tag:blogger.com,1999:blog-3331630668965131132024-03-14T06:35:42.583-07:00Happy Hour MatemáticoSobre as matemáticas que podem ser contadas tomando-se uma xícara de café. Talvez duas, três...Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.comBlogger35125tag:blogger.com,1999:blog-333163066896513113.post-82341454292404939632020-10-06T14:25:00.002-07:002020-10-15T11:16:48.741-07:00Razão cruzada<p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;"><!--[if gte vml 1]><v:line
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<p class=MsoNormal style='margin-bottom:0cm'><b>A<o:p></o:p></b></p>
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<b>A</b> e <b>B </b>definem uma única reta (chamada de <i>reta pelos pontos </i><b>A</b>
e <b>B</b>)<b>.</b><p></p><p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;">Dado um ponto <b>X</b>
na reta pelos pontos <b>A</b> e <b>B</b>, temos que os vetores <b>AX</b> (vetor
com origem no ponto <b>A </b> e extremidade
no ponto <b>X</b>)<b> </b>e <b>BX </b>(vetor com origem no ponto <b>B </b>e
extremidade no ponto <b>X</b>) são paralelos, ou seja, existe um número real <b> r</b> tal que <b>AX = rBX</b> (o vetor <b>BX</b>
é um múltiplo do vetor <b>AX</b>). O número <b>r</b>, como definido agora,
indica a posição relativa em que o ponto <b>X </b>está posicionado na reta com
respeito aos pontos <b>A</b> e <b>B</b>. </p><p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;"><br /></p><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjVx7HFvRtJ3B5T5V1edspX2uXNkQuAf7s_ePsWxYhBkmk_KLp7b6QaEioM65QQk51PvRVZgYRjQmtTIh-3yGNKmVpJVmVNQrzh0_CzGM2oB7-CHJRbDyvujPzFTmiEJfzc_HpOZw1u3fI/" style="margin-left: 1em; margin-right: 1em; text-align: center; text-indent: 47.2px;"><img alt="" data-original-height="185" data-original-width="618" height="120" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjVx7HFvRtJ3B5T5V1edspX2uXNkQuAf7s_ePsWxYhBkmk_KLp7b6QaEioM65QQk51PvRVZgYRjQmtTIh-3yGNKmVpJVmVNQrzh0_CzGM2oB7-CHJRbDyvujPzFTmiEJfzc_HpOZw1u3fI/w400-h120/image.png" width="400" /></a></div><p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;"><span style="text-indent: 35.4pt;">Neste exemplo acima,</span><b style="text-indent: 35.4pt;"> AX = 2BX </b><span style="text-indent: 35.4pt;">(</span><b style="text-indent: 35.4pt;">r = 2</b><span style="text-indent: 35.4pt;">), tem-se que o ponto </span><b style="text-indent: 35.4pt;">B</b><span style="text-indent: 35.4pt;"> vem como ponto médio do segmento com extremidades </span><b style="text-indent: 35.4pt;">A</b><span style="text-indent: 35.4pt;"> e </span><b style="text-indent: 35.4pt;">X.</b></p>
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<td><div class="separator" style="clear: both; text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuRekeVlEQnTtASsLAdfe3GkcL6ggN_oyrmMfU7DtPK3oklOCt8XPR7C7__-dNlX9UwXXL4HBxoyy1QOLaO7bBlzYIhjkvK6NShgPja_KwkSvUQQjVrNuEu6OdJDqKr9DhKuOsGDy-DCs/" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="144" data-original-width="735" height="79" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhuRekeVlEQnTtASsLAdfe3GkcL6ggN_oyrmMfU7DtPK3oklOCt8XPR7C7__-dNlX9UwXXL4HBxoyy1QOLaO7bBlzYIhjkvK6NShgPja_KwkSvUQQjVrNuEu6OdJDqKr9DhKuOsGDy-DCs/w400-h79/image.png" width="400" /></a></div><br /><br /></div></td>
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</tbody></table><span style="text-align: justify; text-indent: 35.4pt;"> </span>
<p class="MsoNormal" style="text-align: justify;"><o:p><br /> </o:p><span style="text-align: left; text-indent: 35.4pt;"> </span></p><p class="MsoNormal" style="text-align: justify;"><span style="text-align: left; text-indent: 35.4pt;"><br /></span></p><p class="MsoNormal" style="text-align: justify;"><span style="text-align: left; text-indent: 35.4pt;"><br /></span></p><p class="MsoNormal" style="text-align: justify;"><span style="text-align: left; text-indent: 35.4pt;"><span> </span><span> </span><span> </span>Neste outro exemplo acima,<b> AX = -1/2BX </b></span><span style="text-align: left; text-indent: 35.4pt;">(</span><b style="text-align: left; text-indent: 35.4pt;">r = -1/2</b><span style="text-align: left; text-indent: 35.4pt;">), tem-se que o ponto </span><b style="text-align: left; text-indent: 35.4pt;">X</b><span style="text-align: left; text-indent: 35.4pt;"> está posicionado entre </span><span style="text-align: left; text-indent: 35.4pt;">os pontos </span><b style="text-align: left; text-indent: 35.4pt;">A</b><span style="text-align: left; text-indent: 35.4pt;"> e </span><b style="text-align: left; text-indent: 35.4pt;"> B. </b><span style="text-align: left; text-indent: 35.4pt;"> </span></p><p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;"><br /></p><p class="MsoNormal">Quando temos <b>AX = r BX</b>, usamos a notação<b> AX/BX</b> para
representar o número real <b>r</b>.</p><p class="MsoNormal"><o:p></o:p></p>
<p class="MsoNormal" style="margin-left: 35.4pt;"> </p><p class="MsoNormal" style="margin-left: 35.4pt;">Considere, agora, os pontos <b>A</b> e <b>B</b> e na
reta pelos pontos <b>A</b> e <b>B</b> considere dois pontos <b>X</b> e <b>Y</b>. Definimos a <b>razão cruzada (A,B:X,Y) </b> por <b>(AX/BX)<span style="font-size: 14pt; line-height: 107%; mso-fareast-font-family: "Times New Roman"; mso-fareast-theme-font: minor-fareast;"> : (</span></b><b>AY/BY</b><b><span style="font-size: 14pt; line-height: 107%; mso-fareast-font-family: "Times New Roman"; mso-fareast-theme-font: minor-fareast;">)</span></b><span style="font-size: 14pt; line-height: 107%; mso-fareast-font-family: "Times New Roman"; mso-fareast-theme-font: minor-fareast;">.</span></p>
<p class="MsoNormal"><b> </b></p>
<p class="MsoNormal"><b>Teorema. </b>A
razão cruzada <b>(A,B:X,Y)</b> de quatro pontos colineares é um invariante
projetivo. Em outras palavras, se <b>A’, B’, X’, Y’ </b> são outros quatro pontos colineares e tais que
as retas pelos pontos <b>A</b> e <b> A’</b>;
<b>B</b> e <b>B’</b>; <b>X</b> e <b>X’</b>; <b>Y</b> e <b>Y’</b> concorrem em
um ponto <b>O</b>, então <b>(A,B:X,Y) = (A’,B’:X’,Y’)<o:p></o:p></b></p><p class="MsoNormal"><b></b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLBzkYCGXAWGV2iQ3ND1JH7wRc8Ptw86nQnknuxBZ1nDcTzRw6zUbKeshiXBUpoiD3d-VC2WyxFNwVR85rg0rMjg3NXBW9weyaDgQnVU0zG455zO7R0r2wRyzxUevpTNvn-EQFC9isCL4/" style="margin-left: 1em; margin-right: 1em;"><img alt="" data-original-height="434" data-original-width="733" height="236" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLBzkYCGXAWGV2iQ3ND1JH7wRc8Ptw86nQnknuxBZ1nDcTzRw6zUbKeshiXBUpoiD3d-VC2WyxFNwVR85rg0rMjg3NXBW9weyaDgQnVU0zG455zO7R0r2wRyzxUevpTNvn-EQFC9isCL4/w400-h236/image.png" width="400" /></a></b></div><b><br /><br /></b><p></p>
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<p align="center" class="MsoNormal" style="text-align: center; text-indent: 35.4pt;"><b><span style="font-size: 14pt; line-height: 107%;">Aplicação ao estudo de Metrologia em
Imagens<o:p></o:p></span></b></p>
<p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;">Na imagem
abaixo, à esquerda, temos uma rodovia com medidas exatas de sinalização nos
pontos <b>A</b> e <b>B</b>; um carro no ponto <b>C</b> e um cruzamento
sinalizado pelo ponto <b>D</b>. À direita, temos uma fotografia da mesma rodovia,
agora visualizada por uma câmera em um ângulo não perpendicular ao plano da
pista com os respectivos pontos de marcação <b>A’,B’,C’,D’</b> (representando
os pontos <b>A,B,C,D</b> na fotografia). Com a foto posta na tela de um
computador, tem-se facilmente as medidas (em pixels) das distâncias entre os
pontos <b>A’ </b>e <b>D’ </b>(no caso 300 pixels), <b>A’ </b>e <b>C’ </b>(no
caso 275 pixels), <b>B’ </b>e <b>C’</b> (no caso 50 pixels). O problema é:
encontrar a real distância do carro que está no ponto <b>C</b> ao cruzamento
que está no ponto <b>D</b>. Confira a solução na imagem abaixo.<o:p></o:p></p>
<p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;"><b> </b></p><div class="separator" style="clear: both; text-align: center;"><b><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg5P1t5Jo-Hznt0EzDQ1VGK_GcHZtfhs-YUnYDPu5uQku9ROeS7wjzyRWy2YKnILRFTUMzMw819wUj4OQkYPVoVukx471DOUE3EZ5pUj7ASMyXbbtNLJLX2HXac5s9euGB4j7zv0HBCXu4/s1102/cross_ratio.png" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" data-original-height="622" data-original-width="1102" height="303" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEg5P1t5Jo-Hznt0EzDQ1VGK_GcHZtfhs-YUnYDPu5uQku9ROeS7wjzyRWy2YKnILRFTUMzMw819wUj4OQkYPVoVukx471DOUE3EZ5pUj7ASMyXbbtNLJLX2HXac5s9euGB4j7zv0HBCXu4/w535-h303/cross_ratio.png" width="535" /></a></b></div><p></p><p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;">A imagem acima
e a aplicação foram extraídas do vídeo “Robotics – 4.2.4 – Projective Transformations
– Cross Ratios and Single View Metrology” https://www.youtube.com/watch?v=-5B7llKIq10&t=31s.<o:p></o:p></p>
<p class="MsoNormal" style="text-align: justify; text-indent: 35.4pt;"><b> </b></p>
<p class="MsoNormal" style="text-indent: 35.4pt;"><b> </b></p><script type="text/x-mathjax-config">
MathJax.Hub.Config({tex2jax: {inlineMath: [['$','$'], ['\\(','\\)']]}});
</script> <script src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML" type="text/javascript">
</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-45575926602469374062020-08-29T11:57:00.002-07:002020-08-30T07:17:11.272-07:00O que é ... multiplicidade de pontos singulares?<div style="text-align: justify;"><span style="font-size: large;"><b>N</b></span>esta nota, em linhas gerais, além de apresentar o conceito de multiplicidade de pontos singulares, trazemos, em perspectiva, a famosa Conjectura da Multiplicidade de Zariski e o contexto em que nossas principais contribuições sobre o assunto foram desenvolvidas e nossas parcerias científicas nesse projeto. Este texto foi escrito para fornecer um embasamento para algumas de minhas palestras sobre multiplicidade de pontos singulares.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Estrutura Analítica Local</b></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Estamos no contexto de subconjuntos analíticos fechados $X\subset\mathbb{C}^n$. Dizemos que $X\subset\mathbb{C}^n$ possui a mesma <span style="color: #2b00fe;"><b>estrutura analítica</b></span> em $p\in X$ que $Y\subset\mathbb{C}^m$ em $q\in Y$ se existem vizinhanças $U\subset\mathbb{C}^n$ de $p$, $V\subset\mathbb{C}^m$ de $q$ e uma aplicação analítica $F\colon X\cap U\rightarrow Y\cap V$ com inversa $G\colon Y\cap V\rightarrow X\cap U$ também analítica. A definição acima define uma classe de equivalência no conjunto dos pares $(X,p)$. Uma classe de equivalência da relação acima é o que chamamos de uma <b><span style="color: #2b00fe;">estrutura analítica local</span></b>. A linha de pesquisa intitulada Geometria Analítica Local tem como principal objetivo descrever e reconhecer todas as estruturas analíticas locais. Estabelecemos a classe de $(\mathbb{C}^n,0)$ como a estrutural analítica local mais simples e a denominamos por <b><span style="color: #2b00fe;">estrutura analítica local trivial</span></b>.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Denotamos por $\mathcal{O}_{X,p}$ o conjunto das funções analíticas complexas definidas numa vizinhança do ponto $p$ sobre $X$ equipado naturalmente com as operações de adição e multiplicação provenientes do espaço das funções com valores em $\mathbb{C}$. Verifica-se que $\mathcal{O}_{X,p}$ é um anel local com ideal maximal $\mathcal{M}_{X,p}$ constituído das funções que aplicam $p$ no valor $0\in\mathbb{C}$.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><span style="color: red;"><b>Teorema</b></span> <i>Os pares $(X,p)$ e $(Y,q)$ definem as mesmas estruturas analíticas locais se, e somente se, $\mathcal{O}_{X,p}$ e $\mathcal{O}_{Y,q}$ são isomorfos como $\mathbb{C}$-álgebras locais.</i></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Vale observar que o teorema acima oferece uma alternativa de estudo da Geometria Analítica Local via uma abordagem completamente algébrica. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Multiplicidade</b></div><div style="text-align: justify;"><b><br /></b></div><div style="text-align: justify;">A seguir, apresentamos o conceito de multiplicidade o qual aparece naturalmente como uma medida de não-trivialidade de estruturas analíticas locais.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><span style="color: red;"><b>Proposição </b></span>(Definição) <i>Seja $X\subset\mathbb{C}^n$ tal que sua dimensão numa vizinhança do ponto $p\in X$ é $d$. Se $L\colon\mathbb{C}^n\rightarrow\mathbb{C}^d$ é uma projeção linear genérica, então a sua restrição a $X\cap U$ define uma aplicação finita de grau $m$ para toda vizinhança suficientemente pequena $U\subset\mathbb{C}^n$ do ponto $p$. Além disso, o número inteiro $m$ não depende da projeção genérica escolhida. O número inteiro $m$ é chamado de a <span style="color: #2b00fe;"><b>multiplicidade</b></span> de $X$ em $p$ e é denotado por <span style="color: #2b00fe;">$\mu(X,p)$</span>.