{"id":191498,"date":"2025-02-15T07:23:59","date_gmt":"2025-02-15T07:23:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191498"},"modified":"2025-02-15T07:24:02","modified_gmt":"2025-02-15T07:24:02","slug":"find-the-taylor-series-for-fx-ln-x-centered-at-a-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/15\/find-the-taylor-series-for-fx-ln-x-centered-at-a-2\/","title":{"rendered":"Find the taylor series for f(x) = ln x centered at a = 2"},"content":{"rendered":"\n
    \n
  1. Find the taylor series for f(x) = ln x centered at a = 2.[Assumethat f has a power series expansion. Do not show thatRn(x)–>0.]<\/li>\n\n\n\n
  2. Solve each of the following equations: (a) (b) 5 – 2x = 17 6x +1 = 4(3x – 7) 2(x – 4) + = 7 5 10 2. \u2013 71 = 12 (c) (d)<\/li>\n<\/ol>\n\n\n\n

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

    Let’s break down the tasks one by one:<\/p>\n\n\n\n

    1. Find the Taylor Series for ( f(x) = \\ln(x) ) centered at ( a = 2 ):<\/h3>\n\n\n\n

    To find the Taylor series of a function ( f(x) ) around a point ( a ), we use the general formula for a Taylor series:<\/p>\n\n\n\n

    [
    f(x) = f(a) + f'(a)(x – a) + \\frac{f”(a)}{2!}(x – a)^2 + \\frac{f^{(3)}(a)}{3!}(x – a)^3 + \\dots
    ]<\/p>\n\n\n\n

    For ( f(x) = \\ln(x) ), the derivatives of ( f(x) ) are:<\/p>\n\n\n\n

      \n
    1. ( f(x) = \\ln(x) )<\/li>\n\n\n\n
    2. ( f'(x) = \\frac{1}{x} )<\/li>\n\n\n\n
    3. ( f”(x) = -\\frac{1}{x^2} )<\/li>\n\n\n\n
    4. ( f^{(3)}(x) = \\frac{2}{x^3} )<\/li>\n\n\n\n
    5. ( f^{(4)}(x) = -\\frac{6}{x^4} )
      (and so on)<\/li>\n<\/ol>\n\n\n\n

      Now, we evaluate these derivatives at ( x = 2 ):<\/p>\n\n\n\n

        \n
      1. ( f(2) = \\ln(2) )<\/li>\n\n\n\n
      2. ( f'(2) = \\frac{1}{2} )<\/li>\n\n\n\n
      3. ( f”(2) = -\\frac{1}{4} )<\/li>\n\n\n\n
      4. ( f^{(3)}(2) = \\frac{2}{8} = \\frac{1}{4} )<\/li>\n\n\n\n
      5. ( f^{(4)}(2) = -\\frac{6}{16} = -\\frac{3}{8} )<\/li>\n<\/ol>\n\n\n\n

        Therefore, the Taylor series expansion of ( \\ln(x) ) centered at ( a = 2 ) is:<\/p>\n\n\n\n

        [
        \\ln(x) = \\ln(2) + \\frac{1}{2}(x – 2) – \\frac{1}{8}(x – 2)^2 + \\frac{1}{24}(x – 2)^3 – \\frac{3}{192}(x – 2)^4 + \\dots
        ]<\/p>\n\n\n\n

        This series can be truncated depending on the desired level of accuracy.<\/p>\n\n\n\n

        \n\n\n\n

        2. Solve the following equations:<\/h3>\n\n\n\n

        (a) ( 5 – 2x = 17 )<\/h4>\n\n\n\n

        We solve for ( x ):<\/p>\n\n\n\n

        [
        5 – 2x = 17
        ]
        [
        -2x = 17 – 5
        ]
        [
        -2x = 12
        ]
        [
        x = -6
        ]<\/p>\n\n\n\n

        Thus, the solution to the first equation is ( x = -6 ).<\/p>\n\n\n\n

        (b) ( 6x + 1 = 4(3x – 7) )<\/h4>\n\n\n\n

        Distribute the ( 4 ) on the right side:<\/p>\n\n\n\n

        [
        6x + 1 = 12x – 28
        ]<\/p>\n\n\n\n

        Now, move all terms involving ( x ) to one side and constants to the other:<\/p>\n\n\n\n

        [
        6x – 12x = -28 – 1
        ]
        [
        -6x = -29
        ]
        [
        x = \\frac{29}{6}
        ]<\/p>\n\n\n\n

        Thus, the solution to the second equation is ( x = \\frac{29}{6} ).<\/p>\n\n\n\n

        (c) ( 2(x – 4) + 5 = 7 )<\/h4>\n\n\n\n

        Distribute the ( 2 ):<\/p>\n\n\n\n

        [
        2x – 8 + 5 = 7
        ]
        [
        2x – 3 = 7
        ]
        [
        2x = 7 + 3
        ]
        [
        2x = 10
        ]
        [
        x = 5
        ]<\/p>\n\n\n\n

        Thus, the solution to the third equation is ( x = 5 ).<\/p>\n\n\n\n

        (d) ( -71 = 12 )<\/h4>\n\n\n\n

        This equation seems incorrect or incomplete as it stands. If you meant to solve for ( x ), or if there was an error in the equation provided, please clarify.<\/p>\n\n\n\n

        \n\n\n\n

        I will now generate the image with the equations and solutions visually represented.<\/p>\n\n\n\n

        Here is the image showing the equations along with their solutions. Let me know if you’d like further clarification or adjustments!<\/p>\n\n\n\n

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

        The Correct Answer and Explanation is : Let’s break down the tasks one by one: 1. Find the Taylor Series for ( f(x) = \\ln(x) ) centered at ( a = 2 ): To find the Taylor series of a function ( f(x) ) around a point ( a ), we use the general formula […]<\/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-191498","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191498","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=191498"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191498\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191498"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191498"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191498"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}