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Theorem 1.1 The integers \( {a}_{r},{p}_{r} \), and \( {q}_{r} \) for the continued fraction expansion of a rational number \( \alpha = a/b \) are the same as the corresponding numbers generated by the gcd algorithm given input \( \left( {a, b}\right) \) . For both rational and irrational \( \alpha \), and for all rele...
The proofs are straightforward induction.
No
Theorem 1.2 Given an eventually periodic (infinite) continued fraction\n\n\[ \n{u}_{0} + \left\lbrack {\mathbf{u}\overline{\mathbf{v}}}\right\rbrack = {u}_{0} + \left\lbrack {\mathbf{u}\mathbf{v}\mathbf{v}\ldots }\right\rbrack = \left\lbrack {{u}_{0};{u}_{1},{u}_{2},\ldots {u}_{s},{v}_{1},\ldots {v}_{t},{v}_{1},\ldots ...
Proof. Let \( \alpha = {u}_{0} + \left\lbrack \mathbf{u\widetilde{v}}\right\rbrack \) . Since every rational number has a terminating continued fraction expansion, \( \alpha \) is irrational. Now \( \alpha = {u}_{0} + \left\lbrack {\mathbf{u} + \beta }\right\rbrack \) where \( \beta = \left\lbrack {\mathbf{v} + \beta }...
No
Proposition 1.2 The purely periodic expansions are those in which the loop begins immediately. That is, if \( {x}_{0} \) is the quadratic irrational to be expanded, and \( {x}_{j + 1} \mathrel{\text{:=}} \left\{ {1/{x}_{j}}\right\} ,{a}_{j + 1} \mathrel{\text{:=}} \left\lfloor {1/{x}_{j}}\right\rfloor \), then \( {a}_{...
Proof. If \( x \) satisfies the conditions, then \( {x}_{j + 1} = 1/{x}_{j} - {a}_{j + 1} \) so that \( \overline{{x}_{j + 1}} = 1/\overline{{x}_{j}} - {a}_{j + 1} \) . Inductively, then, \( {x}_{j} \) satisfies the same conditions. Now because \( {x}_{0} \) is a quadratic irrational, the sequence \( \left( {x}_{j}\rig...
Yes
Theorem 1.3 Let \( \alpha \) be an irrational number, with continued fraction expansion \( \alpha = \left\lbrack {{a}_{0};{a}_{,}{a}_{2},\ldots }\right\rbrack \) . Let \( \left( {p}_{n}\right) \) and \( \left( {q}_{n}\right) \) be the numerator and denominator sequences of the convergents to \( \alpha \) . If either \(...
It may come as a bit of a surprise that this was not known since classical times, but the only known proof requires a difficult result of van der Poorten, known as the Hadamard Quotient Theorem. [vdP2; vdP3].
No
Proposition 1.3 For all positive integers \( Q \), there exists integer \( q,1 \leq \) \( q \leq Q \) so that \( \parallel {q\alpha }\parallel < 1/Q \) .
Proof. The sequence \( {q\alpha },0 \leq q \leq Q \) has \( Q + 1 \) elements. Some two entries, say \( {q}_{1}\alpha \) and \( {q}_{2}\alpha \), must fall within the same interval \( \lbrack k/Q,\left( {k + 1}\right) /Q) \) mod 1, and these two will differ by strictly less than \( 1/Q \) . Take \( q = \left| {{q}_{1} ...
Yes
Proposition 1.4 Let \( \alpha \in \left( {0,1/2}\right) \) be a real number, and \( p \) and \( q \) be positive integers. Let \( \left( {0/1,{p}_{1}/{q}_{1},{p}_{2}/{q}_{2},\ldots }\right) \) be the convergents of \( \alpha \) . If \( q > {q}_{1} \) and \( \left| {\alpha - p/q}\right| \leq 1/\left( {2{q}^{2}}\right) \...
Proof. Assume \( p/q \) is not a convergent, yet \( \left| {\alpha - p/q}\right| \leq 1/\left( {2{q}^{2}}\right) \) . Choose \( j \geq 2 \) such that \( {q}_{j - 1} < q < {q}_{j} \) . Since \( q > {q}_{1} \), we can do this. Consider the open interval \( A \) with endpoints \( {p}_{j - 1}/{q}_{j - 1} \) and \( {p}_{j -...
Yes
Theorem 1.4 For \( 0 < \alpha < 1/2 \) and \( j \geq 1 \) ,\n\n(i) \( \begin{Vmatrix}{{q}_{j}\alpha }\end{Vmatrix} = {\left( -1\right) }^{j}{\rho }_{j} \) ,\n\n(ii) If \( 1 \leq q < {q}_{j} \) then \( \parallel {q\alpha }\parallel > \left| {\rho }_{j}\right| \) ,\n\n(iii) \( {\rho }_{j} = {a}_{j}{\rho }_{j - 1} + {\rho...
Proof. We begin with (iii). The recurrence follows from the fact that \( \left. \left\langle {p}_{j}\right\rangle \right\rangle \) and \( \left\langle {q}_{j}\right\rangle \) obey that recurrence, and \( {\rho }_{j} \) is a linear combination of these. Now (vi) follows from (iii), and (vii) from (vi).\n\nFor (ii), we h...
Yes
Lemma 3.2 Let \( u = \left( {{\mathbf{v}}^{ - }, - 1 + {a}_{r},1 + {a}_{r},{\mathbf{v}}^{-\prime }}\right) \) . Then \( \left| \mathbf{u}\right| = {\left| \mathbf{v}\right| }^{2} \).
Proof. We have\n\n\[ \left| \mathbf{u}\right| = \left| {{\mathbf{v}}^{ - },{a}_{r} - 1}\right| \left| {{\mathbf{v}}^{ - },{a}_{r} + 1}\right| + {\left| {\mathbf{v}}^{ - }\right| }^{2} \]\n\n
No
Theorem 3.3 If \( \alpha \) is an irrational real number, and \( {s}_{n}\left( \alpha \right) = \lfloor \left( {n + 1}\right) \alpha \rfloor - \lfloor {n\alpha }\rfloor \), then there are sequences \( \left( {A}_{k}\right) ,\left( {B}_{k}\right) \) and \( \left( {C}_{k}\right) \) of words, and sequences \( \left( {\alp...
Proof. We first establish a lemma.\n\nLemma 3.3 Suppose \( 0
No
Proposition 4.2 The probability measure \( {\nu }_{M} \) on \( {E}_{M} \) is invariant under \( T : x \rightarrow 1/x - \lfloor 1/x\rfloor ;{\nu }_{M}\left( {{T}^{-1}A}\right) = {\nu }_{M}\left( A\right) \) for all \( \mu \) -measurable subsets \( A \) of \( \left\lbrack {0,1}\right\rbrack \) .
Proof. It is sufficient to show that for \( r \geq 1 \) and \( \mathbf{v} \in {M}^{r},{\nu }_{M}\left( {{T}^{-1}{I}_{\mathbf{v}}}\right) = \) \( {\nu }_{M}\left( {I}_{\mathbf{v}}\right) \) . Since each \( {I}_{\mathbf{v}} \) can be written as a disjoint union, apart from common endpoints,(which can occur if \( M \) is ...
Yes
Lemma 4.1 For \( \gamma \in {F}_{N}^{ \bot },\begin{Vmatrix}{{G}^{2}\gamma }\end{Vmatrix} \leq \left( {1 - {2}^{-N}}\right) \parallel \gamma \parallel \) .
Proof. First, we note that \( G{\psi }_{\varnothing, T} = {\psi }_{T,\varnothing } \) . Thus, it will be sufficient to show that for \( S \neq \varnothing ,\begin{Vmatrix}{G{\psi }_{\left( S, T\right) }}\end{Vmatrix} \leq 1 - {2}^{-N} \) . For \( n \in \mathbb{N} \), let \( S - n = \) \( \{ s - n : s \in S\} \cap \math...
Yes
Theorem 5.2 If \( z \in B \) has a Hurwitz continued fraction expansion to depth \( n + 2 \), then\n\n\[ \left| \frac{{q}_{n + 2}}{{q}_{n}}\right| \geq 3/2 \]
Proof. Recall that\n\n\[ {q}_{n + 1} = {a}_{n + 1}{q}_{n} + {q}_{n - 1} \]\n\n\[ {z}_{n + 1} = \frac{1}{{z}_{n}} - {a}_{n + 1} \]\n\n\[ {w}_{n + 1} = \frac{1}{{a}_{n + 1} + {w}_{n}} \]\n\n\[ {a}_{n + 1} = \left\lbrack \frac{1}{{z}_{n}}\right\rbrack \]\n\n\[ \frac{{q}_{n}}{{q}_{n + 1}} = {w}_{n + 1} \]\n\nLet \( {G}_{a}...