</i></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Como mencionado acima, no sentido que precisaremos abaixo, a multiplicidade $\mu(X,p)$ mede a não-trivialidade da estrutura analítica local do par $(X,p)$ quando $X$ tem dimensão pura numa vizinhança do ponto $p$, a saber: <span style="color: #990000;">o par $(X,p)$ define uma estrutura analítica local trivial se, e somente se, $\mu(X,p)=1$</span>.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b><span style="color: red;">Exemplo </span></b>(Cúspide) $X=\{(x,y) \ : \ x^2=y^3\}$ em $\mathbb{C}^2$. Temos que a projeção ortogonal no eixo-$x$ \'e uma projeção linear genérica para $X$ no ponto $0\in X$ (pois não contém em seu núcleo vetores tangentes a $X$ no ponto $0$). Segue-se que $\mu(X,0)=2$.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Do ponto de vista algébrico, a multiplicidade de $X$ no ponto $p$ é introduzida via o <i>Polinômio de Hilbert-Samuel</i> do anel local $\mathcal{O}_{X,p}$ o qual apresentamos na proposição a seguir. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><span style="color: red;"><b>Proposição </b></span>(Definição) <i>Seja $X\subset\mathbb{C}^n$ tal que sua dimensão numa vizinhança do ponto $p\in X$ é $d$. Então, existe um polinômio $P_{X,p}$ de grau $d$ tal que $P_{X,p}(k)$ coincide com a dimensão do $\mathbb{C}$-espaço vetorial $\displaystyle\frac{\mathcal{O}_{X,p}}{\mathcal{M}_{X,p}^k}$ para todo inteiro $k$ suficientemente grande. Além disso, $d!$ multiplicado pelo coeficiente líder do polinômio $P_{X,p}$ é um número inteiro, o qual chamamos de <b><span style="color: #2b00fe;">multiplicidade de Hilbert-Samuel</span></b> de $X$ em $p$.</i></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">De volta ao Exemplo da Cúspide acima, verifica-se que o polinômio de Hilbert-Samuel de $X,0$ é $P_{X,0}(t)=2t-1$, donde segue que a multiplicidade de Hilbert-Samuel de $X$ no ponto $0$ vale 2, isto é, coincide com $\mu(X,0)$. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">De uma forma geral, como resultado não trivial da Geometria Analítica Local, <span style="color: #990000;">a multiplicidade $\mu(X.p)$ coincide com a multiplicidade de Hilbert-Samuel de $X$ no ponto $p$</span>. Assim sendo, recorrendo ao teorema, enunciado alguns parágrafos acima, o qual possibilita o estudo da Geometria Analítica Local via uma abordagem totalmente algébrica, recebemos que $\mu(X,p)$ é um invariante da estrutura analítica local de $(X,p)$ no seguinte sentido: <span style="color: #990000;">se os pares $(X,p)$ e $(Y,q)$ definem a mesma estrutura analítica local, então $\mu(X,p)=\mu(Y,q)$</span>.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Existência de moduli</b></div><div style="text-align: justify;"><b><br /></b></div><div style="text-align: justify;">Embora a multiplicidade seja bastante para identificar as estruturas analíticas locais triviais, o exemplo a seguir nos mostra que invariantes discretos não são bastante para descrever todas as estruturas analíticas locais. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b><span style="color: red;">Exemplo </span></b>(4 retas) Seja $X_{t} \ : \ xy(x-y)(x-ty)=0$ em $\mathbb{C}^2$. Tem-se que para $t\neq s$ genéricos, os pares $(X_{t},0)$ e $(X_s,0)$ não possuem a mesma estrutura analítica local.</div><div style="text-align: justify;"><br /></div><div class="separator" style="clear: both; text-align: justify;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjTifF3rfciZ2XQeewg9fkA244kVS887z_hftHcOApXV3h8bgghwhQ1_C85_4XsQxf9uEhiT3GD_sOzf-GvDxrf9AUPQGNFbpZwIx4hbeXjY7ZUVvByZGZx5AgZsXoMAumQN9vsXwcFV6I/s588/4_retas+%25282%2529.jpg" style="margin-left: 1em; margin-right: 1em;"><img alt="4 retas pela origem" border="0" data-original-height="366" data-original-width="588" height="199" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjTifF3rfciZ2XQeewg9fkA244kVS887z_hftHcOApXV3h8bgghwhQ1_C85_4XsQxf9uEhiT3GD_sOzf-GvDxrf9AUPQGNFbpZwIx4hbeXjY7ZUVvByZGZx5AgZsXoMAumQN9vsXwcFV6I/w320-h199/4_retas+%25282%2529.jpg" width="320" /></a></div><div style="text-align: justify;"><br /></div><div class="separator" style="clear: both; text-align: justify;"><br /></div><div style="text-align: justify;">Em outras palavras, o exemplo acima demonstra que <span style="color: #990000;">existe uma quantidade não enumerável de estruturas analíticas locais</span>. Os exemplos do tipo acima foram obtidos em meados da década de 60. Especificamente, o Exemplo das 4 retas é devido a H. Whitney (1965).</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Topologia e Geometria Local</b></div><div style="text-align: justify;"><b><br /></b></div><div style="text-align: justify;">Ainda na década de 60, resultados que apontaram para a possibilidade de descrever a topologia local de conjuntos analíticos somente com invariantes discretos muito impulsionaram a pesquisa no tema e atraíram eminentes matemáticos para questões ainda hoje em aberto.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Nos parágrafos seguintes, definiremos o conceito de topologia local de conjuntos analíticos empregado no parágrafo acima e explicaremos com precisão as afirmações lá enunciadas. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Conjectura da Multiplicidade de Zariski</b></div><div style="text-align: justify;"><b><br /></b></div><div style="text-align: justify;">Sejam $X$ e $Y$ subconjuntos analíticos fechados de um mesmo ambiente $\mathbb{C}^n$, e sejam $p\in X$, $q\in Y$. Dizemos que os pares $(X,p)$ e $(Y,q)$ são <b><span style="color: #2b00fe;">topologicamente equivalentes</span></b> ou que possuem a mesma <b><span style="color: #2b00fe;">topologia local mergulhada</span></b> em $\mathbb{C}^n$ se existem vizinhanças $U\subset\mathbb{C}^n$ de $p$ e $V\subset\mathbb{C}^n$ de $q$, e um homeomorfismo $F\colon U\rightarrow V$ tal que $F(X\cap U)=Y\cap V$ e $F(p)=q$.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Na década de 60, foi provado que <span style="color: #990000;">existe uma quantidade enumerável de topologias locais mergulhadas de conjuntos analíticos</span>; não sei a referência exata da primeira prova deste resultado, de todo modo, S. Lojasiewicz é uma boa referência para o tema.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Era conhecido por O. Zariski que a multiplicidade era um invariante da topologia local mergulhada de curvas analíticas no plano complexo, de modo que, em 1971, O. Zariski formulou a seguinte pergunta: <span style="color: #990000;">se os pares $(X,p)$ e $(Y,q)$ de hipersuperfícies em $\mathbb{C}^n$ possuem a mesma topologia local mergulhada, então $\mu(X,p)=\mu(Y,q)$?</span> </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Estrutura Lipschitz Local</b></div><div style="text-align: justify;"><b><br /></b></div><div style="text-align: justify;">Dizemos que $X\subset\mathbb{C}^n$ possui a mesma <b><span style="color: #2b00fe;">estrutura Lipschitz</span></b> em $p\in X$ que $Y\subset\mathbb{C}^m$ em $q\in Y$ se existem vizinhanças $U\subset\mathbb{C}^n$ de $p$, $V\subset\mathbb{C}^m$ de $q$ e uma aplicação Lipschitz $F\colon X\cap U\rightarrow Y\cap V$ com inversa $G\colon Y\cap V\rightarrow X\cap U$ também Lipschitz. A definição acima define uma classe de equivalência no conjunto dos pares $(X,p)$. Uma classe de equivalência da relação acima é o que chamamos de uma <b><span style="color: #2b00fe;">estrutura Lipschitz local</span></b>.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Ressaltamos a importância de citar algumas das principais referências de resultados pioneiros na proposta de investigação da multiplicidade como invariante de estruturas locais mais rígidas do que topológicas e menos do que analíticas, a saber: no artigo [6], Gau e Lipman provaram a invariância da multiplicidade para estruturas locais diferenciáveis (Ephraim abordou o caso de estruturas locais continuamente diferenciáveis [5]) e, no artigo [4], assumindo restrições severas sobre as constantes de Lipschitz que conjungam dois paraes $(X,p)$ e $(Y,q)$, Comte mostrou que $\mu(X,p)=\mu(Y,q)$. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Em meados da década de 80, Tadeusz Mostowski provou que <span style="color: #990000;">existe uma quantidade enumerável de estruturas Lipschitz locais analíticos</span>; esse resultado estimulou sobremaneira a pesquisa sobre a Geometria Lipschitz Local de conjuntos analíticos. Estabelecemos a classe de $(\mathbb{C}^n,0)$ como a estrutural Lipschitz local mais simples e a denominamos por <span style="color: #2b00fe;"><b>estrutura Lipschitz local trivial</b></span>.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">Em trabalhos recentes escritos em parceria com Lev Birbrair (UFC), Javier de Bobadilla (BCAM), Edson Sampaio (UFC), Lê Dung Trang (Aix-Marseille) e Misha Verbitsky (IMPA) provamos os seguintes resultados.</div><div style="text-align: justify;"><ol><li>Estrutura Lipschitz local trivial é equivalente a estrutura analítica local trivial [1] (Sampaio provou a versão mais completa deste resultado em [7]);</li><li>Em dimensão 2, a multiplicidade é um invariante da estrutura Lipschitz local [2];</li><li>Em dimensão maior do que 2, a multiplicidade não é invariante da estrutura Lipschitz local [3].</li></ol></div><div style="text-align: justify;">Em outras palavras, o primeiro resultado listado acima diz que multiplicidade igual a 1 é um invariante da estrutura Lipschitz Local. Quanto ao terceiro resultado acima, de fato, provamos que para qualquer dimensão $d$ maior do que 2 existem pares $(X,p)$ e $(Y,q)$ de dimensão $d$ com mesma estrutura Lipschitz local, porém com multiplicidades distintas. Portanto, os resultados 1 e 2 acima, juntamente com os resultados de Pham e Teissier que, no fim da década de 60, demonstraram que a multiciplicidade é invariante da estrutura Lipschitz local em dimensão 1, podem ser resumidos da seguinte forma: <span style="color: #990000;">a multiplicidade é invariante da estrutura Lipschitz local somente em dimensão 1 e 2</span>.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><b>Referências bibliográficas</b></div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[1] BIRBRAIR, L.; FERNANDES, A.; LÊ D. T. and SAMPAIO, J. E. <i>Lipschitz regular complex algebraic sets are smooth</i>. <b>Proc. Amer. Math. Soc 144</b> (2016), pp 983--987.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[2] J. DE BOBADILLA; FERNANDES, A. and SAMPAIO, J. E. <i>Multiplicity and degree as bi-Lipschitz invariants for complex sets.</i> <b>Journal of Topology 11 </b>(2018), pp 957--965. </div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[3] BIRBRAIR, L; FERNANDES, A. and SAMPAIO, J. E., Verbtisky, M. <i>Multiplicity of singularities is not a bi-Lipschitz invariant.</i> <b>Math. Annalen 377 </b>(2020), pp 115--121.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[4] COMTE, G. <i>Multiplicity of complex analytic sets and bi-Lipschitz maps</i>.</div><div style="text-align: justify;">Real analytic and algebraic singularities (Nagoya/Sapporo/Hachioji, 1996 <b>Pitman Res. Notes Math. Ser., vol. 381</b>, pp.182--188, 1998.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[5] EPHRAIM, R. <i>$C^1$ preservation of multiplicity</i>. Duke Math. 43 (1976) pp. 797--803.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[6] GAU, Y.-N. and LIPMAN, J. <i>Differential invariance of multiplicity on analytic varieties.</i></div><div style="text-align: justify;"><b>Inventiones Mathematicae 73</b> (1983), pp. 165--188.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[7] SAMPAIO, J. E. <i>Bi-Lipschitz homeomorphic subanalytic sets have bi-lipschitz homeomorphic tangent cones. </i><b>Selecta Math. (N.S.) 22</b>(2), pp 553--559, (2016).</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;">[8] ZARISKI, O. <i>Some open questions in the theory of singularities</i>. <b>Bull. of the Amer. Math. Soc. 77</b> (1971), pp 481--491.</div><div style="text-align: justify;"><br /></div><div style="text-align: justify;"><br /></div><script type="text/x-mathjax-config">
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Estou muito satisfeito com o resultado da organização do evento 10th Miniworkshop on Singularities, Geometry and Differential Equations. Esta edição do Miniworkshop foi uma iniciativa do Grupo de Singualaridades da Pós-Graduação em Matemática da UFC e do Grupo de Singularidades do ICMC-USP e foi realizada no período de 10 a 13 de maio de 2015 no Hotel Maredomus em Fortaleza-CE. Muita gente apareceu por lá e prestigiou nosso trabalho! Vejam as fotos postadas na minha coleção do google+ <a href="https://plus.google.com/u/0/collection/Q57Nc" target="_blank">Miniworkshop on Singularities 2015</a>. Para mais informações sobre evento visite o nosso <a href="http://www.mat.ufc.br/~sing2014" style="line-height: 18.2000007629395px;" target="_blank">website</a></div>
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-49117940371964121612015-01-25T14:07:00.002-08:002015-01-25T16:57:38.983-08:00Medalha de Ouro na OBI 2014<div style="text-align: justify;">
