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		<title>Integral definido de um radical imbricado</title>
		<link>http://problemasteoremas.wordpress.com/2013/05/24/integral-definido-de-um-radical-imbricado/</link>
		<comments>http://problemasteoremas.wordpress.com/2013/05/24/integral-definido-de-um-radical-imbricado/#comments</comments>
		<pubDate>Fri, 24 May 2013 15:56:03 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Cálculo]]></category>
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		<description><![CDATA[Na questão Integral with 4 radicals-hat, no MSE, Chris&#8217;s wise sister apresenta o seguinte integral com radicais, pedindo o seu cálculo, bem como uma generalização Tradução da minha resposta: 1. Seja Visto que tem-se Cada um dos termos pode integrar-se &#8230; <a href="http://problemasteoremas.wordpress.com/2013/05/24/integral-definido-de-um-radical-imbricado/">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&#038;blog=1866481&#038;post=16927&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Na questão <a href="http://math.stackexchange.com/questions/151665/integral-with-4-radicals-hat">Integral with 4 radicals-hat</a>, no MSE, <a href="http://math.stackexchange.com/users/32016/chriss-wise-sister">Chris&#8217;s wise sister</a> apresenta o seguinte integral com <img src='http://s0.wp.com/latex.php?latex=k%3D4&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k=4' title='k=4' class='latex' /> radicais, pedindo o seu cálculo, bem como uma generalização</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Csqrt%7B1%2B%5Csqrt%7B1+%2B+%7B%5Csqrt%7B1%2B+%5Csqrt%7Bx%7D%7D%7D%7D%7D%5C%2C+dx+.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{1}&#92;sqrt{1+&#92;sqrt{1 + {&#92;sqrt{1+ &#92;sqrt{x}}}}}&#92;, dx .' title='&#92;displaystyle&#92;int_{0}^{1}&#92;sqrt{1+&#92;sqrt{1 + {&#92;sqrt{1+ &#92;sqrt{x}}}}}&#92;, dx .' class='latex' /></p>
<p>Tradução da minha <a href="http://math.stackexchange.com/a/154305/752">resposta</a>:</p>
<p>1. Seja</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7Du%26%3D%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7Bx%7D%7D%7D+%5CLeftrightarrow+%26x%3D%5Cleft%28+%5Cleft%28+u%5E%7B2%7D-1%5Cright%29+%5E%7B2%7D-1%5Cright%29++%5E%7B2%7D%3Du%5E%7B8%7D-4u%5E%7B6%7D%2B4u%5E%7B4%7D.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}u&amp;=&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}} &#92;Leftrightarrow &amp;x=&#92;left( &#92;left( u^{2}-1&#92;right) ^{2}-1&#92;right)  ^{2}=u^{8}-4u^{6}+4u^{4}.&#92;end{aligned}' title='&#92;begin{aligned}u&amp;=&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}} &#92;Leftrightarrow &amp;x=&#92;left( &#92;left( u^{2}-1&#92;right) ^{2}-1&#92;right)  ^{2}=u^{8}-4u^{6}+4u^{4}.&#92;end{aligned}' class='latex' /></p>
<p>Visto que</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=dx%3D%5Cleft%28+8u%5E%7B7%7D-24u%5E%7B5%7D%2B16u%5E%7B3%7D%5Cright%29+du&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='dx=&#92;left( 8u^{7}-24u^{5}+16u^{3}&#92;right) du' title='dx=&#92;left( 8u^{7}-24u^{5}+16u^{3}&#92;right) du' class='latex' /></p>
<p>tem-se</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7DI%26%3A%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7Bx%7D%7D%7D%7D%5C%2C+dx%5C%5C%26%3D%5Cdisplaystyle%5Cint_%7B%5Csqrt%7B2%7D%7D%5E%7B%5Csqrt%7B1%2B%5Csqrt%7B2%7D%7D%7D%5Csqrt%7B1%2Bu%7D%5Cleft%288u%5E%7B7%7D-24u%5E%7B5%7D%2B16u%5E%7B3%7D%5Cright%29%5C%2C+du.%5Cquad%5Ctext%7B%28c%5C%27%7Ba%7Dlculo+abaixo%29%7D%5E%5Cdag%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}I&amp;:=&#92;displaystyle&#92;int_{0}^{1}&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}}}&#92;, dx&#92;&#92;&amp;=&#92;displaystyle&#92;int_{&#92;sqrt{2}}^{&#92;sqrt{1+&#92;sqrt{2}}}&#92;sqrt{1+u}&#92;left(8u^{7}-24u^{5}+16u^{3}&#92;right)&#92;, du.&#92;quad&#92;text{(c&#92;&#039;{a}lculo abaixo)}^&#92;dag&#92;end{aligned}' title='&#92;begin{aligned}I&amp;:=&#92;displaystyle&#92;int_{0}^{1}&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}}}&#92;, dx&#92;&#92;&amp;=&#92;displaystyle&#92;int_{&#92;sqrt{2}}^{&#92;sqrt{1+&#92;sqrt{2}}}&#92;sqrt{1+u}&#92;left(8u^{7}-24u^{5}+16u^{3}&#92;right)&#92;, du.&#92;quad&#92;text{(c&#92;&#039;{a}lculo abaixo)}^&#92;dag&#92;end{aligned}' class='latex' /></p>
<p>Cada um dos termos pode integrar-se usando a substituição <img src='http://s0.wp.com/latex.php?latex=t%3D%5Csqrt%7B1%2Bu%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='t=&#92;sqrt{1+u}' title='t=&#92;sqrt{1+u}' class='latex' /></p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7Ba%7D%5E%7Bb%7D%5Csqrt%7B1%2Bu%7Du%5E%7Bn%7Ddu%3D2%5Cdisplaystyle%5Cint_%7B%5Csqrt%7B1%2Ba%7D%7D%5E%7B%5Csqrt%7B1%2Bb%7D%7Dt%5E%7B2%7D%5Cleft%28++t%5E%7B2%7D-1%5Cright%29%5E%7Bn%7D%5C%2Cdt%2C%5Cquad+a%3D%5Csqrt%7B2%7D%2Cb%3D%5Csqrt%7B1%2B%5Csqrt%7B2%7D%7D.++&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{a}^{b}&#92;sqrt{1+u}u^{n}du=2&#92;displaystyle&#92;int_{&#92;sqrt{1+a}}^{&#92;sqrt{1+b}}t^{2}&#92;left(  t^{2}-1&#92;right)^{n}&#92;,dt,&#92;quad a=&#92;sqrt{2},b=&#92;sqrt{1+&#92;sqrt{2}}.  ' title='&#92;displaystyle&#92;int_{a}^{b}&#92;sqrt{1+u}u^{n}du=2&#92;displaystyle&#92;int_{&#92;sqrt{1+a}}^{&#92;sqrt{1+b}}t^{2}&#92;left(  t^{2}-1&#92;right)^{n}&#92;,dt,&#92;quad a=&#92;sqrt{2},b=&#92;sqrt{1+&#92;sqrt{2}}.  ' class='latex' /></p>
<p>2. <em>Generalização a <img src='http://s0.wp.com/latex.php?latex=k%3D5&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='k=5' title='k=5' class='latex' /> radicais</em></p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=J%3A%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7Bx%7D%7D%7D%7D%7D%5C%2C+dx.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='J:=&#92;displaystyle&#92;int_{0}^{1}&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}}}}&#92;, dx.' title='J:=&#92;displaystyle&#92;int_{0}^{1}&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}}}}&#92;, dx.' class='latex' /></p>
<p>De forma semelhante à de cima a substituição é agora</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=v%3D%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7Bx%7D%7D%7D%7D%5CLeftrightarrow+x%3D%5Cleft%28%5Cleft%28%5Cleft%28v%5E%7B2%7D-1%5Cright%29%5E%7B2%7D-1%5Cright%29%5E%7B2%7D-1%5Cright%29%5E%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='v=&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}}}&#92;Leftrightarrow x=&#92;left(&#92;left(&#92;left(v^{2}-1&#92;right)^{2}-1&#92;right)^{2}-1&#92;right)^{2}' title='v=&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{x}}}}&#92;Leftrightarrow x=&#92;left(&#92;left(&#92;left(v^{2}-1&#92;right)^{2}-1&#92;right)^{2}-1&#92;right)^{2}' class='latex' /></p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=x%3Dv%5E%7B16%7D-8v%5E%7B14%7D%2B24v%5E%7B12%7D-32v%5E%7B10%7D%2B14v%5E%7B8%7D%2B8v%5E%7B6%7D-8v%5E%7B4%7D-1%2C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x=v^{16}-8v^{14}+24v^{12}-32v^{10}+14v^{8}+8v^{6}-8v^{4}-1,' title='x=v^{16}-8v^{14}+24v^{12}-32v^{10}+14v^{8}+8v^{6}-8v^{4}-1,' class='latex' /></p>
