a set of nonnegative integers, \( 0 \in W \) . If \( h\left( {n, k;W}\right) \) is the number of hands of \( k \) cards of weight \( n \), such that each card appears with a multiplicity that belongs... | Observe that the theorem reduces to theorem 3.14.1 in the case where \( W = {Z}^{ + } \), the set of all nonnegative integers.\n\nA noteworthy special case is \( W = \{ 0,1\} \), which means that we can choose a card for our hand or not, but we can't take more than one copy of it. In that case (3.14.9) gives\n\n\[ \mat... | Yes |
Theorem 3.15.1. Let \( a \) and \( b \) be relatively prime positive integers. Then\n\n(a) every integer \( n \geq \kappa = \left( {a - 1}\right) \left( {b - 1}\right) \) is of the form \( n = {xa} + {yb} \) , \( x, y \geq 0 \), and\n\n(b) the integer \( \kappa - 1 \) is not of that form, and\n\n(c) of the integers \( ... | Proof. (Our proof follows [NW]) Since \( {gcd}\left( {a, b}\right) = 1 \), we can certainly write every integer \( m \) as \( {xa} + {yb} \) if \( x, y \) can have either sign. The representation is unique if we require that \( 0 \leq x < b \) . Then \( m \in \mathcal{S} \) if \( y \geq 0 \), and \( m \notin \mathcal{S... | Yes |
Let \( W = \{ 0,1\}, R = \{ 1,2,\ldots \} \) . Then \( p\left( {n, k;W, R}\right) \) is the number of partitions of \( n \) into \( k \) distinct parts, and we have\n\n\[ \mathop{\sum }\limits_{{n, k}}p\left( {n, k;\{ 0,1\} ,{\mathbf{Z}}^{ + }}\right) {x}^{n}{y}^{k} = \mathop{\prod }\limits_{{r \geq 1}}\left( {1 + y{x}... | If we let \( y = 1 \) we obtain\n\n\[ \mathop{\sum }\limits_{n}p\left( {n;\{ 0,1\} ,{\mathbf{Z}}^{ + }}\right) {x}^{n} = \mathop{\prod }\limits_{{r \geq 1}}\left( {1 + {x}^{r}}\right) \]\n\n\[ = \mathop{\prod }\limits_{{r \geq 1}}\frac{1 - {x}^{2r}}{1 - {x}^{r}} \]\n\n\[ = \frac{\left( {1 - {x}^{2}}\right) \left( {1 - ... | Yes |
Theorem 3.16.2. Fix \( q \) and let \( W = \{ 0,1,\ldots, q\} \) and \( R = {\mathbf{Z}}^{ + } \). Then, the following identity holds:\n\[\n\mathop{\prod }\limits_{{r \geq 1}}\left( {1 + {t}^{r} + \cdots + {t}^{qr}}\right) = \mathop{\prod }\limits_{{r \geq 1}}\left( \frac{1 - {t}^{r\left( {q + 1}\right) }}{1 - {t}^{r}}... | Each factor in the numerator of this product cancels one in the denominator, leaving in the denominator only those factors in which \( r \) is not divisible by \( q + 1 \). This proves the following result, which reduces to theorem 3.16.1 when \( q = 1 \). | Yes |
In an exponential family \( \mathcal{F} \), the average number of cards in hands of weight \( n \) is | \[ \mu \left( n\right) = \left\lbrack \frac{h\left( n\right) {x}^{n}}{n!}\right\rbrack \mathcal{D}\left( x\right) \mathcal{H}\left( x\right) \]\n\[ = \frac{1}{h\left( n\right) }\mathop{\sum }\limits_{r}\left( \begin{array}{l} n \\ r \end{array}\right) {d}_{r}h\left( {n - r}\right) \]\n\n(4.1.6)\n\n## Example 1. Cycles ... | Yes |
Consider the sum\n\n\[ \mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} k \\ n - k \end{matrix}\right) \;\left( {n = 0,1,2,\ldots }\right) . | The free variable is \( n \), so let’s call the sum \( f\left( n\right) \) . Write it out like this:\n\n\[ f\left( n\right) = \mathop{\sum }\limits_{{k \geq 0}}\left( \begin{matrix} k \\ n - k \end{matrix}\right) \]\n\nOK, now multiply both sides by \( {x}^{n} \) and sum over \( n \) . You have now arrived at step (c) ... | Yes |
Consider the sum\n\n\\[ \n\\mathop{\\sum }\\limits_{k}\\left( \\begin{matrix} n + k \\\\ m + {2k} \\end{matrix}\\right) \\left( \\begin{matrix} {2k} \\\\ k \\end{matrix}\\right) \\frac{{\\left( -1\\right) }^{k}}{k + 1}\\;\\left( {m, n \\geq 0}\\right) .\n\\]\n\n(4.3.4)\n\nCan it be that the same method will do this sum... | Indeed; just pour enough Snake Oil on it and it will be cured. Let \\( f\\left( n\\right) \\) denote the sum in question, and let \\( F\\left( x\\right) \\) be its opsgf. Dive in immediately by multiplying by \\( {x}^{n} \\) and summing over \\( n \\geq 0 \\), to get\n\n\\[ \nF\\left( x\\right) = \\mathop{\\sum }\\limi... | Yes |