No
Lemma 5.1 Suppose \( z \in B, n \geq 1 \), and \( z \) has a Hurwitz continued fraction to depth \( n + 2 \) . If \( \left| {w}_{n}\right| \geq 2/3 \), then \( \left| {w}_{n + 1}\right| < 2/3 \) . Furthermore, either \( \left| {w}_{n}\right| < 2/3 \), or \( \frac{2}{3} \leq \left| {w}_{n}\right| < 1 \) and one of the f...
The proof of the lemma is inductive and begins with the observation that the claim is true at the outset because \( {w}_{0} = 0 \) . We use a fact about reciprocals of disks: If \( D \) is a disk in \( \mathbb{C} \) with center \( z \) and radius \( r \), and if \( \left| z\right| > r \), then with the notation \( D\le...
Yes
Theorem 5.4 Let \( u \) and \( v \) be positive integers. Let \( \alpha = \sqrt{u} + i\sqrt{v} \) . Let \( p, q \) be Gaussian integers with \( {\left| q\right| }^{2} \geq 9 \), and assume \( \left| {p - {q\alpha }}\right| \leq 2\sqrt{2}/\left| q\right| \) . Let \( e\left( {q, p,\alpha }\right) \) be the scaled error \...
Proof. Let \( p = {p}_{1} + i{p}_{2}, q = {q}_{1} + i{q}_{2} \) with \( {p}_{1},{p}_{2},{q}_{1},{q}_{2} \in \mathbb{Z} \) . Let \( c = {p}_{1}{q}_{1} + \) \( {p}_{2}{q}_{2}, d = {p}_{2}{q}_{1} - {p}_{1}{q}_{2} \), so that \( {c}^{2} + {d}^{2} = {\left| p\right| }^{2}{\left| q\right| }^{2} \) . Let \( {\theta }_{1} + i{...
Yes
Theorem 5.5 There is a unique measure \( \mu \) on \( B \) that is absolutely continuous with respect to Lebesgue measure and invariant under \( T \) . This measure has a density function \( \rho \) ; the density is continuous except perhaps along the arcs \( \left| {z \pm 1}\right| = 1,\left| {z \pm i}\right| = 1 \), ...
The general plan of the proof is this: We show that if \( f \) is a continuous probability density function on \( B \) and \( X \) a random variable on \( B \) with density \( f \), then there exists a positive integer \( n \), depending on \( f \), and \( \epsilon > 0 \) , such that the density of \( {T}^{n}X \) is gr...
Yes
Theorem 7.1 Let \( \alpha \) be a real algebraic number of degree \( n > 1 \). Let \( \mu > 1 \) be a unit of \( \mathbb{Q}\left( \alpha \right) \) of degree \( n \). Let \( p = \left| {\det {P}_{\mu }}\right| \). Suppose \( \left( \nu \right) \) is a sequence of units of degree \( n \) in \( \mathbb{Q}\left( \alpha \r...
Proof. Suppose \( \left( {\sigma, q,\alpha }\right) \) is good. Then the integers nearest \( q{\alpha }_{j} \) respectively, call them \( \left( {{p}_{1},\ldots ,{p}_{n - 1}}\right) \), satisfy \( \left| {{p}_{j} - q{\alpha }^{j}}\right| \leq \sigma {q}^{-1/\left( {n - 1}\right) } \) for \( 1 \leq j \leq n - 1 \). Choo...
Yes
Theorem 7.2 Let \( \\alpha \) be a real algebraic integer of degree \( n > 1 \) . Then for all sufficiently large \( \\sigma \), there exists a finite set \( B\\left( {\\sigma ,\\alpha }\\right) \\subset \\mathbb{Q}\\left( \\alpha \\right) \\cap \\left( {1,\\infty }\\right) \), and a constant \( C = C\\left( {\\\sigma ...
Proof. Choose \( \\sigma \) so that for all \( {q}_{k} \) associated with \( {\\nu }_{k},\\left( {\\\sigma ,{q}_{k},\\alpha }\\right) \) is good. Choose \( N \) so that, for all \( q \) such that \( \\left( {\\\sigma, q,\\alpha }\\right) \) is good, there exists a \( \\mathbf{c} \\in \) \( {\\left\\lbrack -N, N\\right\...
Yes
Theorem 7.3 Let \( \alpha \) be an algebraic integer of degree \( n > 1 \) and signature \( \left( {{r}_{1},{r}_{2}}\right) \) . Then for all \( \sigma > 1 \) there is a finite set \( R\left( {\sigma ,\alpha }\right) \) of positive elements of \( \mathbb{Q}\left( \alpha \right) \), and a nonsingular matrix \( {W}_{\alp...
Proof. Without loss of generality we may take \( \sigma \) arbitrarily large, because if \( \left( {{\sigma }_{1}, q,\alpha }\right) \) is good and \( {\sigma }_{2} > {\sigma }_{1} \), then \( \left( {{\sigma }_{2}, q,\alpha }\right) \) is good. We may also take \( q \) arbitrarily large. We recall that \( p = \left| {...
Yes
Lemma 9.1 For \( \alpha > 1,{G}_{M, s} \) maps \( {H}_{\alpha } \) into \( {H}_{9/4} \) .
Proof. Suppose \( \alpha > 1 \) and choose \( \epsilon ,0 < \epsilon < \min \left\lbrack {1/{10},1/2 - 1/\left( {1 + \sqrt{\alpha }}\right) }\right\rbrack \) . Let \( b = 9/4 + \epsilon \) . The disk \( \left| {z - 1}\right| < \left( {3/2}\right) + \epsilon \) is mapped by \( z \rightarrow 1/\left( {k + z}\right) \) on...
Yes
Lemma 9.2 If \( f\left( z\right) = \mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{\left( z - 1\right) }^{n} \in {H}_{\alpha } \) then \( {G}_{M, s}\left\lbrack f\right\rbrack \left( z\right) = \) \( \mathop{\sum }\limits_{{m = 0}}^{\infty }{b}_{m}{\left( z - 1\right) }^{m} \) where \[ {b}_{m} = \mathop{\sum }\limits_...
Proof. We have already seen that \( {G}_{M, s}\left\lbrack f\right\rbrack \in {H}_{\left( {9/4}\right) + \epsilon } \) for some \( \epsilon > 0 \) , so that there exist \( {b}_{0},{b}_{1},\ldots \) for which \( {G}_{M, s}\left\lbrack f\right\rbrack \left( z\right) = \mathop{\sum }\limits_{{m = 0}}^{\infty }{b}_{m}{\lef...
Yes
Lemma 9.3 If \( M \subseteq {\mathbb{Z}}^{ + } \) with \( \left| M\right| \geq 2 \) (including infinite \( M \) ), and if \( \mathop{\sum }\limits_{{k \in M}}{k}^{-\sigma } < \infty ,\; \) then \( \; \) for \( \; \) all \( \;m, n \geq 0, \)\n\n\[ \left| {{\gamma }_{M, s}\left\lbrack {m, n}\right\rbrack }\right| \leq {2...
Proof. Let \( C = C\left\lbrack {M, s}\right\rbrack \mathrel{\text{:=}} {2}^{\sigma }\mathop{\sum }\limits_{{k \in M}}{k}^{-\sigma } \) . (For the moment, \( M \) and \( s \) are fixed so we drop the subscripts.) We must show \( \left| {\gamma \left\lbrack {m, n}\right\rbrack }\right| \leq C{\left( 3/2\right) }^{-m} \)...
Yes
Lemma 9.4 The spectrum of \( {G}_{M, s} \) is independent of \( \alpha \) .
Proof. Suppose \( \lambda \neq 0 \) belongs to the spectrum of \( {G}_{M, s} \) seen as an operator on \( {H}_{\alpha },1 < \alpha \leq 9/4 \) . Then there exists \( f \in {H}_{\alpha } \) so that \( \left( {{G}_{M, s} - {\lambda I}}\right) \left\lbrack f\right\rbrack = 0 \) . Thus \( f = {\lambda }^{-1}{G}_{M, s}f \in...
Yes
Lemma 9.5 If \( P \) is an orthogonal projection of \( H \) onto a subspace of \( H \) of dimension \( n \), then \( {s}_{n + 1}\left( A\right) \leq \mathop{\sup }\limits_{{\parallel x\parallel = 1}}\parallel {Ax} - {PAx}\parallel \)
This is a corollary to Theorem 2.2, p. 31 of [GK].
No
Lemma 9.7 For all \( M \) and \( \sigma \) satisfying the standard conditions, and for all \( x \) with \( 0 \leq x \leq 1 \), we have\n\n\[ \n{\mu }_{M,\sigma }\lbrack 0, x) = \mathop{\lim }\limits_{{r \rightarrow \infty }}{\lambda }_{M}{\left( \sigma \right) }^{-r}\mathop{\sum }\limits_{{\mathbf{v} \in {M}^{r},\left\...
