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		<title>Desigualdade complexa útil na majoração de certos integrais de contorno</title>
		<link>http://problemasteoremas.wordpress.com/2012/01/21/desigualdade-complexa-util-na-majoracao-de-certos-integrais-de-contorno/</link>
		<comments>http://problemasteoremas.wordpress.com/2012/01/21/desigualdade-complexa-util-na-majoracao-de-certos-integrais-de-contorno/#comments</comments>
		<pubDate>Sat, 21 Jan 2012 16:57:51 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Desigualdades matemáticas]]></category>
		<category><![CDATA[Exercícios Matemáticos]]></category>
		<category><![CDATA[Matemática]]></category>
		<category><![CDATA[Problemas]]></category>
		<category><![CDATA[Exercícios]]></category>
		<category><![CDATA[Problema]]></category>

		<guid isPermaLink="false">http://problemasteoremas.wordpress.com/?p=15475</guid>
		<description><![CDATA[No post anterior, ao aplicar o teorema dos resíduos, utilizei a desigualdade que é uma aplicação de duas desigualdades particulares do tipo em que e são dois complexos quaisquer cuja soma não seja nula. Como exercício, proponho a demonstração de &#8230; <a href="http://problemasteoremas.wordpress.com/2012/01/21/desigualdade-complexa-util-na-majoracao-de-certos-integrais-de-contorno/">Continuar a ler <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&amp;blog=1866481&amp;post=15475&amp;subd=problemasteoremas&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p style="text-align:justify;">No post <a href="http://problemasteoremas.wordpress.com/2011/10/13/integral-real-calculado-pelo-metodo-dos-residuos-com-aplicacao-do-lema-de-jordan/">anterior</a>, ao aplicar o teorema dos resíduos, utilizei a<br />
desigualdade</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdfrac%7B1%7D%7B%5Cleft%28+z%5E%7B2%7D%2B1%5Cright%29+%5E%7B2%7D%7D%3D%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+z-i%5Cright%5Cvert%5E%7B2%7D%5Cleft%5Cvert+z%2Bi%5Cright%5Cvert+%5E%7B2%7D%7D%5Cleq+%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+%5Cleft%5Cvert+z%5Cright%5Cvert+-%5Cleft%5Cvert+i%5Cright%5Cvert+%5Cright%5Cvert+%5E%7B2%7D%7D%5Ctimes+%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+%5Cleft%5Cvert+z%5Cright%5Cvert+-%5Cleft%5Cvert+-i%5Cright%5Cvert+%5Cright%5Cvert+%5E%7B2%7D%7D%3D%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+%5Cleft%5Cvert+z%5Cright%5Cvert+-1%5Cright%5Cvert+%5E%7B4%7D%7D%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;dfrac{1}{&#92;left( z^{2}+1&#92;right) ^{2}}=&#92;dfrac{1}{&#92;left&#92;vert z-i&#92;right&#92;vert^{2}&#92;left&#92;vert z+i&#92;right&#92;vert ^{2}}&#92;leq &#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert i&#92;right&#92;vert &#92;right&#92;vert ^{2}}&#92;times &#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert -i&#92;right&#92;vert &#92;right&#92;vert ^{2}}=&#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -1&#92;right&#92;vert ^{4}},' title='&#92;dfrac{1}{&#92;left( z^{2}+1&#92;right) ^{2}}=&#92;dfrac{1}{&#92;left&#92;vert z-i&#92;right&#92;vert^{2}&#92;left&#92;vert z+i&#92;right&#92;vert ^{2}}&#92;leq &#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert i&#92;right&#92;vert &#92;right&#92;vert ^{2}}&#92;times &#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert -i&#92;right&#92;vert &#92;right&#92;vert ^{2}}=&#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -1&#92;right&#92;vert ^{4}},' class='latex' /></p>
<p>que é uma aplicação de duas desigualdades particulares do tipo</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+z%2Bw%5Cright%5Cvert%7D%5Cleq%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+%5Cleft%5Cvert+z%5Cright%5Cvert+-%5Cleft%5Cvert+w%5Cright%5Cvert+%5Cright%5Cvert+%7D%5Cqquad+%28%5Cast%29%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;dfrac{1}{&#92;left&#92;vert z+w&#92;right&#92;vert}&#92;leq&#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert w&#92;right&#92;vert &#92;right&#92;vert }&#92;qquad (&#92;ast),' title='&#92;dfrac{1}{&#92;left&#92;vert z+w&#92;right&#92;vert}&#92;leq&#92;dfrac{1}{&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert w&#92;right&#92;vert &#92;right&#92;vert }&#92;qquad (&#92;ast),' class='latex' /></p>
<p style="text-align:justify;">em que <img src='http://s0.wp.com/latex.php?latex=z&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z' title='z' class='latex' /> e <img src='http://s0.wp.com/latex.php?latex=w&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='w' title='w' class='latex' /> são dois complexos quaisquer cuja soma não seja nula. Como exercício, proponho a demonstração de <img src='http://s0.wp.com/latex.php?latex=%28%5Cast%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(&#92;ast)' title='(&#92;ast)' class='latex' />,  sugerindo que parta da desigualdade triangular</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cleft%5Cvert+z%2Bw%5Cright%5Cvert+%5Cleq+%5Cleft%5Cvert+z%5Cright%5Cvert+%2B%5Cleft%5Cvert+w%5Cright%5Cvert&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;left&#92;vert z+w&#92;right&#92;vert &#92;leq &#92;left&#92;vert z&#92;right&#92;vert +&#92;left&#92;vert w&#92;right&#92;vert' title='&#92;left&#92;vert z+w&#92;right&#92;vert &#92;leq &#92;left&#92;vert z&#92;right&#92;vert +&#92;left&#92;vert w&#92;right&#92;vert' class='latex' /></p>
<p>e obtenha a desigualdade</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cleft%5Cvert+z%2Bw%5Cright%5Cvert+%5Cgeq+%5Cleft%5Cvert+%5Cleft%5Cvert+z%5Cright%5Cvert+-%5Cleft%5Cvert+w%5Cright%5Cvert+%5Cright%5Cvert+.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;left&#92;vert z+w&#92;right&#92;vert &#92;geq &#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert w&#92;right&#92;vert &#92;right&#92;vert .' title='&#92;left&#92;vert z+w&#92;right&#92;vert &#92;geq &#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert w&#92;right&#92;vert &#92;right&#92;vert .' class='latex' /></p>