Evaluate the sums\n\n\\[ \n{f}_{n} = \\mathop{\\sum }\\limits_{k}\\left( \\begin{matrix} n + k \\ {2k} \\end{matrix}\\right) {2}^{n - k}\\;\\left( {n \\geq 0}\\right) .\n\\]\n\n(4.3.10) | Without stopping to think, let \\( F \\) be the opsgf of the sequence, multiply both sides of (4.3.10) by \\( {x}^{n} \\), sum over \\( n \\geq 0 \\), and interchange the two sums on the right. This produces\n\n\\[ \nF = \\mathop{\\sum }\\limits_{k}{2}^{-k}\\mathop{\\sum }\\limits_{{n \\geq 0}}\\left( \\begin{matrix} n... | Yes |
Suppose we want to prove the identity\n\n\\[ \n\\mathop{\\sum }\\limits_{k}\\left( \\begin{array}{l} n \\\\ k \\end{array}\\right) = {2}^{n}\\;\\left( {n \\geq 0}\\right) \n\\]\n\n(4.4.5) | If we divide by the right hand side we find that the function \\( F\\left( {n, k}\\right) \\) of \\( \\left( {4.4.1}\\right) \\) is\n\n\\[ \nF\\left( {n, k}\\right) = \\left( \\begin{array}{l} n \\\\ k \\end{array}\\right) /{2}^{n}\\;\\left( {n \\geq 0}\\right) .\n\\]\n\n(4.4.6)\n\nNow we need to find the mate \\( G\\l... | Yes |
Theorem 4.4.2. (The Pfaff-Saalschütz identity)\n\n\[ \mathop{\sum }\limits_{k}\frac{\left( {a + k}\right) !\left( {b + k}\right) !\left( {c - a - b + n - 1 - k}\right) !}{\left( {k + 1}\right) !\left( {n - k}\right) !\left( {c + k}\right) !} = \]\n\n\[ \frac{\left( {a - 1}\right) !\left( {b - 1}\right) !\left( {c - a -... | Proof: Take\n\n\[ R\left( {n, k}\right) = - \frac{\left( {b + k}\right) \left( {a + k}\right) }{\left( {c - b + n + 1}\right) \left( {c - a + n + 1}\right) }.\] | No |
Theorem 4.4.3. (Dixon's identity)\n\n\\[ \n\\mathop{\\sum }\\limits_{k}{\\left( -1\\right) }^{k}\\left( \\begin{array}{l} n + b \\\\ n + k \\end{array}\\right) \\left( \\begin{array}{l} n + c \\\\ c + k \\end{array}\\right) \\left( \\begin{array}{l} b + c \\\\ b + k \\end{array}\\right) = \\frac{\\left( {n + b + c}\\ri... | Proof: Take \\( R\\left( {n, k}\\right) = \\left( {c + 1 - k}\\right) \\left( {b + 1 - k}\\right) /\\left( {2\\left( {n + k}\\right) \\left( {n + b + c + 1}\\right) }\\right) \\) . | No |
Theorem 4.5.2. Let \( p\left( x\right) = {c}_{0} + {c}_{1}x + {c}_{2}{x}^{2} + \cdots + {c}_{n}{x}^{n} \) be a polynomial all of whose zeros are real and negative. Then the coefficient sequence \( {\left\{ {c}_{r}\right\} }_{0}^{n} \) is strictly log concave. | To prove the theorem we need to recall Rolle's theorem of elementary calculus. It holds that if \( f\left( x\right) \) is continuously differentiable in \( \left( {a, b}\right) \), and if \( f\left( a\right) = f\left( b\right) \), then somewhere between \( a \) and \( b \) the derivative \( {f}^{\prime } \) must vanish... | Yes |
Corollary 4.5.1. The binomial coefficient sequence \( {\left\{ \left( \begin{array}{l} n \\ k \end{array}\right) \right\} }_{k = 0}^{n} \) is log concave, and therefore unimodal. | Proof. The zeros of the generating polynomial \( {\left( 1 + x\right) }^{n} \) are evidently real and negative. | No |
Corollary 4.5.2. The sequence of Stirling numbers of the first kind\n\n\[ \n{\\left\\{ \\left\\lbrack \\begin{array}{l} n \\ k \\end{array}\\right\\rbrack \\right\\} }_{k = 1}^{n}\n\]\n\nis log concave, and therefore unimodal. | Proof. According to (3.5.2), the opsgf of these Stirling numbers is the polynomial\n\n\[ \n\\mathop{\\sum }\\limits_{j}\\left\\lbrack \\begin{array}{l} n \\ j \\end{array}\\right\\rbrack {x}^{j - 1} = \\left( {x + 1}\\right) \\left( {x + 2}\\right) \\cdots \\left( {x + n - 1}\\right) ,\n\]\n\nwhose zeros are clearly re... | Yes |
Corollary 4.5.3. The sequence of Stirling numbers of the second kind \( {\left\{ \left\{ \begin{array}{l} n \\ k \end{array}\right\} \right\} }_{k = 1}^{n} \) is log concave, and therefore unimodal. | Proof. We'll have to work just a little harder for this one, because the zeros of the polynomial\n\n\[ \n{A}_{n}\left( x\right) = \mathop{\sum }\limits_{j}\left\{ \begin{array}{l} n \\ j \end{array}\right\} {x}^{j} \]\n\nare not easy to find. They are, however, real and negative, and here is one way to see that: by (1.... | Yes |
