{"id":47853,"date":"2025-07-02T16:22:06","date_gmt":"2025-07-02T16:22:06","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47853"},"modified":"2025-07-02T16:22:08","modified_gmt":"2025-07-02T16:22:08","slug":"find-a-power-series-for-fxlnx-centered-at-1","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/find-a-power-series-for-fxlnx-centered-at-1\/","title":{"rendered":"Find a power series for f(x)=lnx, centered at 1"},"content":{"rendered":"\n<p>Find a power series for f(x)=lnx, centered at 1<\/p>\n\n\n\n<p>The Correct Answer and Explanation is:<\/p>\n\n\n\n<p>To find the power series expansion of f(x)=ln\u2061(x)f(x) = \\ln(x)f(x)=ln(x) centered at x=1x = 1x=1, we can use the Taylor series formula, which is given by:f(x)=f(a)+f\u2032(a)(x\u2212a)+f\u2032\u2032(a)2!(x\u2212a)2+f(3)(a)3!(x\u2212a)3+\u22eff(x) = f(a) + f'(a)(x &#8211; a) + \\frac{f&#8221;(a)}{2!}(x &#8211; a)^2 + \\frac{f^{(3)}(a)}{3!}(x &#8211; a)^3 + \\cdotsf(x)=f(a)+f\u2032(a)(x\u2212a)+2!f\u2032\u2032(a)\u200b(x\u2212a)2+3!f(3)(a)\u200b(x\u2212a)3+\u22ef<\/p>\n\n\n\n<p>For the function f(x)=ln\u2061(x)f(x) = \\ln(x)f(x)=ln(x), we want the series centered at a=1a = 1a=1. Let&#8217;s first compute the necessary derivatives of f(x)f(x)f(x):<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Zeroth derivative<\/strong>: f(x)=ln\u2061(x)f(x) = \\ln(x)f(x)=ln(x) Evaluating at x=1x = 1x=1: f(1)=ln\u2061(1)=0f(1) = \\ln(1) = 0f(1)=ln(1)=0<\/li>\n\n\n\n<li><strong>First derivative<\/strong>: f\u2032(x)=1xf'(x) = \\frac{1}{x}f\u2032(x)=x1\u200b Evaluating at x=1x = 1x=1: f\u2032(1)=11=1f'(1) = \\frac{1}{1} = 1f\u2032(1)=11\u200b=1<\/li>\n\n\n\n<li><strong>Second derivative<\/strong>: f\u2032\u2032(x)=\u22121x2f&#8221;(x) = -\\frac{1}{x^2}f\u2032\u2032(x)=\u2212x21\u200b Evaluating at x=1x = 1x=1: f\u2032\u2032(1)=\u22121f&#8221;(1) = -1f\u2032\u2032(1)=\u22121<\/li>\n\n\n\n<li><strong>Third derivative<\/strong>: f(3)(x)=2x3f^{(3)}(x) = \\frac{2}{x^3}f(3)(x)=x32\u200b Evaluating at x=1x = 1x=1: f(3)(1)=2f^{(3)}(1) = 2f(3)(1)=2<\/li>\n<\/ol>\n\n\n\n<p>Now, we can plug these values into the Taylor series expansion:ln\u2061(x)=0+1(x\u22121)+\u221212!(x\u22121)2+23!(x\u22121)3+\u22ef\\ln(x) = 0 + 1(x &#8211; 1) + \\frac{-1}{2!}(x &#8211; 1)^2 + \\frac{2}{3!}(x &#8211; 1)^3 + \\cdotsln(x)=0+1(x\u22121)+2!\u22121\u200b(x\u22121)2+3!2\u200b(x\u22121)3+\u22ef<\/p>\n\n\n\n<p>This simplifies to:ln\u2061(x)=(x\u22121)\u221212(x\u22121)2+13(x\u22121)3\u221214(x\u22121)4+\u22ef\\ln(x) = (x &#8211; 1) &#8211; \\frac{1}{2}(x &#8211; 1)^2 + \\frac{1}{3}(x &#8211; 1)^3 &#8211; \\frac{1}{4}(x &#8211; 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<p>Thus, the power series for ln\u2061(x)\\ln(x)ln(x) centered at x=1x = 1x=1 is:ln\u2061(x)=\u2211n=1\u221e(\u22121)n+1n(x\u22121)n\\ln(x) = \\sum_{n=1}^{\\infty} \\frac{(-1)^{n+1}}{n}(x &#8211; 1)^nln(x)=n=1\u2211\u221e\u200bn(\u22121)n+1\u200b(x\u22121)n<\/p>\n\n\n\n<p>This is a valid expansion for ln\u2061(x)\\ln(x)ln(x) for \u2223x\u22121\u2223&lt;1|x &#8211; 1| &lt; 1\u2223x\u22121\u2223&lt;1, i.e., xxx must lie in the interval (0,2)(0, 2)(0,2). The series converges to ln\u2061(x)\\ln(x)ln(x) within this interval.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-286.jpeg\" alt=\"\" class=\"wp-image-47854\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-286.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-286-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-286-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-286-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Find a power series for f(x)=lnx, centered at 1 The Correct Answer and Explanation is: To find the power series expansion of f(x)=ln\u2061(x)f(x) = \\ln(x)f(x)=ln(x) centered at x=1x = 1x=1, we can use the Taylor series formula, which is given by:f(x)=f(a)+f\u2032(a)(x\u2212a)+f\u2032\u2032(a)2!(x\u2212a)2+f(3)(a)3!(x\u2212a)3+\u22eff(x) = f(a) + f'(a)(x &#8211; a) + \\frac{f&#8221;(a)}{2!}(x &#8211; a)^2 + \\frac{f^{(3)}(a)}{3!}(x &#8211; a)^3 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-47853","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47853","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=47853"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47853\/revisions"}],"predecessor-version":[{"id":47855,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47853\/revisions\/47855"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47853"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47853"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47853"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}