{"id":216518,"date":"2025-05-19T11:52:54","date_gmt":"2025-05-19T11:52:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=216518"},"modified":"2025-05-19T11:52:56","modified_gmt":"2025-05-19T11:52:56","slug":"identify-the-function-represented-by-the-following-power-series-sigma-1kxk-3k-click-the-icon-to-view-a-table-of-taylor-series-for-common-functions-fx-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/19\/identify-the-function-represented-by-the-following-power-series-sigma-1kxk-3k-click-the-icon-to-view-a-table-of-taylor-series-for-common-functions-fx-2\/","title":{"rendered":"Identify the function represented by the following power series Sigma (-1)kxk\/3k Click the icon to view a table of Taylor series for common functions F(x) ="},"content":{"rendered":"\n<p>Identify the function represented by the following power series Sigma (-1)kxk\/3k Click the icon to view a table of Taylor series for common functions F(x) =<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are given the following power series: \u2211k=0\u221e(\u22121)kxk3k\\sum_{k=0}^{\\infty} \\frac{(-1)^k x^k}{3^k}<\/p>\n\n\n\n<p>We want to <strong>identify the function<\/strong> represented by this power series.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Recognize the General Form<\/strong><\/h3>\n\n\n\n<p>The given series looks like: \u2211k=0\u221e(\u22121)k(x3)k\\sum_{k=0}^{\\infty} (-1)^k \\left(\\frac{x}{3}\\right)^k<\/p>\n\n\n\n<p>This can be written as: \u2211k=0\u221e(\u2212x3)k\\sum_{k=0}^{\\infty} \\left(-\\frac{x}{3}\\right)^k<\/p>\n\n\n\n<p>This is a geometric series with the common ratio r=\u2212x3r = -\\frac{x}{3}. The geometric series formula is: \u2211k=0\u221erk=11\u2212r,for&nbsp;\u2223r\u2223&lt;1\\sum_{k=0}^{\\infty} r^k = \\frac{1}{1 &#8211; r}, \\quad \\text{for } |r| &lt; 1<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Apply the Geometric Series Formula<\/strong><\/h3>\n\n\n\n<p>Let\u2019s apply the formula with r=\u2212x3r = -\\frac{x}{3}: \u2211k=0\u221e(\u2212x3)k=11\u2212(\u2212x\/3)=11+x\/3\\sum_{k=0}^{\\infty} \\left(-\\frac{x}{3}\\right)^k = \\frac{1}{1 &#8211; (-x\/3)} = \\frac{1}{1 + x\/3}<\/p>\n\n\n\n<p>We can simplify this expression: 11+x\/3=13+x3=3x+3\\frac{1}{1 + x\/3} = \\frac{1}{\\frac{3 + x}{3}} = \\frac{3}{x + 3}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>F(x)=3x+3F(x) = \\frac{3}{x + 3}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300+ words)<\/strong><\/h3>\n\n\n\n<p>The given power series is: \u2211k=0\u221e(\u22121)kxk3k\\sum_{k=0}^{\\infty} \\frac{(-1)^k x^k}{3^k}<\/p>\n\n\n\n<p>This represents a <strong>geometric series<\/strong> because each term is formed by raising a constant ratio to the power kk. Recognizing this allows us to use one of the most important results in calculus: the formula for the sum of a geometric series. The key formula is: \u2211k=0\u221erk=11\u2212r,where&nbsp;\u2223r\u2223&lt;1\\sum_{k=0}^{\\infty} r^k = \\frac{1}{1 &#8211; r}, \\quad \\text{where } |r| &lt; 1<\/p>\n\n\n\n<p>In this series, r=\u2212x3r = -\\frac{x}{3}, so the sum becomes: 11\u2212(\u2212x\/3)=11+x\/3\\frac{1}{1 &#8211; (-x\/3)} = \\frac{1}{1 + x\/3}<\/p>\n\n\n\n<p>This formula is only valid when the series converges, which occurs for \u2223x\/3\u2223&lt;1|x\/3| &lt; 1, or equivalently, \u2223x\u2223&lt;3|x| &lt; 3. So within this <strong>interval of convergence<\/strong>, the series correctly represents the function.<\/p>\n\n\n\n<p>To simplify the expression further: 11+x\/3=13+x3=3x+3\\frac{1}{1 + x\/3} = \\frac{1}{\\frac{3 + x}{3}} = \\frac{3}{x + 3}<\/p>\n\n\n\n<p>Hence, the power series represents the <strong>rational function<\/strong>: F(x)=3x+3F(x) = \\frac{3}{x + 3}<\/p>\n\n\n\n<p>This is a useful technique in calculus, especially in function approximation, where we use power series like Taylor or Maclaurin series to approximate or represent functions in a simplified polynomial form.<\/p>\n\n\n\n<p>This also shows how infinite series can &#8220;build&#8221; common functions \u2014 an important concept for solving problems in analysis, physics, and engineering.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Identify the function represented by the following power series Sigma (-1)kxk\/3k Click the icon to view a table of Taylor series for common functions F(x) = The Correct Answer and Explanation is: We are given the following power series: \u2211k=0\u221e(\u22121)kxk3k\\sum_{k=0}^{\\infty} \\frac{(-1)^k x^k}{3^k} We want to identify the function represented by this power series. Step 1: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-216518","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216518","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=216518"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216518\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216518"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216518"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216518"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}