Theorem 4.7.3. Let \( S \) be a set of positive integers for which \( \mathop{\sum }\limits_{{s \in S}}1/s \) converges, and let \( \mathbf{a} \) be a fixed cycle type vector. The probability that the cycle type vector of a random permutation agrees with \( \mathbf{a} \) in all of its components whose subscripts lie in... | \[ {e}^{-\left( {\mathop{\sum }\limits_{{s \in S}}1/s}\right) }\left\lbrack {\mathbf{x}}^{\mathbf{a}}\right\rbrack \exp \left( {\mathop{\sum }\limits_{{s \in S}}\frac{{x}_{s}}{s}}\right) = \frac{1}{\mathop{\prod }\limits_{{s \in S}}\left( {{e}^{\frac{1}{s}}{s}^{{a}_{s}}{a}_{s}!}\right) }.\] | Yes |
A permutation \( \sigma \) has a \( k \) th root if and only if for every \( m = 1,2,\ldots \) it is true that the number of \( m \) -cycles that \( \sigma \) has is a multiple of \( \left( \left( {m, k}\right) \right) \) . | To prove this, let \( \sigma = {\tau }^{k} \) be a permutation of \( n \) letters, and suppose \( \sigma \) has exactly \( {\nu }_{m} \) cycles of length \( m \), for each \( m = 1,2,\ldots \) Consider a cycle of length \( r \) in \( \tau \) . In \( {\tau }^{k} \) this contributes \( \left( {r, k}\right) \) cycles of l... | Yes |
Theorem 4.9.1. Let \( f\left( n\right) \) be the number of \( n \) -celled HC-polyominoes. Then\n\n\[ \mathop{\sum }\limits_{{n \geq 1}}f\left( n\right) {x}^{n} = \frac{x{\left( 1 - x\right) }^{3}}{1 - {5x} + 7{x}^{2} - 4{x}^{3}} \] | \[ = x + 2{x}^{2} + 6{x}^{3} + {19}{x}^{4} + {61}{x}^{5} + {196}{x}^{6} + {629}{x}^{7} + {2017}{x}^{8}. \]\n\n\[ + {6466}{x}^{9} + {20727}{x}^{10} + {66441}{x}^{11} + {212980}{x}^{12} + \cdots \]\n\n(4.9.9)\n\nWe now will give a preview of the material in chapter 5 , by working out an asymptotic formula for \( f\left( ... | Yes |
For a set of pairs \( \left( {{a}_{i},{b}_{i}}\right) \;\left( {i = 1,\ldots, k}\right) \) to be an exact covering sequence it is necessary and sufficient that the relation (4.10.1) hold. | One conclusion that we can draw immediately is that in an ECS we must have \( \mathop{\sum }\limits_{i}1/{b}_{i} = 1 \) . To see that, just multiply (4.10.1) by \( 1 - x \) and let \( x \rightarrow 1 \) . But we can learn much more by comparing the partial fraction expansions of the left and right sides of (4.10.1).\n\... | Yes |
Theorem 4.10.2. A set of pairs of integers \( \left( {{a}_{1},{b}_{1}}\right) ,\ldots ,\left( {{a}_{k},{b}_{k}}\right) \), in which the \( a \) ’s are nonnegative and the \( b \) ’s are positive, is an exact covering sequence if and only if \( \mathop{\sum }\limits_{j}1/{b}_{j} = 1 \) and for each \( s > 1 \), the poly... | For an example, take the pairs\n\n\[ \left( {0,4}\right) ,\left( {2,4}\right) ,\left( {1,6}\right) ,\left( {3,6}\right) ,\left( {5,{12}}\right) ,\left( {{11},{12}}\right) \text{.} \]\n\nThen \( \mathop{\sum }\limits_{j}1/{b}_{j} = 1/4 + 1/4 + 1/6 + 1/6 + 1/{12} + 1/{12} = 1 \), and the divisibility conditions of the th... | Yes |
Theorem 5.1.1. (The Lagrange Inversion Formula) Let \( f\left( u\right) \) and \( \phi \left( u\right) \) be formal power series in \( u \), with \( \phi \left( 0\right) = 1 \) . Then there is a unique formal power series \( u = u\left( t\right) \) that satisfies (5.1.1). Further, the value \( f\left( {u\left( t\right)... | Proof. First we note that it suffices to prove the theorem in the case where \( f \) and \( \phi \) are polynomials. Indeed, if \( n \) is fixed, and if \( f \) and \( \phi \) are full formal power series, then suppose that we truncate both of those series by discarding all terms that involve powers \( {u}^{k} \) for \... | Yes |
Theorem 5.2.1. Let \( f \) be analytic in a region \( \Re \) containing the origin, except for finitely many poles. Let \( R > 0 \) be the modulus of the pole(s) of smallest modulus, and let \( {z}_{0},\ldots ,{z}_{s} \) be all of the poles of \( f\left( z\right) \) whose modulus is \( R \) . Further, let \( {R}^{\prim... | Proof. By theorem 2.4.3, this theorem will be proved as soon as we establish that if we subtract from \( f\left( z\right) \) the sum of all of its principal parts from singularities on the circle \( \left| z\right| = R \), then the resulting function is analytic in the larger disk \( \left| z\right| < {R}^{\prime } \) ... | Yes |