<p>e</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=dx%3D%5Cleft%2816v%5E%7B15%7D-112v%5E%7B13%7D%2B288v%5E%7B11%7D-320v%5E%7B9%7D%2B112v%5E%7B7%7D%2B48v%5E%7B5%7D-32v%5E%7B3%7D%5Cright%29+dv.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='dx=&#92;left(16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}&#92;right) dv.' title='dx=&#92;left(16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}&#92;right) dv.' class='latex' /></p>
<p>Assim</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=J%3D%5Cdisplaystyle%5Cint_%7B%5Calpha+%7D%5E%7B%5Cbeta+%7D%5Csqrt%7B1%2Bv%7D%5Cleft%28++16v%5E%7B15%7D-112v%5E%7B13%7D%2B288v%5E%7B11%7D-320v%5E%7B9%7D%2B112v%5E%7B7%7D%2B48v%5E%7B5%7D-32v%5E%7B3%7D%5Cright%29dv&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='J=&#92;displaystyle&#92;int_{&#92;alpha }^{&#92;beta }&#92;sqrt{1+v}&#92;left(  16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}&#92;right)dv' title='J=&#92;displaystyle&#92;int_{&#92;alpha }^{&#92;beta }&#92;sqrt{1+v}&#92;left(  16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}&#92;right)dv' class='latex' /></p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Calpha%3D%5Csqrt%7B1%2B%5Csqrt%7B2%7D%7D%2C%5Cbeta+%3D%5Csqrt%7B1%2B%5Csqrt%7B1%2B%5Csqrt%7B2%7D%7D%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;alpha=&#92;sqrt{1+&#92;sqrt{2}},&#92;beta =&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{2}}}.' title='&#92;alpha=&#92;sqrt{1+&#92;sqrt{2}},&#92;beta =&#92;sqrt{1+&#92;sqrt{1+&#92;sqrt{2}}}.' class='latex' /></p>
<p>&#8211;</p>
<p><img src='http://s0.wp.com/latex.php?latex=%5Cdag&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;dag' title='&#92;dag' class='latex' /> Obtive no SWP</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7DI%26%3D-%5Cdfrac%7B26%5C%2C704%7D%7B765%5C%2C765%7D%5Csqrt%7B1%2B%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%7D%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%5Csqrt%7B2%7D%2B%5Cdfrac%7B83%5C%2C584%7D%7B765%5C%2C765%7D%5Csqrt%7B1%2B%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%7D%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%5C%5C%26%2B%5Cdfrac%7B344%5C%2C096%7D%7B765%5C%2C765%7D%5Csqrt%7B1%2B%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%7D%2B%5Cdfrac%7B67%5C%2C328%7D%7B109%5C%2C395%7D%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D-%5Cdfrac%7B256%7D%7B3003%7D%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%5Csqrt%7B2%7D%5C%5C%26-%5Cdfrac%7B17%5C%2C168%7D%7B765%5C%2C765%7D%5Csqrt%7B1%2B%5Csqrt%7B%5Csqrt%7B2%7D%2B1%7D%7D%5Csqrt%7B2%7D%5C%5C%26%5Capprox+1%2C584%5C%2C9.%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}I&amp;=-&#92;dfrac{26&#92;,704}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}&#92;sqrt{&#92;sqrt{2}+1}&#92;sqrt{2}+&#92;dfrac{83&#92;,584}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}&#92;sqrt{&#92;sqrt{2}+1}&#92;&#92;&amp;+&#92;dfrac{344&#92;,096}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}+&#92;dfrac{67&#92;,328}{109&#92;,395}&#92;sqrt{&#92;sqrt{2}+1}-&#92;dfrac{256}{3003}&#92;sqrt{&#92;sqrt{2}+1}&#92;sqrt{2}&#92;&#92;&amp;-&#92;dfrac{17&#92;,168}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}&#92;sqrt{2}&#92;&#92;&amp;&#92;approx 1,584&#92;,9.&#92;end{aligned}' title='&#92;begin{aligned}I&amp;=-&#92;dfrac{26&#92;,704}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}&#92;sqrt{&#92;sqrt{2}+1}&#92;sqrt{2}+&#92;dfrac{83&#92;,584}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}&#92;sqrt{&#92;sqrt{2}+1}&#92;&#92;&amp;+&#92;dfrac{344&#92;,096}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}+&#92;dfrac{67&#92;,328}{109&#92;,395}&#92;sqrt{&#92;sqrt{2}+1}-&#92;dfrac{256}{3003}&#92;sqrt{&#92;sqrt{2}+1}&#92;sqrt{2}&#92;&#92;&amp;-&#92;dfrac{17&#92;,168}{765&#92;,765}&#92;sqrt{1+&#92;sqrt{&#92;sqrt{2}+1}}&#92;sqrt{2}&#92;&#92;&amp;&#92;approx 1,584&#92;,9.&#92;end{aligned}' class='latex' /></p>
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		<title>Yitang Zhang</title>
		<link>http://problemasteoremas.wordpress.com/2013/05/20/yitang-zhang/</link>
		<comments>http://problemasteoremas.wordpress.com/2013/05/20/yitang-zhang/#comments</comments>
		<pubDate>Mon, 20 May 2013 15:01:48 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Matemática]]></category>

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		<description><![CDATA[Reblogged from Pink Iguana: Maggie McKee, Nature, First proof that infinitely many prime numbers come in pairs, here. Better than hitting the Powerball Lottery for Prof. Zhang if this holds. When is the Princeton University Public Lecture scheduled? I'm in &#8230; <a href="http://problemasteoremas.wordpress.com/2013/05/20/yitang-zhang/">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&#038;blog=1866481&#038;post=16918&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<div class="reblog-post"><p class="reblog-from"><img alt='' src='http://1.gravatar.com/avatar/d27eaccc0e0d64e60bb7b3ae15e5de9d?s=25&amp;d=identicon&amp;r=G' class='avatar avatar-25' height='25' width='25' /> <a href="http://jsandber.wordpress.com/2013/05/15/yitang-zhang/">Reblogged from Pink Iguana:</a></p><div class="wpcom-enhanced-excerpt"><div class="wpcom-enhanced-excerpt-content"><p dir='auto'>

</p><p><strong>Maggie McKee</strong>, Nature, First proof that infinitely many prime numbers come in pairs, <a href="http://www.nature.com/news/first-proof-that-infinitely-many-prime-numbers-come-in-pairs-1.12989">here</a>. Better than hitting the Powerball Lottery for Prof. Zhang if this holds. When is the Princeton University Public Lecture scheduled? I'm in for that.</p>
<blockquote><p>The new result, from Yitang Zhang of the University of New Hampshire in Durham, finds that there are infinitely many pairs of primes that are less than 70 million units apart without relying on unproven conjectures.</p></blockquote>

</div> <p class="read-more"><a href="http://jsandber.wordpress.com/2013/05/15/yitang-zhang/" target="_self"><span>Ler mais&hellip;</span> mais 1.140 palavras</a></p></div></div><div class="reblogger-note"><div class='reblogger-note-content'>
Compilação de notícias, em inglês, sobre o seguinte resultado de Yitang Zhang:

Há um número infinito de números primos p e q tais que | p - q | &lt; 70000000.