<br />Filed under: <a href='http://problemasteoremas.wordpress.com/category/matematica/desigualdades-matematicas/'>Desigualdades matemáticas</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/exercicios-matematicos/'>Exercícios Matemáticos</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/'>Matemática</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/problemas/'>Problemas</a> Tagged: <a href='http://problemasteoremas.wordpress.com/tag/exercicios/'>Exercícios</a>, <a href='http://problemasteoremas.wordpress.com/tag/matematica/'>Matemática</a>, <a href='http://problemasteoremas.wordpress.com/tag/problema/'>Problema</a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/gocomments/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/comments/problemasteoremas.wordpress.com/15475/" /></a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/godelicious/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/delicious/problemasteoremas.wordpress.com/15475/" /></a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/gofacebook/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/facebook/problemasteoremas.wordpress.com/15475/" /></a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/gotwitter/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/twitter/problemasteoremas.wordpress.com/15475/" /></a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/gostumble/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/stumble/problemasteoremas.wordpress.com/15475/" /></a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/godigg/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/digg/problemasteoremas.wordpress.com/15475/" /></a> <a rel="nofollow" href="http://feeds.wordpress.com/1.0/goreddit/problemasteoremas.wordpress.com/15475/"><img alt="" border="0" src="http://feeds.wordpress.com/1.0/reddit/problemasteoremas.wordpress.com/15475/" /></a> <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&amp;blog=1866481&amp;post=15475&amp;subd=problemasteoremas&amp;ref=&amp;feed=1" width="1" height="1" />]]></content:encoded>
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		<title>Integral real calculado pelo método dos resíduos com aplicação do lema de Jordan</title>
		<link>http://problemasteoremas.wordpress.com/2011/10/13/integral-real-calculado-pelo-metodo-dos-residuos-com-aplicacao-do-lema-de-jordan/</link>
		<comments>http://problemasteoremas.wordpress.com/2011/10/13/integral-real-calculado-pelo-metodo-dos-residuos-com-aplicacao-do-lema-de-jordan/#comments</comments>
		<pubDate>Thu, 13 Oct 2011 15:10:33 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Análise Complexa]]></category>
		<category><![CDATA[Integrais]]></category>
		<category><![CDATA[Matemática]]></category>
		<category><![CDATA[Mathematics Stack Exchange]]></category>
		<category><![CDATA[Exercícios]]></category>
		<category><![CDATA[MSE]]></category>

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		<description><![CDATA[Nesta questão do MSE EMKA perguntou como calcular o integral real pelos métodos da Análise Complexa. Apresento a tradução da minha resposta. Seja O resíduo de em é Designemos  o contorno da metade superior do disco , percorrido no sentido &#8230; <a href="http://problemasteoremas.wordpress.com/2011/10/13/integral-real-calculado-pelo-metodo-dos-residuos-com-aplicacao-do-lema-de-jordan/">Continuar a ler <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&amp;blog=1866481&amp;post=15432&amp;subd=problemasteoremas&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p style="text-align:justify;"><a href="http://math.stackexchange.com/questions/71932/verify-integrals-with-residue-theorem">Nesta questão</a> do MSE <a href="http://math.stackexchange.com/users/8324/emka">EMKA</a> perguntou como calcular o integral real</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B%5Ccos+ax%7D%7B%5Cleft%28+1%2Bx%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddx%5Cqquad%5Ctext%7Bcom%7D%5Cquad+a%3E0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{&#92;infty }&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx&#92;qquad&#92;text{com}&#92;quad a&gt;0' title='&#92;displaystyle&#92;int_{0}^{&#92;infty }&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx&#92;qquad&#92;text{com}&#92;quad a&gt;0' class='latex' /></p>
<p>pelos métodos da Análise Complexa. Apresento a tradução da minha resposta.</p>
<p>Seja</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=f%28z%29%3D%5Cdfrac%7Be%5E%7Biaz%7D%7D%7B%5Cleft%281%2Bz%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7D%3D%5Cdfrac%7Be%5E%7Biaz%7D%7D%7B%28z-i%29%5E%7B2%7D%28z%2Bi%29%5E%7B2%7D%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f(z)=&#92;dfrac{e^{iaz}}{&#92;left(1+z^{2}&#92;right) ^{2}}=&#92;dfrac{e^{iaz}}{(z-i)^{2}(z+i)^{2}}.' title='f(z)=&#92;dfrac{e^{iaz}}{&#92;left(1+z^{2}&#92;right) ^{2}}=&#92;dfrac{e^{iaz}}{(z-i)^{2}(z+i)^{2}}.' class='latex' /></p>
<p>O <a href="http://en.wikipedia.org/wiki/Residue_%28complex_analysis%29#Limit_formula_for_higher_order_poles">resíduo de <img src='http://s0.wp.com/latex.php?latex=f%28z%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f(z)' title='f(z)' class='latex' /> em <img src='http://s0.wp.com/latex.php?latex=z%3Di&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z=i' title='z=i' class='latex' /></a> é</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Cunderset%7Bz%3Di%7D%7B%5Cmathrm%7Bres%7D%7Df%28z%29+%26%3D%5Cdfrac%7B1%7D%7B%282-1%29%21%7D%5Clim_%7Bz%5Crightarrow+i%7D%5Cdfrac%7Bd%7D%7Bdz%7D%5Cleft%28%28z-i%29%5E2f%28z%29%5Cright%29%5C%5C%26%3D%5Clim_%7Bz%5Crightarrow+i%7D%5Cdfrac%7Bd%7D%7Bdz%7D%5Cleft%28+%5Cdfrac%7Be%5E%7Biaz%7D%7D%7B%28z%2Bi%29%5E%7B2%7D%7D%5Cright%29%3D%5Clim_%7Bz%5Crightarrow+i%7D%5Cdfrac%7Biae%5E%7Biaz%7D%28z%2Bi%29%5E%7B2%7D-e%5E%7Biaz%7D2%5Cleft%28+z%2Bi%5Cright%29+%7D%7B%28z%2Bi%29%5E%7B4%7D%7D+%5C%5C%26%3D-%5Cdfrac%7B1%7D%7B4%7Di%5Cleft%28+a%2B1%5Cright%29+e%5E%7B-a%7D.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;underset{z=i}{&#92;mathrm{res}}f(z) &amp;=&#92;dfrac{1}{(2-1)!}&#92;lim_{z&#92;rightarrow i}&#92;dfrac{d}{dz}&#92;left((z-i)^2f(z)&#92;right)&#92;&#92;&amp;=&#92;lim_{z&#92;rightarrow i}&#92;dfrac{d}{dz}&#92;left( &#92;dfrac{e^{iaz}}{(z+i)^{2}}&#92;right)=&#92;lim_{z&#92;rightarrow i}&#92;dfrac{iae^{iaz}(z+i)^{2}-e^{iaz}2&#92;left( z+i&#92;right) }{(z+i)^{4}} &#92;&#92;&amp;=-&#92;dfrac{1}{4}i&#92;left( a+1&#92;right) e^{-a}.&#92;end{aligned}' title='&#92;begin{aligned}&#92;underset{z=i}{&#92;mathrm{res}}f(z) &amp;=&#92;dfrac{1}{(2-1)!}&#92;lim_{z&#92;rightarrow i}&#92;dfrac{d}{dz}&#92;left((z-i)^2f(z)&#92;right)&#92;&#92;&amp;=&#92;lim_{z&#92;rightarrow i}&#92;dfrac{d}{dz}&#92;left( &#92;dfrac{e^{iaz}}{(z+i)^{2}}&#92;right)=&#92;lim_{z&#92;rightarrow i}&#92;dfrac{iae^{iaz}(z+i)^{2}-e^{iaz}2&#92;left( z+i&#92;right) }{(z+i)^{4}} &#92;&#92;&amp;=-&#92;dfrac{1}{4}i&#92;left( a+1&#92;right) e^{-a}.&#92;end{aligned}' class='latex' /></p>
