{"id":2283,"date":"2023-05-07T19:01:11","date_gmt":"2023-05-07T22:01:11","guid":{"rendered":"http:\/\/localhost\/?page_id=2283"},"modified":"2023-05-08T21:21:45","modified_gmt":"2023-05-09T00:21:45","slug":"o-problema-da-interpolacao","status":"publish","type":"page","link":"http:\/\/localhost\/index.php\/ensino\/geometria-algebrica-e-algebra-linear\/o-problema-da-interpolacao\/","title":{"rendered":"O problema da interpola\u00e7\u00e3o"},"content":{"rendered":"<div style=\"color: black;\">\n<div class=\"example\">Sejam $P_1=(1,1)$, $P_2=(-1,1)$, $P_3=(2,1)$ e $P_3=(-2,-11)$ quatro pontos no plano. Queremos achar uma fun\u00e7\u00e3o polinomial<br \/>\n\\[<br \/>\nf(x)=\\alpha_3 x^3+\\alpha_2 x^2+\\alpha_1 x+\\alpha_0<br \/>\n\\]<br \/>\ntal que o grafo de $f(x)$ passe pelos pontos $P_1$, $P_2$, $P_3$, $P_4$. Em outras palavras,<br \/>\n\\[<br \/>\nf(1)=1,\\quad f(-1)=1,\\quad f(2)=1,\\quad f(-2)=-11.<br \/>\n\\]<br \/>\nSubstituindo em $f(x)$ obtemos o seguinte sistema de equa\u00e7\u00f5es lineares.<br \/>\n\\begin{align*}<br \/>\n\\alpha_0+\\alpha_1+\\alpha_2+\\alpha_3&amp;=1\\\\<br \/>\n\\alpha_0-\\alpha_1+\\alpha_2-\\alpha_3&amp;=1\\\\<br \/>\n\\alpha_0+2\\alpha_1+4\\alpha_2+8\\alpha_3&amp;=1\\\\<br \/>\n\\alpha_0-2\\alpha_1+4\\alpha_2-8\\alpha_3&amp;=-11<br \/>\n\\end{align*}<br \/>\nPara resolver o problema da interpola\u00e7\u00e3o, precisamos resolver o sistema acima.<\/p>\n<pre><code class=\"language-r\">\n\/\/ definimos a matriz do sistema\nA = [ 1 1 1 1; 1 -1 1 -1; 1 2 4 8; 1 -2 4 -8 ]\n A  = \n\n   1.   1.   1.   1.\n   1.  -1.   1.  -1.\n   1.   2.   4.   8.\n   1.  -2.   4.  -8.\n\n\/\/ definimos o vetor b\n--&gt; b = [1;1;1;-11] \n b  = \n\n   1.\n   1.\n   1.\n  -11.\n\n\/\/ calculamos a solu\u00e7\u00e3o do sistema usando \n\/\/ o m\u00e9todo da matriz inversa\n--&gt; A^-1*b\n ans  =\n\n   3.0000000000000004440892\n  -0.999999999999999888978\n  -2.\n   1.\n <\/code><\/pre>\n<p>Obtemos da computa\u00e7\u00e3o que $\\alpha_0=3$, $\\alpha_1=-1$, $\\alpha_2=-2$, $\\alpha_3=1$. Ou seja, o polin\u00f4mio procurado \u00e9<br \/>\n\\[<br \/>\nf(x)=x^3-2x^2-x+3<br \/>\n\\]<br \/>\nPara visualizar o polin\u00f4mio obtido, podemos executar as seguintes linhas.<\/p>\n<pre><code class=\"language-r\">\n\/\/ definimos primeiro o polin\u00f4mio\n--&gt; p = poly( [3,-1,-2,1], &quot;x&quot;, &quot;coeff&quot; )\n p  = \n\n  3 -x -2x^2 +x^3\n\n\/\/ usamos a seguinte instru\u00e7\u00e3o para visualizar a curva\n\/\/ a express\u00e3o &quot;-3:0.1:3&quot; significa a lista de valores \n\/\/ entre -3 e 3 com incremento de 0.1.\n\/\/ \n\/\/ Esta linha deve abrir uma nova janela \n--&gt; plot( -3:0.1:3, p )\n<\/code><\/pre>\n<p>Obtemos uma imagem da seguinte forma.<\/p>\n<p><img decoding=\"async\" class=\"alignnone size-medium wp-image-2288\" src=\"http:\/\/localhost\/wp-content\/uploads\/2023\/05\/poly-300x226.jpeg\" alt=\"\" width=\"300\" height=\"226\" srcset=\"http:\/\/localhost\/wp-content\/uploads\/2023\/05\/poly-300x226.jpeg 300w, http:\/\/localhost\/wp-content\/uploads\/2023\/05\/poly.jpeg 610w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"exercise\">Ache um polin\u00f4mio $f(x)$ de grau $5$ tal que<br \/>\n\\[<br \/>\nf(0)=-1,\\ f(1)=-2,\\ f(-1)=0,\\ f(2)=-57,\\ f(-2)=79,\\ f(3)=-436.<br \/>\n\\]<br \/>\nVizualize o polin\u00f4mio encontrado.<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Sejam $P_1=(1,1)$, $P_2=(-1,1)$, $P_3=(2,1)$ e $P_3=(-2,-11)$ quatro pontos no plano. Queremos achar uma fun\u00e7\u00e3o polinomial \\[ f(x)=\\alpha_3 x^3+\\alpha_2 x^2+\\alpha_1 x+\\alpha_0 \\] tal que o grafo de $f(x)$ passe pelos pontos $P_1$, $P_2$, $P_3$, $P_4$. Em outras palavras, \\[ f(1)=1,\\quad f(-1)=1,\\quad f(2)=1,\\quad f(-2)=-11. \\] Substituindo em $f(x)$ obtemos o seguinte sistema de equa\u00e7\u00f5es lineares. \\begin{align*} \\alpha_0+\\alpha_1+\\alpha_2+\\alpha_3&amp;=1\\\\ &hellip; <a href=\"http:\/\/localhost\/index.php\/ensino\/geometria-algebrica-e-algebra-linear\/o-problema-da-interpolacao\/\" class=\"more-link\">Continue reading <span class=\"screen-reader-text\">O problema da interpola\u00e7\u00e3o<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2265,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":[],"_links":{"self":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/2283"}],"collection":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/comments?post=2283"}],"version-history":[{"count":6,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/2283\/revisions"}],"predecessor-version":[{"id":2320,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/2283\/revisions\/2320"}],"up":[{"embeddable":true,"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/pages\/2265"}],"wp:attachment":[{"href":"http:\/\/localhost\/index.php\/wp-json\/wp\/v2\/media?parent=2283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}