That is, \( {\mu }_{M,\sigma } \) is the limiting case of discrete measures assigning mass\n\n\[ \n{\lambda }_{M}{\left( \sigma \right) }^{-r}{\left| \mathbf{v}\right| }^{-\sigma }g\left( {\{ \mathbf{v}\} }\right) \n\]\n\nto the point \( \left\lbrack \mathbf{v}\right\rbrack \), summed over all \( \mathbf{v} \in {M}^{r}...
Yes
Lemma 9.9 S maps \( {H}_{M} \) into \( {H}_{\alpha } \) and is a bounded linear operator.
Proof. By the Cauchy-Schwarz inequality, if \( z = u + {iv} \) with \( u > - 1/2 \) then\n\n\[ \left| {{S}_{M,\sigma }\left\lbrack \phi \right\rbrack \left( z\right) }\right| \leq {\left( {\int }_{0}^{\infty }{x}^{\sigma - 1}{e}^{-{2ux}}\mathop{\sum }\limits_{{k \in M}}{e}^{-{kx}}dx\right) }^{1/2}\parallel \phi {\paral...
Yes
Lemma 9.10 \( S = {S}_{M,\sigma } \) is nonsingular, that is, if \( \phi \in {H}_{M} \neq 0 \) then \( S\left\lbrack \phi \right\rbrack \in {H}_{\alpha } \neq 0. \)
Proof. \( \;S\left\lbrack \phi \right\rbrack \) is the classical laplace transform of \( {x}^{\left( {\sigma - 1}\right) /2}{e}_{M}\left( x\right) \phi \) . If \( \phi \) is not the zero function in \( {H}_{M} \) (that is, it is not the case that \( \phi \) is zero almost everywhere), then this other function is likewi...
Yes
Lemma 9.11 If \( f\left( z\right) = {\int }_{0}^{\infty }{\left( 1 + \theta z\right) }^{-\sigma }{d\mu }\left( \theta \right) \) where \( \mu \) is a finite signed measure on \( \lbrack 0,\infty ) \) then \( G\left\lbrack f\right\rbrack \) has the same form, and\n\n\[ S\left( {\frac{1}{\Gamma \left( \sigma \right) }{\i...
Proof. It is sufficient to establish the result for the case that \( \mu \) is a probability measure on \( \lbrack 0,\infty ) \) . So suppose \( f\left( z\right) = {\int }_{0}^{\infty }{\left( 1 + \theta z\right) }^{-\sigma }{d\mu }\left( \theta \right) \) . Then\n\n\[ G\left\lbrack f\right\rbrack \left( z\right) = {\i...
Yes
Proposition 9.1 If \( \phi \in {H}_{M} \) and if \( {\zeta }_{M}\left( \sigma \right) = \mathop{\sum }\limits_{{k \in M}}{k}^{-\sigma } < \infty \), then for \( x > 0, K\left\lbrack \phi \right\rbrack \left( x\right) \) is defined. Moreover, the function \( f = K\left\lbrack \phi \right\rbrack \) satisfies \( \parallel...
Proof. By the Cauchy-Schwarz inequality,\n\n\[ \left| {K\left\lbrack \phi \right\rbrack \left( x\right) }\right| \leq \parallel \phi {\parallel }_{M}{\int }_{0}^{\infty }{\left( {J}_{\sigma - 1}^{2}\left( 2\sqrt{xy}\right) {e}_{M}\left( y\right) \right) }^{1/2}{dy}. \]\n\nNow\n\n\[ {\int }_{0}^{\infty }{J}_{\sigma - 1}...
Yes
Theorem 9.3 If \( \sigma > 0, K = {K}_{M,\sigma } \), and \( \phi \in {H}_{M} \), then\n\n\[ K\left\lbrack \phi \right\rbrack = \mathop{\sum }\limits_{{k = 0}}^{\infty }\left\langle {{u}_{k},\phi }\right\rangle {e}_{k} \]\n\nwhere\n\n\[ {u}_{k}\left( w\right) = {w}^{\left( {\sigma - 1}\right) /2}{L}_{k}^{s - 1}\left( w...
Proof. First we establish the identity for \( {J}_{\sigma - 1}\left( {2\sqrt{vw}}\right) \) claimed above.\n\n\[ {J}_{\sigma - 1}\left( {2\sqrt{vw}}\right) = {\left( vw\right) }^{\left( {\sigma - 1}\right) /2}\mathop{\sum }\limits_{{k = 0}}^{\infty }\frac{{w}^{k}{e}^{-w}}{\Gamma \left( {k + \sigma }\right) }{L}_{k}^{\s...
Yes
Lemma 9.12\n\n\[ \n{G}_{M,\sigma }^{2} : {\mathcal{A}}_{M,{2\sigma }} \rightarrow {\mathcal{A}}_{M,\left( {3/2}\right) \sigma } \n\]
Proof. Clearly if \( f \) is positive on \( \left( {0,1}\right) \) then so is \( {G}^{2}f \) . Taking derivatives and simplifying, we see that the claim holds provided that for all \( \left( {j, k}\right) \in M \) ,\n\n\[ \n\frac{\sigma k}{{\left( jk + 1 + tk\right) }^{\sigma + 1}} + \frac{2\sigma }{{\left( jk + 1 + tk...
Yes
Theorem 9.4 Uniformly over \( \sigma \) and \( M \) satisfying the standard conditions, and uniformly over \( f \in {H}_{A} \) ,\n\n\[ \n{\lambda }_{M,\sigma }^{-r}{G}_{M,\sigma }^{r}\left\lbrack f\right\rbrack = \frac{{\int }_{0}^{1}{fd}{\mu }_{M,\sigma }}{{\int }_{0}^{1}{g}_{M,\sigma }\;d{\mu }_{M,\sigma }}{g}_{M,\si...
Proof. We begin with some lemmas.
No
Lemma 9.13 Let \( {C}_{\sigma } \mathrel{\text{:=}} 3{e}^{\sigma }\left( {1 + 1/\sigma }\right) \) . If \( h = h\left\lbrack f\right\rbrack = f + {C}_{\sigma }\parallel f{\parallel }_{A}{g}_{M,\sigma } \) then \( h \in {\mathcal{A}}_{2\sigma } \) .
Proof. We have \( \left| {f\left( t\right) }\right| \leq \parallel f{\parallel }_{A} \) and \( \left| {{f}^{\prime }\left( t\right) }\right| \leq \parallel f{\parallel }_{A} \) for \( 0 \leq t \leq 1 \) . Now \( {g}_{M,\sigma }\left( t\right) = {\int }_{0}^{1}{\left( 1 + \theta t\right) }^{-\sigma }{d\mu }\left( \theta...
Yes
Lemma 9.14 \( {G}_{M,\sigma }^{2} : {\mathcal{A}}_{2\sigma } \rightarrow {\mathcal{A}}_{{3\sigma }/2} \) .
Proof. All that is needed is to expand \( {G}^{2}f \) and check the details:\n\n\[ \n{G}^{2}f\left\lbrack t\right\rbrack = \mathop{\sum }\limits_{{j, k \in M}}{\left( jk + 1 + jt\right) }^{-\sigma }f\left( {1/\left( {j + 1/\left( {k + t}\right) }\right) }\right) \n\] \n\nand \n\n\[ \n{\left( {G}^{2}f\right) }^{\prime }...
No
Lemma 9.15 If \( u \in {\mathcal{A}}_{{3\sigma }/2} \) then\n\n\[ u - \left( {1/6}\right) {e}^{-{3\sigma }}{L}_{M,\sigma }\left\lbrack u\right\rbrack {g}_{M,\sigma } \in {\mathcal{A}}_{2\sigma } \]
Proof. It suffices to show two things: first, that \( u\left( 0\right) \geq {e}^{-{3\sigma }/2}L\left\lbrack u\right\rbrack \), and second, that if \( \zeta \leq \left( {1/6}\right) {e}^{-{3\sigma }/2}u\left( 0\right) \), then \( u - {\zeta g} \in {\mathcal{A}}_{2\sigma } \) .\n\nThe first is simple: since \( u \in {\m...
Yes
For all \( r \geq 1 \), \n\n\[ \n{h}_{r} \in {\mathcal{A}}_{2\sigma } \n\] \n\nand \n\n\[ \n{L}_{M,\sigma }\left\lbrack {h}_{r + 1}\right\rbrack = \left( {1 - \left( {1/6}\right) {e}^{-{3\sigma }}{L}_{M,\sigma }{g}_{M,\sigma }}\right) {L}_{M,\sigma }\left\lbrack {h}_{r}\right\rbrack . \n\]
Proof. By \( {9.13},{h}_{0} \in {\mathcal{A}}_{2\sigma } \) . By 9.14, if \( {h}_{r} \in {\mathcal{A}}_{2\sigma } \) then \( {\lambda }^{-2}{G}^{2}\left\lbrack {h}_{r}\right\rbrack \in \) \( {\mathcal{A}}_{{3\sigma }/2} \) . To apply 9.15 and conclude that \( {h}_{r + 1} \in {\mathcal{A}}_{2\sigma } \), we need to know...