<p><a href="http://problemasteoremas.files.wordpress.com/2011/10/contourjordan-001.jpg"><img class="aligncenter size-medium wp-image-15449" title="contourjordan 001" src="http://problemasteoremas.files.wordpress.com/2011/10/contourjordan-001.jpg?w=490&#038;h=337" alt="" width="490" height="337" /></a></p>
<p>Designemos  o contorno da metade superior do disco <img src='http://s0.wp.com/latex.php?latex=%5Cleft%5Cvert+z%5Cright%5Cvert+%3DR&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;left&#92;vert z&#92;right&#92;vert =R' title='&#92;left&#92;vert z&#92;right&#92;vert =R' class='latex' />, percorrido no sentido contrário ao dos ponteiros do relógio (ver figura) por <img src='http://s0.wp.com/latex.php?latex=C_%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='C_{R}' title='C_{R}' class='latex' />. Pelo <a href="http://en.wikipedia.org/wiki/Residue_theorem">teorema dos resíduos</a></p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Cdisplaystyle%5Cint_%7B-R%7D%5E%7BR%7D%5Cdfrac%7Be%5E%7Biax%7D%7D%7B%5Cleft%28+1%2Bx%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddx%2B%5Cdisplaystyle%5Cint_%7BC_%7BR%7D%7D%5Cdfrac%7Be%5E%7Biaz%7D%7D%7B%5Cleft%28+1%2Bz%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddz+%26%3D2%5Cpi+i%5Cunderset%7Bz%3Di%7D%7B%5C+%5Cmathrm%7B+res%7D%7Df%28z%29e%5E%7Biaz%7D+%5C%5C%26%3D%5Cdfrac%7B1%7D%7B2%7D%5Cpi+%5Cleft%28+a%2B1%5Cright%29+e%5E%7B-a%7D.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;displaystyle&#92;int_{-R}^{R}&#92;dfrac{e^{iax}}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx+&#92;displaystyle&#92;int_{C_{R}}&#92;dfrac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right) ^{2}}dz &amp;=2&#92;pi i&#92;underset{z=i}{&#92; &#92;mathrm{ res}}f(z)e^{iaz} &#92;&#92;&amp;=&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}.&#92;end{aligned}' title='&#92;begin{aligned}&#92;displaystyle&#92;int_{-R}^{R}&#92;dfrac{e^{iax}}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx+&#92;displaystyle&#92;int_{C_{R}}&#92;dfrac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right) ^{2}}dz &amp;=2&#92;pi i&#92;underset{z=i}{&#92; &#92;mathrm{ res}}f(z)e^{iaz} &#92;&#92;&amp;=&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}.&#92;end{aligned}' class='latex' /></p>
<p>Assim,</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Ctext%7BRe%7D%5Cdisplaystyle%5Cint_%7B-R%7D%5E%7BR%7D%5Cfrac%7Be%5E%7Biax%7D%7D%7B%5Cleft%28+1%2Bx%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddx%2B%5Ctext%7BRe%7D%5Cdisplaystyle%5Cint_%7BC_%7BR%7D%7D%5Cdfrac%7Be%5E%7Biaz%7D%7D%7B%5Cleft%28+1%2Bz%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddz+%26%3D%5Ctext%7BRe%7D%5Cdfrac%7B1%7D%7B2%7D%5Cpi+%5Cleft%28+a%2B1%5Cright%29+e%5E%7B-a%7D%2C+%5C%5C+%5C%5C%5Cdisplaystyle%5Cint_%7B-R%7D%5E%7BR%7D%5Cdfrac%7B%5Ccos+ax%7D%7B%5Cleft%28+1%2Bx%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddx%2B%5Ctext%7BRe%7D%5Cdisplaystyle%5Cint_%7BC_%7BR%7D%7D%5Cfrac%7Be%5E%7Biaz%7D%7D%7B%5Cleft%28+1%2Bz%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddz+%26%3D%5Cdfrac%7B1%7D%7B2%7D%5Cpi+%5Cleft%28+a%2B1%5Cright%29+e%5E%7B-a%7D.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;text{Re}&#92;displaystyle&#92;int_{-R}^{R}&#92;frac{e^{iax}}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx+&#92;text{Re}&#92;displaystyle&#92;int_{C_{R}}&#92;dfrac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right) ^{2}}dz &amp;=&#92;text{Re}&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}, &#92;&#92; &#92;&#92;&#92;displaystyle&#92;int_{-R}^{R}&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx+&#92;text{Re}&#92;displaystyle&#92;int_{C_{R}}&#92;frac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right) ^{2}}dz &amp;=&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}.&#92;end{aligned}' title='&#92;begin{aligned}&#92;text{Re}&#92;displaystyle&#92;int_{-R}^{R}&#92;frac{e^{iax}}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx+&#92;text{Re}&#92;displaystyle&#92;int_{C_{R}}&#92;dfrac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right) ^{2}}dz &amp;=&#92;text{Re}&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}, &#92;&#92; &#92;&#92;&#92;displaystyle&#92;int_{-R}^{R}&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx+&#92;text{Re}&#92;displaystyle&#92;int_{C_{R}}&#92;frac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right) ^{2}}dz &amp;=&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}.&#92;end{aligned}' class='latex' /></p>
<p style="text-align:justify;">Quando <img src='http://s0.wp.com/latex.php?latex=%5Cleft%5Cvert+z%5Cright%5Cvert+%3DR&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;left&#92;vert z&#92;right&#92;vert =R' title='&#92;left&#92;vert z&#92;right&#92;vert =R' class='latex' />, verifica-se (veja identidade mais geral <a href="http://problemasteoremas.wordpress.com/2012/01/21/desigualdade-complexa-util-na-majoracao-de-certos-integrais-de-contorno/">aqui</a>.)</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cleft%5Cvert+%5Cdfrac%7B1%7D%7B%281%2Bz%5E%7B2%7D%29%5E%7B2%7D%7D%5Cright%5Cvert+%3D%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert+z%2Bi%5Cright%5Cvert+%5E%7B2%7D%5Cleft%5Cvert+z-i%5Cright%5Cvert+%5E%7B2%7D%7D%5Cleq+%5Cdfrac%7B1%7D%7B%5Cleft%5Cvert%5Cleft%5Cvert+z%5Cright%5Cvert+-%7Ci%5Cright%5Cvert+%5E%7B2%7D%5Cleft%5Cvert+%5Cleft%5Cvert+z%5Cright%5Cvert+-%5Cleft%5Cvert+i%5Cright%5Cvert+%5Cright%5Cvert+%5E%7B2%7D%7D%3D%5Cdfrac%7B1%7D%7B%28R-1%29+%5E%7B4%7D%7D%3D%3AM_R%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;left&#92;vert &#92;dfrac{1}{(1+z^{2})^{2}}&#92;right&#92;vert =&#92;dfrac{1}{&#92;left&#92;vert z+i&#92;right&#92;vert ^{2}&#92;left&#92;vert z-i&#92;right&#92;vert ^{2}}&#92;leq &#92;dfrac{1}{&#92;left&#92;vert&#92;left&#92;vert z&#92;right&#92;vert -|i&#92;right&#92;vert ^{2}&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert i&#92;right&#92;vert &#92;right&#92;vert ^{2}}=&#92;dfrac{1}{(R-1) ^{4}}=:M_R,' title='&#92;left&#92;vert &#92;dfrac{1}{(1+z^{2})^{2}}&#92;right&#92;vert =&#92;dfrac{1}{&#92;left&#92;vert z+i&#92;right&#92;vert ^{2}&#92;left&#92;vert z-i&#92;right&#92;vert ^{2}}&#92;leq &#92;dfrac{1}{&#92;left&#92;vert&#92;left&#92;vert z&#92;right&#92;vert -|i&#92;right&#92;vert ^{2}&#92;left&#92;vert &#92;left&#92;vert z&#92;right&#92;vert -&#92;left&#92;vert i&#92;right&#92;vert &#92;right&#92;vert ^{2}}=&#92;dfrac{1}{(R-1) ^{4}}=:M_R,' class='latex' /></p>