</div></div>]]></content:encoded>
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		<title>De SPM Notícias &#8212; Vídeos Khan Academy: Matemática para todos em português</title>
		<link>http://problemasteoremas.wordpress.com/2013/05/09/de-spm-noticias-videos-khan-academy-matematica-para-todos-em-portugues/</link>
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		<pubDate>Thu, 09 May 2013 21:55:27 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Matemática]]></category>
		<category><![CDATA[Matemática-Básico]]></category>
		<category><![CDATA[Matemática-Secundário]]></category>
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		<category><![CDATA[Vídeo]]></category>

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		<description><![CDATA[Uma centena de vídeos com conteúdos matemáticos certificados pela SPM está disponível gratuitamente em http://fundacao.telecom.pt/. Os vídeos, desenvolvidos pela Khan Academy, estão a ser adaptados para português por iniciativa da Fundação Portugal Telecom, que, em parceria com a SPM, assegura &#8230; <a href="http://problemasteoremas.wordpress.com/2013/05/09/de-spm-noticias-videos-khan-academy-matematica-para-todos-em-portugues/">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&#038;blog=1866481&#038;post=16910&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<blockquote>
<p style="text-align:justify;"><span style="color:#000000;">Uma centena de <b>vídeos</b> com <b>conteúdos matemáticos certificados pela SPM</b> está disponível gratuitamente em <span style="color:#0000ff;"><a href="http://fundacao.telecom.pt/"><span style="color:#0000ff;">http://fundacao.telecom.pt/</span></a></span>. Os vídeos, desenvolvidos pela <b>Khan Academy</b>, estão a ser adaptados para português por iniciativa da <b>Fundação Portugal Telecom</b>, que, em parceria com a SPM, assegura a sua adequação ao currículo escolar nacional.</span></p>
<p style="text-align:justify;"><span style="color:#000000;">Até ao final do ano estarão disponíveis cerca de <b>400 vídeos</b> que abordarão essencialmente matérias dos <b>2.º, 4.º, 6.º, 9.º e 12.º anos</b>. Além do público português, o projeto pretende também chegar aos internautas dos Países Africanos de Língua Oficial Portuguesa e Timor-Leste através do Sapo Internacional.</span></p>
</blockquote>
<p>Exemplo: Resolver equações de 2.º grau completando o quadrado &#8211; Khan Academy em Português, 9.º ano</p>
<p style="text-align:center;"><span class='embed-youtube' style='text-align:center; display: block;'><iframe class='youtube-player' type='text/html' width='640' height='390' src='http://www.youtube.com/embed/VGKwJuO_2w4?version=3&#038;rel=1&#038;fs=1&#038;showsearch=0&#038;showinfo=1&#038;iv_load_policy=1&#038;wmode=transparent' frameborder='0'></iframe></span></p>
<br />Filed under: <a href='http://problemasteoremas.wordpress.com/category/matematica/'>Matemática</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/matematica-basico/'>Matemática-Básico</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/matematica-secundario/'>Matemática-Secundário</a>, <a href='http://problemasteoremas.wordpress.com/category/spm/'>SPM</a>, <a href='http://problemasteoremas.wordpress.com/category/video/'>Vídeo</a> Tagged: <a href='http://problemasteoremas.wordpress.com/tag/matematica/'>Matemática</a>, <a href='http://problemasteoremas.wordpress.com/tag/matematica-basico/'>Matemática-Básico</a>, <a href='http://problemasteoremas.wordpress.com/tag/matematica-secundario/'>Matemática-Secundário</a>, <a href='http://problemasteoremas.wordpress.com/tag/video/'>Vídeo</a> <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&#038;blog=1866481&#038;post=16910&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></content:encoded>
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		<title>Mathematics Stack Exchange trusted user (20000 rep)</title>
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		<pubDate>Mon, 29 Apr 2013 18:56:39 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Mathematics Stack Exchange]]></category>
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		<description><![CDATA[Aos 20000 reputation points atingi ontem o nível de trusted user do fórum Mathematics Stack Exchange (Q&#38;A for people studying mathematics at any level and professionals in related fields.) Iniciei a minha participação em Agosto de 2010, um mês após &#8230; <a href="http://problemasteoremas.wordpress.com/2013/04/29/20000/">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&#038;blog=1866481&#038;post=16846&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Aos 20000 <em><a href="http://math.stackexchange.com/faq#reputation">reputation</a> points</em> atingi ontem o nível de <a href="http://math.stackexchange.com/privileges/trusted-user"><em>trusted user</em></a> do fórum <a href="http://math.stackexchange.com/">Mathematics Stack Exchange</a> (<em>Q&amp;A</em> <em>for people studying mathematics at any level and professionals in related fields.</em>)</p>
<p style="text-align:center;"><a href="http://problemasteoremas.files.wordpress.com/2013/04/mse2013-04-28profile.png"><img class="aligncenter size-full wp-image-16847" alt="mse2013-04-28profile" src="http://problemasteoremas.files.wordpress.com/2013/04/mse2013-04-28profile.png?w=640&#038;h=72" width="640" height="72" /></a></p>
<p style="text-align:justify;">Iniciei a <a href="http://stackexchange.com/users/209651/americo-tavares?tab=reputation">minha participação</a> em Agosto de 2010, um mês após a lançamento do site, <a href="http://math.stackexchange.com/users/752/americo-tavares">coloquei 37 questões e dei 415 respostas</a>, repartidas por várias categorias (<em>tags</em>), sendo as 3 primeiras o<a href="http://math.stackexchange.com/search?q=user:752+[calculus]"><em> calculus</em></a>, a <a href="http://math.stackexchange.com/search?q=user:752+[algebra-precalculus]"><em>algebra-precalculus</em></a> e a <a href="http://math.stackexchange.com/search?q=user:752+[trigonometry]"><em>trigonometry</em></a>. Poderá ver as cerca de 5o  que publiquei neste blog, traduzidas para português,  seleccionando a categoria Mathematics Stack Exchange, na barra lateral, ou clicando <a href="http://problemasteoremas.wordpress.com/category/mathematics-stack-exchange/">aqui</a>.</p>
<p style="text-align:justify;">Com cerca de 61000 (61k) utilizadores, e um número de questões de 132k,  83% das quais foram respondidas, e 205k respostas, este site tem tido um crescimento contínuo, sendo já o 4.º do <a href="http://stackexchange.com/sites#">grupo de sites Stack Exchange</a>. O maior é o StackOverflow (4900k questões), e os dois sequintes, um ano mais velhos, o Super User (171k questões) e o ServerFault (149k questões).</p>
<p style="text-align:justify;"><em>[Título alterado em 1-5-2013]</em></p>
<p style="text-align:left;">ADENDA: acrescento hoje, 1-5-2013, a minha questão  de Agosto 2011 e a resposta de <a href="http://math.stackexchange.com/users/299/pete-l-clark">Pete L. Clark</a>.</p>
<p style="text-align:left;">Questão: <a href="http://math.stackexchange.com/questions/58104/what-is-the-importance-of-calculus-in-todays-mathematics">What is the importance of Calculus in today&#8217;s Mathematics?</a></p>
<blockquote>
<p style="text-align:left;">For engineering (e. g. electrical engineering) and physics, Calculus is important. But for a future mathematician, is the classical approach to Calculus still important? What is normally taught, as a minimum, in most Universities worldwide?</p>
<p style="text-align:left;">Hoping it is useful, I transcribe three comments of mine (to this question):</p>