<span style="font-size: large;"> Depois de um longo tempo sem postar, retorno para parabenizar meu filho, Alexandre Lima Fernandes, por sua grande conquista na última edição da Olimpíada Brasileira de Informática <a href="http://olimpiada.ic.unicamp.br/" target="_blank">OBI</a>; <span style="color: orange;">Medalha de Ouro</span>.</span><script type="text/x-mathjax-config">
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<span style="font-size: large;"> Abaixo, segue uma pequena amostra da prova da segunda (e última) etapa da OBI 2014 Modalidade Iniciação: </span><span style="font-size: large;"> </span></div>
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<span style="font-size: large;">Armário de Troféus</span><br />
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O Centro Acadêmico mandou construir dois armários, um vermelho e um branco (as cores do colégio) para guardar os troféus das equipes de esporte. Há quatro prateleiras em cada armário, numeradas de 1 a 4, de baixo para cima. Há sete tipos de troféus que devem ser colocados nas prateleiras: handebol, vôlei, futebol, futsal, judô, natação e tênis. Cada tipo de troféu deve ser colocado em exatamente uma prateleira, exceto futebol, que ocupa duas prateleiras, uma acima da outra, no mesmo armário. As seguintes condições devem ser obedecidas:</div>
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<li style="text-align: justify;">Os troféus de volei e futsal estão nas prateleiras mais altas do armário.</li>
<li style="text-align: justify;">Os troféus de tênis estão numa prateleira 2.</li>
<li style="text-align: justify;"> Os troféus de natação e judô estão em armários distintos.</li>
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<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<b>Questão 28.</b> Qual das seguintes poderia ser uma lista completa e correta dos troféus, nas prateleiras de 1 a 4, respectivamente, nos dois armários?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(A) vermelho: futebol, futebol, handebol, futsal branco: natação, tênis, judô, volei</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(B) vermelho: futebol, judô, futebol, futsal branco: handebol, tênis, natação, volei</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(C) vermelho: judô, handebol, tênis, volei branco: natação, futebol,futebol,futsal</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(D) vermelho: natação, futebol, futebol, volei branco: handebol, tênis, judô,futsal</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(E) vermelho: judô, futebol, futebol, handebol branco: judô, tênis, volei, futsal</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<b>Questão 29.</b> Se tênis está no armário vermelho, qual das seguintes afirmativas é necessariamente verdadeira?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(A) Natação está na prateleira 1 do armário vermelho.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(B) Futebol está na prateleira 2 do armário branco.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(C) Volei está na prateleira 4 do armário vermelho.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(D) Futebol está na prateleira 3 do armário branco.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(E) Handebol está na prateleira 3 do armário vermelho.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<b>Questão 30.</b> Qual das seguintes afirmativas é necessariamente falsa?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(A) Tênis e natação estão no mesmo armário.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(B) Judô e handebol estão no mesmo armário.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(C) Futsal e futebol estão no mesmo armário.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(D) Futebol e handebol estão no mesmo armário.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(E) Tênis e judô estão no mesmo armário.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<b>Questão 31.</b> Se futebol está na prateleira 1 do armário vermelho, quantos tipos diferentes de troféus poderiam estar na prateleira 1 do armário branco?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(A) 1</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(B) 2</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(C) 3</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(D) 4</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(E) 5</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<b>Questão 32.</b> Se judô está em uma prateleira de número mais baixo do que natação, qual das seguintes</div>
<div style="text-align: justify;">
afirmativas é necessariamente verdadeira?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(A) Judô está numa prateleira 1.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(B) Natação está numa prateleira 2.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(C) Futebol está numa prateleira 1.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(D) Futebol e handebol estão na mesma prateleira.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(E) Judô e handebol estão na mesma prateleira.</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<b>Questão 33.</b> Se tênis está no armário vermelho, em uma prateleira abaixo de natação, quantas diferentes maneiras de guardar os troféus no armário branco existem?</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(A) 2</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(B) 3</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(C) 4</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(D) 5</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
(E) 6</div>
<span style="text-align: justify;">.</span></div>
</div>
<div style="text-align: justify;">
<br /></div>
</div>
<div>
<span style="font-size: large;"> Como vocês podem observar nos problemas acima, a prova da Modalidade Iniciação consiste em problemas de Lógica; uma das matérias favoritas do meu filho.</span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;"> Com a palavra, Alexandre; o filho!</span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div style="text-align: right;">
<span style="color: blue; font-size: large;">Por Alexandre Lima Fernandes</span></div>
<div>
<span style="color: blue; font-size: large;"><br /></span></div>
<div>
<div style="text-align: justify;">
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; font-size: 14px; line-height: 19.3199996948242px;"> </span><span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;">Lógica pode parecer um conceito trivial - porque de fato é -, mas no Brasil é algo de muita demanda. Veja como isso é simples de perceber:</span></div>
<br /><span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"></span>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;">
</span>
<br />
<div style="text-align: justify;">
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"><span style="line-height: 19.3199996948242px;"> Uma pessoa sem lógica não consegue achar relação entre determinados elementos em uma situação quando ela existe. Ou seja, tem interpretação limitada. Posso dar um exemplo simples de frequente diálogo brasileiro:</span></span></div>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;">
</span><span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"></span>
<div style="text-align: justify;">
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"><span style="line-height: 19.3199996948242px;"><br /></span></span></div>
<span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;">
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;">"Falante: - Observei que por aqui sempre chove em domingos.</span></div>
</span><span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"><div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;">Ouvinte: -Mas ontem choveu e ontem foi terça-feira.</span></div>
</span><span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"><div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;">Falante: - É verd</span><span style="line-height: 19.3199996948242px;">ade! Perdoe o equívoco."</span></div>
</span><span style="background-color: white; color: #141823; font-family: Helvetica, Arial, 'lucida grande', tahoma, verdana, arial, sans-serif; line-height: 19.3199996948242px;"><div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"><br /></span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"> Quem se prepara para a Olimpíada Brasileira de Informática na modalidade Iniciação cansa de ouvir expressões como "Se e somente se", "ida e volta" e afins. Basta conhecer esses conceitos para compreender que, se em todo domingo ocorre uma chuva, não necessariamente só chove em domingos. O problema é que muitos brasileiros - principalmente jovens - são desprovidos, seja por erro natural ou biológico, de análises triviais como essa!</span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"> </span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"> Mas qual é o peso disso? O peso disso é não pensar com coerência. O peso disso é incapacidade de crítica. E quem não consegue tecer uma crítica ou um próprio pensamento, dependendo de obedecer cegamente ao que ouve sem contestar, denomina-se o quê? Denomina-se idiota útil.</span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"> </span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"> A consequência disso são manifestações sem ideias fixas contra um suposto "fascismo" cujo significado é desconhecido pela maioria dos manifestantes. A consequência disso são atos de pura guerrilha contra um "sistema" cujos integrantes os próprios guerrilheiros são incapazes de descrever.</span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"><br /></span></div>
<div style="text-align: justify;">
<span style="line-height: 19.3199996948242px;"> Em suma: lógica é um conceito trivial, mas necessário para a inteligência humana e a capacidade de debate.</span></div>
</span></div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com8tag:blogger.com,1999:blog-333163066896513113.post-60656462658381976892013-09-22T07:08:00.000-07:002013-09-22T07:08:18.350-07:00Parabéns pelo blog, Rui!<div>
<span style="font-size: large;"><span style="line-height: 115%;"><a href="http://estudamelhor.blogspot.com.br/" target="_blank">Conversando sobre matemática se estuda melhor</a> </span><span class="apple-converted-space"><span style="font-family: Calibri, sans-serif; line-height: 115%;">é o </span></span><span style="line-height: 115%;">Blog do meu amigo Professor Rui Brasileiro, o qual seguiremos a partir de hoje. Na mais recente postagem do Rui, ele resolve o problema do <a href="http://happyhourmatematico.blogspot.com.br/2012/07/triangulo-russo.html" target="_blank">Triângulo Russo</a> de forma extremamente elegante; vale a pena conferir!</span></span><span style="font-size: large; line-height: 115%;"> </span></div>
<div>
<span style="font-size: large;"><span style="line-height: 115%;"><br /></span></span></div>
<div>
<span style="font-size: large;"><span style="line-height: 115%;">Ao </span></span><a href="http://estudamelhor.blogspot.com.br/" style="line-height: 27px;" target="_blank"><span style="font-size: large;">Conversando sobre matemática se estuda melhor</span></a><span style="font-size: large; line-height: 115%;">,</span></div>
<div>
<span style="font-size: large;"><span style="line-height: 115%;"><br /></span></span></div>
<div>
<span style="font-size: large;"><span style="line-height: 115%;">Sinceros votos de sucesso e parceria! </span></span></div>
<div>
<span style="font-size: large;"><span style="line-height: 115%;"><br /></span></span></div>
<div>
<span style="font-size: large;"><span style="line-height: 115%;">Happy Hour Matematico.</span></span></div>
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Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com1tag:blogger.com,1999:blog-333163066896513113.post-13147330268141259672013-09-16T15:32:00.001-07:002013-09-16T15:32:47.939-07:00Irracionalidade de Raiz de 2<div style="text-align: justify;">
<span style="font-size: large;">Esta semana, recebi um email de meu amigo Professor Diego Marques (UNB) comunicando que iniciara um projeto de postagem dos vídeos de suas palestras em um canal no youtube. Excelente iniciativa!!! Fui lá</span><span style="font-size: large;"> conferir. O primeiro vídeo é uma bela palestra em que o Diego apresenta algumas provas da irracionalidade de $\sqrt{2}$ com o objetivo apresentar técnicas de prova que são utilizadas em outros problemas correlatos. Esse vídeo corresponde à</span><span style="font-size: large;"> primeira aula de um minicurso sobre Teoria dos N<span style="font-family: Calibri, sans-serif; line-height: 115%;">ú</span>meros ministrado no IMPA.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Abaixo, compartilho o link do referido vídeo</span></div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<span style="font-size: large;"><a href="https://www.youtube.com/watch?v=uhZf8Oa2F7Q" target="_blank">https://www.youtube.com/watch?v=uhZf8Oa2F7Q</a> </span><script type="text/x-mathjax-config">