Yes
Theorem 9.5 For all \( M \) and \( \sigma \) satisfying the standard conditions, the operator \( {G}_{M,\sigma } \) has spectral radius \( {\lambda }_{M}\left( \sigma \right) \), and as \( r \rightarrow \infty \) ,
\[ {\lambda }_{M}^{-r}\left( \sigma \right) {G}_{M,\sigma }\left\lbrack f\right\rbrack \left( t\right) = {C}_{f}{g}_{M,\sigma }\left( t\right) + O\left( {{\left( 1 - \frac{1}{12}{e}^{-{7\sigma }/2}\right) }^{r}\parallel f{\parallel }_{A}}\right) \] uniformly over \( f \in {H}_{A} \) and over \( \sigma \) and \( M \) .
No
Theorem 10.2 Let \( \sigma > 0 \) . Let \( F\left( s\right) = \mathop{\sum }\limits_{0}^{\infty }{a}_{n}{s}^{n} \) be a Dirichlet series with nonnegative coefficients \( \left( {a}_{n}\right) \) that converges for \( \Re \left( s\right) > \sigma \) . Assume further that \( F\left( s\right) \) extends to an analytic fun...
\[ \mathop{\sum }\limits_{{n \leq x}}{a}_{n} = \frac{A\left( \sigma \right) }{{\sigma \Gamma }\left( {\sigma + 1}\right) }{x}^{\sigma }{\log }^{\gamma }x\left( {1 + o\left( 1\right) }\right) . \]
Yes
Theorem 10.4 Uniformly over \( y \in \mathbb{R} \), as \( x \rightarrow \infty \), \[ \frac{1}{\left| {\Omega }_{x}\right| }\# \left\{ {\left( {u, v}\right) : \mu \log x + {\delta y}\sqrt{\log x} - 1/2 < l\left( {u, v}\right) }\right. \] \[ \left. { \leq \mu \log x + {\delta y}\sqrt{\log x} + 1/2}\right\} = \frac{1}{\d...
The proof depends heavily on a key innovation: the complex analysis is extended from treating just a neighborhood of the critical point \( (s = 2 \) or \( s = 1 \), depending on how we define \( \lambda \) or \( \Lambda \) ), to treating at least a punctured half plane. One defines again a linear operator \( {H}_{s, w}...
No
Theorem 12.1 If \( \mathop{\sum }\limits_{1}^{\infty }\left| {b}_{j}\right| < \infty \), then the sequence \( \left( {K}_{r}\right) \), given by\n\n\[ \n{K}_{r} = \frac{1}{{b}_{1} + \frac{1}{{b}_{2} \ddots + \frac{1}{{b}_{r}}}}\n\]\n\nfails to converge.
Proof. We give what may be a new proof, which yields en passant an explicit lower bound for the difference \( \left| {{K}_{2r} - {K}_{{2r} + 1}}\right| \) :\n\n\[ \n\left| {{K}_{2r} - {K}_{{2r} + 1}}\right| \geq \exp \left( {-\mathop{\sum }\limits_{1}^{{{2r} + 1}}\left| {b}_{j}\right| }\right)\n\]\n\nWe use the spheric...
Yes
Theorem 12.3 Suppose \( \left( {b}_{n}\right) \) is a sequence of nonzero complex numbers within the wedge \( V\left\lbrack \epsilon \right\rbrack = \{ z \in \mathbb{C}\left| {z \neq 0\text{ and }}\right| \arg \left( z\right) | \leq \pi /2 - \epsilon \} \) . Let \( {f}_{n} = \left\lbrack {{b}_{1},\ldots ,{b}_{n}}\right...
Proof. We first prove the last item. Recall that if \( {M}_{0} = \left( \begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \) and if\n\n\( {M}_{n} = {M}_{n - 1}\left( \begin{array}{ll} 0 & 1 \\ 1 & {b}_{n} \end{array}\right) \) then \( {M}_{n} = \left( \begin{array}{ll} {A}_{n - 1} & {A}_{n} \\ {B}_{n - 1} & {B}_{n} \...
Yes
Theorem 12.5 For all \( n \geq 1 \) and all \( z \) with \( \left| z\right| \leq n/2 \) , \[ \left| {\frac{{p}_{n}\left( z\right) }{{q}_{n}\left( z\right) }{e}^{-z} - 1}\right| \leq {2}^{-n} \]
Proof. We use the classical identity \[ {p}_{n}\left( z\right) - {q}_{n}\left( z\right) {e}^{z} = \frac{{\left( -1\right) }^{n}{z}^{{2n} + 1}}{\left( {2n}\right) !}{\int }_{0}^{1}{e}^{tz}{\left( 1 - t\right) }^{n}{t}^{n}{dt} \] and we use a result of Saff and Varga [SV] concerning the zeros of \( {q}_{n}\left( z\right)...
Yes
17.3 Calculation of a Basis of the Additive Group \( \Gamma \left( A\right) \)
17.3.1 Step 1: Preliminary Statements 241\n17.3.2 Step 2: Application of the LLL-Algorithm 242\n17.3.3 Step 3: Calculation of an Integer Basis Having a Basis of an Integer Sublattice 242
Yes
Let us calculate the above polynomials for the continued fractions \( \left\lbrack {x}_{0}\right\rbrack ,\left\lbrack {{x}_{0};{x}_{1}}\right\rbrack ,\left\lbrack {{x}_{0};{x}_{1} : {x}_{2}}\right\rbrack ,\ldots \) . The polynomials are as follows:
\[ {P}_{0}\left( {x}_{0}\right) = {x}_{0},\;{Q}_{0}\left( {x}_{0}\right) = 1; \] \[ {P}_{1}\left( {{x}_{0},{x}_{1}}\right) = {x}_{0}{x}_{1} + 1,\;{Q}_{1}\left( {{x}_{0},{x}_{1}}\right) = {x}_{1}; \] \[ {P}_{2}\left( {{x}_{0},{x}_{1},{x}_{2}}\right) = {x}_{0}{x}_{1}{x}_{2} + {x}_{2} + {x}_{0},\;{Q}_{2}\left( {{x}_{0},{x...
No
Lemma 1.11 The following holds:\n\n\[ \left\{ \begin{array}{l} {p}_{k} = {a}_{0}{\widehat{p}}_{k} + {\widehat{q}}_{k} \\ {q}_{k} = {\widehat{p}}_{k} \end{array}\right. \]
Proof Since\n\n\[ \frac{{\widehat{p}}_{k}}{{\widehat{q}}_{k}} = \left\lbrack {{a}_{1};{a}_{2} : \cdots : {a}_{k}}\right\rbrack \]\n\nwe get\n\n\[ \frac{{p}_{k}}{{q}_{k}} = {a}_{0} + \frac{1}{{\widehat{p}}_{k}/{\widehat{q}}_{k}} = \frac{{a}_{0}{\widehat{p}}_{k} + {\widehat{q}}_{k}}{{\widehat{p}}_{k}}. \]\n\nTherefore,\n...
Yes
Proposition 1.12 Let \( \left\lbrack {{a}_{0};{a}_{1} : \cdots : {a}_{n}}\right\rbrack \) be a continued fraction with integer elements. Then the corresponding integers \( {p}_{k} \) and \( {q}_{k} \) are relatively prime.
Proof We prove the statement by induction on \( k \) . It is clear that \( {p}_{0} = {a}_{0} \) and \( {q}_{0} = 1 \) are relatively prime. Suppose that the statement holds for \( k - 1 \) . Then \( {\widehat{p}}_{k} \) and \( {\widehat{q}}_{k} \) are relatively prime by the induction assumption. Now the statement ho...
Yes
Proposition 1.13 For every integer \( k \) we get\n\n\[ \left\{ \begin{array}{l} {p}_{k} = {a}_{k}{p}_{k - 1} + {p}_{k - 2} \\ {q}_{k} = {a}_{k}{q}_{k - 1} + {q}_{k - 2} \end{array}\right. \]
Proof We prove the statement by induction on \( k \) .\n\nFor \( k = 2 \) the statement holds, since\n\n\[ \frac{{p}_{0}}{{q}_{0}} = \frac{{a}_{0}}{1}\;\text{ and }\;\frac{{p}_{1}}{{q}_{1}} = \frac{{a}_{0}{a}_{1} + 1}{{a}_{1}} \]\n\nand therefore\n\n\[ \frac{{p}_{2}}{{q}_{2}} = \frac{{a}_{2} + {a}_{0}{a}_{1}{a}_{2} + {...