<p style="text-align:left;">1. I had [have] in mind for instance Tom Apostol&#8217;s books, although learning differentiation before integration. (in response to Qiaochu Yuan&#8217;s &#8220;What is the classical approach to calculus?&#8221;)</p>
<p style="text-align:left;">2. Elementary Calculus, continuous functions, functions of several variables, partial differentiation, implicit-functions, vectors and vector fields, multiple integrals, infinite series, uniform convergence, power series, Fourier series and integrals, etc. (in response to a comment by Geoff Robinson).</p>
<p style="text-align:left;">3. I had [have] in mind calculus for math students, although I am a retired electrical engineer. (in response to Andy&#8217;s comment &#8220;Are you talking about what is usually taught to engineers and physicists, or also about a calculus curriculum for math majors? &#8220;)</p>
</blockquote>
<p style="text-align:left;"><a href="http://math.stackexchange.com/a/58119/752">Resposta</a> de <a href="http://math.stackexchange.com/users/299/pete-l-clark">Pete L. Clark</a></p>
<blockquote>
<p style="text-align:left;"><span style="color:#0000ff;">In a comment to his question, Américo has clarified that by &#8220;classical calculus&#8221; he means something relatively rigorous and theoretical, as for instance in Apostol&#8217;s book (or Spivak&#8217;s).  I think the answer to the question was probably yes no matter what, but when restricted in this way it becomes a big booming <strong>YES</strong>.</span></p>
<p><span style="color:#0000ff;">The methods of rigorous calculus &#8212; may I say elementary real analysis? it seems more specific &#8212; are an indispensable part of the cultural knowledge of all mathematicians, pure and applied.  Not all mathematicians will <em>directly</em> use this material in their work: I for one am a mathematician with relatively broad interests almost to a fault, but I have never written &#8220;by the Fundamental Theorem of Calculus&#8221; or &#8220;by the Mean Value Theorem&#8221; in any of my research papers.  But nevertheless familiarity and even deep understanding of these basic ideas and themes permeates all of modern mathematics.  For instance, as an arithmetic geometer the functions I differentiate are usually polynomials or rational functions, but the idea of differentiation is still there, in fact abstracted in the notion of derivations and modules of differentials.  One of the most important concepts in algebraic / arithmetic geometry is <em>smoothness</em>, and although you could in principle try to swallow this as a piece of pure algebra, I say good luck with that if you have never taken multivariable calculus and understood the inverse and implicit function theorems.</span></p>
<p><span style="color:#0000ff;">Eschewing &#8220;classical&#8221; mathematics in favor of more modern, abstract or specialized topics is one of the biggest traps a bright young student of mathematics can fall into.  (If you spend any time at a place like Harvard, as I did as a graduate student, you see undergraduates falling for this with distressing regularity, almost as if the floor outside your office was carpeted with banana peels.)  The people who created the fancy modern machinery did so by virtue of their knowledge of classical stuff, and are responding to it in ways that are profound even if they are unfortunately not made explicit.  Although I am very far from really knowing what I&#8217;m talking about here, my feeling is that the analogy to the fine arts is rather apt: abstract modern art is very much <em>a response</em> to classical, figurative, realistic (I was tempted to say &#8220;mimetic&#8221;, so I had better end this digression soon!) art: if you decide to forego learning about perspective in favor of arranging black squares on a white canvas, you&#8217;re severely missing the point.</span></p>
<p><span style="color:#0000ff;">The material of elementary real analysis &#8212; and even freshman calculus &#8212; is remarkably rich.  I have taught more or less the same freshman calculus courses about a dozen times, and each time I find something new to think about, sometimes in resonance with my other mathematical thoughts of the moment but sometimes I just find that I have the chance to stop and think about something that never occurred to me before.  Once for instance I was talking about computing volumes of solids of revolution and it occurred to me that I had never thought about <em>proving in general</em> that the method of shells will give the same answer as the method of washers.  It was pretty good fun to do it, and I mentioned it to a couple of my colleagues and they had a similar reaction: &#8220;No, I never thought of that before, but it sounds like fun.&#8221;  There are thousands of little projects and discoveries like this in freshman calculus.</span></p>
<p><span style="color:#0000ff;">I confess though that it would be interesting to hear mathematicians talk about parts of calculus that they never liked and never had any use for.  As for me, I really dread the part of the course where we do related rates problems and min / max problems.  The former seems like an exercise whose only point is to exploit &#8212; sometimes to the point of cruelty &#8212; the shakiness of students&#8217; understanding of implicit differentiation, and the latter was sort of fun for me for the first ten problems but twenty years and thousands of min / max problems later I could hardly imagine something more tedious.  (Moreover I am *<em>not that good*</em> at these problems. I had a couple of embarrassing failures as a graduate student, and ever since I look to make sure I can do the problems before I assign them, something I have stopped needing to do in most other undergraduate courses.)</span></p>
<p><span style="color:#0000ff;"><strong>Added</strong>: Let me be explicit that I am not answering the second part of the question, i.e., what is a minimum that is or should be taught.  It goes hand in hand with the richness of these topics that if you tried to make a list of everything that it would be valuable for students to know, your (surely severely incomplete!) list would contain vastly more material than could be reasonably covered in the allotted courses.  This is one subject where books which aim to be &#8220;comprehensive&#8221; come off as pretty daunting.  For instance I own the first of Courant and John&#8217;s two volumes on advanced calculus: it&#8217;s more than six hundred pages!  Is there anything in there which I am willing to point to as &#8220;dispensable&#8221;?  Not much.  (Not to mention that the second volume of their work comes in two parts, the second part of which is itself 954 pages long!) The challenge of teaching these courses lies in the fact that the potential landscape is almost infinite and virtually none of it manifestly unimportant, so you have to make hard choices about what not to do.</span></p>
<p style="text-align:left;">
</blockquote>
<p>O <a href="http://www.math.uga.edu/~pete/">Professor Pete L. Clark</a>  é autor destas Lições de <a href="http://www.math.uga.edu/~pete/2400full.pdf">Honors Calculus</a></p>
<br />Filed under: <a href='http://problemasteoremas.wordpress.com/category/mathematics-stack-exchange/'>Mathematics Stack Exchange</a> Tagged: <a href='http://problemasteoremas.wordpress.com/tag/mse/'>MSE</a> <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&#038;blog=1866481&#038;post=16846&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></content:encoded>
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		<title>Volume de região delimitada por cone e cilindro de eixos paralelos</title>