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<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Antes de finalizar esta postagem, gostaria de dar a minha <span style="font-family: inherit;">contribuição</span></span><span style="font-size: large;"><span style="font-family: inherit;"> </span>apresentando uma prova da irracionalidade de $\sqrt{2}$ diferente das apresentadas pelo Diego.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Observe que se a soma de dois quadrados de inteiros, digamos $m^2+n^2$, é divisível por 3 então cada inteiro, $m$ e $n$, é divisível por 3. Logo, se supusermos que $\sqrt{2}$ seja um n<span style="font-family: Calibri, sans-serif; line-height: 115%;">ú</span>mero racional, podemos escrever $\sqrt{2}=\frac{m}{n}$, em que $m$ e $n$ são números inteiros primos entre si. Então, $3n^2=m^2+n^2$, daí $m$ e $n$ necessariamente são múltiplos de 3. Absurdo!</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Ah! Quase me esqueci que havia prometido que apresentaria, </span><span style="font-size: large;">nesta postagem,</span><span style="font-size: large;"> uma solução do problema abaixo . </span><br />
<span style="font-size: large;"><br /></span>
<span style="color: blue; font-size: large;">Problema.</span><span style="font-size: large;"> Seja $f:\mathbb{R}\rightarrow\mathbb{R}$ uma função contínua tal que $f(f(f(x)))=x^2+1$ para todo $x$. Mostre que $f$ é uma função par.</span><br />
<span style="font-size: large;"><br /></span>
<span style="color: blue; font-size: large;">Solução. </span><span style="font-size: large;">Desde que $x^2+1>x$ para todo $x$, temos que $f$ não possui ponto fixo. Em particular, $f(0)\neq 0$. Sejam</span><br />
<span style="font-size: large;">$P=\{x : f(x)=f(-x)\}$ e $I=\{x : f(x)=-f(-x)\}$. Como $f$ é contínua, Temos que os subconjuntos acima são subconjuntos fechados de $\mathbb{R}$. Mostraremos que:</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;"><span style="color: red;">1</span>. $\mathbb{R}=P\cup I$;</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;"><span style="color: red;">2</span>. $P\cap I=\emptyset$.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Uma vez provados 1 e 2,desde que $\mathbb{R}$ é conexo e $0\in P$, segue que $P=\mathbb{R}$, isto é, $f$ é uma função par.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;"><i><span style="color: red;">Prova de 1</span></i>. </span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Dado $x$, temos que $f(x)^2+1=f(f(f(f(x))))=f(x^2+1)$. Portanto, $f(-x)^2+1=f(x)^2+1$, isto é, $x\in P\cup I$.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;"><i><span style="color: red;">Prova de 2</span></i>. </span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Por contradição, suponha que $x\in P\cap I$. Logo $f(x)=0$ e $f(-x)=0$. Como $f(0)\neq 0$, garantimos a existência de um y positivo tal que $f(y)=0$.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Temos $f(1)=f(0^2+1)=f(f(f(f(0))))=f(0)^2+1>1$. Assim, a função $g(z)=f(z)-z$ é positiva no ponto 1 e negativa no ponto y. Pelo Teorema do Valor Intermediário, a função $g$ tem um zero, isto é, a função $f$ tem um ponto fixo. O que é um absurdo.</span><br />
<span style="font-size: large;"><br /></span>
<div style="text-align: right;">
<span style="color: blue; font-size: large;">C.Q.D.</span></div>
</div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-57659400450819098722013-07-01T18:47:00.000-07:002013-07-02T08:28:47.967-07:00Equações funcionais e Eu<div style="text-align: justify;">
<br />
<br />
<span style="font-size: large;">Semana passada, um aluno muito crítico observou algo muito interessante que me deixou meio reflexivo; minhas soluções nas aulas de exercícios de E.D.O. são ad hoc. Em principio, não recebi a critica suavemente, todavia a crítica fez-me lembrar e motivou-me a pesquisar sobre solu</span><span style="font-size: large;">ções</span><span style="font-size: large;"> que eu considero ad hoc. N</span><span style="font-size: large;">ã</span><span style="font-size: large;">o pude escapar de um reencontro com uma classe de problemas que, realmente, eu nunca consegui achar um padrão (ou sistemática se preferir) para resolv</span><span style="font-size: large;">ê</span><span style="font-size: large;">-los. Nessa classe, cada problema é uma história diferente. Estou falando dos problemas sobre equações funcionais. </span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Em minhas pesquisas pela web, encontrei um artigo do Professor Eduardo Tengan, sobre Eq. Funcionais, publicado na Revista Eureka (<a href="http://www.obm.org.br/opencms/revista_eureka/lista.html" target="_blank">aqui</a>). Nesse artigo, o Professor Tengan esbo</span><span style="font-size: large;">ça </span><span style="font-size: large;">uma tentativa de domar tais problemas e obt</span><span style="font-size: large;">é</span><span style="font-size: large;">m como produto um artigo que vale a pena ser lido; uma iniciativa excelente.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Durante minhas reflexões resolvi matar a saudade de resolver um desses problemas de equações funcionais que você não imagina por onde começar a resolvê-los. Abaixo o problema.</span><br />
<br />
<span style="color: blue; font-size: large;">Problema</span><br />
<span style="font-size: large;">Seja $f\colon\mathbb{R}\rightarrow\mathbb{R}$ uma fun</span><span style="font-size: large;">ção </span><span style="font-size: large;">cont</span><span style="font-size: large;">ínua </span><span style="font-size: large;"> tal que $f(f(f(x)))=1+x^2$ para todo $x\in\mathbb{R}$. Mostre que $f(x)$ é uma fun</span><span style="font-size: large;">ção</span><span style="font-size: large;"> par.</span><br />
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;">Na próxima postagem a solução.</span></div>
<div>
<span style="font-size: large;"><br /></span></div>
<div>
<span style="font-size: large;">Agradecimentos. Ao meu aluno!</span></div>
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-61765773539742942912013-05-29T12:21:00.000-07:002013-05-29T12:21:05.429-07:00A reta possui raiz: um ponto de vista algébrico<span style="color: blue; font-size: large;">Por Rafael A. da Ponte</span><script type="text/x-mathjax-config">
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<div>
<span style="color: blue; font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><div>
A convite do prof. Alexandre, estou aqui para continuar a série de postagens sobre raízes. Agradeço o chamado, professor! </div>
<div>
<br /></div>
<div>
Nesse post "contrariaremos" a resposta da pergunta que originou toda a discussão:</div>
<div>
<br /></div>
<div style="text-align: center;">
"$\mathbb{R}$ possui raiz?" </div>
<div>
<br /></div>
<div>
Na verdade, consideraremos agora o grupo aditivo dos reais, e não mais a reta como espaço topológico (naturalmente, o conceito de raiz é o algébrico, já apresentado pelo Bill em duas oportunidades). Vale lembrar que essa pergunta é a primeira da postagem anterior (<a href="http://happyhourmatematico.blogspot.com.br/2013/05/raizes-de-grupos-e-um-outro-problema.html" target="_blank">aqui</a>).</div>
<div>
<br /></div>
<div>
Então, vamos à solução:</div>
<div>
<br /></div>
<div>
Suponhamos primeiro que ($\mathbb{R}$,+) possui uma raiz ($X$, digamos). Consideraremo-la um subgrupo aditivo de $\mathbb{R}$, da maneira canônica. É simples de ver que $X$, com as operações usuais, é um subespaço de $\mathbb{R}$ sobre $\mathbb{Q}$ (dado qualquer natural $q$ e $x$ em $X$, existe $(a,b)$ em $X\times X$ tal que $q.(a,b) = (x,0)$, logo $b=0$, e assim $a \in X$ é tal que $a=x/q$, daí segue da estrutura de grupo que $X$ é fechado por produto por escalar). </div>
<div>
Disso segue que $X$ é raiz da reta na estrutura de espaço vetorial sobre $\mathbb{Q}$ (o que nos dá mais um conceito de raiz, dessa vez no contexto de espaços vetoriais).</div>
<div>
<br /></div>
<div>
Feito isso, note que se $B$ é base de Hamel de $X$, $B\times \{ 0 \} \cup \{ 0 \}\times B$ é base de $X\times X$ e suas imagens serão uma base de Hamel de $\mathbb{R}$. Daí, o problema de achar uma raiz de ($\mathbb{R}$,+) resume-se a, dada uma base de Hamel $H$ de $\mathbb{R}$, encontrarmos um subconjunto $B$ contido em $H$ em bijeção com $H\backslash B$.</div>
<div>
Para isso, defina o conjunto dos pares $(C,f)$, $C$ contido em $H$ e $f: C\rightarrow H\C$ injetivas ordenado com $(C,f)\leq(D,g) \Leftrightarrow C \subset D$ e $g$ estende $f$. Esse conjunto satisfaz as condições do Lema de Zorn, logo tem um elemento maximal $(B,h)$. Agora, verifique que $(H\backslash B) \backslash h(B)$ tem, no máximo, um elemento, e em ambos os casos de cardinalidades de $(H\backslash B) \backslash h(B)$, $B$ é um subconjunto como pedimos no parágrafo acima.</div>
<div>
<br /></div>
<div style="text-align: right;">
<span style="color: blue;">Fim da solução.</span></div>
<div>
<br /></div>
<div>
<br /></div>
<div>
É isso aí. Até a próxima vez!</div>
</span></div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-44468653479645848402013-05-22T11:09:00.001-07:002013-05-22T11:09:56.576-07:00Raízes de Grupos e um outro Problema não relacionado!<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="color: blue; font-size: large;">Por Bill Bastos</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="color: blue; font-size: large;"><br /></span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Bem, promessa feita &
promessa cumprida! Tive uma pequena conversação com o Prof.
Alexandre e combinamos fazer um pequeno post de Teoria dos Grupos.
Peço desculpas, a priori, por ter incluído um problema não
relacionado ao Tema: “raízes de grupos”. Tive boas intenções!
Mas quero ver para onde a conversa converge!</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><u>Raízes
de Grupos</u></span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Pequeno(a) repeteco
(retrospectiva):
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">“Dizemos que um
espaço topológico $Y$ é uma raiz de um outro espaço topológico
$X$ se: $X$ é homeomorfo a $Y \times Y$.”
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Ops, os espaços
topológicos $\mathbb{R}$, $\mathbb{R}^3$ e $\mathbb{S}^2$ não possuem raízes!!! Veja os posts
anteriores (<a href="http://happyhourmatematico.blogspot.com.br/2013/05/a-reta-nao-possui-raiz.html" target="_blank">aqui</a> e <a href="http://happyhourmatematico.blogspot.com.br/2013/05/a-reta-nao-possui-raiz-topologica.html" target="_blank">aqui</a>).</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><u>Raízes de Grupos:</u></span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Nada nos impede de
pensar esse problema numa versão para grupos:</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Dizemos que um grupo $G
= (G, *)$ tem raiz se existe um outro grupo $H = (H, . )$ de tal sorte
que G é isomorfo a $H \times H$.</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Nessa perspectiva, já
conseguimos mostrar que $(\mathbb{Q},+)$, $(K_{p},+)$ (Um grupo de
Prüfer, i.e., tome um primo $p$ e considere $K_p$ a união de todas
as raízes $p^a$-ésimas da unidade, para todo $a \in \mathbb{N}$)
não possuem raízes (ver seção de comentários do post anterior <a href="http://happyhourmatematico.blogspot.com.br/2013/05/a-reta-nao-possui-raiz.html" target="_blank">aqui</a>). Ficam as seguintes perguntas:</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">P.1) o grupo aditivo
dos reais admite raiz?</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">P.2) Existe um grupo
$G$ tal que $G \times G$ é isomorfo a $G \times G \times G$ & G
não é isomorfo a $G \times G$?</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Não disponho de um
exemplo ELEGANTE satisfazendo as propriedades requisitadas em P.2),
i.e. bonito, mas elementar. Mas tenho uma referência onde é
apresentado um; caso alguém se interesse, posso repassar!
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">P.3) A questão P.2) dá
para ser pensada em espaços topológicos? :) Existe?</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Desculpem a minha
indecisão! Num consegui evitar o retorno! :(</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><u>Um
Problema não relacionado:</u></span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">P.4) (Problema bem
conhecido na Teoria dos Grupos) Dado um grupo solúvel G, no qual
todos os seus subgrupos abelianos são finitos. Então G é finito?
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">Grande Abraço!</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">ps. Gostaria de
agradecer ao meu amigo Ismael Lins por me lembrar do Problema P.2), o
qual está proposto no livro ``Examples of groups, Michael
Weinstein''.