Lemma 5.3.1. Let \( \\left\\{ {a}_{n}\\right\\} ,\\left\\{ {b}_{n}\\right\\} \) be two sequences that satisfy (a) \( {a}_{n} = \) \( O\\left( {n}^{-\\gamma }\\right) \) and (b) \( {b}_{n} = O\\left( {\\theta }^{n}\\right) \\left( {0 < \\theta < 1}\\right) \) . Then\n\n\[ \n\\mathop{\\sum }\\limits_{k}{a}_{k}{b}_{n - k}... | Proof. We have first (the \( C \) ’s are not all the same constant)\n\n\[ \n\\left| {\\mathop{\\sum }\\limits_{{0 \\leq k \\leq n/2}}{a}_{k}{b}_{n - k}}\\right| \\leq \\left\\{ {\\mathop{\\max }\\limits_{{0 \\leq k \\leq n/2}}\\left| {a}_{k}\\right| }\\right\\} \\left\\{ {\\mathop{\\sum }\\limits_{{0 \\leq k \\leq n/2}... | Yes |
Lemma 5.3.2. If \( \beta \notin \{ 0,1,2,\ldots \} \), then\n\n\[ \left\lbrack {z}^{n}\right\rbrack {\left( 1 - z\right) }^{\beta } \sim \frac{{n}^{-\beta - 1}}{\Gamma \left( {-\beta }\right) }.\] | Proof. We have\n\n\[ \left\lbrack {z}^{n}\right\rbrack {\left( 1 - z\right) }^{\beta } = \left( \begin{array}{l} \beta \\ n \end{array}\right) {\left( -1\right) }^{n} \]\n\n\[ = \left( \begin{matrix} n - \beta - 1 \\ n \end{matrix}\right) \]\n\n\[ = \frac{\Gamma \left( {n - \beta }\right) }{\Gamma \left( {-\beta }\righ... | Yes |
Lemma 5.3.3. Let \( u\left( z\right) = {\left( 1 - z\right) }^{\gamma }v\left( z\right) \), where \( v\left( z\right) \) is analytic in some disk \( \left| z\right| < 1 + \eta ,\left( {\eta > 0}\right) \) . Then\n\n\[ \left\lbrack {z}^{n}\right\rbrack u\left( z\right) = O\left( {n}^{-\gamma - 1}\right) . \] | Proof. Apply lemma 5.3.1 with \( {a}_{n} = \left\lbrack {z}^{n}\right\rbrack {\left( 1 - z\right) }^{\gamma } \) and \( {b}_{n} = \left\lbrack {z}^{n}\right\rbrack v\left( z\right) \) . Since \( v \) is analytic in a disk \( \left| z\right| < 1 + \eta \), we have \( {b}_{n} = O\left( {\theta }^{n}\right) \) . The resul... | Yes |
Theorem 5.3.1. (Darboux) Let \( v\left( z\right) \) be analytic in some disk \( \left| z\right| < 1 + \eta \) , and suppose that in a neighborhood of \( z = 1 \) it has the expansion \( v\left( z\right) = \) \( \sum {v}_{j}{\left( 1 - z\right) }^{j} \) . Let \( \beta \notin \{ 0,1,2,\ldots \} \) . Then\n\n\[ \left\lbra... | Proof. We have\n\n\[ {\left( 1 - z\right) }^{\beta }v\left( z\right) - \mathop{\sum }\limits_{{j = 0}}^{m}{v}_{j}{\left( 1 - z\right) }^{\beta + j} = \mathop{\sum }\limits_{{j > m}}{v}_{j}{\left( 1 - z\right) }^{\beta + j} \]\n\n\[ = {\left( 1 - z\right) }^{\beta + m + 1}\widetilde{v}\left( z\right) \]\n\nwhere the reg... | Yes |
The pair \( \left( {\mathbb{Z}, + }\right) \) is a group: \( \mathbb{Z} = \{ \ldots , - 2, - 1,0,1,2,\ldots \} \) is the set and the associative operation is addition. | - The element \( 0 \in \mathbb{Z} \) is an identity: \( a + 0 = 0 + a = a \) for any \( a \). \n- Every element \( a \in \mathbb{Z} \) has an additive inverse: \( a + \left( {-a}\right) = \left( {-a}\right) + a = 0 \. | Yes |
Let \( {\mathbb{Q}}^{ \times } \) be the set of nonzero rational numbers. The pair \( \left( {{\mathbb{Q}}^{ \times }, \cdot }\right) \) is a group: the set is \( {\mathbb{Q}}^{ \times } \) and the associative operation is multiplication. | - The element \( 1 \in {\mathbb{Q}}^{ \times } \) is an identity: for any rational number, \( a \cdot 1 = 1 \cdot a = a \). \n- For any rational number \( x \in {\mathbb{Q}}^{ \times } \), we have an inverse \( {x}^{-1} \), such that \n\n\[ \nx \cdot {x}^{-1} = {x}^{-1} \cdot x = 1. \n\] | No |
The pair \( \left( {\mathbb{Q}, \cdot }\right) \) is NOT a group. (Here \( \mathbb{Q} \) is rational numbers.) | While there is an identity element, the element \( 0 \in \mathbb{Q} \) does not have an inverse. | Yes |
