{"id":248127,"date":"2025-07-08T06:59:04","date_gmt":"2025-07-08T06:59:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=248127"},"modified":"2025-07-08T06:59:06","modified_gmt":"2025-07-08T06:59:06","slug":"find-the-power-series-expansion-of-yln%e2%81%a1x-at-x1-that-is-as-s-power","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/find-the-power-series-expansion-of-yln%e2%81%a1x-at-x1-that-is-as-s-power\/","title":{"rendered":"Find the power series expansion of\u00a0y=ln\u2061x\u00a0at\u00a0x=1, that is, as s power"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The problem asks to find the power series expansion of y=ln\u2061(x)y = \\ln(x)y=ln(x) around x=1x = 1x=1 in terms of a series involving (x\u22121)(x – 1)(x\u22121), i.e., in the form:ln\u2061(x)=\u2211n=0\u221ean(x\u22121)n\\ln(x) = \\sum_{n=0}^{\\infty} a_n (x – 1)^nln(x)=n=0\u2211\u221e\u200ban\u200b(x\u22121)n<\/p>\n\n\n\n

We will derive the coefficients ana_nan\u200b by differentiating y=ln\u2061(x)y = \\ln(x)y=ln(x) and evaluating each derivative at x=1x = 1x=1. Here are the steps to find these coefficients:<\/p>\n\n\n\n

Step 1: Define the function and its derivatives.<\/h3>\n\n\n\n

The function is f(x)=ln\u2061(x)f(x) = \\ln(x)f(x)=ln(x).<\/p>\n\n\n\n

    \n
  • First derivative: f\u2032(x)=1x=x\u22121f'(x) = \\frac{1}{x} = x^{-1}f\u2032(x)=x1\u200b=x\u22121.<\/li>\n\n\n\n
  • Second derivative: f\u2032\u2032(x)=\u2212x\u22122f”(x) = -x^{-2}f\u2032\u2032(x)=\u2212x\u22122.<\/li>\n\n\n\n
  • Third derivative: f(3)(x)=2x\u22123f^{(3)}(x) = 2x^{-3}f(3)(x)=2x\u22123.<\/li>\n\n\n\n
  • Fourth derivative: f(4)(x)=\u22126x\u22124f^{(4)}(x) = -6x^{-4}f(4)(x)=\u22126x\u22124.<\/li>\n\n\n\n
  • Continue this process for higher derivatives.<\/li>\n<\/ul>\n\n\n\n

    Step 2: Evaluate the derivatives at x=1x = 1x=1.<\/h3>\n\n\n\n

    For each derivative, we will evaluate it at x=1x = 1x=1 to find an=f(n)(1)n!a_n = \\frac{f^{(n)}(1)}{n!}an\u200b=n!f(n)(1)\u200b:<\/p>\n\n\n\n

      \n
    • f(1)=ln\u2061(1)=0f(1) = \\ln(1) = 0f(1)=ln(1)=0.<\/li>\n\n\n\n
    • f\u2032(1)=1f'(1) = 1f\u2032(1)=1.<\/li>\n\n\n\n
    • f\u2032\u2032(1)=\u22121f”(1) = -1f\u2032\u2032(1)=\u22121.<\/li>\n\n\n\n
    • f(3)(1)=2f^{(3)}(1) = 2f(3)(1)=2.<\/li>\n\n\n\n
    • f(4)(1)=\u22126f^{(4)}(1) = -6f(4)(1)=\u22126.<\/li>\n\n\n\n
    • f(5)(1)=24f^{(5)}(1) = 24f(5)(1)=24, and so on.<\/li>\n<\/ul>\n\n\n\n

      Step 3: Write the power series.<\/h3>\n\n\n\n

      Now we can write the power series expansion. We will use the formula for the coefficients ana_nan\u200b:an=f(n)(1)n!a_n = \\frac{f^{(n)}(1)}{n!}an\u200b=n!f(n)(1)\u200b<\/p>\n\n\n\n

      Substitute the values of the derivatives evaluated at x=1x = 1x=1:<\/p>\n\n\n\n

        \n
      • a0=0a_0 = 0a0\u200b=0<\/li>\n\n\n\n
      • a1=1a_1 = 1a1\u200b=1<\/li>\n\n\n\n
      • a2=\u221212a_2 = -\\frac{1}{2}a2\u200b=\u221221\u200b<\/li>\n\n\n\n
      • a3=13a_3 = \\frac{1}{3}a3\u200b=31\u200b<\/li>\n\n\n\n
      • a4=\u221214a_4 = -\\frac{1}{4}a4\u200b=\u221241\u200b<\/li>\n\n\n\n
      • a5=15a_5 = \\frac{1}{5}a5\u200b=51\u200b<\/li>\n\n\n\n
      • And so on.<\/li>\n<\/ul>\n\n\n\n

        Thus, the power series expansion of ln\u2061(x)\\ln(x)ln(x) at x=1x = 1x=1 is:ln\u2061(x)=(x\u22121)\u221212(x\u22121)2+13(x\u22121)3\u221214(x\u22121)4+\u22ef\\ln(x) = (x – 1) – \\frac{1}{2}(x – 1)^2 + \\frac{1}{3}(x – 1)^3 – \\frac{1}{4}(x – 1)^4 + \\cdotsln(x)=(x\u22121)\u221221\u200b(x\u22121)2+31\u200b(x\u22121)3\u221241\u200b(x\u22121)4+\u22ef<\/p>\n\n\n\n

        This is the desired power series expansion of y=ln\u2061(x)y = \\ln(x)y=ln(x) at x=1x = 1x=1.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        The Correct Answer and Explanation is: The problem asks to find the power series expansion of y=ln\u2061(x)y = \\ln(x)y=ln(x) around x=1x = 1x=1 in terms of a series involving (x\u22121)(x – 1)(x\u22121), i.e., in the form:ln\u2061(x)=\u2211n=0\u221ean(x\u22121)n\\ln(x) = \\sum_{n=0}^{\\infty} a_n (x – 1)^nln(x)=n=0\u2211\u221e\u200ban\u200b(x\u22121)n We will derive the coefficients ana_nan\u200b by differentiating y=ln\u2061(x)y = \\ln(x)y=ln(x) and evaluating […]<\/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-248127","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248127","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=248127"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248127\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248127"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248127"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248127"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}