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;"><br />
</span></div>
<div class="western" style="margin-bottom: 0cm; text-align: justify;">
<span style="font-size: large;">pps. Fico esperando a
solução do Rafael para o P.1).</span></div>
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com4tag:blogger.com,1999:blog-333163066896513113.post-61341101031508475502013-05-21T14:47:00.001-07:002013-05-21T14:50:57.494-07:00A reta não possui raiz (topológica). Solucão.<div style="text-align: justify;">
<span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">Na postagem anterior, propusemos o seguinte problema.</span></div>
<div style="text-align: justify;">
<span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;"><i>Não existe um espa</i></span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; line-height: 18px;"><span style="font-family: inherit;">ço</span></span><i style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;"> topológico $X$ tal que $\mathbb{R}$ seja homeomorfo a $X\times X$.</i></span></div>
<div style="text-align: justify;">
<span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"><i><br /></i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">Abaixo, segue a solu</span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; line-height: 18px;"><span style="font-family: inherit;">ç</span></span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">ão</span><span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;"> apresentada pelo leitor Rodrigo Mendes. Eu tinha prometido postar a minha solu</span></span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; line-height: 18px;"><span style="font-family: inherit;">ç</span></span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">ão, mas mudei de ideia!</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="color: blue; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;"><br /></span></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="color: blue; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">Solu</span><span style="color: blue;"><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; line-height: 18px;"><span style="font-family: inherit;">ç</span></span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">ão</span></span></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="color: blue;"><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;"><br /></span></span></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">Se um tal espa</span></span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; line-height: 18px;"><span style="font-family: inherit;"><span style="font-size: large;">ço $X$ existe, ent</span></span></span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">ão $X$ deve ser de Hausdorff e conexo. Mais ainda, desde que $X$ </span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; line-height: 25px;"><span style="font-size: large;">é </span></span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"> homeomorfo ao subconjunto diagonal de $X\times X$, o qual </span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; line-height: 25px;">é</span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; line-height: 25px;"> </span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"> um fechado de $X\times X$, devemos ter $X$ homeomorfo a um intervalo fechado de $\mathbb{R}$. Absurdo pois $\mathbb{R}$ n</span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">ão </span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; line-height: 25px;">é</span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; line-height: 25px;"> </span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"> homeomorfo a um produto de dois intervalos.</span></div>
<div style="text-align: justify;">
<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"><br /></span></div>
<div style="text-align: right;">
<span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"><span style="color: blue;"> Final da Solu</span></span><span style="background-color: white; color: blue; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; line-height: 18px; text-align: justify;"><span style="font-family: inherit;">ç</span></span><span style="background-color: white; color: blue; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; text-align: justify;">ão</span></div>
<div style="text-align: justify;">
<span style="background-color: white; color: blue; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; text-align: justify;"><br /></span></div>
<div style="text-align: justify;">
<span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">Antes de finalizar esta postagem, gostaria de agradecer a todos</span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"> que comentaram na postagem anterior e observar que aquela discuss</span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">ão sobre raizes de grupos foi bem interessante. Em primeira m</span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">ão anuncio que </span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;">Rafael e Bill prometeram postagens sobre esse assunto; j</span><span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; line-height: 25px;"><span style="font-size: large;">á</span></span><span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large;"> estamos esperando! </span></div>
<div style="text-align: left;">
<span style="background-color: white; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: large; text-align: justify;"><br /></span></div>
<div style="text-align: left;">
<br /></div>
<div style="text-align: right;">
<span style="background-color: white; text-align: justify;"></span><br />
<div style="text-align: justify;">
<span style="background-color: white; text-align: justify;"><br /></span></div>
<span style="background-color: white; text-align: justify;">
</span></div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<span style="color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: medium;"><br /></span></div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-45829240508464216412013-05-03T18:08:00.001-07:002013-05-10T10:26:52.057-07:00A reta não possui raiz <div style="text-align: justify;">
<span style="font-size: large;">Iniciei meu final de semana lendo o artigo intitulado "$\mathbb{R}^3$ <i><b>has no root</b></i></span><span style="font-size: large;"><i>"</i> escrito por Robbert Fokkink e publicado em 2002 no peri</span><span style="font-size: large;">ó</span><span style="font-size: large;">dico <b>American Math. Monthly</b>. Fokkink mostra que n</span><span style="font-size: large;">ão</span><span style="font-size: large;"> existe um espa</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">o topol</span><span style="font-size: large;">ó</span><span style="font-size: large;">gico $X$ tal que $\mathbb{R}^3$ seja</span><i style="font-size: x-large;"> </i><span style="font-size: large;">homeomorfo a $X\times X$, isto </span><span style="font-size: large;">é</span><span style="font-size: large;"> exatamente o que Fokkink entende por $\mathbb{R}^ 3$ n</span><span style="font-size: large;">ão</span><span style="font-size: large;"> possuir uma raiz. Como Fokkink observa no artigo, </span><span style="font-size: large;">é</span><span style="font-size: large;"> poss</span><span style="font-size: large;">í</span><span style="font-size: large;">vel mostrar que $\mathbb{R}^3$ n</span><span style="font-size: large;">ão</span><span style="font-size: large;"> possui raiz utilizando a f</span><span style="font-size: large;">ó</span><span style="font-size: large;">rmula de Kunneth para grupos de homologia local de espa</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">os euclidianos, contudo a ideia de Fokkink </span><span style="font-size: large;">é</span><span style="font-size: large;"> trazer uma prova elementar daquele fato. A prova de Fokkink </span><span style="font-size: large;">é</span><span style="font-size: large;"> elementar e se baseia somente na seguinte observa</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span></div>
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<i><span style="font-size: large;"><br /></span></i></div>
<div style="text-align: center;">
<span style="font-size: large;"><i>- Se </i>$h\colon\mathbb{R}^ n\rightarrow\mathbb{R}^ n$ </span><span style="font-size: large; text-align: justify;"><i>é</i></span><span style="font-size: large;"><i> um homeomorfismo, ent</i></span><span style="font-size: large; text-align: justify;"><i>ã</i></span><span style="font-size: large;"><i>o </i>$h^ 2=h\circ h$ </span><span style="font-size: large; text-align: justify;"><i>é</i></span><span style="font-size: large;"><i> um homeomorfismo que preserva orienta</i></span><i><span style="line-height: 18px; text-align: justify;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large; text-align: justify;">ão</span></i><i style="font-size: x-large;">.</i></div>
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<i><span style="font-size: large;"><br /></span></i></div>
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<span style="font-size: large;">Resolvi fazer uma pergunta mais b</span><span style="background-color: white; color: #222222; font-size: large; line-height: 25px; text-align: justify;">á</span><span style="font-size: large;">sica; </span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: center;">
<span style="font-size: large;">- $\mathbb{R}$ <i>possui raiz ?</i></span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Utilizando somente Topologia Geral, cheguei a uma resposta negativa da pergunta acima. Ent</span><span style="font-size: large;">ão</span><span style="font-size: large;">, fica prometido que na pr</span><span style="font-size: large;">ó</span><span style="font-size: large;">xima postagem trarei a minha demonstra</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> de que $\mathbb{R}$ n</span><span style="font-size: large;">ão</span><span style="font-size: large;"> possui raiz.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Bom final de semana!</span></div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com20tag:blogger.com,1999:blog-333163066896513113.post-69476371907217210052013-03-17T07:42:00.000-07:002013-05-03T18:10:16.335-07:00Positiva vezes simétrica é diagonalizável. Solução.<div style="text-align: justify;">
<span style="font-size: large;">Hoje, trago as solu</span><span style="font-size: large; line-height: 18px;">çõ</span><span style="font-size: large;">es que conhe</span><span style="font-size: large; line-height: 18px;">ç</span><span style="font-size: large;">o para o problema de álgebra linear proposto na postagem anterior. Abaixo, o problema e as solu</span><span style="font-size: large; line-height: 18px;">ç</span><span style="font-size: large;">ões.</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><b>Problema.</b> </span><span style="background-color: white; color: #222222; line-height: 18px;"><span style="font-family: inherit; font-size: large;"><i>Sejam A e B duas matrizes simétricas. Se A é positiva definida, então AB é diagonalizável.</i></span></span><br />
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<span style="font-size: large;">A primeira solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> que apresento é devida a Rafael A. da Ponte</span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="color: blue; font-size: large;">Solucão 1. </span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">Mostra-se, primeiro, que os autovalores de $AB$ são reais. </span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Seja $\alpha$ um autovalor de $AB$, seja $v$ um autovetor correspondente a $\alpha$ (eventualmente com entradas não-reais) e $\beta$ o conjugado de $\alpha$. Então, teremos:</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">$Bv\cdot v = v \cdot Bv $ <=> $ A^{-1}ABv\cdot v = v\cdot A^{-1}ABv $ <=> $A^{-1}\alpha v\cdot v = v\cdot A^{-1}\alpha v$ <=> $\alpha A^{-1}v\cdot v = \beta v \cdot A^{-1}v. $</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Daí, teremos $\beta = \alpha$ ou $Av\cdot v = 0.$ A última possibilidade não acontece, pois $A^{-1}$ é positiva e $v$ é não nulo. Logo $\alpha$ é real. </span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Uma vez mostrado que os autovalores de $AB$ são reais, supomos que $AB$ não é diagonalizável, isto é, existem $\alpha$ real e vetores $v$ e $w$ não-nulos tais que </span><span style="font-size: large;">$ABv = \alpha v + w$ e $ABw = \alpha w$ (tais vetores aparecem naturalmente no Teorema de Jordan). Então:</span></div>
<div style="text-align: justify;">
<span style="font-size: large;">$Bv\cdot w = Bw\cdot v$ <=> $A^{-1}ABv\cdot w = A^{-1}ABw\cdot v$ <=> $\alpha A^{-1}v\cdot w + A^{-1}w\cdot w = \alpha A^{-1}w\cdot v$ <=> $A^{-1}w\cdot w = 0.$ </span></div>
<div style="text-align: justify;">
<span style="font-size: large;">A última igualdade não pode ocorrer, uma vez que $A^{-1}$ é positiva e $w$ é não-nulo. Logo, a matriz de Jordan de $AB$ é diagonal, logo $AB$ é diagonalizável. </span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="color: blue;"><span style="font-size: large;">Final da <span style="font-family: inherit;">solu</span></span><span style="line-height: 115%;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;"><span style="font-family: inherit;">ão </span>1.</span></span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">A solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> seguinte conheci através do site <a href="http://math.stackexchange.com/questions/267109/t-u-self-adjoint-u-positive-definite-then-tu-has-only-real-eigenvalues" target="_blank">http://math.stackexchange.com</a> (user 1551) e coincide com a solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> enviada por Diego Sousa (o autor do blog <a href="http://gigamatematica.blogspot.com.br/" target="_blank">gigamatematica</a>)</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="color: blue;"><span style="font-size: large;">Solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> 2.</span></span></div>
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<span style="color: blue;"><span style="font-size: large;"><br /></span></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Seja $M$ matriz simétrica e invertível tal que $M^2=A.$ A existência de $M$ vem do fato de $A$ ser uma matriz positiva. Ent</span><span style="font-size: large;">ã</span><span style="font-size: large;">o, $AB= M (MBM) M^{-1}$, isto é, $AB$ é semelhante </span></div>
<div style="text-align: justify;">
<span style="font-size: large;">a $MBM$ que é matriz sim</span><span style="font-size: large;">é</span><span style="font-size: large;">trica, logo $AB$ é diagonalizável.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"> </span></div>
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<span style="color: blue;"><span style="font-size: large;">Final da solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> 2.</span></span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Agora, a solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> que o <span style="font-family: inherit;">Professor <span style="background-color: white; color: #222222; line-height: 25px;">Flávio</span><span style="background-color: white; color: #222222; line-height: 25px;"> </span></span></span><span style="font-size: large;"><span style="font-family: inherit;">Cruz</span> me apresentou. </span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="color: blue;"><span style="font-size: large;">Solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span></span><span style="font-size: large;"><span style="color: blue;"> 3.</span> </span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Desde que $A$ é uma matriz sim</span><span style="font-size: large;">é</span><span style="font-size: large;">trica positiva, a seguinte fun</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão</span><span style="font-size: large;"> $(v,w)\rightarrow A^{-1}v\cdot w$ define um produto </span><span style="font-size: large;">interno em $\mathbb{R}^n$ que faz da matriz $AB$ um operador auto-adjunto, donde diagonalizável.</span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="color: blue; font-size: large;">Final da solucão 3.</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Finalmente, gostaria de observar que a questão levantada por Diego Sousa no campo de comentários da postagem anterior, a saber: </span><br />
<div style="text-align: center;">
<i><span style="font-size: large;"><br /></span></i></div>
<div style="text-align: center;">
<i><span style="font-size: large;">"qual seria a vers</span><span style="font-size: large;">ão complexa do problema acima ?</span></i><span style="font-size: large;"><i>"</i> </span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">foi respondida por Rafael A. da Ponte no mesmo campo de coment</span><span style="background-color: white; color: #222222; font-size: large; line-height: 25px;">ários fazendo referência à Solu</span><span style="line-height: 18px;"><span style="font-family: inherit; font-size: large;">ç</span></span><span style="font-size: large;">ão 1 apresentada aqui.</span></div>
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com4tag:blogger.com,1999:blog-333163066896513113.post-23389426486592323612013-02-27T13:39:00.001-08:002013-03-01T09:41:16.796-08:00Positiva vezes simétrica é diagonalizável<div class="separator" style="clear: both; text-align: center;">
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<div style="text-align: justify;">
<span style="font-size: large;">A cena acima descreve parte do meu encontro, na última segunda-feira, com meu amigo Flávio. Flávio é um matemático que trabalha sobre imersões isométricas. </span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">Pois bem! Após a saudação, minhas próximas palavras foram: </span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">- Rafael, aluno do bacharelado em matemática da UFC (e leitor deste blog!), resolveu o problema que você propôs.</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">O problema proposto pelo Professor Flávio era um problema de Álgebra Linear que apareceu como subproduto de sua pesquisa.</span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">Eis o dito cujo!</span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;"><b>Problema.</b> <i>Sejam A e B duas matrizes simétricas. Se A é positiva definida, então AB é diagonalizável.</i></span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Ops! Quase esquecia de falar qual é a relação do Professor Flávio com o Icasa. Icasa é um time de futebol aqui no Ceará que vive um momento de distinção entre os times cearenses porque será um dos nossos representantes na série B (isso mesmo, série B) do campeonato brasileiro de 2013. E, obviamente, o professor Flávio é um dos esperançosos torcedores do Icasa. Salve Icasa!</span></div>
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<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Sobre o problema acima, hoje conheço 3 soluções para ele: a solução do Rafael, a do Professor Flávio e uma outra que encontrei no site <a href="http://math.stackexchange.com/" target="_blank">http://math.stackexchange.com/</a></span></div>
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<br /></div>
<div style="text-align: justify;">
<span style="font-size: large;">Na próxima postagem, mostrarei as 3 soluções!</span></div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com4tag:blogger.com,1999:blog-333163066896513113.post-84166677447937559432013-01-22T18:28:00.001-08:002013-11-20T05:03:27.483-08:00Solução do problema sobre distorções de curvas<div class="MsoNormal" style="text-align: justify;">
<span style="font-size: large;"><span style="font-family: 'Times New Roman', serif; line-height: 115%;">Como prometido, apresentaremos uma prova, devida a Michael Gromov, de que a distorção
de uma curva fechada </span><span style="font-family: 'Times New Roman', serif; line-height: 115%;">é </span><span style="font-family: 'Times New Roman', serif; line-height: 115%;">maior do que ou igual à metade de $\pi$. Para ver uma definição
de distorção de curvas fechadas e entender melhor o problema que estamos
resolvendo, consulte a postagem anterior (<a href="http://happyhourmatematico.blogspot.com.br/2012/12/um-problema-sobre-distorcoes-de-curvas.html" target="_blank">aqui</a>).</span></span></div>
<div class="MsoNormal" style="text-align: justify;">
<span style="font-family: 'Times New Roman', serif; line-height: 115%;"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify;">
<span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;"><b>Prova</b>. Seja $\gamma$
curva fechada de comprimento 2L. Suponhamos que $\gamma$ </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 27px;">está</span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;"> parametrizada
pelo comprimento de arco. Para cada
ponto p em $\gamma$, denotemos por p* o único ponto da curva tal que p e p*
dividem a curva em dois arcos de comprimento L. Então, para provar a afirmação,
e suficiente provar que existe um ponto p em $\gamma$ tal que |p-p*| </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 27px;">é</span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;"> menor do que ou igual a 2L/$\pi$. Para tanto, suponhamos o contrario e definamos a
curva $\alpha(t)=\gamma(t)-\gamma(t+\mbox{L})$ com $t$ variando entre 0 e L. Vale observar que $\alpha(t)$ </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 27px;">é definido como a diferenca entre o ponto $\gamma(t)$ e o seu ponto * correspondente. </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;">Facilmente, vemos que $\alpha$ tem comprimento no máximo 2L. Por outro lado, $\alpha$ </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 27px;">é</span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;"> uma
curva cujos pontos final e inicial são antipodais e, por suposição, $\alpha$ está
contida no exterior da bola euclidiana de centro na origem e raio 2L/$\pi$.