Let \( {S}^{1} \) denote the set of complex numbers \( z \) with absolute value one; that is\n\n\[ \n{S}^{1} \mathrel{\text{:=}} \{ z \in \mathbb{C}\left| \right| z \mid = 1\} \n\]\n\nThen \( \left( {{S}^{1}, \times }\right) \) is a group because | - The complex number \( 1 \in {S}^{1} \) serves as the identity, and\n\n- Each complex number \( z \in {S}^{1} \) has an inverse \( \frac{1}{z} \) which is also in \( {S}^{1} \), since \( \left| {z}^{-1}\right| = {\left| z\right| }^{-1} = 1 \n\nThere is one thing I ought to also check: that \( {z}_{1} \times {z}_{2} \)... | Yes |
Let \( n \) be a positive integer. Then \( {\mathrm{{GL}}}_{n}\left( \mathbb{R}\right) \) is defined as the set of \( n \times n \) real matrices which have nonzero determinant. It turns out that with this condition, every matrix does indeed have an inverse, so \( \left( {{\mathrm{{GL}}}_{n}\left( \mathbb{R}\right) , \... | (The fact that \( {\mathrm{{GL}}}_{n}\left( \mathbb{R}\right) \) is closed under \( \times \) follows from the linear algebra fact that \( \det \left( {AB}\right) = \det A\det B \), proved in later chapters.) | No |
Proposition 1.2.4 (Inverse of products)\n\nLet \( G \) be a group, and \( a, b \in G \) . Then \( {\left( ab\right) }^{-1} = {b}^{-1}{a}^{-1} \) . | Proof. Direct computation. We have\n\n\[\n\left( {ab}\right) \left( {{b}^{-1}{a}^{-1}}\right) = a\left( {b{b}^{-1}}\right) {a}^{-1} = a{a}^{-1} = {1}_{G}.\n\]\n\nHence \( {\left( ab\right) }^{-1} = {b}^{-1}{a}^{-1} \) . Similarly, \( \left( {{b}^{-1}{a}^{-1}}\right) \left( {ab}\right) = {1}_{G} \) as well. | Yes |
Lemma 1.2.5 (Left multiplication is a bijection)\n\nLet \( G \) be a group, and pick a \( g \in G \) . Then the map \( G \rightarrow G \) given by \( x \mapsto {gx} \) is a bijection. | Exercise 1.2.6. Check this by showing injectivity and surjectivity directly. (If you don't know what these words mean, consult Appendix E.) | No |
Example 1.3.3 (Primitive roots modulo 7)\n\nAs a nontrivial example, we claim that \( \mathbb{Z}/6\mathbb{Z} \cong {\left( \mathbb{Z}/7\mathbb{Z}\right) }^{ \times } \) . The bijection is\n\n\[ \phi \left( {a{\;\operatorname{mod}\;6}}\right) = {3}^{a}{\;\operatorname{mod}\;7}. \]\n\nTo check that this is an isomorphism... | - First, we need to check this map actually makes sense: why is it the case that if \( a \equiv b\left( {\;\operatorname{mod}\;6}\right) \), then \( {3}^{a} \equiv {3}^{b}\left( {\;\operatorname{mod}\;7}\right) \) ? The reason is that Fermat’s little theorem guarantees that \( {3}^{6} \equiv 1\left( {\;\operatorname{mo... | Yes |
We can make \( {\mathbb{R}}^{2} \) into a metric space by imposing the Euclidean distance function | \[ d\left( {\left( {{x}_{1},{y}_{1}}\right) ,\left( {{x}_{2},{y}_{2}}\right) }\right) = \sqrt{{\left( {x}_{1} - {x}_{2}\right) }^{2} + {\left( {y}_{1} - {y}_{2}\right) }^{2}}. \] | Yes |
Theorem 2.3.3 (Sequential continuity)\n\nA function \( f : M \rightarrow N \) of metric spaces is continuous at a point \( p \in M \) if and only if the following property holds: if \( {x}_{1},{x}_{2},\ldots \) is a sequence in \( M \) converging to \( p \), then the sequence \( f\left( {x}_{1}\right), f\left( {x}_{2}\... | Proof. One direction is not too hard:\n\nExercise 2.3.4. Show that \( \varepsilon - \delta \) continuity implies sequential continuity at each point.\n\nConversely, we will prove if \( f \) is not \( \varepsilon - \delta \) continuous at \( p \) then it does not preserve convergence.\n\nIf \( f \) is not continuous at ... | No |
Proposition 2.3.5 (Composition of continuous functions is continuous)\n\nLet \( f : M \rightarrow N \) and \( g : N \rightarrow L \) be continuous maps of metric spaces. Then their composition \( g \circ f \) is continuous. | Proof. Dead simple with sequences: Let \( p \in M \) be arbitrary and let \( {x}_{n} \rightarrow p \) in \( M \) . Then \( f\left( {x}_{n}\right) \rightarrow f\left( p\right) \) in \( N \) and \( g\left( {f\left( {x}_{n}\right) }\right) \rightarrow g\left( {f\left( p\right) }\right) \) in \( L \), QED. | Yes |