<p style="text-align:justify;">o que significa que <img src='http://s0.wp.com/latex.php?latex=M_R%3E0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='M_R&gt;0' title='M_R&gt;0' class='latex' /> e</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Clim_%7BR%5Crightarrow+%5Cinfty+%7DM_R%3D+%5Clim_%7BR%5Crightarrow+%5Cinfty+%7D%5Cdfrac%7B1%7D%7B%28R-1%29+%5E%7B4%7D%7D%3D0.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;lim_{R&#92;rightarrow &#92;infty }M_R= &#92;lim_{R&#92;rightarrow &#92;infty }&#92;dfrac{1}{(R-1) ^{4}}=0.' title='&#92;lim_{R&#92;rightarrow &#92;infty }M_R= &#92;lim_{R&#92;rightarrow &#92;infty }&#92;dfrac{1}{(R-1) ^{4}}=0.' class='latex' /></p>
<p style="text-align:justify;">Podemos aplicar o <a href="http://en.wikipedia.org/wiki/Jordan_lemma">lema de Jordan</a>, qualquer que seja a constante positiva <img src='http://s0.wp.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' />, e concluir que</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Clim_%7BR%5Crightarrow+%5Cinfty+%7D%5Cdisplaystyle%5Cint_%7BC_%7BR%7D%7D%5Cdfrac%7Be%5E%7Biaz%7D%7D%7B%5Cleft%28+1%2Bz%5E%7B2%7D%5Cright%29%5E%7B2%7D%7Ddz%3D0.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;lim_{R&#92;rightarrow &#92;infty }&#92;displaystyle&#92;int_{C_{R}}&#92;dfrac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right)^{2}}dz=0.' title='&#92;lim_{R&#92;rightarrow &#92;infty }&#92;displaystyle&#92;int_{C_{R}}&#92;dfrac{e^{iaz}}{&#92;left( 1+z^{2}&#92;right)^{2}}dz=0.' class='latex' /></p>
<p style="text-align:justify;">Em consequência,</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B-%5Cinfty+%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B%5Ccos+ax%7D%7B%5Cleft%28+1%2Bx%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddx%3D%5Cdfrac%7B1%7D%7B2%7D%5Cpi+%5Cleft%28+a%2B1%5Cright%29+e%5E%7B-a%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{-&#92;infty }^{&#92;infty }&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx=&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}' title='&#92;displaystyle&#92;int_{-&#92;infty }^{&#92;infty }&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx=&#92;dfrac{1}{2}&#92;pi &#92;left( a+1&#92;right) e^{-a}' class='latex' /></p>
<p style="text-align:justify;">e</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B%5Ccos+ax%7D%7B%5Cleft%28+1%2Bx%5E%7B2%7D%5Cright%29+%5E%7B2%7D%7Ddx%3D%5Cdfrac%7B1%7D%7B4%7D%5Cpi+%5Cleft%28+a%2B1%5Cright%29+e%5E%7B-a%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{&#92;infty }&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx=&#92;dfrac{1}{4}&#92;pi &#92;left( a+1&#92;right) e^{-a}.' title='&#92;displaystyle&#92;int_{0}^{&#92;infty }&#92;dfrac{&#92;cos ax}{&#92;left( 1+x^{2}&#92;right) ^{2}}dx=&#92;dfrac{1}{4}&#92;pi &#92;left( a+1&#92;right) e^{-a}.' class='latex' /></p>
<p style="text-align:justify;"><em>Nota</em>: O exercício 3 da página 265 de<em> Complex Variables and Applications</em> por James Brown e Ruell Churchill generaliza este integral para</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B%5Ccos+ax%7D%7B%28+b%5E2%2Bx%5E%7B2%7D%29+%5E%7B2%7D%7Ddx%3D%5Cdfrac%7B1%7D%7B4b%5E3%7D%5Cpi+%5Cleft%28+ab%2B1%5Cright%29+e%5E%7B-ab%7D%5Cqquad+%28a%3E0%2Cb%3E0%29.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{&#92;infty }&#92;dfrac{&#92;cos ax}{( b^2+x^{2}) ^{2}}dx=&#92;dfrac{1}{4b^3}&#92;pi &#92;left( ab+1&#92;right) e^{-ab}&#92;qquad (a&gt;0,b&gt;0).' title='&#92;displaystyle&#92;int_{0}^{&#92;infty }&#92;dfrac{&#92;cos ax}{( b^2+x^{2}) ^{2}}dx=&#92;dfrac{1}{4b^3}&#92;pi &#92;left( ab+1&#92;right) e^{-ab}&#92;qquad (a&gt;0,b&gt;0).' class='latex' /></p>
<p style="text-align:justify;">
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			<media:title type="html">ATavares</media:title>
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			<media:title type="html">contourjordan 001</media:title>
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		<title>Determinação de uma matriz que transforma um vector noutro</title>
		<link>http://problemasteoremas.wordpress.com/2011/10/03/determinacao-de-uma-matriz-que-transforma-um-vector-noutro/</link>
		<comments>http://problemasteoremas.wordpress.com/2011/10/03/determinacao-de-uma-matriz-que-transforma-um-vector-noutro/#comments</comments>
		<pubDate>Mon, 03 Oct 2011 12:08:55 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Álgebra linear]]></category>
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		<guid isPermaLink="false">http://problemasteoremas.wordpress.com/?p=15386</guid>
		<description><![CDATA[Nesta questão de Miro, no MSE, é dado um vector e pretende-se determinar uma matriz tal que . A matriz deverá rodar . Tradução da minha resposta: o seu problema é equivalente a determinar a transformação entre as coordenadas de &#8230; <a href="http://problemasteoremas.wordpress.com/2011/10/03/determinacao-de-uma-matriz-que-transforma-um-vector-noutro/">Continuar a ler <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&amp;blog=1866481&amp;post=15386&amp;subd=problemasteoremas&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p style="text-align:justify;"><a href="http://math.stackexchange.com/q/63503/752">Nesta questão</a> de <a href="http://math.stackexchange.com/users/11319/miro">Miro</a>, no MSE, é dado um vector <img src='http://s0.wp.com/latex.php?latex=u%3D%28x%2Cy%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='u=(x,y)' title='u=(x,y)' class='latex' /> e pretende-se determinar uma matriz <img src='http://s0.wp.com/latex.php?latex=M&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='M' title='M' class='latex' /> tal que <img src='http://s0.wp.com/latex.php?latex=Mu%3D%281%2C0%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='Mu=(1,0)' title='Mu=(1,0)' class='latex' />. A matriz deverá rodar <img src='http://s0.wp.com/latex.php?latex=u&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='u' title='u' class='latex' />.</p>