Por isso, $\alpha$ deve ter comprimento maior do que o de uma semicircunferência
de raio 2L/$\pi$, ou seja, $\alpha$ tem comprimento maior do que 2L. O
que </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 27px;">é </span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;">o desejado absurdo.</span></div>
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<span style="font-family: 'Times New Roman', serif; line-height: 115%;"><span style="font-size: large;"><b>Final da Prova</b>.<o:p></o:p></span></span></div>
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<br /></div>
<div class="MsoNormal" style="text-align: justify;">
<span style="font-family: 'Times New Roman', serif; line-height: 115%;"><span style="font-size: large;">Fica como desafio
provar que se a distorção de $\gamma$ vale a metade de $\pi$, então $\gamma$
deve ser um circulo.<o:p></o:p></span></span></div>
<div class="MsoNormal" style="text-align: justify;">
<span style="font-family: 'Times New Roman', serif; line-height: 115%;"><span style="font-size: large;"><br /></span></span></div>
<div class="MsoNormal" style="text-align: justify;">
<span style="font-family: 'Times New Roman', serif; line-height: 115%;"><span style="font-size: large;">At</span></span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 27px;">é</span><span style="font-family: 'Times New Roman', serif; font-size: large; line-height: 115%;"> a próxima. </span></div>
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-37143224628061775992012-12-11T18:55:00.001-08:002012-12-11T19:02:17.163-08:00Um problema sobre distorções de curvas<div style="text-align: justify;">
<span style="font-size: large;">Seja $\gamma$ uma curva simples e fechada em $\mathbb{R}^n$ de comprimento finito L. Dados quaisquer dois pontos distintos $p$ e $q$ sobre $\gamma$, existem exatamente dois arcos sobre a curva que conectam $p$ e $q$ e, pelo menos um deles tem comprimento $D_{\gamma}(p,q)$ menor do que ou igual à metade de L. Definimos a distância em $\gamma$ entre p e q como sendo $D_{\gamma}(p,q)$. De fato, a função $D_{\gamma}$, assim definida, é uma distância sobre a curva $\gamma$. Pois bem, agora definimos a distorção de $\gamma$ como o supremo dos quocientes $\frac{D_{\gamma}(p,q)}{|p-q|}$ quando $p\neq q$ variam em $\gamma$.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Eis o nosso problema de hoje!</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><span style="color: blue;">Problema. </span><i>A distorção de $\gamma$ é sempre maior do que ou igual à metade de $\pi$ e, ocorre a igualdade se, e somente se, $\gamma$ é um círculo.</i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><i><br /></i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Na próxima postagem, trarei uma belíssima solução para este problema. </span></div>
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Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-42140784594788175542012-11-22T20:45:00.003-08:002012-11-23T06:32:41.173-08:00Renan, Medalha de Ouro<div style="text-align: justify;">
Renan da Silva Santos, aquele camarada do blog <a href="http://pontodeacumulacao.blogspot.com/">ponto de acumulação</a>, aluno do Bacharelado em Matemática da UFC, conquistou a medalha de ouro no <a href="http://www.impa.br/opencms/pt/eventos/store_old/evento_1205?link=25">VI SIMPÓSIO NACIONAL / JORNADAS DE INICIAÇÃO CIENTÍFICA</a> que aconteceu no IMPA, Rio de Janeiro, de 04 a 10 de novembro de 2012. O Renan, sob a minha orientação, apresentou uma monografia sobre endomorfismos injetivos de variedades algébricas. O resultado mais conhecido a respeito do tópico acima é o Teorema de Ax-Grothendick o qual diz que, no caso em que temos variedades algébricas sobre corpos algebricamente fechados, tais endomorfismos são necessariamente sobrejetivos. Além de trazer uma prova do Teorema de Ax-Grothendick, a monografia do Renan aborda resultados do tipo Ax-Grothendick quando consideradas variedades algébricas sobre corpos não necessariamente algebricamente fechados, vale a pena conferir.</div>
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<br /></div>
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Parabéns, Renan!</div>
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Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-49424636899270785082012-09-18T16:54:00.000-07:002012-12-11T18:05:27.936-08:00Complementar de conjuntos algébricos em $\mathbb{R}^n$<div style="text-align: justify;">
<div style="text-align: center;">
<span style="font-size: large;"><i>"Seja $V\subset\mathbb{R}^n$ um subconjunto algébrico não vazio. Então, $\mathbb{R}^n$ e $\mathbb{R}^n-V$ não são homeomorfos"</i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Abaixo, segue uma prova do resultado acima</span></div>
<div style="text-align: justify;">
<br /></div>
<span style="color: blue;"><span style="font-size: large;">Por Rodrigo Mendes Pereira</span></span></div>
<div style="text-align: justify;">
<span style="color: blue;"><span style="font-size: large;"><br /></span></span></div>
<div style="text-align: justify;">
<div>
Primeiramente, meus agradecimentos ao professor Alexandre pelo convite! </div>
<div>
<br /></div>
<div>
A prova que vou apresentar aqui usa uma versão semialgébrica da Homologia de Borel-Moore. Ressalto que o interesse em trabalhar com uma teoria de homologia dentro da estrutura semialgébrica é, dentre outros fatores, por ter a disposição boas ferramentas (como decomposição celular, descrição por fórmulas e existência de uma boa triangulação). De posse disto, é garantida a existência de uma classe fundamental sobre variedades algébricas que é essencial, em particular, para esta prova. Vamos lá então!</div>
<div>
<br /></div>
<div>
Gostaria de iniciar uma sequência observações sobre conjuntos semialgébricos apresentando o conceito de dimensão de conjuntos semialgébricos. Temos um teorema de decomposição de conjuntos semialgébricos que garante o seguinte: cada $X$ semialgébrico pode ser escrito como união finita e disjunta (a menos de um homeomorfismo semialgébrico) de cubos unitários abertos $(0,1)^{d_i}, \ d_i \in \mathbb{N} \cup \left\{0\right\}, i = 1,2 \ldots, k$. Com isto temos uma definição natural para dimensão de $X$, a saber, $dim X = max\left\{d_1, \ldots, d_k \right\}$ </div>
<div>
<br /></div>
<div>
Antes de iniciarmos a prova propriamente dita, vamos definir e fixar mais alguns fatos importantes referentes à <i>teoria de homologia semialgébrica.</i></div>
<div>
<br /></div>
<div>
<ul>
<li><b> Sobre triangulação</b></li>
</ul>
</div>
<div>
Se $X$ é um semialgébrico compacto em $\mathbb{R}^n$, então existem $|\mathcal{K}| \subset \mathbb{R}^n$ complexo simplicial finito e um homemorfismo $\varphi: |\mathcal{K}| \rightarrow X$ tal que o gráfico de $\varphi$ é um conjunto semialgébrico em $\mathbb{R}^{2n}$ (definição de um aplicação semialgébrica) e tal triangulação pode ser escolhida compatível com uma família $X_1, X_2, \ldots, X_n$ de conjuntos semialgébricos em $X$. Isto significa que $\varphi^{-1}\left(X_i\right) = L_i$ subcomplexo de $\mathcal{K}$. Por outro lado, se X é ilimitado ou não contém todos os seus pontos de fronteira podemos obter uma versão semialgébrica da compactificação de Alexandrov:</div>
<div>
<span class="Apple-tab-span" style="white-space: pre;"> </span></div>
<div style="text-align: center;">
<i><span class="Apple-tab-span" style="white-space: pre;"> </span> Existem, para X localmente compacto, um subconjunto semialgébrico $X^* \subset \mathbb{R}^{n+1}$ compacto, um mergulho semialgébrico de $X$ em $X^*$</i><i> de forma que $X^*-X = \left\{p\right\} \in \mathbb{R}^{n+1}$.</i></div>
<div style="text-align: center;">
<br /></div>
<div>
Portanto, de posse de uma triangulação de $X$ (localmente compacto) podemos definir a </div>
<div>
<br /></div>
<div>
<ul>
<li><b>Homologia semialgébrica de Borel-Moore</b></li>
</ul>
</div>
<div style="text-align: center;">
<span class="Apple-tab-span" style="white-space: pre;"> </span> $H_i^{BM}(X,\mathbb{Z}_2 )$, se $X$ for compacto, é definido por $H_i(X, \mathbb{Z}_2)$ e, se $X$ não for compacto, é definido por $H_i(X^*,X^*-X, \mathbb{Z}_2)$.</div>
<div style="text-align: center;">
<br /></div>
<div>
$i = 0,1, \ldots, dimX$.</div>
<div>
<br /></div>
<div>
Um comentário: $H_i(X, \mathbb{Z}_2)$ trata-se de um grupo de homologia simplicial (e um espaço vetorial!), induzido pelo quociente entre ciclos e bordos sobre complexo de cadeias $\mathcal{C}_i(X,\mathbb{Z}_2)$, que não depende da triangulação, ou seja, é invariante topológico. Por conseguinte, também independem da triangulação, grupos de homologia relativa induzidos pelo quociente natural entre os respectivos complexos de cadeias (considerando uma escolha compatível de triangulação).</div>
<div>
<br /></div>
<div>
Vamos agora a 3 resultados que serão usados na demonstração (todos os grupos de homologia considerados são com coeficientes em $\mathbb{Z}_2$):</div>
<div>
<br /></div>
<div>
<b>1.</b> Sejam $A \subset B$ semialgébricos compactos em $\mathbb{R}^n$. Temos que</div>
<div>
$H_q(A,B) = H_q^{BM}(A - B)$.<br />
<br /></div>
<div>
<br /></div>
<div>
<b>2.</b> Sejam $U \subset V$, $V$ localmente compacto e $U$ subconjunto semialgébrico fechado de $V$. Então existe uma sequência exata longa<br />
<div style="text-align: center;">
<br /></div>
</div>
<div>
<div style="text-align: center;">
$\ldots H_{q+1}^{BM} (V - U) \rightarrow H_q^{BM}(U) \rightarrow H_q^{BM}(V) \rightarrow H_{q}^{BM}(V - U) \rightarrow \ldots $</div>
</div>
<div>
<div style="text-align: center;">
</div>
<div style="text-align: justify;">
<br /></div>
</div>
<div>
<b>3. (</b>Existência da classe fundamental<b>)</b> Seja $X$ um conjunto algébrico compacto de dimensão $d$ e considere $\phi:|\mathcal{K}| \rightarrow X$ uma triangulação semialgébrica de $X$. A soma de todos os $d$-simplexos de $\mathcal{K}$ é um ciclo não nulo com coeficientes em $\mathbb{Z}_2$, elemento de $H_d(X,\mathbb{Z}_2)$. Este elemento (que denotaremos por $[X]$) é independente da escolha da triangulação e é dito a classe fundamental de $X$.</div>
<div>
<br /></div>
<div>
Vamos agora à prova do teorema: Ah! Nessa altura, vale lembrar o que se deseja provar:</div>
<div>
<br /></div>
<div>
<b>Teorema principal.</b></div>
<div>
O complemento de um subconjunto algébrico de $\mathbb{R}^n$ não é homeomorfo a $\mathbb{R}^n$.</div>
<div>
<br /></div>
<div>
Prova:</div>
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<br /></div>
<div>
Para nossos propósitos vamos tomar grupos de homologia reduzida $\tilde{H}$:<br />
<br /></div>
<div>
$H_0(|\mathcal{K}|) = \mathbb{Z}\oplus \tilde{H}_0(|\mathcal{K}|)$ e $\tilde{H}_i(|\mathcal{K}|) = H_i(|\mathcal{K}|)$ se $i > 0$.</div>
<div>
<br /></div>
<div>
Considere $V$ algébrico com $dimV < n$ e a seguinte sequência exata:</div>
<div>
<br /></div>
<div>
<div style="text-align: center;">
$\ldots \rightarrow H_i^{BM}(V) \rightarrow H_i^{BM}(\mathbb{R}^n) \rightarrow H_{i}^{BM}(\mathbb{R}^n - V) \rightarrow H_{i-1}^{BM}(V) \ldots $</div>
</div>
<div>
<div style="text-align: center;">
<br /></div>
</div>
<div>
<br /></div>
<div>
Sabemos que $H_n^{BM}(\mathbb{R}^n) = \mathbb{Z}_2$ e $H_i^{BM}(\mathbb{R}^n)=0$ para $i < n$. Além disso,</div>
<div>
<br /></div>
<div>
$H_{dimV}^{BM}(V)$ é não trivial e $H_i^{BM}(V) = 0, \ i > dimV$.</div>
<div>
<br /></div>
<div>
Pela exatidão da sequência acima, obtemos ($0 < i < n$)</div>
<div>
<br /></div>
<div>
<div style="text-align: center;">
$0 \rightarrow H_i^{BM}(\mathbb{R}^n-V) \rightarrow H_{i-1}^{BM}(V) \rightarrow 0$</div>
</div>
<div>
<div style="text-align: center;">
<br /></div>
</div>
<div>
E, portanto, $H_i^{BM}(\mathbb{R}^n - V) = H_{i-1}^{BM}(V)$, \ $0< i < n$ (I)</div>
<div>
<br />
Para $i = n$, ficamos com o seguinte:</div>
<div>
<br /></div>
<div>
<div style="text-align: center;">
$0 \rightarrow \mathbb{Z}_2 \rightarrow H_n^{BM}(\mathbb{R}^n-V) \rightarrow H_{n-1}^{BM}(V) \rightarrow 0$</div>
</div>
<div>
<br /></div>
<div>
Como os elementos dessa sequência podem ser vistos como módulos sobre um corpo (espaços vetoriais) temos que: $H_n^{BM}(\mathbb{R}^n-V) = \mathbb{Z}_2 \oplus H_{n-1}^{BM}(V)$ (II).</div>
<div>
<br /></div>
<div>
<br /></div>
<div>
Nesse ponto, olhando para a equação (II), se fossem $\mathbb{R}^n$ e $\mathbb{R}^n-V$ homeomorfos, não poderia ser $dim V = n-1$ pois</div>
<div>
<br /></div>
<div>
<div style="text-align: center;">
$ \mathbb{Z}_2 = H_n^{BM}(\mathbb{R}^n) = \mathbb{Z}_2 \oplus H_{n-1}^{BM}(V)$</div>
</div>
<div>
<br /></div>
<div>
E nem poderia ser $0\leq p = dimV < n-1$ pois usando a equação (I) teríamos</div>
<div>
<br /></div>
<div>
<div style="text-align: center;">
$0 = H_{p+1}^{BM}(\mathbb{R}^n) = H_{p}^{BM}(V)$</div>
</div>
<div>
<br /></div>
<div>
Logo, $V$ deve ser $\emptyset$ e isto conclui a prova.</div>
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Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-49423166390684918992012-09-01T20:48:00.000-07:002012-09-01T20:49:47.389-07:00Complementar de curvas algébricas em $\mathbb{R}^2$ II<div style="text-align: justify;">
<span style="font-size: large;">A postagem de hoje é destinada a apresentação de uma solução para o seguinte problema.</span></div>