It may have seemed strange that our metric function on \( {S}^{1} \) was the one inherited from \( {\mathbb{R}}^{2} \), meaning the distance between two points on the circle was defined to be the length of the chord. Wouldn't it have made more sense to use the circumference of the smaller arc joining the two points? | In fact, it doesn’t matter: if we consider \( {S}^{1} \) with the \ | No |
Example 2.4.5 (Homeomorphisms really don't preserve size)\n\nSurprisingly, the open interval \( \left( {-1,1}\right) \) is homeomorphic to the real line \( \mathbb{R} \) ! | One bijection is given by\n\n\[ x \mapsto \tan \left( {\pi /{2x}}\right) \]\n\nwith the inverse being given by \( t \mapsto \frac{2}{\pi }\arctan \left( t\right) \) . | Yes |
Proposition 2.5.4 (Convergence in the product metric is by component)\n\nWe have \( \left( {{x}_{n},{y}_{n}}\right) \rightarrow \left( {x, y}\right) \) if and only if \( {x}_{n} \rightarrow x \) and \( {y}_{n} \rightarrow y \) . | Proof. We have \( {d}_{\max }\left( {\left( {x, y}\right) ,\left( {{x}_{n},{y}_{n}}\right) }\right) = \max \left\{ {{d}_{M}\left( {x,{x}_{n}}\right) ,{d}_{N}\left( {y,{y}_{n}}\right) }\right\} \) and the latter approaches zero as \( n \rightarrow \infty \) if and only if \( {d}_{M}\left( {x,{x}_{n}}\right) \rightarrow ... | Yes |
Proposition 2.5.5 (Addition and multiplication are continuous)\n\nThe addition and multiplication maps are continuous maps \( \\mathbb{R} \\times \\mathbb{R} \\rightarrow \\mathbb{R} \) . | Proof. For multiplication: for any \( n \) we have\n\n\[ \n{x}_{n}{y}_{n} = \\left( {x + \\left( {{x}_{n} - x}\\right) }\\right) \\left( {y + \\left( {{y}_{n} - y}\\right) }\\right)\n\]\n\n\[ \n= {xy} + y\\left( {{x}_{n} - x}\\right) + x\\left( {{y}_{n} - y}\\right) + \\left( {{x}_{n} - x}\\right) \\left( {{y}_{n} - y}... | Yes |
Theorem 2.6.11 (Open set condition)\n\nA function \( f : M \rightarrow N \) of metric spaces is continuous if and only if the pre-image of every open set in \( N \) is open in \( M \) . | Proof. I'll just do one direction...\n\nNow assume \( f \) is continuous. First, suppose \( V \) is an open subset of the metric space \( N \) ; let \( U = {f}^{\text{pre }}\left( V\right) \) . Pick \( x \in U \), so \( y = f\left( x\right) \in V \) ; we want an open neighborhood of \( x \) inside \( U \) .\n\n![1dfd45... | No |
The same group can have very different presentations. For instance consider\n\n\[ \n{D}_{2n} = \left\langle {x, y \mid {x}^{2} = {y}^{2} = 1,{\left( xy\right) }^{n} = 1.}\right\rangle .\n\] | (To see why this is equivalent, set \( x = s, y = {rs} \) .) | No |
Theorem 3.4.1 (Lagrange's theorem)\n\nLet \( G \) be a finite group, and let \( H \) be any subgroup. Then \( \left| H\right| \) divides \( \left| G\right| \) . | The proof is very simple: note that the cosets of \( H \) all have the same size and form a partition of \( G \) (even when \( H \) is not necessarily normal). Hence if \( n \) is the number of cosets, then \( n \cdot \left| H\right| = \left| G\right| \) . | Yes |
Example 3.5.2 (Example of a non-normal subgroup)\n\nLet \( {D}_{12} = \left\langle {r, s \mid {r}^{6} = {s}^{2} = 1,{rs} = s{r}^{-1}}\right\rangle \) . Consider the subgroup of order two \( H = \) \( \{ 1, s\} \) and notice that | \[ {rs}{r}^{-1} = r\left( {s{r}^{-1}}\right) = r\left( {rs}\right) = {r}^{2}s \notin H. \]\n\nHence \( H \) is not normal, and cannot be the kernel of any homomorphism. | Yes |
Theorem 3.5.4 (Algebraic condition for normal subgroups)\n\nLet \( H \) be a subgroup of \( G \) . Then the following are equivalent:\n\n- \( H \trianglelefteq G \) .\n\n- For every \( g \in G \) and \( h \in H,{gh}{g}^{-1} \in H \) . | Proof. We already showed one direction.\n\nFor the other direction, we need to build a homomorphism with kernel \( H \) . So we simply define the group \( G/H \) as the cosets. To put a group operation, we need to verify:\n\nClaim 3.5.5. If \( {g}_{1}^{\prime }{ \sim }_{H}{g}_{1} \) and \( {g}_{2}^{\prime }{ \sim }_{H}... | Yes |