<p style="text-align:justify;">Tradução da <a href="http://math.stackexchange.com/questions/63503/how-to-create-2x2-matrix-to-rotate-vector-to-right-side/63536#63536">minha resposta</a>: o seu problema é equivalente a determinar a transformação entre as coordenadas <img src='http://s0.wp.com/latex.php?latex=x%2Cy&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x,y' title='x,y' class='latex' /> de um ponto e as coordenadas <img src='http://s0.wp.com/latex.php?latex=x%27%2Cy%27&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x&#039;,y&#039;' title='x&#039;,y&#039;' class='latex' /> do mesmo ponto num sistema de coordenadas rodado em relação ao inicial, seguida da multiplicação pelo factor  <img src='http://s0.wp.com/latex.php?latex=k%3D1%2F%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k=1/&#92;sqrt{x^{2}+y^{2}}' title='k=1/&#92;sqrt{x^{2}+y^{2}}' class='latex' />, de tal maneira que  <img src='http://s0.wp.com/latex.php?latex=x%27%27%3Dkx%27%3D1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x&#039;&#039;=kx&#039;=1' title='x&#039;&#039;=kx&#039;=1' class='latex' /> and <img src='http://s0.wp.com/latex.php?latex=y%27%27%3Dkx%27%3D0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y&#039;&#039;=kx&#039;=0' title='y&#039;&#039;=kx&#039;=0' class='latex' />. O ângulo de rotação deve ser <img src='http://s0.wp.com/latex.php?latex=%5Ctheta+%3D%5Carctan+%5Cdfrac%7By%7D%7Bx%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;theta =&#92;arctan &#92;dfrac{y}{x}' title='&#92;theta =&#92;arctan &#92;dfrac{y}{x}' class='latex' /> (ver figura).</p>
<p><img class="aligncenter" src="http://i.stack.imgur.com/fMxsi.jpg" alt="" width="542" height="400" /></p>
<p>Da trigonometria sabemos que</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7Dx%27%3Dx%5Ccos+%5Ctheta+%2By%5Csin+%5Ctheta+%3D%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%5C%5C+%5C%5Cy%27%3D-x%5Csin+%5Ctheta+%2By%5Ccos+%5Ctheta+%3D0%5Cend%7Barray%7D%5Cright.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;left&#92;{&#92;begin{array}{c}x&#039;=x&#92;cos &#92;theta +y&#92;sin &#92;theta =&#92;sqrt{x^{2}+y^{2}}&#92;&#92; &#92;&#92;y&#039;=-x&#92;sin &#92;theta +y&#92;cos &#92;theta =0&#92;end{array}&#92;right.&#92;end{aligned}' title='&#92;begin{aligned}&#92;left&#92;{&#92;begin{array}{c}x&#039;=x&#92;cos &#92;theta +y&#92;sin &#92;theta =&#92;sqrt{x^{2}+y^{2}}&#92;&#92; &#92;&#92;y&#039;=-x&#92;sin &#92;theta +y&#92;cos &#92;theta =0&#92;end{array}&#92;right.&#92;end{aligned}' class='latex' /></p>
<p>e como</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Ccos%5Cleft%28%5Carctan+%5Cdfrac%7By%7D%7Bx%7D%5Cright%29+%3D%5Cdfrac%7Bx%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;cos&#92;left(&#92;arctan &#92;dfrac{y}{x}&#92;right) =&#92;dfrac{x}{&#92;sqrt{x^{2}+y^{2}}}' title='&#92;cos&#92;left(&#92;arctan &#92;dfrac{y}{x}&#92;right) =&#92;dfrac{x}{&#92;sqrt{x^{2}+y^{2}}}' class='latex' /></p>
<p style="text-align:center;"> <img src='http://s0.wp.com/latex.php?latex=%5Csin%5Cleft%28%5Carctan%5Cdfrac%7By%7D%7Bx%7D%5Cright%29+%3D%5Cdfrac%7By%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7D%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sin&#92;left(&#92;arctan&#92;dfrac{y}{x}&#92;right) =&#92;dfrac{y}{&#92;sqrt{x^{2}+y^{2}}},' title='&#92;sin&#92;left(&#92;arctan&#92;dfrac{y}{x}&#92;right) =&#92;dfrac{y}{&#92;sqrt{x^{2}+y^{2}}},' class='latex' /></p>
<p>tem-se</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bc%7Dx%27%27%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7D%5C+x%27%3D%5Cdfrac%7Bx%5E%7B2%7D%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%2B%5Cdfrac%7By%5E%7B2%7D%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%3D1%5C%5Cy%27%27%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7D%5C+y%27%3D-%5Cdfrac%7Bxy%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%2B%5Cdfrac%7Bxy%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%3D0.%5Cend%7Barray%7D%5Cright.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;left&#92;{&#92;begin{array}{c}x&#039;&#039;=&#92;dfrac{1}{&#92;sqrt{x^{2}+y^{2}}}&#92; x&#039;=&#92;dfrac{x^{2}}{x^{2}+y^{2}}+&#92;dfrac{y^{2}}{x^{2}+y^{2}}=1&#92;&#92;y&#039;&#039;=&#92;dfrac{1}{&#92;sqrt{x^{2}+y^{2}}}&#92; y&#039;=-&#92;dfrac{xy}{x^{2}+y^{2}}+&#92;dfrac{xy}{x^{2}+y^{2}}=0.&#92;end{array}&#92;right.&#92;end{aligned}' title='&#92;begin{aligned}&#92;left&#92;{&#92;begin{array}{c}x&#039;&#039;=&#92;dfrac{1}{&#92;sqrt{x^{2}+y^{2}}}&#92; x&#039;=&#92;dfrac{x^{2}}{x^{2}+y^{2}}+&#92;dfrac{y^{2}}{x^{2}+y^{2}}=1&#92;&#92;y&#039;&#039;=&#92;dfrac{1}{&#92;sqrt{x^{2}+y^{2}}}&#92; y&#039;=-&#92;dfrac{xy}{x^{2}+y^{2}}+&#92;dfrac{xy}{x^{2}+y^{2}}=0.&#92;end{array}&#92;right.&#92;end{aligned}' class='latex' /></p>
<p>Em notação matricial</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Bpmatrix%7Dx%27%27%5C%5Cy%27%27%5Cend%7Bpmatrix%7D++%3D%5Cdfrac%7B1%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7D%5Cbegin%7Bpmatrix%7Dx%27%5C%5Cy%27++%5Cend%7Bpmatrix%7D%3D%5Cbegin%7Bpmatrix%7D++%5Cdfrac%7Bx%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%26%5Cdfrac%7By%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%5C%5C++-%5Cdfrac%7By%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%26%5Cdfrac%7Bx%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D++%5Cend%7Bpmatrix%7D%5Cbegin%7Bpmatrix%7Dx%5C%5Cy%5Cend%7Bpmatrix%7D%3D%5Cbegin%7Bpmatrix%7D1%5C%5C+0%5Cend%7Bpmatrix%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{pmatrix}x&#039;&#039;&#92;&#92;y&#039;&#039;&#92;end{pmatrix}  =&#92;dfrac{1}{&#92;sqrt{x^{2}+y^{2}}}&#92;begin{pmatrix}x&#039;&#92;&#92;y&#039;  &#92;end{pmatrix}=&#92;begin{pmatrix}  &#92;dfrac{x}{x^{2}+y^{2}}&amp;&#92;dfrac{y}{x^{2}+y^{2}}&#92;&#92;  -&#92;dfrac{y}{x^{2}+y^{2}}&amp;&#92;dfrac{x}{x^{2}+y^{2}}  &#92;end{pmatrix}&#92;begin{pmatrix}x&#92;&#92;y&#92;end{pmatrix}=&#92;begin{pmatrix}1&#92;&#92; 0&#92;end{pmatrix}' title='&#92;begin{pmatrix}x&#039;&#039;&#92;&#92;y&#039;&#039;&#92;end{pmatrix}  =&#92;dfrac{1}{&#92;sqrt{x^{2}+y^{2}}}&#92;begin{pmatrix}x&#039;&#92;&#92;y&#039;  &#92;end{pmatrix}=&#92;begin{pmatrix}  &#92;dfrac{x}{x^{2}+y^{2}}&amp;&#92;dfrac{y}{x^{2}+y^{2}}&#92;&#92;  -&#92;dfrac{y}{x^{2}+y^{2}}&amp;&#92;dfrac{x}{x^{2}+y^{2}}  &#92;end{pmatrix}&#92;begin{pmatrix}x&#92;&#92;y&#92;end{pmatrix}=&#92;begin{pmatrix}1&#92;&#92; 0&#92;end{pmatrix}' class='latex' /></p>