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<span style="font-size: large;"><b>Problema. </b>O complemento de <i>um subconjunto algébrico de $\mathbb{R}^2$ </i></span><span style="font-size: large;"><i>não é homeomorfo a $\mathbb{R}^2$.</i></span></div>
</div>
<div style="text-align: justify;">
<span style="font-size: large;"><i><br /></i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><b>Solução.</b> Seja <i>p(x,y)</i> um polinômio irredutível que não muda de sinal em $\mathbb{R}^2$. Mostraremos que o conjuntos dos zeros deste polinômio é finito. De fato, dado <i>(a,b)</i> ponto que anula aquele polinômio, temos que, para qualquer que seja o vetor <i>(u,v)</i>, a função real de uma variável real $t$ dada por</span></div>
<div style="text-align: justify;">
<div style="text-align: justify;">
<i><span style="font-size: large;"><br /></span></i></div>
<div style="text-align: center;">
<i><span style="font-size: large;">f(t)=p(a+tu,b+tv)</span></i></div>
<div style="text-align: justify;">
<i><span style="font-size: large;"><br /></span></i></div>
<div style="text-align: justify;">
<span style="font-size: large;">possui um ponto de mínimo em <i>t=0</i>, portanto, <i>f'(0)=0, </i>ou seja, o gradiente de <i> p(x,y)</i> no ponto <i>(a,b)</i> <i> </i>é ortogonal ao vetor <i>(u,v). </i>Desde que escolhemos o vetor <i>(u,v)</i> arbitrariamente, constatamos que o gradiente de <i>p(x,y)</i> se anula no ponto <i> (a,b). </i> Segue do Teorema de Bezout que <i> </i>o conjunto dos zeros de <i>p(x,y)</i> é finito.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
</div>
<div>
<span style="font-size: large;">Suponhamos que <i>X</i> seja definido por um polinômio <i>p(x,y)</i>. Isto é, <i>X</i> é o conjunto dos pontos <i>(x,y)</i> em que <span style="text-align: center;"><i>p(x,y)=0</i>. Se os fatores irredutíveis de <i>p(x,y) </i>não mudam </span>de sinal em $\mathbb{R}^2$ temos que <i>X</i> é um conjunto finito (foi o que vimos acima). Assim, o complemento de <i>X</i> não é simplesmente conexo.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Então, resta-nos supor que <i>p(x,y) </i>possui um fator irredutível <i>q(x,y) </i>que muda de sinal em $\mathbb{R}^2$<i>.</i> Desta forma, os conjuntos abaixo</span></div>
<div>
<div>
<ul>
<li><span style="font-size: large;">pontos <i>(x,y)</i> em que q<i>(x,y)</i> é positivo</span></li>
<li><span style="font-size: large;">pontos <i>(x,y)</i> em que q<i>(x,y)</i> é negativo</span></li>
</ul>
</div>
<div>
<span style="font-size: large;">são abertos, disjuntos e não-vazios. Como o complemento de <i>X</i> é um subconjunto denso do plano que está contido na reunião dos conjuntos acima, temos que ele não é conexo<i>.</i></span></div>
<div style="text-align: right;">
<br /></div>
<div style="text-align: right;">
C.Q.D.</div>
</div>
<div style="text-align: justify;">
<br /></div>
<div style="text-align: justify;">
<span style="font-size: large;">Antes de finalizar esta postagem, gostaria de informar que o problema acima possui uma versão em dimensão <i>n</i>, quero dizer, temos o seguinte resultado em $\mathbb{R}^n$.</span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><b>Teorema.</b><i> O complemento de um subconjunto algébrico de $\mathbb{R}^n$ não é homeomorfo a $\mathbb{R}^n$.</i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span style="font-size: large;">Recentemente, meu aluno de mestrado, Rodrigo Mendes Pereira, escreveu sua dissertação sobre Homologia Semialgébrica sobre corpos reais fechados e provou uma versão do teorema acima para $\mathbb{K}^n$, em que $\mathbb{K}$ é um corpo real fechado. Por isto, convidei o Rodrigo a escrever uma postagem para este blog trazendo uma demonstração para o teorema acima. Já estamos esperando, Rodrigo!</span></div>
</div>
Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-13567844298564258622012-08-11T10:36:00.000-07:002012-09-01T18:36:02.962-07:00Complementar de curvas algébricas em $\mathbb{R}^2$<script type="text/x-mathjax-config">
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<span style="font-size: large;">Eu e Renan
(meu aluno de iniciação científica), recentemente, quando estudávamos o Teorema
de Newman (ver abaixo), deparamo-nos com o seguinte problema: <i>se removermos um subconjunto algébrico,
próprio, do plano euclidiano o resultado não será homeomorfo ao plano.</i></span></div>
<div style="text-align: justify;">
<span style="font-size: large;"><i><br /></i></span></div>
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<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;">Antes de
tecer mais alguns comentários sobre o problema acima, gostaria de enunciar o
Teorema de Newman<o:p></o:p></span></div>
</div>
<div class="MsoNormal">
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;"><b>Teorema </b>(Donald J. Newman 1969). <i>Se uma aplicação polinomial $\mathbb{R}^2\rightarrow\mathbb{R}^2$ é injetiva, então ela também é sobrejetiva.</i><o:p></o:p></span></div>
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;"><i><br /></i></span></div>
</div>
<div class="MsoNormal">
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;">O teorema
acima é parte de uma belíssima história na matemática que tem como
palavra-chave “Teorema de Ax-Grothendick”. Se você quiser conhecer um pouco dessa
história, sugiro o post publicado por Terence Tao
em seu blog <i>What’s new</i> (<a href="http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/">aqui</a>).</span></div>
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;"><br /></span></div>
</div>
<div class="MsoNormal">
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;">De volta ao
primeiro problema descrito acima, gostaria de observar que existe subconjunto
semi-algébrico, próprio, do plano euclidiano de sorte que seu complementar
continua homeomorfo ao plano. Com efeito, o complementar de uma semi-reta no
plano é homeomorfo ao plano. Ora, se não fosse assim, o Teorema de Newman
seria consequência imediata desse fato.
Infelizmente, a vida não é tão fácil! Ou não, como diria o Renan.<o:p></o:p></span></div>
</div>
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<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;"><br /></span></div>
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;">Então,
finalizo esta postagem prometendo trazer uma solução bonita para o seguinte<o:p></o:p></span></div>
<div style="text-align: justify;">
<span lang="PT-BR" style="font-size: large;"><br /></span></div>
</div>
<div class="MsoNormal">
<div style="text-align: justify;">
<span lang="PT-BR"><span style="font-size: large;"><b>Problema</b>:
Se $X\subset\mathbb{R}^2$ é subconjunto algébrico, próprio, então $\mathbb{R}^2\setminus X$ </span></span><span style="font-size: large;">não é homeomorfo a </span><span style="font-size: large;">$\mathbb{R}^2$.</span></div>
<div style="text-align: justify;">
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com2tag:blogger.com,1999:blog-333163066896513113.post-26027024349967563192012-07-15T21:45:00.000-07:002012-07-18T13:29:00.985-07:00Triângulo Russo<div style="text-align: justify;">
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<span style="font-size: large;">Hoje encontrei um amigo que há muito tempo não via, Professor Ivair. Realmente, foi uma grande satisfação </span><span style="background-color: white; font-size: large;">encontrá-lo novamente. Vocês podem estar perguntando por que eu relataria aqui um reencontro com um amigo. </span><span style="background-color: white; font-size: large;">Já explico o porquê de um reencontro com o Professor Ivair gerar uma postagem no <span style="color: blue;">Happy Hour Matemático</span>. O negócio </span><span style="background-color: white; font-size: large;">é o seguinte: ele conhece muitos e bons problemas de matemática, coleciona-os em apostilas e tem dezenas </span><span style="background-color: white; font-size: large;">delas. Ele é, de fato, uma boa fonte para este blog. </span></div>
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<span style="font-size: large;">Não há como eu me encontrar com ele e não lembrar de um problema casca grossa. Confesso que hoje não foi diferente! </span><span style="background-color: white; font-size: large;">Lembrei-me de um famoso problema de geometria plana conhecido como </span><span style="background-color: white; font-size: large;">o Triângulo Russo, o qual, também, foi-me apresentado pelo Ivair em uma de nossas conversas na UFC. </span></div>
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<span style="font-size: large;">Abaixo segue o <b>Problema do Triângulo Russo</b>.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOpJprPaaAcyk0-UTfhLw3jf7Z1vGpPUgBcETrD_oAemd7YFqGGfeOTW_mK9_xIyGUjIbfBqMsDMAixfQqaYJBM6L2WbehwUFcBcX8aMfEDsCI3j1s07zUB9dZsWFhIuGZGHtKmTdiiRg/s1600/triangulo-russo.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOpJprPaaAcyk0-UTfhLw3jf7Z1vGpPUgBcETrD_oAemd7YFqGGfeOTW_mK9_xIyGUjIbfBqMsDMAixfQqaYJBM6L2WbehwUFcBcX8aMfEDsCI3j1s07zUB9dZsWFhIuGZGHtKmTdiiRg/s400/triangulo-russo.jpg" width="398" /></a></div>
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<i><span style="font-size: large;">Sabendo que AB=AC, encontre o valor do ângulo x.</span></i></div>
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<span style="font-size: large;">Até a próxima! </span><br />
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com13tag:blogger.com,1999:blog-333163066896513113.post-12486736094532146692012-07-06T13:05:00.000-07:002012-07-06T19:56:14.754-07:0012th International Workshop on Real and Complex Singularities<div class="separator" style="clear: both; text-align: center;">
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<span style="font-size: large;"><i><span style="color: red;"><a href="http://www.icmc.usp.br/~sing/main_site/"><span style="color: red;">International Workshop on Real and Complex Singularities</span></a> </span></i>é um congresso de matemática promovido pelo Instituto de Ciências Matemáticas e Computação da USP que ocorre a cada dois anos na cidade de São Carlos (SP). O tradicional evento é marcado <span style="background-color: white;">pela presença de renomados pesquisadores </span></span><span style="background-color: white; font-size: large;">das áreas de Singularidades, Geometria Algébrica e Teoria de Bifurcações, assim como pela presença maciça de jovens pesquisadores/alunos brasileiros e estrangeiros. A edição deste ano celebrará o 60<strike>°</strike> aniversário do Professor Shyuichi Izumiya e ocorrerá na semana 22-27 de julho (<a href="http://www.icmc.usp.br/~sing/main_site/2012/index.php">link</a>).</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNF6INRrDlO_BVtexAh8PNAYmplrpD0uNBvXAhS_vYB4JWo8XsDKKEpg__mcruJl8cCJ5RDZ3PKZ7gZ4w8hVIT0HObhw0wf7BJZwPVHshXxY8AfLxtVG5s93QcqMv7jynRb01M6agCH6E/s1600/banner.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhNF6INRrDlO_BVtexAh8PNAYmplrpD0uNBvXAhS_vYB4JWo8XsDKKEpg__mcruJl8cCJ5RDZ3PKZ7gZ4w8hVIT0HObhw0wf7BJZwPVHshXxY8AfLxtVG5s93QcqMv7jynRb01M6agCH6E/s640/banner.jpg" width="451" /></a></div>
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<span style="background-color: white; font-size: large;">Vale a pena conferir!</span></div>
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</script>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-4386105284378711962012-06-09T08:36:00.001-07:002012-08-11T09:38:52.463-07:00Desigualdade de Weyl<script type="text/x-mathjax-config">
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<span style="font-size: large;">Em 2007, um conhecido stick, um stick que de certa forma é observado pelos seus pares, </span>
<span style="font-size: large;">um tipo que se expressa </span><span style="font-size: large;">com uma mão no bolso de sua calça e a outra livre fazendo movimentos como de um maestro que comanda uma orquestra sinfônica (acho que não sei defini-lo!), apresentou-me o seguinte problema:</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXR1x1oJ2VXHGf8LPlclzFQIQdDsNW9YCJ4NIi3JhOtxufPadpz8HEJkJsnAFXjrCNHlzz7UuLZQNAOjXcUK2RmCgLUa9ACN7lZOfuwZqVA-yHe0k134qXjppFCg5aiJHvPowqq6ECblg/s1600/WEYL.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXR1x1oJ2VXHGf8LPlclzFQIQdDsNW9YCJ4NIi3JhOtxufPadpz8HEJkJsnAFXjrCNHlzz7UuLZQNAOjXcUK2RmCgLUa9ACN7lZOfuwZqVA-yHe0k134qXjppFCg5aiJHvPowqq6ECblg/s320/WEYL.jpg" width="319" /></a></div>
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<span style="font-size: large;"><i><br /></i></span></div>
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<span style="font-size: large;"><i>É bem conhecido que os autovalores de uma matriz simétrica variam continuamente com a matriz. </i></span><i style="font-size: x-large;">Podemos afirmar que essa variação é lipschitziana?</i></div>
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<span style="font-size: large;">Aceitei o desafio e, juntamente com um colega de trabalho, obtive uma resposta positiva para a pergunta acima. Contudo, a </span><span style="font-size: large;">solução obtida era muito feia e, para mim, aquela prova merecia ser esquecida. Antes de esquecê-la, refleti bastante sobre a seguinte frase de Michel Atiyah</span></div>
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<span style="font-size: large;"><i>"Existe teorema bonito com prova feia, mas não existe teorema feio com prova bonita".</i></span></div>