Consider again the product group \( G \times H \). Earlier we identified a subgroup \[ {G}^{\prime } = \left\{ {\left( {g,{1}_{H}}\right) \mid g \in G}\right\} \cong G. \] You can easily see that \( {G}^{\prime } \trianglelefteq G \times H \). (Easy calculation.) Moreover, just as the notation would imply, you can chec... | Indeed, we have \( \left( {g, h}\right) { \sim }_{{G}^{\prime }}\left( {{1}_{G}, h}\right) \) for all \( g \in G \) and \( h \in H \). | Yes |
Example 4.4.4 (First examples of fields)\n\n(a) \( \mathbb{Q},\mathbb{R},\mathbb{C} \) are fields, since the notion \( \frac{1}{c} \) makes sense in them.\n\n(b) If \( p \) is a prime, then \( \mathbb{Z}/p\mathbb{Z} \) is a field, which we denote will usually denote by \( {\mathbb{F}}_{p} \) . | The trivial ring 0 is not considered a field, since we require fields to be nontrivial. | No |
Theorem 4.6.6 (Ring analog of normal subgroups)\n\nLet \( R \) be a ring and \( I \varsubsetneq R \) . Then \( I \) is the kernel of some homomorphism if and only if it's an ideal. | Proof. It's quite similar to the proof for the normal subgroup thing, and you might try\n\nit yourself as an exercise.\n\nObviously the conditions are necessary. To see they're sufficient, we define a ring by \ | No |
Proposition 4.7.1 (Proper ideal \( \Leftrightarrow \) no units)\n\nLet \( R \) be a ring and \( I \subseteq R \) an ideal. Then \( I \) is proper (i.e. \( I \neq R \) ) if and only if it contains no units of \( R \) . | Proof. Suppose \( I \) contains a unit \( u \), i.e. an element \( u \) with an inverse \( {u}^{-1} \) . Then it contains \( u \cdot {u}^{-1} = 1 \), and thus \( I = R \) . Conversely, if \( I \) contains no units, it is obviously proper. | Yes |
In \( \mathbb{Z}\left\lbrack x\right\rbrack, I = \left( {x,{2015}}\right) \) is not a principal ideal. | For if \( I = \left( f\right) \) for some polynomial \( f \in I \) then \( f \) divides \( x \) and 2015. This can only occur if \( f = \pm 1 \), but then \( I \) contains \( \pm 1 \), which it does not. | Yes |
(c) It turns out that for a field \( k \) the ring \( k\left\lbrack x\right\rbrack \) is always a PID. For example, \( \mathbb{Q}\left\lbrack x\right\rbrack \) , \( \mathbb{R}\left\lbrack x\right\rbrack ,\mathbb{C}\left\lbrack x\right\rbrack \) are PID’s. | If you want to try and prove this, first prove an analog of Bezout's lemma, which implies the result. | No |
Example 4.9.3 (Non-Noetherian ring breaks ACC)\n\nIn the ring \( R = \mathbb{Z}\left\lbrack {{x}_{1},{x}_{2},{x}_{3},\ldots }\right\rbrack \) we have an infinite ascending chain\n\n\[ \left( {x}_{1}\right) \varsubsetneq \left( {{x}_{1},{x}_{2}}\right) \varsubsetneq \left( {{x}_{1},{x}_{2},{x}_{3}}\right) \varsubsetneq ... | From the example, you can kind of see why the proposition is true: from an infinitely generated ideal you can extract an ascending chain by throwing elements in one at a time. I’ll leave the proof to you if you want to do it. \( {}^{3} \) | No |
Example 5.4.2 (Examples of maximal ideals)\n\n(a) The ideal \( I = \\left( 7\\right) \) of \( \\mathbb{Z} \) is maximal, because if an ideal \( J \) contains 7 and an element \( n \) not in \( I \) it must contain \( \\gcd \\left( {7, n}\\right) = 1 \), and hence \( J = \\mathbb{Z} \) . | because if an ideal \( J \) contains 7 and an element \( n \) not in \( I \) it must contain \( \\gcd \\left( {7, n}\\right) = 1 \), and hence \( J = \\mathbb{Z} \) | Yes |
Theorem 5.4.4 ( \( I \) maximal \( \Leftrightarrow R/I \) field)\n\nAn ideal \( I \) is maximal if and only if \( R/I \) is a field. | Proof. A ring is a field if and only if (0) is the only maximal ideal. So this follows by Problem \( 4{\mathrm{D}}^{ \star } \) . | No |
Corollary 5.4.5 (Maximal ideals are prime)\n\nIf \( I \) is a maximal ideal of a ring \( R \), then \( I \) is prime. | Proof. If \( I \) is maximal, then \( R/I \) is a field, hence an integral domain, so \( I \) is prime. | Yes |