<p>Assim</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=M%3D%5Cbegin%7Bpmatrix%7D%5Cdfrac%7Bx%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%26%5Cdfrac%7By%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%5C%5C-%5Cdfrac%7By%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%26%5Cdfrac%7Bx%7D%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D++%5Cend%7Bpmatrix%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='M=&#92;begin{pmatrix}&#92;dfrac{x}{x^{2}+y^{2}}&amp;&#92;dfrac{y}{x^{2}+y^{2}}&#92;&#92;-&#92;dfrac{y}{x^{2}+y^{2}}&amp;&#92;dfrac{x}{x^{2}+y^{2}}  &#92;end{pmatrix}.' title='M=&#92;begin{pmatrix}&#92;dfrac{x}{x^{2}+y^{2}}&amp;&#92;dfrac{y}{x^{2}+y^{2}}&#92;&#92;-&#92;dfrac{y}{x^{2}+y^{2}}&amp;&#92;dfrac{x}{x^{2}+y^{2}}  &#92;end{pmatrix}.' class='latex' /></p>
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			<media:title type="html">ATavares</media:title>
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		<title>Interpretação geométrica de ∬dx dy = ∬ r dα dr na transformação de coordenadas cartesianas em polares</title>
		<link>http://problemasteoremas.wordpress.com/2011/09/14/interpretacao-geometrica-de-%e2%88%acdx-dy-%e2%88%ac-r-d%ce%b1-dr-na-transformacao-de-coordenadas-cartesianas-em-polares/</link>
		<comments>http://problemasteoremas.wordpress.com/2011/09/14/interpretacao-geometrica-de-%e2%88%acdx-dy-%e2%88%ac-r-d%ce%b1-dr-na-transformacao-de-coordenadas-cartesianas-em-polares/#comments</comments>
		<pubDate>Wed, 14 Sep 2011 14:39:29 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Cálculo]]></category>
		<category><![CDATA[Integrais]]></category>
		<category><![CDATA[Matemática]]></category>
		<category><![CDATA[Mathematics Stack Exchange]]></category>
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		<description><![CDATA[Nesta minha resposta a uma questão de hhh, no MSE, apresentei a seguinte interpretação geométrica da transformação de coordenadas cartesianas em polares, no cálculo do integral duplo  do elemento de área, que traduzo. Por definição do integral duplo de uma &#8230; <a href="http://problemasteoremas.wordpress.com/2011/09/14/interpretacao-geometrica-de-%e2%88%acdx-dy-%e2%88%ac-r-d%ce%b1-dr-na-transformacao-de-coordenadas-cartesianas-em-polares/">Continuar a ler <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&amp;blog=1866481&amp;post=15338&amp;subd=problemasteoremas&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p style="text-align:justify;"><a href="http://math.stackexchange.com/questions/37044/explain-iint-mathrm-dx-mathrm-dy-iint-r-mathrm-d-alpha-mathrm-dr/37071#37071">Nesta minha resposta</a> a <a href="http://math.stackexchange.com/q/37044/752">uma questão</a> de <a href="http://math.stackexchange.com/users/5902/hhh">hhh</a>, no MSE, apresentei a seguinte interpretação geométrica da transformação de coordenadas cartesianas em polares, no cálculo do integral duplo  do elemento de área, que traduzo.</p>
<p style="text-align:justify;">Por definição do integral duplo de uma função contínua numa região  <img src='http://s0.wp.com/latex.php?latex=R&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='R' title='R' class='latex' /> fechada e limitada <img src='http://s0.wp.com/latex.php?latex=R&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='R' title='R' class='latex' />, no plano <img src='http://s0.wp.com/latex.php?latex=xy&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='xy' title='xy' class='latex' />, tem-se</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Ciint_R+f%28x%2Cy%29%5C%3B%5Cmathrm%7Bd%7Dx%5C%3B%5Cmathrm%7Bd%7Dy%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7Bn%7Df%28x_%7Bi%7D%2Cy_%7Bi%7D%29%5CDelta+A_%7Bi%7D%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;iint_R f(x,y)&#92;;&#92;mathrm{d}x&#92;;&#92;mathrm{d}y=&#92;displaystyle&#92;lim_{n&#92;to&#92;infty}&#92;displaystyle&#92;sum_{i=1}^{n}f(x_{i},y_{i})&#92;Delta A_{i},' title='&#92;displaystyle&#92;iint_R f(x,y)&#92;;&#92;mathrm{d}x&#92;;&#92;mathrm{d}y=&#92;displaystyle&#92;lim_{n&#92;to&#92;infty}&#92;displaystyle&#92;sum_{i=1}^{n}f(x_{i},y_{i})&#92;Delta A_{i},' class='latex' /></p>
<p>em que <img src='http://s0.wp.com/latex.php?latex=%5CDelta+A_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;Delta A_{i}' title='&#92;Delta A_{i}' class='latex' /> é a área de uma célula rectangular genérica e <img src='http://s0.wp.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> o número de células.</p>
<p>Se <img src='http://s0.wp.com/latex.php?latex=f%28x%2Cy%29%3D1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='f(x,y)=1' title='f(x,y)=1' class='latex' />, obtém-se a área de <img src='http://s0.wp.com/latex.php?latex=R&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='R' title='R' class='latex' /></p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Ciint_R%5Cmathrm%7Bd%7Dx%5C%3B%5Cmathrm%7Bd%7Dy%3D%5Cdisplaystyle%5Clim_%7Bn%5Cto%5Cinfty%7D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7Bn+%7D%5CDelta+A_%7Bi%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;iint_R&#92;mathrm{d}x&#92;;&#92;mathrm{d}y=&#92;displaystyle&#92;lim_{n&#92;to&#92;infty}&#92;displaystyle&#92;sum_{i=1}^{n }&#92;Delta A_{i}.' title='&#92;displaystyle&#92;iint_R&#92;mathrm{d}x&#92;;&#92;mathrm{d}y=&#92;displaystyle&#92;lim_{n&#92;to&#92;infty}&#92;displaystyle&#92;sum_{i=1}^{n }&#92;Delta A_{i}.' class='latex' /></p>