		<link>http://problemasteoremas.wordpress.com/2013/04/21/volume-de-regiao-limitada-por-cone-e-cilindro-com-eixos-paralelos/</link>
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		<pubDate>Sun, 21 Apr 2013 23:07:26 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Cálculo]]></category>
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		<description><![CDATA[Na questão do MSE Find volume inside the cone and inside the cylinder , de user68203, pede-se, como o título indica, para determinar o volume da região do espaço interior ao cilindro e limitada superiormente pelo cone e inferiormente pelo plano &#8230; <a href="http://problemasteoremas.wordpress.com/2013/04/21/volume-de-regiao-limitada-por-cone-e-cilindro-com-eixos-paralelos/">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&#038;blog=1866481&#038;post=16785&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Na questão do MSE <em><a href="http://math.stackexchange.com/q/339195/752">Find volume inside the cone <img src='http://s0.wp.com/latex.php?latex=z%3D+2a-%5Csqrt%7Bx%5E2%2By%5E2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z= 2a-&#92;sqrt{x^2+y^2}' title='z= 2a-&#92;sqrt{x^2+y^2}' class='latex' /> and inside the cylinder <img src='http://s0.wp.com/latex.php?latex=x%5E2%2By%5E2%3D2ay&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x^2+y^2=2ay' title='x^2+y^2=2ay' class='latex' /></a></em>, de <a href="http://math.stackexchange.com/users/68203/user68203">user68203,</a> pede-se, como o título indica, para determinar o volume da região do espaço interior ao cilindro <em><img src='http://s0.wp.com/latex.php?latex=x%5E2%2By%5E2%3D2ay&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x^2+y^2=2ay' title='x^2+y^2=2ay' class='latex' /></em> e limitada superiormente pelo cone <em><img src='http://s0.wp.com/latex.php?latex=z%3D2a-%5Csqrt%7Bx%5E2%2By%5E2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z=2a-&#92;sqrt{x^2+y^2}' title='z=2a-&#92;sqrt{x^2+y^2}' class='latex' /> </em>e inferiormente pelo plano <img src='http://s0.wp.com/latex.php?latex=z%3D0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z=0' title='z=0' class='latex' /><em>.</em></p>
<p>Tradução da minha resolução:</p>
<p style="text-align:justify;">Convertendo a equação cartesiana da base (<img src='http://s0.wp.com/latex.php?latex=z%3D0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z=0' title='z=0' class='latex' />) do cilindro vertical dado, a qual tem o centro em <img src='http://s0.wp.com/latex.php?latex=%280%2Ca%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(0,a)' title='(0,a)' class='latex' /> e raio <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' />,</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=x%5E%7B2%7D%2By%5E%7B2%7D%3D2ay%5CLeftrightarrow+x%5E%7B2%7D%2B%5Cleft%28+y-a%5Cright%29+%5E%7B2%7D%3Da%5E%7B2%7D%5Cqquad+%281%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x^{2}+y^{2}=2ay&#92;Leftrightarrow x^{2}+&#92;left( y-a&#92;right) ^{2}=a^{2}&#92;qquad (1)' title='x^{2}+y^{2}=2ay&#92;Leftrightarrow x^{2}+&#92;left( y-a&#92;right) ^{2}=a^{2}&#92;qquad (1)' class='latex' /></p>
<p style="text-align:justify;">a <em>coordenadas polares</em> <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' />, obtive</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=r%3D2a%5Csin+%5Ctheta+%2C%5Cqquad+%281%5Cmathrm%7Ba%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r=2a&#92;sin &#92;theta ,&#92;qquad (1&#92;mathrm{a})' title='r=2a&#92;sin &#92;theta ,&#92;qquad (1&#92;mathrm{a})' class='latex' /></p>
<p style="text-align:justify;">porque <img src='http://s0.wp.com/latex.php?latex=x%5E%7B2%7D%2By%5E%7B2%7D%3Dr%5E%7B2%7D&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='x^{2}+y^{2}=r^{2}' title='x^{2}+y^{2}=r^{2}' class='latex' />, <img src='http://s0.wp.com/latex.php?latex=r%3E0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r&gt;0' title='r&gt;0' class='latex' />, e</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7Dx%5E%7B2%7D%2By%5E%7B2%7D-2ay%26%3D0%5CLeftrightarrow+r%5E%7B2%7D-2ar%5Csin%5Ctheta+%3D0%5C%5C%5CLeftrightarrow+r%5Cleft%28+r-2a%5Csin+%5Ctheta%5Cright%29%26%3D0%5CLeftrightarrow+r%3D2a%5Csin%5Ctheta%5Cend%7Baligned%7D%5Cqquad+%281%5Cmathrm%7Bb%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}x^{2}+y^{2}-2ay&amp;=0&#92;Leftrightarrow r^{2}-2ar&#92;sin&#92;theta =0&#92;&#92;&#92;Leftrightarrow r&#92;left( r-2a&#92;sin &#92;theta&#92;right)&amp;=0&#92;Leftrightarrow r=2a&#92;sin&#92;theta&#92;end{aligned}&#92;qquad (1&#92;mathrm{b})' title='&#92;begin{aligned}x^{2}+y^{2}-2ay&amp;=0&#92;Leftrightarrow r^{2}-2ar&#92;sin&#92;theta =0&#92;&#92;&#92;Leftrightarrow r&#92;left( r-2a&#92;sin &#92;theta&#92;right)&amp;=0&#92;Leftrightarrow r=2a&#92;sin&#92;theta&#92;end{aligned}&#92;qquad (1&#92;mathrm{b})' class='latex' /></p>
<p style="text-align:justify;">É de salientar que <img src='http://s0.wp.com/latex.php?latex=%281%5Cmathrm%7Ba%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(1&#92;mathrm{a})' title='(1&#92;mathrm{a})' class='latex' /> poderia ser calculada directamente da figura abaixo, usando trigonometria simples.</p>
<p><img class="aligncenter" alt="" src="http://i.stack.imgur.com/HsGaa.jpg" width="600" height="335" /></p>
<p style="text-align:center;">Disco <img src='http://s0.wp.com/latex.php?latex=R%3D%5C%7B%28x%2Cy%29%5Cin%5Cmathbb%7BR%7D%5E2%3A0%5Cle+x%5E%7B2%7D%2B%5Cleft%28y-a%5Cright%29%5E%7B2%7D%5Cle+a%5E%7B2%7D+%5C%7D%3B%5C%2C+0%5Cleq+r%5Cleq+2a%5Csin%5Ctheta%2C0%5Cleq%5Ctheta+%5Cleq%5Cpi&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='R=&#92;{(x,y)&#92;in&#92;mathbb{R}^2:0&#92;le x^{2}+&#92;left(y-a&#92;right)^{2}&#92;le a^{2} &#92;};&#92;, 0&#92;leq r&#92;leq 2a&#92;sin&#92;theta,0&#92;leq&#92;theta &#92;leq&#92;pi' title='R=&#92;{(x,y)&#92;in&#92;mathbb{R}^2:0&#92;le x^{2}+&#92;left(y-a&#92;right)^{2}&#92;le a^{2} &#92;};&#92;, 0&#92;leq r&#92;leq 2a&#92;sin&#92;theta,0&#92;leq&#92;theta &#92;leq&#92;pi' class='latex' /></p>
<p style="text-align:justify;">O volume da região interior ao cilindro e limitada inferiormente pelo plano <img src='http://s0.wp.com/latex.php?latex=z%3D0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z=0' title='z=0' class='latex' /> e superiormente pelo cone dado</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=z%3D2a-%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%3D2a-r%5Cqquad+%282%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z=2a-&#92;sqrt{x^{2}+y^{2}}=2a-r&#92;qquad (2)' title='z=2a-&#92;sqrt{x^{2}+y^{2}}=2a-r&#92;qquad (2)' class='latex' /></p>
<p style="text-align:justify;">pode ser expressa pelo integral duplo</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=V%3D%5Cdisplaystyle%5Ciint_%7BR%7D2a-%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%5C%2CdA%2C%5Cqquad+%283%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='V=&#92;displaystyle&#92;iint_{R}2a-&#92;sqrt{x^{2}+y^{2}}&#92;,dA,&#92;qquad (3)' title='V=&#92;displaystyle&#92;iint_{R}2a-&#92;sqrt{x^{2}+y^{2}}&#92;,dA,&#92;qquad (3)' class='latex' /></p>