Example 5.5.4 (Gaussian rationals) | Just like we defined \( \mathbb{Z}\left\lbrack i\right\rbrack \) by abusing notation, we can also write \( \mathbb{Q}\left( i\right) = \operatorname{Frac}\left( {\mathbb{Z}\left\lbrack i\right\rbrack }\right) \) . Officially, it should consist of\n\n\[ \mathbb{Q}\left( i\right) = \left\{ {\left. \frac{f\left( i\right) ... | Yes |
Example 6.2.7 ( \( \mathbb{Q} \) completes to \( \mathbb{R} \) ) | The completion of \( \mathbb{Q} \) is \( \mathbb{R} \) . | Yes |
Question 7.2.3. Show that this is equivalent to \( f \) and its inverse both being continuous. | Therefore, any property defined only in terms of open sets is preserved by homeomorphism. Such a property is called a topological property. However, the later adjectives we define (\ | No |
The open intervals form a basis of \( \mathbb{R} \) . | In fact, more generally we have: | No |
Theorem 7.8.4 (Basis of metric spaces)\n\nThe \( r \) -neighborhoods form a basis of any metric space \( M \) . | Proof. Kind of silly - given an open set \( U \) draw an \( {r}_{p} \) -neighborhood \( {U}_{p} \) contained entirely inside \( U \) . Then \( \mathop{\bigcup }\limits_{p}{U}_{p} \) is contained in \( U \) and covers every point inside it.\n\nHence, an open set in \( {\mathbb{R}}^{2} \) is nothing more than a union of ... | Yes |
Proposition 8.1.6 (Closed subsets of compacts)\n\nClosed subsets of sequentially compact sets are compact. | Question 8.1.7. Prove this. (It should follow easily from definitions.) | No |
Theorem 8.2.1 (Tychonoff's theorem)\n\nIf \( X \) and \( Y \) are compact spaces, then so is \( X \times Y \) . | Proof. Problem \( 8\mathrm{E} \) . | No |
Theorem 8.2.2 (The interval is compact)\n\n\\( \\left\\lbrack {0,1}\\right\\rbrack \\) is compact. | Proof. Killed by Problem \\( 8{\\mathrm{\\;F}}^{ \\dagger } \\) ; however, here is a sketch of a direct proof. Split \\( \\left\\lbrack {0,1}\\right\\rbrack \\) into \\( \\left\\lbrack {0,\\frac{1}{2}}\\right\\rbrack \\cup \\left\\lbrack {\\frac{1}{2},1}\\right\\rbrack \\) . By Pigeonhole, infinitely many terms of the ... | No |
A subset of \( {\mathbb{R}}^{n} \) is compact if and only if it is closed and bounded. | Proof. Well, look at a closed and bounded \( S \subseteq {\mathbb{R}}^{n} \) . Since it’s bounded, it lives inside some box \( \left\lbrack {{a}_{1},{b}_{1}}\right\rbrack \times \left\lbrack {{a}_{2},{b}_{2}}\right\rbrack \times \cdots \times \left\lbrack {{a}_{n},{b}_{n}}\right\rbrack \) . By Tychonoff’s theorem, sinc... | No |
Suppose we cover the unit square \( M = {\left\lbrack 0,1\right\rbrack }^{2} \) by putting an open disk of diameter 1 centered at every point (trimming any overflow). This is clearly an open cover because, well, every point lies in many of the open sets, and in particular is the center of one. | But this is way overkill - we only need about four of these circles to cover the whole square. That's what is meant by a \ | No |
Theorem 8.3.5 (Sequentially compact \( \Leftrightarrow \) compact)\n\nA metric space \( M \) is sequentially compact if and only if it is compact. | We defer the proof to the last section. | No |
Proposition 8.4.1 (Compact \( \Rightarrow \) totally bounded)\n\nLet \( M \) be compact. Then \( M \) is totally bounded. | Proof using covers. For every point \( p \in M \), take an \( \varepsilon \) -neighborhood of \( p \), say \( {U}_{p} \) . These cover \( M \) for the horrendously stupid reason that each point \( p \) is at the very least covered by its open neighborhood \( {U}_{p} \) . Compactness then lets us take a finite subcover. | No |
Theorem 8.4.2 (Images of compacts are compact)\n\nLet \( f : X \rightarrow Y \) be a continuous function, where \( X \) is compact. Then the image\n\n\[{f}^{\text{img }}\left( X\right) \subseteq Y\]\n\nis compact. | Proof using covers. Take any open cover \( \left\{ {V}_{\alpha }\right\} \) in \( Y \) of \( {f}^{\mathrm{{img}}}\left( X\right) \) . By continuity of \( f \), it pulls back to an open cover \( \left\{ {U}_{\alpha }\right\} \) of \( X \) . Thus some finite subcover of this covers \( X \) . The corresponding \( V \) ’s ... | Yes |
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