Yes
Corollary 1.14 For regular continued fractions the following estimates hold:\n\n\[ \n\left| {p}_{k}\right| \geq {F}_{k}\;\text{ and }\;{q}_{k} \geq {F}_{k + 1} \n\]
Proof We prove the statement by induction on \( k \) . Direct calculations show that\n\n\[ \n\left| {p}_{0}\right| \geq 0,\;\left| {p}_{1}\right| \geq 1,\;\text{ and }\;\left| {p}_{2}\right| \geq 1, \n\]\n\n\[ \n\left| {q}_{0}\right| \geq 1,\;\left| {q}_{1}\right| \geq 1,\;\text{ and }\;\left| {q}_{2}\right| \geq 2. \n...
Yes
Proposition 1.15 For every \( k \geq 1 \), the following holds:\n\n\[ \frac{{p}_{k - 1}}{{q}_{k - 1}} - \frac{{p}_{k}}{{q}_{k}} = \frac{{\left( -1\right) }^{k}}{{q}_{k - 1}{q}_{k}} \]
Proof Let us multiply both sides by \( {q}_{k - 1}{q}_{k} \) . We get\n\n\[ {p}_{k - 1}{q}_{k} - {p}_{k}{q}_{k - 1} = {\left( -1\right) }^{k}. \]\n\nWe prove this by induction on \( k \).\n\nFor \( k = 1 \) we have\n\n\[ {p}_{0}{q}_{1} - {p}_{1}{q}_{0} = {a}_{1}{a}_{0} - \left( {{a}_{1}{a}_{0} + 1}\right) = - 1. \]\n\n...
Yes
Theorem 1.16 The sequence of \( k \) -convergents of an arbitrary regular continued fraction converges.
Proof of Theorem 1.16 Let \( \left\lbrack {{a}_{0};{a}_{1} : \cdots }\right\rbrack \) be an infinite regular continued fraction. From Proposition 1.15 and Corollary 1.14 we have\n\n\[ \left| {\frac{{p}_{k - 1}}{{q}_{k - 1}} - \frac{{p}_{k}}{{q}_{k}}}\right| \leq \frac{1}{{F}_{k}{F}_{k + 1}} \]\n\nSince the sum\n\n\[ \m...
No
Proposition 1.19 The sequence of real numbers\n\n\[ \n\\left| {\\alpha - \\frac{{p}_{k}}{{q}_{k}}}\\right| ,\\;k = 0,1,2,\\ldots ,\n\]\n\nis strongly decreasing, except for the case of \( \\alpha = \\left\\lbrack {{a}_{0},1,1}\\right\\rbrack \), where this sequence consists of the following three equivalent elements: \...
Proof Recall that \( {r}_{k} \) denotes the reminder \( \\left\\lbrack {{a}_{k + 1};{a}_{k + 2} : \\cdots }\\right\\rbrack \).\n\nLet us first prove that\n\n\[ \n\\left| {\\alpha -\\lfloor \\alpha \\rfloor }\\right| > \\left| {\\alpha - \\frac{{p}_{1}}{{q}_{1}}}\\right| \n\]\n\nThis inequality is equivalent to\n\n\[ \n...
Yes
Lemma 1.21 Theorem 1.20 is true if for every irrational \( \alpha \), there are infinitely many integer solutions \( \left( {p, q}\right) \) of\n\n\[ \left| {\alpha - \frac{p}{q}}\right| \leq \frac{1}{\sqrt{5}{q}^{2}} \]
Proof It is clear that for rational \( \alpha = p/q \) the integer pairs \( \left( {{np},{nq}}\right) \) are solutions. In the irrational case it is enough to prove the theorem for \( c = \frac{1}{\sqrt{5}} \) . Let us prove that the equality can occur at most twice. Suppose we have\n\n\[ \alpha - \frac{p}{q} = \pm \fr...
Yes
Lemma 1.22 Consider a real number \( \alpha \) with regular continued fraction \( \left\lbrack {{a}_{0};{a}_{1}}\right. \) : \( \cdots \rbrack \) . Let \( {p}_{k}/{q}_{k} \) be its \( k \) -convergents. Then for an infinite sequence of integers we have\n\n\[ \frac{{q}_{k}}{{q}_{k - 1}} \geq \frac{1 + \sqrt{5}}{2} \]
Proof By Proposition 1.13 we have\n\n\[ \frac{{q}_{k}}{{q}_{k - 1}} = {a}_{k} + \frac{{q}_{k - 2}}{{q}_{k - 1}}. \]\n\nHence if \( {a}_{k} \geq 2 \), then\n\n\[ \frac{{q}_{k}}{{q}_{k - 1}} \geq {a}_{k} \geq 2 > \frac{1 + \sqrt{5}}{2}. \]\n\nSuppose now that \( {a}_{k} = 1 \) and let\n\n\[ \frac{{q}_{k}}{{q}_{k - 1}} < ...
Yes
Proposition 1.23 Let \( \alpha \) be the golden ratio, i.e., \n\n\[ \n\alpha = \frac{1 + \sqrt{5}}{2} = \left\lbrack {1;1 : 1 : 1 : 1 : \cdots }\right\rbrack \n\] \n\nIf \( c < \frac{1}{\sqrt{5}} \), then inequality (1.5) has only finitely many solutions.
Proof of Proposition 1.23 First of all, let us show that it is enough to check only all the convergents \( {p}_{k}/{q}_{k} \) . From Theorem 1.16 it follows that best approximations are convergents to a number. Let \( p/q \) be a rational number such that \( {q}_{k} \leq q < {q}_{k + 1} \) . Therefore, \n\n\[ \n{q}^{2}...
Yes
The index of a sublattice generated by a pair of integer vectors \( v \) and \( w \) in \( {\mathbb{Z}}^{2} \) equals the number of all integer points \( P \) satisfying\n\n\[ \n{AP} = {\alpha v} + {\beta w}\;\text{ with }0 \leq \alpha < 1;0 \leq \beta < 1, \n\]\n\nwhere \( A \) is an arbitrary integer point.
Proof Let \( H \) be the subgroup of \( {\mathbb{Z}}^{2} \) generated by \( v \) and \( w \) . Define\n\n\[ \n\operatorname{Par}\left( {v, w}\right) = \{ {\alpha v} + {\beta w} \mid 0 \leq \alpha ,\beta < 1\} . \n\]\n\nFirst, we show that for every integer vector \( g \) there exists an integer point \( P \in \) \( \op...
Yes
Proposition 2.5 Two integer segments are congruent if and only if they have the same integer length.
Proof Let \( {AB} \) be an integer segment of length \( k \) . Let us prove that it is integer congruent to the integer segment with endpoints \( O\left( {0,0}\right) \) and \( K\left( {k,0}\right) \) . Consider an integer translation sending \( A \) to the origin \( O \) . Let this translation send \( B \) to an integ...
Yes
Proposition 2.11 Consider an integer triangle \( \bigtriangleup {ABC} \) . Then the following statements are equivalent:\n\n(a) \( \bigtriangleup {ABC} \) is empty;\n\n(b) \( \operatorname{IS}\left( {\bigtriangleup {ABC}}\right) = 1 \) ;\n\n(c) \( S\left( {\bigtriangleup {ABC}}\right) = 1/2 \) .
Proof (a) \( \Rightarrow \) (b). Let an integer triangle \( \bigtriangleup {ABC} \) be empty. Then by symmetry, a parallelogram with edges \( {AB} \) and \( {AC} \) does not contain integer points except for its vertices as well. Therefore, by Proposition 2.2 there is only one coset for the subgroup generated by \( {AB...
Yes
Corollary 2.15 The integer area of polygons in the plane is twice the Euclidean area.
Proof The statement holds for empty triangles by Proposition 2.11. Now the corollary follows directly from Proposition 2.13 and the definition of integer area.
No
Example 2.18 For instance, for the following pentagon\n\n![1c600fd7-25c8-4af1-a5fa-ba617d01167c_45_0.jpg](images/1c600fd7-25c8-4af1-a5fa-ba617d01167c_45_0.jpg)\n\nthe number of inner integer points equals 4 , the number of integer points on the edges equals 6 , and the area equals 6 . So,\n\n\[ 6 = 4 + 6/2 - 1\text{.} ...
Proof of Theorem 2.17 We prove Pick's formula by the induction on the area.\n\nBase of induction. If the Euclidean area of an integer polygon is \( 1/2 \), then by Proposition 2.11 the polygon is an empty triangle. For the empty triangle we have\n\n\[ S = 0 + 3/2 - 1 = 1/2 \]\n\nInduction step. Fix \( k \) . Suppose th...