<p style="text-align:justify;">em que <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' /> é o disco limitado por <img src='http://s0.wp.com/latex.php?latex=%281%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(1)' title='(1)' class='latex' />. Em coordenadas polares <img src='http://s0.wp.com/latex.php?latex=dA%3Ddx%5C%2C+dy%3Dr%5C%2Cdr%5C%2Cd%5Ctheta+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='dA=dx&#92;, dy=r&#92;,dr&#92;,d&#92;theta ' title='dA=dx&#92;, dy=r&#92;,dr&#92;,d&#92;theta ' class='latex' /> e a região de integração <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' /> é definida por <img src='http://s0.wp.com/latex.php?latex=0%5Cleq+r%5Cleq+2a%5Csin%5Ctheta+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='0&#92;leq r&#92;leq 2a&#92;sin&#92;theta ' title='0&#92;leq r&#92;leq 2a&#92;sin&#92;theta ' class='latex' />, com <img src='http://s0.wp.com/latex.php?latex=0%5Cleq+%5Ctheta+%5Cleq+%5Cpi+&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='0&#92;leq &#92;theta &#92;leq &#92;pi ' title='0&#92;leq &#92;theta &#92;leq &#92;pi ' class='latex' />. Logo  <img src='http://s0.wp.com/latex.php?latex=%283%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(3)' title='(3)' class='latex' /> transforma-se em</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7DV%26%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cleft%28%5Cint_%7B0%7D%5E%7B2a%5Csin%5Ctheta+%7D%5Cleft%28+2a-r%5Cright%29%5C%2Cr%5C%2Cdr%5Cright%29+%5C%2Cd%5Ctheta%5C%5C%26%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cleft.+ar%5E%7B2%7D-%5Cdfrac%7B1%7D%7B3%7Dr%5E%7B3%7D%5Cright%5Cvert+_%7B0%7D%5E%7B2a%5Csin%5Ctheta+%7D%5C%2Cd%5Ctheta%5C%5C%26%3Da%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cleft%28+2a%5Csin%5Ctheta%5Cright%29+%5E%7B2%7D%5C%2Cd%5Ctheta+-%5Cdfrac%7B1%7D%7B3%7D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cleft%282a%5Csin%5Ctheta%5Cright%29%5E%7B3%7D%5C%2Cd%5Ctheta%5C%5C%26%3D2a%5E%7B3%7D%5Cpi+-%5Cdfrac%7B32%7D%7B9%7Da%5E%7B3%7D%3D%5Cleft%282%5Cpi+-%5Cdfrac%7B32%7D%7B9%7D%5Cright%29a%5E%7B3%7D%2C%5Cend%7Baligned%7D%5Cqquad+%284%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}V&amp;=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left(&#92;int_{0}^{2a&#92;sin&#92;theta }&#92;left( 2a-r&#92;right)&#92;,r&#92;,dr&#92;right) &#92;,d&#92;theta&#92;&#92;&amp;=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left. ar^{2}-&#92;dfrac{1}{3}r^{3}&#92;right&#92;vert _{0}^{2a&#92;sin&#92;theta }&#92;,d&#92;theta&#92;&#92;&amp;=a&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left( 2a&#92;sin&#92;theta&#92;right) ^{2}&#92;,d&#92;theta -&#92;dfrac{1}{3}&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left(2a&#92;sin&#92;theta&#92;right)^{3}&#92;,d&#92;theta&#92;&#92;&amp;=2a^{3}&#92;pi -&#92;dfrac{32}{9}a^{3}=&#92;left(2&#92;pi -&#92;dfrac{32}{9}&#92;right)a^{3},&#92;end{aligned}&#92;qquad (4)' title='&#92;begin{aligned}V&amp;=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left(&#92;int_{0}^{2a&#92;sin&#92;theta }&#92;left( 2a-r&#92;right)&#92;,r&#92;,dr&#92;right) &#92;,d&#92;theta&#92;&#92;&amp;=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left. ar^{2}-&#92;dfrac{1}{3}r^{3}&#92;right&#92;vert _{0}^{2a&#92;sin&#92;theta }&#92;,d&#92;theta&#92;&#92;&amp;=a&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left( 2a&#92;sin&#92;theta&#92;right) ^{2}&#92;,d&#92;theta -&#92;dfrac{1}{3}&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left(2a&#92;sin&#92;theta&#92;right)^{3}&#92;,d&#92;theta&#92;&#92;&amp;=2a^{3}&#92;pi -&#92;dfrac{32}{9}a^{3}=&#92;left(2&#92;pi -&#92;dfrac{32}{9}&#92;right)a^{3},&#92;end{aligned}&#92;qquad (4)' class='latex' /></p>
<p style="text-align:justify;">em que os integrais de <img src='http://s0.wp.com/latex.php?latex=%5Csin%5E2%5Ctheta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sin^2&#92;theta' title='&#92;sin^2&#92;theta' class='latex' /> e <img src='http://s0.wp.com/latex.php?latex=%5Csin%5E3%5Ctheta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sin^3&#92;theta' title='&#92;sin^3&#92;theta' class='latex' /> foram calculados como segue:</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Csin+%5E%7B2%7D%5Ctheta+%5C%2Cd%5Ctheta%26%3D%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cdfrac%7B1-%5Ccos+2%5Ctheta+%7D%7B2%7D%5C%2Cd%5Ctheta+%3D%5Cdfrac%7B%5Cpi+%7D%7B2%7D%2C%5C%5C%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Csin+%5E%7B3%7D%5Ctheta%5C%2Cd%5Ctheta%26%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cdfrac%7B3%5Csin++%5Ctheta+-%5Csin+3%5Ctheta+%7D%7B4%7D%5C%2Cd%5Ctheta+%3D%5Cdfrac%7B4%7D%7B3%7D.++%5Cend%7Baligned%7D%5Cqquad+%285%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;sin ^{2}&#92;theta &#92;,d&#92;theta&amp;=&#92;int_{0}^{&#92;pi }&#92;dfrac{1-&#92;cos 2&#92;theta }{2}&#92;,d&#92;theta =&#92;dfrac{&#92;pi }{2},&#92;&#92;&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;sin ^{3}&#92;theta&#92;,d&#92;theta&amp;=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;dfrac{3&#92;sin  &#92;theta -&#92;sin 3&#92;theta }{4}&#92;,d&#92;theta =&#92;dfrac{4}{3}.  &#92;end{aligned}&#92;qquad (5)' title='&#92;begin{aligned}&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;sin ^{2}&#92;theta &#92;,d&#92;theta&amp;=&#92;int_{0}^{&#92;pi }&#92;dfrac{1-&#92;cos 2&#92;theta }{2}&#92;,d&#92;theta =&#92;dfrac{&#92;pi }{2},&#92;&#92;&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;sin ^{3}&#92;theta&#92;,d&#92;theta&amp;=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;dfrac{3&#92;sin  &#92;theta -&#92;sin 3&#92;theta }{4}&#92;,d&#92;theta =&#92;dfrac{4}{3}.  &#92;end{aligned}&#92;qquad (5)' class='latex' /></p>
<p style="text-align:justify;"><em>Comentário final</em>: talvez esteja mais próximo do espírito da pergunta, escrever os integrais duplos <img src='http://s0.wp.com/latex.php?latex=%283%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(3)' title='(3)' class='latex' /> e <img src='http://s0.wp.com/latex.php?latex=%284%29&amp;bg=ffffff&amp;fg=333333&amp;s=0' alt='(4)' title='(4)' class='latex' /> como triplos, respectivamente,</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=V%3D%5Cdisplaystyle%5Ciint_%7BR%7D%5Cleft%28+%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B2a-%5Csqrt%7Bx%5E%7B2%7D%2By%5E%7B2%7D%7D%7Ddz%5Cright%29+%5C%2CdA%5Cqquad+%283%5E%5Cprime%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='V=&#92;displaystyle&#92;iint_{R}&#92;left( &#92;displaystyle&#92;int_{0}^{2a-&#92;sqrt{x^{2}+y^{2}}}dz&#92;right) &#92;,dA&#92;qquad (3^&#92;prime)' title='V=&#92;displaystyle&#92;iint_{R}&#92;left( &#92;displaystyle&#92;int_{0}^{2a-&#92;sqrt{x^{2}+y^{2}}}dz&#92;right) &#92;,dA&#92;qquad (3^&#92;prime)' class='latex' /></p>
<p style="text-align:justify;">e</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=V%3D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B%5Cpi+%7D%5Cleft%28+%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B2a%5Csin+%5Ctheta+%7D%5Cleft%28++%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B2a-r%7Ddz%5Cright%29+%5C%2Cr%5C%2Cdr%5Cright%29+%5C%2Cd%5Ctheta+.%5Cqquad+%284%5E%5Cprime%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='V=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left( &#92;displaystyle&#92;int_{0}^{2a&#92;sin &#92;theta }&#92;left(  &#92;displaystyle&#92;int_{0}^{2a-r}dz&#92;right) &#92;,r&#92;,dr&#92;right) &#92;,d&#92;theta .&#92;qquad (4^&#92;prime)' title='V=&#92;displaystyle&#92;int_{0}^{&#92;pi }&#92;left( &#92;displaystyle&#92;int_{0}^{2a&#92;sin &#92;theta }&#92;left(  &#92;displaystyle&#92;int_{0}^{2a-r}dz&#92;right) &#92;,r&#92;,dr&#92;right) &#92;,d&#92;theta .&#92;qquad (4^&#92;prime)' class='latex' /></p>
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		<title>Três quadrados e dois triângulos equiláteros &#8212; relação entre lados</title>
		<link>http://problemasteoremas.wordpress.com/2013/03/09/tres-quadrados-e-dois-triangulos-equilateros-relacao-entre-lados/</link>
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		<pubDate>Sat, 09 Mar 2013 21:32:10 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Exercícios Matemáticos]]></category>
		<category><![CDATA[Geometria]]></category>
		<category><![CDATA[Matemática]]></category>
