{"id":220119,"date":"2025-05-27T14:01:54","date_gmt":"2025-05-27T14:01:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220119"},"modified":"2025-05-27T14:01:56","modified_gmt":"2025-05-27T14:01:56","slug":"identify-the-function-represented-by-the-following-power-series-sigma-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/identify-the-function-represented-by-the-following-power-series-sigma-2\/","title":{"rendered":"Identify the function represented by the following power series Sigma"},"content":{"rendered":"\n<p>Identify the function represented by the following power series Sigma (-1)<sup>k<\/sup>x<sup>k<\/sup>\/3<sup>k<\/sup>&nbsp;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-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\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>We can rewrite this series 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 of the form: \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<p>Here, the common ratio r=\u2212x3r = \\frac{-x}{3}. Since this is a geometric series, we can express it as a function: F(x)=11\u2212(\u2212x\/3)=11+x\/3F(x) = \\frac{1}{1 &#8211; (-x\/3)} = \\frac{1}{1 + x\/3}<\/p>\n\n\n\n<p>To simplify: F(x)=11+x3=13+x3=33+xF(x) = \\frac{1}{1 + \\frac{x}{3}} = \\frac{1}{\\frac{3 + x}{3}} = \\frac{3}{3 + x}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>F(x)=33+xF(x) = \\frac{3}{3 + x}<\/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<\/strong><\/h3>\n\n\n\n<p>The given series is: \u2211k=0\u221e(\u22121)kxk3k\\sum_{k=0}^{\\infty} \\frac{(-1)^k x^k}{3^k}<\/p>\n\n\n\n<p>To identify the function represented by this power series, we recognize it as a geometric series. A geometric series has the general form: \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<p>We manipulate the original expression: (\u22121)kxk3k=(\u2212x3)k\\frac{(-1)^k x^k}{3^k} = \\left(\\frac{-x}{3}\\right)^k<\/p>\n\n\n\n<p>Thus, the series becomes: \u2211k=0\u221e(\u2212x3)k\\sum_{k=0}^{\\infty} \\left( \\frac{-x}{3} \\right)^k<\/p>\n\n\n\n<p>This fits the form of a geometric series with ratio r=\u2212x3r = \\frac{-x}{3}. The series converges for \u2223x\u2223&lt;3|x| &lt; 3, because \u2223\u2212x3\u2223&lt;1\\left|\\frac{-x}{3}\\right| &lt; 1. Applying the geometric series sum formula, we get: \u2211k=0\u221e(\u2212x3)k=11\u2212(\u2212x3)=11+x3\\sum_{k=0}^{\\infty} \\left( \\frac{-x}{3} \\right)^k = \\frac{1}{1 &#8211; \\left( \\frac{-x}{3} \\right)} = \\frac{1}{1 + \\frac{x}{3}}<\/p>\n\n\n\n<p>To simplify further: 11+x3=33+x\\frac{1}{1 + \\frac{x}{3}} = \\frac{3}{3 + x}<\/p>\n\n\n\n<p>Hence, the power series represents the function F(x)=33+xF(x) = \\frac{3}{3 + x}, valid within the radius of convergence \u2223x\u2223&lt;3|x| &lt; 3. Recognizing and manipulating power series into geometric form is a powerful technique in calculus and mathematical analysis.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner7-45.jpeg\" alt=\"\" class=\"wp-image-220120\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Identify the function represented by the following power series Sigma (-1)kxk\/3k&nbsp;Click the icon to view a table of Taylor series for common functions F(x) = The Correct Answer and Explanation is: The given power series is: \u2211k=0\u221e(\u22121)kxk3k\\sum_{k=0}^{\\infty} \\frac{(-1)^k x^k}{3^k} We can rewrite this series as: \u2211k=0\u221e(\u2212x3)k\\sum_{k=0}^{\\infty} \\left( \\frac{-x}{3} \\right)^k This is a geometric series of [&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-220119","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220119","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=220119"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220119\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220119"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220119"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220119"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}