Yes
Theorem 3.4 Consider \( 0 < \alpha < 1 \) . Let \( {A}_{0}{A}_{1}{A}_{2}\ldots \) and \( {B}_{0}{B}_{1}{B}_{2}\ldots \) be the principal parts of the corresponding sails (finite or infinite). Then\n\n\[ \n{A}_{i} = \left( {{p}_{2i},{q}_{2i}}\right) \;\text{ and }\;{B}_{i} = \left( {{p}_{{2i} - 1},{q}_{{2i} - 1}}\right)...
Proof The proof repeats the proof of Theorem 3.1 except for the following difference. Since \( {a}_{0} = 0 \), the point \( {A}_{0} \) should coincide with \( {A}_{1} \) . This is the explanation for the shift in indices.
Yes
Corollary 3.5 Consider \( \alpha \geq 1 \) . Let \( {A}_{0}{A}_{1}{A}_{2}\ldots \) and \( {B}_{0}{B}_{1}{B}_{2}\ldots \) be the principal parts of the corresponding sails (finite or infinite). Then\n\n\[ \n\operatorname{l}\ell \left( {{A}_{i}{A}_{i + 1}}\right) = {a}_{2i}\;\text{ and }\;\operatorname{l}\ell \left( {{B}...
Proof of Corollary 3.5 The corollary follows directly from the explicit formulas for \( {A}_{k} \) and \( {B}_{k} \) of Theorem 3.1, after applying Proposition 1.13.
No
Theorem 3.6 Consider \( \alpha \geq 1 \) . Let \( {A}_{0}{A}_{1}{A}_{2}\ldots \) and \( {B}_{0}{B}_{1}{B}_{2}\ldots \) be the principal parts of the corresponding sails (finite or infinite). Then\n\n\[ \operatorname{l}\alpha \left( {\angle {A}_{i}{A}_{i + 1}{A}_{i + 2}}\right) = {a}_{{2i} + 1}\;\text{ and }\;\operatorn...
Proof Let us calculate the index of an integer angle at some vertex of the principal part of one of the two sails. By Theorem 3.1 it is equivalent to calculate the index of the angle between a pair of vectors \( \left( {{p}_{i - 1},{q}_{i - 1}}\right) \) and \( \left( {{p}_{i + 1},{q}_{i + 1}}\right) \) :\n\n\[ \operat...
Yes
Proposition 4.3 The integer sine of a rational angle coincides with the index of the angle.
Proof of Proposition 4.3 Consider a rational angle \( \angle {ABC} \) with \( A, B, C \) not contained on one line. Let \( {A}^{\prime } \) and \( {C}^{\prime } \) be the nearest integer points to \( B \) in the open rays \( {BA} \) and \( {BC} \), respectively. Then from the definition of integer length, we get\n\n\[ ...
Yes
Proposition 4.8 Integer length, integer area, index and integer sine of a rational angle are invariant under the action of the group of integer affine transformations \( \operatorname{Aff}\left( {2,\mathbb{Z}}\right) \) .
Proof Every integer linear transformation sends subgroups to subgroups and the corresponding cosets to the corresponding cosets; hence all integer indices are invariant under integer linear transformations. Thus, every integer affine transformation preserves indexes of the corresponding integer lattice subgroups in \( ...
Yes
Corollary 4.9 The LLS sequence is an invariant of lattice angles with respect to \( \operatorname{Aff}\left( {2,\mathbb{Z}}\right) \) .
Proof First, note that convex hulls are preserved by the elements of \( \operatorname{Aff}\left( {2,\mathbb{Z}}\right) \) (we leave this as an easy exercise for the reader). Then the statement follows directly from the fact that the integer length and the index are invariants of \( \operatorname{Aff}\left( {2,\mathbb{Z...
No
Theorem 4.11 For every sequence of positive integers (odd finite or infinite on one or both sides) there exists an angle with vertex at the origin whose LLS sequence is this sequence.
Proof First we consider the case of a sequence \( {a}_{0},{a}_{1},{a}_{2},\ldots \) (odd finite or infinite to the right). Let \( A = \left( {1,0}\right), B = \left( {0,0}\right) \), and \( C = \left( {1,\alpha }\right) \), where \( \left\lbrack {{a}_{0};{a}_{1} : {a}_{2} : \cdots }\right\rbrack \) is the continued fra...
Yes
Theorem 5.7 (Trigonometric relations for transpose angles) Let an integer angle \( \alpha \) be not contained in a line. Then\n\n(1) If \( \alpha \cong \operatorname{larctan}\left( 1\right) \), then \( {\alpha }^{t} \cong \operatorname{larctan}\left( 1\right) \).\n\n(2) If \( \alpha ≆ \operatorname{larctan}\left( 1\rig...
Proof Consider an integer angle \( \alpha \) . Let \( \operatorname{ltan}\alpha = p/q \), where \( \gcd \left( {p, q}\right) = 1 \) . By Proposition 5.4(ii), we have \( \alpha \cong \operatorname{larctan}\left( {p/q}\right) \) . The case \( p/q = 1 \) is trivial. Consider the case \( p/q > 1 \) .\n\nLet \( A = \left( {...
Yes
Theorem 5.9 (Trigonometric relations for adjacent angles) Let \( \alpha \) be some integer angle. Then one of the following holds:\n\n(1) If \( \alpha \) is the zero angle, then \( \pi - \alpha \cong \pi \) .\n\n(2) If \( \alpha \) is the straight angle, then \( \pi - \alpha \cong 0 \) .\n\n(3) If \( \alpha \cong \oper...
Proof of Theorem 5.9 Consider an integer angle \( \alpha \) . Directly from the definitions it follows that if \( \alpha \cong 0 \), then \( \pi - \alpha \cong \pi \), and, if \( \alpha \cong \pi \), then \( \pi - \alpha \cong 0 \) .\n\nSuppose that \( \operatorname{ltan}\alpha = p/q > 0 \), where \( \gcd \left( {p, q}...
Yes
Proposition 5.14 Every integer right angle is integer-congruent to exactly one of the following two angles: \( \operatorname{larctan}\left( 1\right) ,\operatorname{larctan}\left( 2\right) \) .
Proof Let \( \alpha \) be an integer right angle.\n\nSince \( \left( {\pi - 0}\right) ≆ 0 \) and \( \left( {\pi - \pi }\right) \ncong \pi \), we have \( \operatorname{ltan}\alpha > 0 \) .\n\nBy the definition of integer right angles and Theorem 5.7, we obtain\n\n\[ \operatorname{lcos}\left( \alpha \right) \equiv \opera...
Yes
Proposition 5.16 Two integer angles opposite interior to each other are integer-congruent.
Proof Consider two distinct integer parallel lines \( {AB} \) and \( {CD} \) . Let the points \( A \) and \( D \) be in distinct open half-planes with respect to the line \( {BC} \) . Let us prove that \( \angle {ABC} \cong \angle {DCB} \) .\n\nConsider the central symmetry \( S \) of the two-dimensional plane at the m...
Yes
Proposition 6.1 (The sine formula for integer triangles) For any integer triangle \( \bigtriangleup {ABC} \) the following holds:\n\n\[ \n\frac{1\ell \left( {AB}\right) }{\operatorname{lsin}\left( {\angle {BCA}}\right) } = \frac{1\ell \left( {BC}\right) }{\operatorname{lsin}\left( {\angle {CAB}}\right) } = \frac{1\ell ...
Proof We have\n\n\[ \n{1S}\left( {\bigtriangleup {ABC}}\right) = 1\ell \left( {AB}\right) 1\ell \left( {AC}\right) \operatorname{lsin}\left( {\angle {CAB}}\right) = 1\ell \left( {BA}\right) 1\ell \left( {BC}\right) \operatorname{lsin}\left( {\angle {BCA}}\right) \n\]\n\n\[ \n= 1\ell \left( {CB}\right) 1\ell \left( {CA}...
Yes
Proposition 6.2 (The first criterion of integer triangle integer congruence) Consider integer triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . Suppose that\n\n\[ \n{AB} \cong {A}^{\prime }{B}^{\prime },\;{AC} \cong {A}^{\prime }{C}^{\prime },\;\text{ and }\;\angle ...
Proof By definition there exists an integer affine transformation taking \( \angle {CAB} \) to \( \angle {C}^{\prime }{A}^{\prime }{B}^{\prime } \) . Since the integer lengths of the corresponding segments are the same, the transformation takes \( B \) and \( C \) to \( {B}^{\prime } \) and \( {C}^{\prime } \) .
No
Example 6.3 (The second criterion of triangle integer congruence does not hold in integer geometry) In Fig. 6.1 we show two integer triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) . We have\n\n\[ \n{AB} \cong {A}^{\prime }{B}^{\prime },\;\angle {ABC} \cong \angle {A...