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		<category><![CDATA[Problema]]></category>

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		<description><![CDATA[Na figura: - os vértices do triângulo equilátero interior situam-se sobre as bissectrizes do exterior; - cada um dos quadrados partilha um lado com o triângulo menor e dois dos seus vértices estão sobre os lados do triângulo maior. Mostre &#8230; <a href="http://problemasteoremas.wordpress.com/2013/03/09/tres-quadrados-e-dois-triangulos-equilateros-relacao-entre-lados/">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&#038;blog=1866481&#038;post=16763&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Na figura:</p>
<p style="text-align:justify;">- os vértices do triângulo equilátero interior situam-se sobre as bissectrizes do exterior;</p>
<p style="text-align:justify;">- cada um dos quadrados partilha um lado com o triângulo menor e dois dos seus vértices estão sobre os lados do triângulo maior.</p>
<p style="text-align:justify;">Mostre que a relação entre os lados do triângulo maior e menor é igual a <img src='http://s0.wp.com/latex.php?latex=2%2B%5Csqrt%7B3%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2+&#92;sqrt{3}' title='2+&#92;sqrt{3}' class='latex' />.</p>
<p style="text-align:center;"><a href="http://problemasteoremas.files.wordpress.com/2013/03/3quadrados3triangulos.jpg"><img class="aligncenter size-full wp-image-16764" alt="3quadrados3triangulos" src="http://problemasteoremas.files.wordpress.com/2013/03/3quadrados3triangulos.jpg?w=640&#038;h=420" width="640" height="420" /></a></p>
<p style="text-align:center;">
<br />Filed under: <a href='http://problemasteoremas.wordpress.com/category/matematica/exercicios-matematicos/'>Exercícios Matemáticos</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/geometria/'>Geometria</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/'>Matemática</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/matematica-secundario/'>Matemática-Secundário</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/matematica-secundario/'>Matemática-Secundário</a>, <a href='http://problemasteoremas.wordpress.com/tag/problema/'>Problema</a> <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&#038;blog=1866481&#038;post=16763&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></content:encoded>
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		<title>Integrais impróprios &#8212; exercício sobre integração de séries</title>
		<link>http://problemasteoremas.wordpress.com/2013/03/06/integrais-improprios-exercicio-sobre-integracao-de-series/</link>
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		<pubDate>Wed, 06 Mar 2013 19:26:10 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Cálculo]]></category>
		<category><![CDATA[Exercícios Matemáticos]]></category>
		<category><![CDATA[Integrais]]></category>
		<category><![CDATA[Integrais impróprios]]></category>
		<category><![CDATA[Matemática]]></category>
		<category><![CDATA[Mathematics Stack Exchange]]></category>
		<category><![CDATA[Séries]]></category>
		<category><![CDATA[Exercícios]]></category>
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		<description><![CDATA[Para obter o desenvolvimento em série de integrais impróprios, muitas vezes pode desenvolver-se em série a função integranda e integrar termo a termo essa série. Vou ilustrar o método com o seguinte exemplo adaptado da minha resposta, no MSE, à &#8230; <a href="http://problemasteoremas.wordpress.com/2013/03/06/integrais-improprios-exercicio-sobre-integracao-de-series/">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&#038;blog=1866481&#038;post=16740&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Para obter o desenvolvimento em série de integrais impróprios, muitas vezes pode desenvolver-se em série a função integranda e integrar termo a termo essa série. Vou ilustrar o método com o seguinte exemplo adaptado da <a href="http://math.stackexchange.com/a/108254/752">minha resposta</a>, no MSE, à <a href="http://math.stackexchange.com/q/108248/752">questão</a> de <a href="http://math.stackexchange.com/users/14829/jozef">Jozef</a>:</p>
<blockquote><p>Mostre que</p></blockquote>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Cdfrac%7B%5Clog+x%7D%7Bx-1%7D%5C%2C+dx%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B1%7D%7Bn%5E%7B2%7D%7D.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log x}{x-1}&#92;, dx=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n^{2}}.' title='&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log x}{x-1}&#92;, dx=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n^{2}}.' class='latex' /></p>
<p>Podemos começar por fazer a substituição  <img src='http://s0.wp.com/latex.php?latex=u%3Dx-1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='u=x-1' title='u=x-1' class='latex' />:</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Cdfrac%7B%5Clog+x%7D%7Bx-1%7D%2C+dx%3D-%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Cdfrac%7B%5Clog+%281-u%29%7D%7Bu%7D%5C%2C+du.&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log x}{x-1}, dx=-&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log (1-u)}{u}&#92;, du.' title='&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log x}{x-1}, dx=-&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log (1-u)}{u}&#92;, du.' class='latex' /></p>
<p>De seguida consideramos o desenvolvimento em série de MacLaurin</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Clog+%281-u%29%3D-x-%5Cdfrac%7B1%7D%7B2%7Dx%5E%7B2%7D-%5Cdfrac%7B1%7D%7B3%7Dx%5E%7B3%7D-%5Cldots+-%5Cdfrac%7Bx%5E%7Bn%7D%7D%7Bn%7D-%5Cldots%5Cqquad%5Cleft%5Cvert+u%5Cright%5Cvert+%3C1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;log (1-u)=-x-&#92;dfrac{1}{2}x^{2}-&#92;dfrac{1}{3}x^{3}-&#92;ldots -&#92;dfrac{x^{n}}{n}-&#92;ldots&#92;qquad&#92;left&#92;vert u&#92;right&#92;vert &lt;1' title='&#92;log (1-u)=-x-&#92;dfrac{1}{2}x^{2}-&#92;dfrac{1}{3}x^{3}-&#92;ldots -&#92;dfrac{x^{n}}{n}-&#92;ldots&#92;qquad&#92;left&#92;vert u&#92;right&#92;vert &lt;1' class='latex' /></p>
<p>do qual obtemos o da função integranda</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdfrac%7B%5Clog+%281-u%29%7D%7Bu%7D%3D-1-%5Cdfrac%7B1%7D%7B2%7Du-%5Cdfrac%7B1%7D%7B3%7Du%5E%7B2%7D-%5Cldots+-%5Cdfrac%7Bu%5E%7Bn%7D%7D%7Bn%7D-%5Cldots+%3D-%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7Bu%5E%7Bn-1%7D%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;dfrac{&#92;log (1-u)}{u}=-1-&#92;dfrac{1}{2}u-&#92;dfrac{1}{3}u^{2}-&#92;ldots -&#92;dfrac{u^{n}}{n}-&#92;ldots =-&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{u^{n-1}}{n}' title='&#92;dfrac{&#92;log (1-u)}{u}=-1-&#92;dfrac{1}{2}u-&#92;dfrac{1}{3}u^{2}-&#92;ldots -&#92;dfrac{u^{n}}{n}-&#92;ldots =-&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{u^{n-1}}{n}' class='latex' /></p>