<p>Se decompusermos <img src='http://s0.wp.com/latex.php?latex=R&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='R' title='R' class='latex' /> em células com a forma de <a href="http://en.wikipedia.org/wiki/Circular_sector">sector circulares</a> definidos por dois arcos cuja diferença de raios é <img src='http://s0.wp.com/latex.php?latex=%5CDelta+r_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;Delta r_{i}' title='&#92;Delta r_{i}' class='latex' /> para a <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' />-ésima célula genérica e por  dois raios que fazem um ângulo <img src='http://s0.wp.com/latex.php?latex=%5CDelta+%5Ctheta+_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;Delta &#92;theta _{i}' title='&#92;Delta &#92;theta _{i}' class='latex' /> entre eles, a área da célula, utilizando a respectiva fórmula, é</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdfrac%7B1%7D%7B2%7D%5Cleft%5B+%5Cleft%28+r_%7Bi%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5CDelta+r_%7Bi%7D%5Cright%29+%5E%7B2%7D-%5Cleft%28+r_%7Bi%7D-%5Cdfrac%7B1%7D%7B2%7D%5CDelta+r_%7Bi%7D%5Cright%29+%5E%7B2%7D%5Cright%5D+%5CDelta+%5Ctheta_%7Bi%7D%3Dr_%7Bi%7D%5CDelta+r_%7Bi%7D%5CDelta%5Ctheta+_%7Bi%7D%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;dfrac{1}{2}&#92;left[ &#92;left( r_{i}+&#92;dfrac{1}{2}&#92;Delta r_{i}&#92;right) ^{2}-&#92;left( r_{i}-&#92;dfrac{1}{2}&#92;Delta r_{i}&#92;right) ^{2}&#92;right] &#92;Delta &#92;theta_{i}=r_{i}&#92;Delta r_{i}&#92;Delta&#92;theta _{i},' title='&#92;dfrac{1}{2}&#92;left[ &#92;left( r_{i}+&#92;dfrac{1}{2}&#92;Delta r_{i}&#92;right) ^{2}-&#92;left( r_{i}-&#92;dfrac{1}{2}&#92;Delta r_{i}&#92;right) ^{2}&#92;right] &#92;Delta &#92;theta_{i}=r_{i}&#92;Delta r_{i}&#92;Delta&#92;theta _{i},' class='latex' /></p>
<p>em que <img src='http://s0.wp.com/latex.php?latex=r_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r_{i}' title='r_{i}' class='latex' /> é o raio do ponto médio da célula. A mesma área de  <img src='http://s0.wp.com/latex.php?latex=R&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='R' title='R' class='latex' /> pode ser expressa por <img src='http://s0.wp.com/latex.php?latex=%5Clim_%7Bn%5Cto%5Cinfty%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dr_%7Bi%7D%5CDelta+r_%7Bi%7D%5CDelta+%5Ctheta+_%7Bi%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;lim_{n&#92;to&#92;infty}&#92;sum_{i=1}^{n}r_{i}&#92;Delta r_{i}&#92;Delta &#92;theta _{i}' title='&#92;lim_{n&#92;to&#92;infty}&#92;sum_{i=1}^{n}r_{i}&#92;Delta r_{i}&#92;Delta &#92;theta _{i}' class='latex' />, o que, por definição de um integral duplo, é igual a</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Ciint_R+r%5C%3B%5Cmathrm%7Bd%7Dr%5C%3B%5Cmathrm%7Bd%7D%5Ctheta.+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;iint_R r&#92;;&#92;mathrm{d}r&#92;;&#92;mathrm{d}&#92;theta. ' title='&#92;displaystyle&#92;iint_R r&#92;;&#92;mathrm{d}r&#92;;&#92;mathrm{d}&#92;theta. ' class='latex' /></p>
<p><img class="aligncenter" src="http://i.stack.imgur.com/8zoCC.jpg" alt="" width="400" height="365" /></p>
<p style="text-align:center;">Figura: <img src='http://s0.wp.com/latex.php?latex=i&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='i' title='i' class='latex' />-ésima célula genérica em coordenadas polares com a forma de um sector circular</p>
<p>Esta transformação é definida rigorosamente pelo valor absoluto do  <a href="http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant">jacobiano da transformação</a>  <img src='http://s0.wp.com/latex.php?latex=%5Cleft%5Cvert%5Cdfrac%7B%5Cpartial+%28x%2Cy%29%7D%7B%5Cpartial+%28r%2C%5Ctheta+%29%7D%5Cright%5Cvert+%3Dr&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;left&#92;vert&#92;dfrac{&#92;partial (x,y)}{&#92;partial (r,&#92;theta )}&#92;right&#92;vert =r' title='&#92;left&#92;vert&#92;dfrac{&#92;partial (x,y)}{&#92;partial (r,&#92;theta )}&#92;right&#92;vert =r' class='latex' /> do sistema de coordenadas cartesianas no de polares (<img src='http://s0.wp.com/latex.php?latex=x%3Dr%5Ccos+%5Ctheta+%2Cy%3Dr%5Csin%5Ctheta+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x=r&#92;cos &#92;theta ,y=r&#92;sin&#92;theta ' title='x=r&#92;cos &#92;theta ,y=r&#92;sin&#92;theta ' class='latex' />):</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Ciint_R%5Cmathrm%7Bd%7Dx%5C%3B%5Cmathrm%7Bd%7Dy%3D%5Cdisplaystyle%5Ciint_R%5Cleft%5Cvert%5Cdfrac%7B%5Cpartial+%28x%2Cy%29%7D%7B%5Cpartial+%28r%2C%5Ctheta+%29%7D%5Cright%5Cvert+%5C%3B%5Cmathrm%7Bd%7Dr%5C%3B%5Cmathrm%7Bd%7D%5Ctheta+%3D%5Cdisplaystyle%5Ciint_R+r%5C%3B%5Cmathrm%7Bd%7Dr%5C%3B%5Cmathrm%7Bd%7D%5Ctheta.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;iint_R&#92;mathrm{d}x&#92;;&#92;mathrm{d}y=&#92;displaystyle&#92;iint_R&#92;left&#92;vert&#92;dfrac{&#92;partial (x,y)}{&#92;partial (r,&#92;theta )}&#92;right&#92;vert &#92;;&#92;mathrm{d}r&#92;;&#92;mathrm{d}&#92;theta =&#92;displaystyle&#92;iint_R r&#92;;&#92;mathrm{d}r&#92;;&#92;mathrm{d}&#92;theta.' title='&#92;displaystyle&#92;iint_R&#92;mathrm{d}x&#92;;&#92;mathrm{d}y=&#92;displaystyle&#92;iint_R&#92;left&#92;vert&#92;dfrac{&#92;partial (x,y)}{&#92;partial (r,&#92;theta )}&#92;right&#92;vert &#92;;&#92;mathrm{d}r&#92;;&#92;mathrm{d}&#92;theta =&#92;displaystyle&#92;iint_R r&#92;;&#92;mathrm{d}r&#92;;&#92;mathrm{d}&#92;theta.' class='latex' /></p>
<p>Cálculo do determinante jacobiano:</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Cdfrac%7B%5Cpartial+%28x%2Cy%29%7D%7B%5Cpartial+%28r%2C%5Ctheta+%29%7D+%26%3D%5Cdet%5Cbegin%7Bpmatrix%7D%5Cpartial+x%2F%5Cpartial+r%26%5Cpartial+x%2F%5Cpartial+%5Ctheta%5C%5C%5Cpartial+y%2F%5Cpartial+r%26%5Cpartial+y%2F%5Cpartial%5Ctheta%5Cend%7Bpmatrix%7D%5C%5C%26%3D%5Cdet%5Cbegin%7Bpmatrix%7D%5Ccos+%5Ctheta+%26+-r%5Csin+%5Ctheta%5C%5C%5Csin%5Ctheta+%26+r%5Ccos+%5Ctheta%5Cend%7Bpmatrix%7D%5C%5C%26%3Dr%5Ccos+%5E%7B2%7D%5Ctheta+%2Br%5Csin+%5E%7B2%7D%5Ctheta%5C%5C%26%3Dr%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;dfrac{&#92;partial (x,y)}{&#92;partial (r,&#92;theta )} &amp;=&#92;det&#92;begin{pmatrix}&#92;partial x/&#92;partial r&amp;&#92;partial x/&#92;partial &#92;theta&#92;&#92;&#92;partial y/&#92;partial r&amp;&#92;partial y/&#92;partial&#92;theta&#92;end{pmatrix}&#92;&#92;&amp;=&#92;det&#92;begin{pmatrix}&#92;cos &#92;theta &amp; -r&#92;sin &#92;theta&#92;&#92;&#92;sin&#92;theta &amp; r&#92;cos &#92;theta&#92;end{pmatrix}&#92;&#92;&amp;=r&#92;cos ^{2}&#92;theta +r&#92;sin ^{2}&#92;theta&#92;&#92;&amp;=r&#92;end{aligned}' title='&#92;begin{aligned}&#92;dfrac{&#92;partial (x,y)}{&#92;partial (r,&#92;theta )} &amp;=&#92;det&#92;begin{pmatrix}&#92;partial x/&#92;partial r&amp;&#92;partial x/&#92;partial &#92;theta&#92;&#92;&#92;partial y/&#92;partial r&amp;&#92;partial y/&#92;partial&#92;theta&#92;end{pmatrix}&#92;&#92;&amp;=&#92;det&#92;begin{pmatrix}&#92;cos &#92;theta &amp; -r&#92;sin &#92;theta&#92;&#92;&#92;sin&#92;theta &amp; r&#92;cos &#92;theta&#92;end{pmatrix}&#92;&#92;&amp;=r&#92;cos ^{2}&#92;theta +r&#92;sin ^{2}&#92;theta&#92;&#92;&amp;=r&#92;end{aligned}' class='latex' /></p>