The triangle \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) is not integer congruent to the triangle \( \bigtriangleup {ABC} \), since\n\n\[ \n\operatorname{lS}\left( {\bigtriangleup {ABC}}\right) = 4\;\text{ and }\;\operatorname{lS}\left( {\bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime }}\right) = ...
Yes
Proposition 6.5 (An additional criterion of integer triangle integer congruence) Consider two integer triangles \( \bigtriangleup {ABC} \) and \( \bigtriangleup {A}^{\prime }{B}^{\prime }{C}^{\prime } \) of the same integer area. Suppose that\n\n\[ \angle {ABC} \cong \angle {A}^{\prime }{B}^{\prime }{C}^{\prime },\;\an...
Proof All the integer lengths of the corresponding edges are the same by the sine formula. Hence the triangles are integer congruent by the first criterion.
No
Theorem 6.8 (On sums of integer tangents of angles in integer triangles)\n\n(i) Let \( \\left( {{\\alpha }_{1},{\\alpha }_{2},{\\alpha }_{3}}\\right) \) be an ordered triple of angles. There exists a triangle with consecutive angles integer congruent to \( {\\alpha }_{1},{\\alpha }_{2} \), and \( {\\alpha }_{3} \) if a...
We are not yet ready to prove the first statement of this theorem; we shall do it later, in Chap. 12. We prove the second statement of the theorem after a small remark.\n\nProof of the second statement of Theorem 6.8 Consider a triangle \\(\\bigtriangleup {ABC}\\) with rational angles \\(\\alpha ,\\beta \\), and \\(\\g...
No
Theorem 6.11 Consider some triangle \( \bigtriangleup {ABC} \) . Let\n\n\[1\ell \left( {AB}\right) = c,\;1\ell \left( {AC}\right) = b,\;\text{ and }\;\angle {CAB} \cong \alpha .\n\]\n\nThen the angles \( \angle {BCA} \) and \( \angle {ABC} \) are defined in the following way:\n\n![1c600fd7-25c8-4af1-a5fa-ba617d01167c_7...
Proof We start with proving the formula for the angle \( \angle {BCA} \) . Let \( \alpha \cong \operatorname{larctan}(p/ \) \( q) \), where \( \gcd \left( {p, q}\right) = 1 \) . Then \( \bigtriangleup {CAB} \cong \bigtriangleup {DOE} \), where \( D = \left( {b,0}\right), O = \left( {0,0}\right) \), and \( E = \left( {{...
Yes
Proposition 6.13 We have\n\n\[ \frac{d}{3} \leq N\left( d\right) \leq \frac{d\left( {d + 1}\right) }{2}. \]\n
Proof First, let us show that \( N\left( d\right) \leq d\left( {d + 1}\right) /2 \) . From Theorem 6.8(ii) we know that every integer triangle is equivalent to some \( \bigtriangleup {A}_{0}{B}_{0}{C}_{0} \) where\n\n\[ {A}_{0} = \left( {0,0}\right) ,\;{B}_{0} = \left( {{\lambda }_{2}q,{\lambda }_{2}p}\right) ,\;{C}_{0...
Yes
Proposition 7.5 Let \( A \) be a real spectrum matrix with no integer eigenvectors. Then the sails of the opposite octants are congruent. The sails of the adjacent octants are dual.
Proof The sails of opposite octants are congruent, since they are taken one to another by the symmetry about the origin.\n\nLet us prove the duality. Let one of the sails for the geometric continued fraction be a broken line \( \left( {A}_{i}\right) \) . Without loss of generality we may fix coordinates such that \( {A...
Yes
Proposition 7.8 Every algebraic sail has a periodic LLS sequence.
Proof of Proposition 7.8 Let \( A \) be a real spectrum matrix in \( \mathrm{{SL}}\left( {2,\mathbb{Z}}\right) \). The matrix \( A \) preserves its invariant lines and the lattice \( {\mathbb{Z}}^{2} \). Hence it acts on each of the sails by shifting the vertices of the corresponding broken line along the broken line. ...
Yes
Proposition 7.11 Two real spectrum \( \mathrm{{SL}}\left( {2,\mathbb{Z}}\right) \) matrices \( {M}_{1} \) and \( {M}_{2} \) with positive eigenvalues are integer conjugate if and only if their LLS periods coincide.
Proof Let \( {M}_{1} \) and \( {M}_{2} \) be integer conjugate. Then they have integer congruent geometric continued fractions, and they define the same shift of these continued fractions. Therefore the LLS periods of \( {M}_{1} \) and \( {M}_{2} \) coincide.\n\nSuppose now that the LLS periods of the matrices \( {M}_{...
Yes
Corollary 7.16 (On enumeration of reduced matrices) The set of real spectrum reduced matrices is in one-to-one correspondence (defined in Remark 7.15) with the set of finite sequences consisting of an even number of positive integer elements.
Proof It is clear that two distinct reduced matrices have distinct sequences as defined in Remark 7.15. Let us show that the sequence\n\n\\[ \n\\left( {{a}_{1},{a}_{2},\\ldots ,{a}_{{2n} - 1},\\lambda }\\right) \n\\]\n\n is realizable for some reduced matrix. If this sequence is \\( \\left( {1,\\alpha }\\right) \\), th...
Yes
Corollary 7.17 (On almost normal forms) The number of reduced matrices in an integer conjugacy class coincides with the number of elements in the minimal period of the corresponding LLS sequence.
Proof In Proposition 7.11 we showed that two matrices are integer conjugate if their LLS periods are conjugate. From Corollary 7.16 it follows that reduced matrices are in one-to-one correspondence with finite sequences of even lengths. The number of sequences that are cyclic permutations of each other exactly coincide...
Yes
Theorem 7.18 (On complete invariant of integer conjugacy classes)\n\n(i) Every LLS period is a complete invariant of an integer conjugacy class for \( \mathrm{{SL}}\left( {2,\mathbb{Z}}\right) \) matrices with distinct positive real eigenvalues.\n\n(ii) An arbitrary finite sequence of an even length with positive integ...
Proof Proposition 7.11 is a reformulation of the first part of the theorem. Corollary 7.16 implies that all sequences are realized.
No
Example 7.19 Consider\n\n\[ \nM = \left( \begin{matrix} {1519} & {1164} \\ - {1964} & - {1505} \end{matrix}\right) \n\]\n\nThe LLS period of \( M \) is \( \left( {1,2,1,2}\right) \) . Hence there are exactly two reduced matrices represented by the sequences \( \left( {1,2,1,2}\right) \) and \( \left( {2,1,2,1}\right) \...
We find the elements \( c \) and \( d \) of the reduced matrices from conditions \( \lambda = \left\lfloor \frac{d - 1}{b}\right\rfloor \) and \( {ad} - {bc} = 1 \) . Finally, we get both reduced matrices integer conjugate to \( M \) :\n\n\[ \n\left( \begin{matrix} 3 & 8 \\ 4 & {11} \end{matrix}\right) \text{ and }\lef...
Yes
Corollary 7.20 A sail with a periodic LLS sequence is algebraic (i.e., a sail of some algebraic real spectrum matrix).
Proof In Theorem 7.14 we constructed the algebraic matrices for all finite sequences as periods. Then in Proposition 7.7 we showed that sails with equivalent LLS sequences are either integer congruent or dual. Therefore, every sail with a periodic LLS sequence is algebraic.
Yes
Proposition 7.33 Every Markov triple is obtained from \( \left( {1,1,1}\right) \) by applying a sequence operations of the following two types:\n\n- permute the numbers \( x, y \), and \( z \) in the triple \( \left( {x, y, z}\right) \) ;\n\n- if \( \left( {x, y, z}\right) \) is a Markov triple, then \( \left( {x, y,{3...
Without loss of generality we consider only solutions with \( x \leq y \leq z \) . The structure of such solutions forms a tree, whose vertices are almost all of valence 3 (except for two vertices). An edge between two of vertices corresponds to the operation described in Proposition 7.33. We show several of the first ...
No
Theorem 8.4 Consider an irreducible real spectrum matrix \( A \in \mathrm{{SL}}\left( {2,\mathbb{Z}}\right) \) . Then its Dirichlet group \( \Xi \left( A\right) \) is homeomorphic to \( \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \) .
Proof The proof is straightforward. Any matrix commuting with \( A \) has the same eigenlines, so it either acts as a shift on the sail, or exchanges the sails. The group of shifts is homeomorphic to \( \mathbb{Z} \) . Suppose it is generated by \( \widehat{A} \) . If the period is even, then the dual sails are not con...