<p style="text-align:justify;">Finalmente integramos esta série, termo a termo, admitindo sem o justificar a validade deste procedimento:</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cbegin%7Baligned%7D-%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Cdfrac%7B%5Clog+%281-u%29%7D%7Bu%7Ddu+%26%3D-%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D-%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D++%5Cdfrac%7Bu%5E%7Bn-1%7D%7D%7Bn%7Ddu%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B1%7D%7Bn%7D%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7Du%5E%7Bn-1%7Ddu%5C%5C%26%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B1%7D%7Bn%7D%5Ctimes%5Cdfrac%7B1%7D%7Bn%7D%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B1%7D%7Bn%5E%7B2%7D%7D++%5Cend%7Baligned%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;begin{aligned}-&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log (1-u)}{u}du &amp;=-&#92;displaystyle&#92;int_{0}^{1}-&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }  &#92;dfrac{u^{n-1}}{n}du=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n}&#92;displaystyle&#92;int_{0}^{1}u^{n-1}du&#92;&#92;&amp;=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n}&#92;times&#92;dfrac{1}{n}=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n^{2}}  &#92;end{aligned}' title='&#92;begin{aligned}-&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log (1-u)}{u}du &amp;=-&#92;displaystyle&#92;int_{0}^{1}-&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }  &#92;dfrac{u^{n-1}}{n}du=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n}&#92;displaystyle&#92;int_{0}^{1}u^{n-1}du&#92;&#92;&amp;=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n}&#92;times&#92;dfrac{1}{n}=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{1}{n^{2}}  &#92;end{aligned}' class='latex' /></p>
<p>Da mesma forma podemos obter</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Cdfrac%7B%5Clog+x%7D%7B1%2Bx%7Ddx%3D-%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5Cdfrac%7B%28-1%29%5E%7Bn-1%7D%7D%7Bn%5E%7B2%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log x}{1+x}dx=-&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{(-1)^{n-1}}{n^{2}}' title='&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{&#92;log x}{1+x}dx=-&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }&#92;dfrac{(-1)^{n-1}}{n^{2}}' class='latex' /></p>
<p>por integração, por partes, da série</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdfrac%7B%5Clog+x%7D%7B1%2Bx%7D%3D%5Cdfrac%7B1%7D%7B1%2Bx%7D%5Clog+x%3D%5Clog+x%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%28-x%29%5E%7Bn-1%7D%3D%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%28-1%29%5E%7Bn-1%7Dx%5E%7Bn-1%7D%5Clog+x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;dfrac{&#92;log x}{1+x}=&#92;dfrac{1}{1+x}&#92;log x=&#92;log x&#92;displaystyle&#92;sum_{n=1}^{&#92;infty}(-x)^{n-1}=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }(-1)^{n-1}x^{n-1}&#92;log x' title='&#92;dfrac{&#92;log x}{1+x}=&#92;dfrac{1}{1+x}&#92;log x=&#92;log x&#92;displaystyle&#92;sum_{n=1}^{&#92;infty}(-x)^{n-1}=&#92;displaystyle&#92;sum_{n=1}^{&#92;infty }(-1)^{n-1}x^{n-1}&#92;log x' class='latex' /></p>
<p>termo a termo</p>
<p style="text-align:center;"><img src='http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7Dx%5E%7Bn-1%7D%5Clog+x%5C%2Cdx%3D-%5Cdisplaystyle%5Cint_%7B0%7D%5E%7B1%7D%5Cdfrac%7B1%7D%7Bx%7D%5Cdfrac%7Bx%5E%7Bn%7D%7D%7Bn%7D%5C%2Cdx%3D-%5Cdfrac%7B1%7D%7Bn%5E%7B2%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;displaystyle&#92;int_{0}^{1}x^{n-1}&#92;log x&#92;,dx=-&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{1}{x}&#92;dfrac{x^{n}}{n}&#92;,dx=-&#92;dfrac{1}{n^{2}}' title='&#92;displaystyle&#92;int_{0}^{1}x^{n-1}&#92;log x&#92;,dx=-&#92;displaystyle&#92;int_{0}^{1}&#92;dfrac{1}{x}&#92;dfrac{x^{n}}{n}&#92;,dx=-&#92;dfrac{1}{n^{2}}' class='latex' />.</p>
<br />Filed under: <a href='http://problemasteoremas.wordpress.com/category/matematica/calculo/'>Cálculo</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/exercicios-matematicos/'>Exercícios Matemáticos</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/integrais/'>Integrais</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/integrais/integrais-improprios/'>Integrais impróprios</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/'>Matemática</a>, <a href='http://problemasteoremas.wordpress.com/category/mathematics-stack-exchange/'>Mathematics Stack Exchange</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/series/'>Séries</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/mse/'>MSE</a> <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&#038;blog=1866481&#038;post=16740&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></content:encoded>
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		<title>Como ler um texto matemático segundo o Matemática Nua &amp; Crua</title>
		<link>http://problemasteoremas.wordpress.com/2013/01/16/como-ler-um-texto-matematico-segundo-o-matematica-nua-crua/</link>
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		<pubDate>Wed, 16 Jan 2013 00:47:54 +0000</pubDate>
		<dc:creator>Américo Tavares</dc:creator>
				<category><![CDATA[Citações]]></category>
		<category><![CDATA[Matemática]]></category>

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		<description><![CDATA[Do Aviso aos Navegantes do blog(ue) Matemática Nua &#38; Crua de Luiz Francisco, Licenciatura Plena em Matemática pelo Instituto Municipal de Ensino Superior de Assis: « Não se deve estudar em um texto matemático ( livro, artigo resolução de problema, &#8230; ) &#8230; <a href="http://problemasteoremas.wordpress.com/2013/01/16/como-ler-um-texto-matematico-segundo-o-matematica-nua-crua/">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&#038;blog=1866481&#038;post=16721&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></description>
				<content:encoded><![CDATA[<p style="text-align:justify;">Do <em>Aviso aos Navegantes</em> do blog(ue) <a href="http://mathluiz.blogspot.pt/">Matemática Nua &amp; Crua</a> de Luiz Francisco, Licenciatura Plena em Matemática pelo Instituto Municipal de Ensino Superior de Assis:</p>
<blockquote>
<p style="text-align:justify;"><span style="color:#0000ff;"><strong>«</strong> Não se deve estudar em um texto matemático ( livro, artigo resolução de problema, &#8230; ) como se estivesse lendo um romance. Também não faz qualquer sentido ler um texto matemático marcando as passagens que você considera importantes. Ler um texto matemático é um processo ativo. Você tem que participar. Leia devagar, com cuidado e sabendo que uma grande parte dos detalhes é em geral omitida quando o texto é escrito.</span></p>
<p style="text-align:justify;"><span style="color:#0000ff;">Qualquer texto matemático que contivesse todos os detalhes seria imenso e seria impossível de ser lido. É normal encontrar nestes textos frases do tipo &#8220;evidentemente&#8221; ou &#8220;é fácil ver que&#8221;. Elas não significam que o que vem em seguida deve ser imediatamente entendido pelo leitor, mas que neste ponto alguns detalhes foram suprimidos e que você deve usar papel e lápis para preencher estes detalhes que estão faltando.<i> </i></span></p>
<p style="text-align:justify;"><span style="color:#0000ff;">Quando for estudar matemática tenha à mão lápis, papel de rascunho e borracha (não tenha medo de errar). Leia o texto com atenção e escreva (e não apenas leia) os exemplos que aparecem no livro. Faça você mesmo os cálculos. Invente seus próprios exemplos a respeito do que esta sendo explicado. <strong>»</strong><br />
</span></p>
</blockquote>
<p style="text-align:justify;">Concordo, excepto em não se dever assinalar as passagens que se consideram importantes. Neste pormenor sempre segui o método contrário, mas pelo menos metade do texto é sempre importante, logo o autor acaba por ter razão.</p>
<br />Filed under: <a href='http://problemasteoremas.wordpress.com/category/citacoes/'>Citações</a>, <a href='http://problemasteoremas.wordpress.com/category/matematica/'>Matemática</a> Tagged: <a href='http://problemasteoremas.wordpress.com/tag/matematica/'>Matemática</a> <img alt="" border="0" src="http://stats.wordpress.com/b.gif?host=problemasteoremas.wordpress.com&#038;blog=1866481&#038;post=16721&#038;subd=problemasteoremas&#038;ref=&#038;feed=1" width="1" height="1" />]]></content:encoded>
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