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<span style="font-size: large;"><i><br /></i></span></div>
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<span style="font-size: large;">Certamente, a dependência Lipschitz dos autovalores de uma matriz simétrica é um teorema bonito e, embora Sir Michael Atiyah trouxesse a permissão de termos provas feias para teoremas bonitos, eu sentia que aquele não seria o caso em que eu deveria me conformar com aquela prova, porque ela era demasiadamente feia. </span></div>
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<span style="font-size: large;">Pouco tempo depois, ministrando uma disciplina de álgebra linear, conheci uma prova adequada para a variação lipschitziana dos autovalores de matrizes simétricas.</span></div>
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<span style="font-size: large;">O objetivo de hoje é mostrar de uma </span><span style="font-size: large;">forma muito elegante a Desigualdade de Weyl, desigualdade que responde positivamente à pergunta daquele stick legal.</span></div>
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<span style="font-size: large;">A partir daqui, passo à apresentação direta da Desigualdade de Weyl e sua prova. </span></div>
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<span style="font-size: large;">Inicialmente, lembramos que o espaço das matrizes de ordem $n$ admite uma norma definida da seguinte maneira: dada uma matriz $A$ de ordem $n$, $$\|A \|= max\{|Ax| \ : \ x\in\mathbb{R}^n, |x|=1\}.$$<ax :="" font="" in="in" mathbb="mathbb" n="n" r="r" x="x"></ax></span><br />
<span style="font-size: large;">No caso em que $A$ é uma matriz simétrica é muito fácil mostrar que $\|A\|$ coincide com o maior valor absoluto dos autovalores de $A$.</span><br />
<span style="font-size: large;"><br /></span>
<span style="font-size: large;">Antes de enunciarmos a desigualdade de Weyl, consideremos mais algumas notações. Para cada matriz simétrica $M$ de de ordem $n$, estabelecemos que o i-ésimo autovalor de $M$, denotado por $\lambda_i(M)$, respeita a seguinte ordem $$\lambda_1(M)\geq\cdots\geq\lambda_n(M).$$</span><br />
<span style="font-size: large;">Utilizamos $x\cdot y$ para representar o produto interno euclidiano entre os vetores $x$ e $y$ em $\mathbb{R}^n$.</span>
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<span style="color: blue; font-size: large;">Desigualdade de Weyl</span><br />
<span style="font-size: large;"><i>Sejam $A$ e $B$ matrizes simétricas de ordem $n$. </i><i>Então, vale a seguinte desigualdade: </i></span><i style="text-align: center;"><span style="font-size: large;">$|\lambda_i(A)-\lambda_i(B) | \leq \| A-B \|, \ i=1,\dots,n$.</span></i><br />
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<span style="font-size: large;"><span style="color: blue;">Prova</span>. Sejam $\{x_1,\dots,x_n\}$ e $\{y_1,\dots,y_n\}$ bases ortonormais de $\mathbb{R}^n$ em que $x_i$ é autovetor de $A$ associado ao i-ésimo autovalor de $A$ e $y_i$ é autovetor de $B$ associado ao i-ésimo autovalor de $B$. Para cada $j=1,\dots,n$, seja $z_j$ vetor unitário na interseção dos subespaços $span\{x_1,\dots,x_j\}$ e $span\{y_j,\dots,y_n\}$. Como $\lambda_j(A)\leq Az_j\cdot z_j$ e $\lambda_j(B)\geq Bz_j\cdot z_j$, obtemos:</span><br />
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<span style="font-size: large;">$\lambda_j(A)-\lambda_j(B)\leq \lambda_1(A-B)$.</span></div>
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<span style="font-size: large;">Substituíndo $A$ por $-A$ e $B$ por $-B$ na desigualdade acima, obtemos:</span></div>
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<span style="font-size: large;">$\lambda_j(A)-\lambda_j(B)\geq \lambda_n(A-B)$.</span>
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<span style="font-size: large;">Juntando as duas últimas desigualdades, obtemos a almejada desigualdade de Weyl.</span></div>
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<span style="font-size: large;"><span style="color: blue;">Final da Prova</span></span></div>
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<span style="font-size: large;">O objetivo da postagem de hoje é apresentar uma prova para a proposição abaixo.</span></div>
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<span style="font-size: large;"><i>"Se uma curva divide o quadrado unitário em dois pedaços, então o comprimento da curva é maior do que ou igual ao valor da área de um desses pedaços."</i></span></div>
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<span style="font-size: large;"><b>Prova. </b>Seja $\gamma$ uma curva que divide o quadrado unitário em pedaços A e B. Mostremos que o comprimento comp($\gamma$) dessa curva é maior do que ou igual ao mínimo das áreas {area(A),area(B)}.</span></div>
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<span style="font-size: large;">Desde que o quadrado tem área unitária, podemos supor que o comprimento da curva é </span><span style="font-size: large;">menor do que 1. </span></div>
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<span style="font-size: large;">Suponhamos que area(B) seja maior do que comp($\gamma$). </span><span style="font-size: large;">Nesse caso, considerando G como sendo a </span><span style="font-size: large;">projeção vertical do quadrado sobre a sua base (veja figura abaixo), temos que </span><span style="font-size: large;">area(B) é menor do que ou igual à área do cilindro de base G(B) e altura 1 e, portanto, </span><span style="font-size: large;">area(B) é menor do que o comprimento do intervalo G(B). </span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5SyrBZrly-NN1NFCmLMxywh871WqO8QU7CpaVorx6GjNY8Ju2FxGcGRbgVLmalfIOInR0Zy4eqOVW3Rgamt-fak-vWhv9SUdOFlXsLoRaprrUKTVGtcN1QHR6xdSsiRoBOjr3jAAfPLs/s1600/PROJECOES-QUADRADO.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5SyrBZrly-NN1NFCmLMxywh871WqO8QU7CpaVorx6GjNY8Ju2FxGcGRbgVLmalfIOInR0Zy4eqOVW3Rgamt-fak-vWhv9SUdOFlXsLoRaprrUKTVGtcN1QHR6xdSsiRoBOjr3jAAfPLs/s400/PROJECOES-QUADRADO.jpg" width="398" /></a></div>
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<span style="font-size: large;">Por outro lado, desde que G é uma projeção ortogonal, </span><span style="font-size: large;">comp($\gamma$)</span><span style="font-size: large;"> é maior do que ou igual ao comprimento do intervalo G($\gamma$). Então, existe uma reta vertical </span><span style="font-size: large;">que não intersecta $\gamma$ e intersecta B.</span></div>
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<span style="font-size: large;">Se, também, temos que area(A) é maior do que </span><span style="font-size: large;">comp($\gamma$), u</span><span style="font-size: large;">tilizando a projeção ortogonal F do quadrado sobre a sua lateral </span><span style="font-size: large;">direita (veja figura acima), com uma análise semelhante à que fizemos acima, concluímos que existe uma </span><span style="font-size: large;">reta horizontal no quadrado que não intersecta $\gamma$ e intersecta A. Desde que retas verticais sempre intersectam retas horizontais no quadrado, temos pontos de A e B que podem ser conectados por um arco poligonal no quadrado que não cruza a curva $\gamma$, resultando que A e B são partes do quadrado que não estão </span><span style="font-size: large;">separadas pela curva $\gamma$.</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgf5qIyqolXpovi3JZ6B4Bebt91hENFWl8vME3GIK2meWtiJAG0xaSjLjJFCRbQK4n_TRjrKxw_KfS6JDwuqdpU7Xevy77OQV8qFPMHCccStSrvSFAhlyyGJtjozvjCfnO51iVGjAGfYro/s1600/isoperimetria.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="227" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgf5qIyqolXpovi3JZ6B4Bebt91hENFWl8vME3GIK2meWtiJAG0xaSjLjJFCRbQK4n_TRjrKxw_KfS6JDwuqdpU7Xevy77OQV8qFPMHCccStSrvSFAhlyyGJtjozvjCfnO51iVGjAGfYro/s400/isoperimetria.jpg" width="400" /></a></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;">Hoje, trago um problema do tipo "desigualdade isoperimétrica".</span></div>
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<span style="font-size: large;"><br /></span></div>
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<span style="font-size: large;"><b>Problema.</b><i> Se uma curva divide o quadrado unitário em dois pedaços, então o comprimento da curva é maior do que ou igual ao valor da área de um desses pedaços.</i></span></div>
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<span style="font-size: large;"><i><br /></i></span></div>
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<span style="font-size: large;">De fato, apesar de apresentar a proposição acima como um problema do tipo "desigualdade isoperimétrica", ainda não é claro para mim que essa proposição decorre de algum teorema clássico de desigualdade isoperimétrica. Isso, eu realmente gostaria de saber. Ideias são muito bem-vindas!</span></div>
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<span style="font-size: large;">Na próxima postagem, trarei uma solução bacaninha para o nosso problema de hoje. Espero! </span></div>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0tag:blogger.com,1999:blog-333163066896513113.post-82400622996855187132012-04-30T20:42:00.000-07:002012-05-01T11:05:48.560-07:00Teorema de Sylvester e Gallai<br />
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<i><span style="font-size: large;">"Seja S um conjunto finito de pontos no plano euclidiano. Suponha que cada reta que passa por dois pontos de S, necessariamente, contém um terceiro ponto de S. Então, os pontos de S estão sobre uma mesma reta."</span></i></div>
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<span style="font-size: large;">O problema acima foi proposto na postagem anterior e, como prometido, hoje trago uma solução para ele. De fato, o primeiro </span><span style="font-size: large;">matemático a propor esse problema foi James Joseph Sylvester em 1893. </span><br />
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<tr><td style="text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvZ0ZtO4vq_6ptXaW3yEj_oLTyHVJbd5mptmQeXha6rza7RS4DrREwdlSDHURkpp4cMS97dfSrGxIxDHOcB8MiFLqF0O_3lOVFDEZO-vou7qo13-7ERvfpJBxXQYCOXhLLmOaYa5D5QiM/s1600/fig1-25.gif" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjvZ0ZtO4vq_6ptXaW3yEj_oLTyHVJbd5mptmQeXha6rza7RS4DrREwdlSDHURkpp4cMS97dfSrGxIxDHOcB8MiFLqF0O_3lOVFDEZO-vou7qo13-7ERvfpJBxXQYCOXhLLmOaYa5D5QiM/s1600/fig1-25.gif" /></a></td></tr>
<tr><td class="tr-caption" style="text-align: center;">James Joseph Sylvester</td></tr>
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<span style="font-size: large;">Acredita-se que Sylvester </span><span style="font-size: large;">não conhecia uma solução desse problema. A primeira prova desse resultado veio algumas décadas depois que Sylvester o </span><span style="font-size: large;">propôs e é devida a Tibor Gallai. Abaixo, apresentamos a prova de Leroy Milton Kelly para o Teorema de Sylvester e Gallai.</span></div>
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<span style="font-size: large;"><b>Prova. </b></span><span style="font-size: large; text-align: justify;">Suponha que os pontos de <i>S</i> não estejam sobre uma única reta. Seja <i>L(S)</i> o conjunto de todas as retas que passam por dois pontos de <i>S</i>. Sejam <i>p</i> ponto de <i>S</i> e <i>r</i> reta de <i>L(S)</i> tais que a reta <i>r</i> não contém <i>p</i> e a </span><span style="font-size: large; text-align: justify;">distância de <i>p</i> a <i>r</i> seja a menor possível dentre tais pares. Por hipótese, a reta <i>r</i> contém 3 pontos de S. Digamos A, B e C (com B entre A e C). Sem perda de generalidade, podemos </span><span style="font-size: large; text-align: justify;">supor que o pé da perpendicular por <i>p</i> e <i>r</i>, o qual denotamos por Q, não está entre C e B (como na figura abaixo). </span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioPLOAIbRlK1pjpNc5zOyocwkYzhguCxeV_BYUXlnkQBhBGk7o7ibdhyphenhyphenJKrCECY_gkOVJsQhiV49MlpR97SGf5hgcEt-810WJ8-g42MC9R39eUjGSacFF0N2SVkNkEepzVpJLhJx43Y3Q/s1600/kelly-proof+(2).jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEioPLOAIbRlK1pjpNc5zOyocwkYzhguCxeV_BYUXlnkQBhBGk7o7ibdhyphenhyphenJKrCECY_gkOVJsQhiV49MlpR97SGf5hgcEt-810WJ8-g42MC9R39eUjGSacFF0N2SVkNkEepzVpJLhJx43Y3Q/s1600/kelly-proof+(2).jpg" /></a></div>
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<span style="font-size: large;">Seja <i>t</i> a reta que passa por P e C (reta azul na figura acima). Denotemos por D o pé da perpendicular por B e a reta <i>t</i>. Assim, <i>t </i> é uma reta de <i>L(S)</i> e B é um ponto de <i>S </i>que não pertence à reta <i>t.</i> Por outro lado, desde que os triângulos PQC e BDC são semelhantes, vemos que a distância de B à reta <i>t</i> é menor do que a </span><span style="font-size: large;">distância de P à reta <i>r</i>. O que é um absurdo.</span><br />
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<span style="font-size: large;">(Fonte: "<i>As provas estão n'O LIVRO</i>" de Aigner-Ziegler ed. Edgar Blucher LTDA).</span></div>
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</div>Alexandre Fernandeshttp://www.blogger.com/profile/11783442988444002636noreply@blogger.com0