Yes
Example 8.7 Consider the reduced matrix\n\n\[ \nA = \left( \begin{array}{ll} {11} & {30} \\ {15} & {41} \end{array}\right) = \left( \begin{matrix} {11} & 8 + 2 \cdot {11} \\ {15} & {11} + 2 \cdot {15} \end{matrix}\right) \n\]\n\nThe period of the LLS sequence related to this continued fraction is \( \left( {1,2,1,2,1,2...
\[ \n\left( \begin{array}{ll} 1 & 0 + 2 \cdot 1 \\ 1 & 1 + 2 \cdot 1 \end{array}\right) = \left( \begin{array}{ll} 1 & 2 \\ 1 & 3 \end{array}\right) \n\]\n\nDenote this matrix by \( \widehat{A} \) . The group \( \Xi \left( A\right) \) is generated by \( \widehat{A} \) and \( - E \) .
Yes
Proposition 8.8 The matrix \( {A}_{3} \) constructed in Step 3 is an nth root of the matrix \( {A}_{2} \) .
Proof There exists a matrix \( M \in \Xi \left( {A}_{2}\right) \) mapping \( \left( {1,0}\right) \) to \( \left( {{a}_{3},{b}_{3}}\right) \) . This matrix corresponds to the sail integer self-congruence related to the shift for the period \( \left( {{a}_{1},\ldots ,{a}_{k}}\right) \) if \( k \) is even. If \( k \) is o...
Yes
Lemma 8.12 For every quadratic irrationality \( \xi \) there exists an \( \mathrm{{SL}}\left( {2,\mathbb{Z}}\right) \) matrix such that one of its eigenvectors is \( \left( {1,\xi }\right) \) .
Proof Let \( \xi \) be a root of the equation \( {c}_{2}{x}^{2} + {c}_{1}x + {c}_{0} = 0 \) with integer coefficients \( {c}_{0},{c}_{1},{c}_{2} \) . Consider an arbitrary matrix\n\n\[ A = \left( \begin{array}{ll} {a}_{11} & {a}_{12} \\ {a}_{21} & {a}_{22} \end{array}\right) \]\n\nIts eigenvectors are\n\n\[ \left( {1,\...
Yes
Proposition 8.13 Let \( {\alpha }_{1} > {\alpha }_{2} > {\alpha }_{3} \) be distinct numbers and let \( {\alpha }_{1} \) be irrational. Consider two angles \( \alpha \) and \( \beta \) defined by pairs of rays in the lines \( \left( {y = {\alpha }_{1}x, y = }\right. \) \( \left. {{\alpha }_{2}x}\right) \) and \( \left(...
Proof Denote by \( \gamma \) the angle defined by rays \( \left( {y = {\alpha }_{2}x, y = {\alpha }_{3}x}\right), x > 0 \) . It is clear that every integer point of the angle \( \beta \) is either in \( \alpha \) or in \( \gamma \) . Then the convex hull of all integer points of \( \beta \) is the convex hull of the un...
Yes
Lemma 9.8 Let \( x = \left\lbrack {0;{a}_{1} : {a}_{2} : \cdots }\right\rbrack \) . Then\n\n\[ \n{T}^{-1}\left( x\right) = \left\{ {\left\lbrack {0;k : {a}_{1} : {a}_{2} : \cdots }\right\rbrack \mid k \in {\mathbb{Z}}_{ + }}\right\} = \left\{ {\left. \frac{1}{x + k}\right| \;k \in {\mathbb{Z}}_{ + }}\right\} .\n\]
Proof The first equality follows directly from the fact that\n\n\[ \nT\left( \left\lbrack {0;{b}_{1} : {b}_{2} : \cdots }\right\rbrack \right) = \left\lbrack {0;{b}_{2} : {b}_{3}\cdots }\right\rbrack \n\]\n\nand the fact that every real number has a unique regular continued fraction expansion with the last element not ...
Yes
Proposition 9.10 The set \( \Psi \) is measurable (i.e., \( \Psi \in \sum \) ).
Proof Denote by \( {\Upsilon }_{n} \) the set of all irrational numbers that contain the element 1 exactly at place \( n \) . Notice that\n\n\[ \n{\Upsilon }_{1} = \left\lbrack {1/2,1}\right\rbrack \n\]\n\nand therefore, it is measurable. Hence for every \( n \) the set\n\n\[ \n{\Upsilon }_{n + 1} = {T}^{n}\left( {\Ups...
Yes
Lemma 9.12 The measure of a segment \( {I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) } \) satisfies the following inequality:\n\n\[ \widehat{\mu }\left( {I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) }\right) < \frac{1}{\ln 2\left( {{q}_{n} + {q}_{n - 1}}\right) \left( {{p}_{n} + {q}_{n}}\right) }.\]
Proof We have\n\n\[ \widehat{\mu }\left( {I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) }\right) = \frac{1}{\ln 2}{\int }_{{I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) }}\frac{dx}{1 + x} = \frac{1}{\ln 2}\left| {{\int }_{\left\lbrack 0;{a}_{1} : \cdots : {a}_{n}\right\rbrack }^{\left\lbrack 0;{a}_{1} : \cdots : {a}_{n}\right...
Yes
Lemma 9.13 For any invariant set \( S \) of positive measure and any interval \( {I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) } \) ,
\[ \widehat{\mu }\left( {S \cap {I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) }}\right) \geq \frac{\ln 2}{2}\widehat{\mu }\left( S\right) \widehat{\mu }\left( {I}_{\left( {a}_{1},\ldots ,{a}_{n}\right) }\right) . \] Proof Since the map \( T \) is surjective, we also have \[ T\left( S\right) = S. \] Let \( \widehat{\mu }\l...
Yes
Theorem 9.14 For every positive integer \( k \) and almost every \( x \) (i.e., in the complement of a set of zero measure) the following holds:\n\n\[ \n{\widehat{P}}_{k}\left( x\right) = \frac{1}{\ln 2}\ln \left( {1 + \frac{1}{k\left( {k + 2}\right) }}\right) .\n\]
Proof of Theorem 9.14 Consider a subset \( S \in I \) . Let \( {\chi }_{S} \) be the characteristic function of \( S \), i.e.,\n\n\[ \n{\chi }_{S}\left( x\right) = \left\{ \begin{array}{ll} 1, & \text{ if }x \in S \\ 0, & \text{ otherwise. } \end{array}\right.\n\]\n\nThen\n\n\[ \n{\widehat{P}}_{n, k}\left( x\right) = \...
Yes
Theorem 9.15 (Gauss-Kuzmin) For \( 0 \leq x \leq 1 \) the following holds:\n\n\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}{m}_{n}\left( x\right) = \frac{\ln \left( {1 + x}\right) }{\ln 2}. \]
This theorem is technically complicated. For the proof we refer to the original manuscripts of R.O. Kuzmin [121] and [122] (see also A.Ya. Khinchin [105]).
No
Corollary 9.16 For every positive integer \( k \), the following holds:\n\n\[ P\left( k\right) = \frac{1}{\ln 2}\ln \left( {1 + \frac{1}{k\left( {k + 2}\right) }}\right) . \]
Proof Notice that \( {P}_{n}\left( k\right) = {m}_{n}\left( \frac{1}{k}\right) - {m}_{n}\left( \frac{1}{k + 1}\right) \) . Now the statement of the corollary follows from the Gauss-Kuzmin theorem.
No
Corollary 9.19 The infinitesimal cross-ratio is an invariant of projective transformations of \( {\mathbb{R}}^{n - 1} \) .
Proof The density of the infinitesimal cross-ratio coincides with the asymptotic coefficient at \( {\varepsilon }^{2} \) of the cross-ratios of 4-tuples of points:\n\n\[ \n x, y, x + {\varepsilon dx}, y + {\varepsilon dy}, \n\] \n\nas \( \varepsilon \) tends to 0 . Therefore, the infinitesimal cross-ratio is a projecti...
No
Proposition 9.22 All Möbius forms of the manifolds \( {\mathrm{{CF}}}_{1} \) and \( {\mathrm{{FCF}}}_{1} \) are proportional.
Proof Transitivity of the action of PGL \( \left( {2,\mathbb{R}}\right) \) implies that all Möbius forms of the manifolds \( C{F}_{1} \) and \( {FC}{F}_{1} \) are proportional.
Yes
Proposition 9.26 The measure \( {\mu }_{{\omega }_{l}} \) coincides with the restriction of some Möbius measures to \( {FC}{F}_{1, l} \) .
Proof Notice that the form \( {\omega }_{l}\left( {{x}_{l},{y}_{l}}\right) \) coincides with the infinitesimal cross-ratio on the line \( l \) . Hence it is invariant under projective transformations of \( l \) (on an everywhere dense subset) in the chart \( {FC}{F}_{1, l} \) . Therefore, the measure \( {\mu }_{{\omega...
Yes