<p>&#8211;</p>
<p>Nota: Um raciocínio idêntico que o generaliza  permite intepretar geometricamente a transformação de coordenadas cartesianas em coordenadas esféricas, ao calcular o integral triplo  do elemento de volume.</p>
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			<media:title type="html">ATavares</media:title>
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		<title>Clube SPM &#8212; Entrevista ao Presidente da Sociedade Portuguesa de Matemática Miguel Abreu</title>
		<link>http://problemasteoremas.wordpress.com/2011/09/13/clube-spm-entrevista-ao-presidente-da-sociedade-portuguesa-de-matematica-miguel-abreu/</link>
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		<pubDate>Tue, 13 Sep 2011 12:27:37 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Divulgação]]></category>
		<category><![CDATA[Matemática]]></category>
		<category><![CDATA[Notícia]]></category>
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		<description><![CDATA[(Fonte) Entrevista a Miguel Abreu (Professor Catedrático  do Departamento de Matemática IST ): « (&#8230;) O gosto pela matemática começou quando? A partir daí foi em exponencial… Começou na escola. Tinha jeito e é fácil gostar daquilo que conseguimos fazer &#8230; <a href="http://problemasteoremas.wordpress.com/2011/09/13/clube-spm-entrevista-ao-presidente-da-sociedade-portuguesa-de-matematica-miguel-abreu/">Continuar a ler <span class="meta-nav">&#8594;</span></a><img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&amp;blog=1866481&amp;post=15322&amp;subd=problemasteoremas&amp;ref=&amp;feed=1" width="1" height="1" />]]></description>
			<content:encoded><![CDATA[<p style="text-align:center;"><img class="aligncenter" src="http://www.spm.pt/files/album/miguelabreu.jpg" alt="" width="150" height="150" /></p>
<p style="text-align:center;">(<a href="http://www.spm.pt/miguel_abreu/">Fonte</a>)</p>
<p><a href="http://www.clube.spm.pt/arquivo/641/">Entrevista</a> a <a href="http://www.spm.pt/miguel_abreu/">Miguel Abreu</a> (<a href="http://www.math.ist.utl.pt/%7Emabreu/">Professor Catedrático  do Departamento de Matemática IST </a>):</p>
<blockquote>
<p style="text-align:justify;"><span style="color:#ff0000;"><strong>«</strong></span> <a href="http://www.clube.spm.pt/arquivo/641/">(&#8230;)</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
<strong>O gosto pela matemática começou quando? A partir daí foi em exponencial…</strong></a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
Começou na escola. Tinha jeito e é fácil gostar daquilo que conseguimos fazer bem. Fui aluno no Colégio Militar e tive lá, durante os três anos do ensino secundário, um professor de matemática fantástico: o Dr. José Sena Neves. Marcou-me muito, tanto a nível pessoal como matemático. Ficámos amigos e continuamos a falar regularmente. Depois entrei para engenharia electrotécnica no Instituto Superior Técnico (IST) e ao fim de um ano e meio percebi que só gostava das cadeiras de matemática. Tive a sorte de poder mudar para o curso de matemática que tinha acabado de abrir no IST e a partir daí &#8220;foi em exponencial&#8221;&#8230;</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
(&#8230;)</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
<strong>Lecciona no IST. O que ensina em concreto?</strong></a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
Ensino as cadeiras de cálculo e álgebra comuns à quase totalidade dos cursos do IST, bem como cadeiras de geometria para os alunos de matemática. Nos últimos anos tenho dado frequentemente a cadeira de Cálculo Diferencial e Integral I. Tenho tido assim o privilégio de ensinar aos caloiros do IST o Teorema Fundamental do Cálculo, certamente um dos teoremas mais bonitos e importantes, tanto para a matemática como para toda a ciência em geral.</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
<strong>Faz investigação Matemática em Geometria e Topologia Simplética. Consegue explicar devagarinho o que é ou será melhor passarmos à próxima pergunta?</strong></a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
A Geometria e Topologia Simplética tem origem na Física, mais precisamente nos espaços de fase e transformações canónicas da Mecânica Clássica. Estuda propriedades de generalizações desses espaços e transformações. No caso mais simples de espaços de dimensão 2, que inclui por exemplo o plano e a superfície de uma esfera, a Geometria e Topologia Simpléctica estuda propriedades das transformações destes espaços que preservam área.</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
(&#8230;)</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
<strong>O que faz a &#8220;SPM &#8212; Sociedade Portuguesa de Matemática&#8221; em concreto?</strong></a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
A SPM faz o que está especificado nos seus estatutos: promover o ensino, investigação e divulgação da matemática. Exemplos concretos na área do ensino são a formação de professores, acreditação de manuais escolares e análise das provas nacionais de avaliação. Na investigação, a SPM promove, apoia e divulga a organização de encontros científicos e contribui para a representação da comunidade matemática portuguesa em organizações internacionais, como a União Matemática Internacional e a Sociedade Europeia de Matemática. Na divulgação, temos por exemplo as Tardes de Matemática e o Clube de Matemática da SPM. Temos também actividades transversais a mais do que uma destas vertentes, como as Olimpíadas de Matemática, os Encontros Nacionais e as Escolas de Verão. Temos 3 publicações periódicas (Boletim, Gazeta e Jornal de Matemática Elementar) e lançamos regularmente livros. Apoiamos também as actividades do Seminário Nacional de História de Matemática que é uma secção autónoma da SPM.</a></p>
<p style="text-align:justify;"><a href="http://www.clube.spm.pt/arquivo/641/"><br />
(&#8230;)</a> <span style="color:#ff0000;"><strong>»</strong></span></p>
</blockquote>
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