{"id":945,"date":"2025-06-20T17:24:23","date_gmt":"2025-06-20T17:24:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=945"},"modified":"2025-08-25T16:25:05","modified_gmt":"2025-08-25T16:25:05","slug":"taylor-and-maclaurin-series-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-learn-it-1\/","title":{"raw":"Taylor and Maclaurin Series: Learn It 1","rendered":"Taylor and Maclaurin Series: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Learn how to find Taylor polynomials of a given order for a function<\/li>\r\n \t<li>Estimate the remainder when using a Taylor series to approximate a function<\/li>\r\n \t<li>Determine when a Taylor series converges to the original function<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Overview of Taylor\/Maclaurin Series<\/h2>\r\n<p class=\"whitespace-normal break-words\">In the previous sections, you learned to find power series representations for functions related to geometric series. Now we'll tackle a broader question: <strong>Which functions can be represented by power series, and how do we find these representations?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">We'll also address a crucial follow-up: If we find a power series for a function [latex]f[\/latex] that converges on some interval, how do we prove the series actually converges to [latex]f[\/latex]?<\/p>\r\n\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding the Right Coefficients<\/h3>\r\n<p class=\"whitespace-normal break-words\">Consider a function [latex]f[\/latex] that has a power series representation at [latex]x = a[\/latex]. The series takes the form:<\/p>\r\n\r\n<div id=\"fs-id1167025118355\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+\\cdots [\/latex].<\/div>\r\nThe key question is: <strong>What should the coefficients be?<\/strong> For now, we'll focus on determining the coefficients and address convergence later.\r\n<p class=\"whitespace-normal break-words\">If this series represents [latex]f[\/latex] at [latex]x = a[\/latex], we want the series to equal [latex]f(a)[\/latex] when [latex]x = a[\/latex]. Evaluating the series at [latex]x = a[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1167024866801\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}&amp; ={c}_{0}+{c}_{1}\\left(a-a\\right)+{c}_{2}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; ={c}_{0}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p class=\"whitespace-normal break-words\">Therefore, for the series to equal [latex]f(a)[\/latex], we need [latex]c_0 = f(a)[\/latex].<\/p>\r\n\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Matching Derivatives<\/h3>\r\n<p class=\"whitespace-normal break-words\">We also want the first derivative of our power series to equal [latex]f'(a)[\/latex] at [latex]x = a[\/latex]. Differentiating term-by-term:<\/p>\r\n\r\n<div id=\"fs-id1167024875132\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots [\/latex].<\/div>\r\nAt [latex]x = a[\/latex], this becomes:\r\n<div id=\"fs-id1167024999505\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ \\hfill \\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)&amp; ={c}_{1}+2{c}_{2}\\left(a-a\\right)+3{c}_{3}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; ={c}_{1}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p class=\"whitespace-normal break-words\">So for the derivative to match, we need [latex]c_1 = f'(a)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Continuing this pattern, the second and third derivatives are:<\/p>\r\n\r\n<div id=\"fs-id1167024955379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+4\\cdot 3{c}_{4}{\\left(x-a\\right)}^{2}+\\cdots [\/latex]<\/div>\r\n<p id=\"fs-id1167024962562\">and<\/p>\r\n\r\n<div id=\"fs-id1167025089047\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(x-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(x-a\\right)}^{2}+\\cdots [\/latex].<\/div>\r\nAt [latex]x = a[\/latex], these derivatives evaluate to:\r\n<div id=\"fs-id1167025091472\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)&amp; =2{c}_{2}+3\\cdot 2{c}_{3}\\left(a-a\\right)+4\\cdot 3{c}_{4}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; =2{c}_{2}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167025018088\">and<\/p>\r\n\r\n<div id=\"fs-id1167025099708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)&amp; =3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(a-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ &amp; =3\\cdot 2{c}_{3}\\hfill \\end{array}[\/latex]<\/div>\r\n<p class=\"whitespace-normal break-words\">For these to equal [latex]f''(a)[\/latex] and [latex]f'''(a)[\/latex] respectively, we need:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]c_2 = \\frac{f''(a)}{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]c_3 = \\frac{f'''(a)}{3 \\cdot 2} = \\frac{f'''(a)}{3!}[\/latex]<\/li>\r\n<\/ul>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">The General Pattern<\/h3>\r\n<p class=\"whitespace-normal break-words\">Following this pattern, if [latex]f[\/latex] has a power series representation at [latex]x = a[\/latex], the coefficients must be:<\/p>\r\n<p id=\"fs-id1167025239876\" style=\"text-align: center;\">[latex]{c}_{n}=\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This gives us the complete series:<\/p>\r\n\r\n<div id=\"fs-id1167024875578\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}=f\\left(a\\right)+{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)+\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots [\/latex].<\/div>\r\n<p id=\"fs-id1167025239902\">This power series for [latex]f[\/latex] is known as the <strong>Taylor series<\/strong> for [latex]f[\/latex] at [latex]a[\/latex]. If [latex]a=0[\/latex], then this series is known as the <strong>Maclaurin series<\/strong> for [latex]f[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Taylor and Maclaurin series<\/h3>\r\n<p id=\"fs-id1167025101458\">If [latex]f[\/latex] has derivatives of all orders at [latex]x=a[\/latex], then the <strong>Taylor series<\/strong> for the function [latex]f[\/latex] at [latex]a[\/latex] is<\/p>\r\n\r\n<div id=\"fs-id1167025228678\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}=f\\left(a\\right)+{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)+\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1167024875404\">The Taylor series for [latex]f[\/latex] at [latex]0[\/latex] is known as the <strong>Maclaurin series<\/strong> for [latex]f[\/latex].<\/p>\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Uniqueness of Taylor Series<\/h3>\r\n<p class=\"whitespace-normal break-words\">Power series representations are unique, which leads to an important result: if a function [latex]f[\/latex] has a power series at [latex]a[\/latex], then it must be the Taylor series for [latex]f[\/latex] at [latex]a[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>theorem: uniqueness of Taylor series<\/h3>\r\nIf a function [latex]f[\/latex] has a power series at [latex]a[\/latex] that converges to [latex]f[\/latex] on some open interval containing [latex]a[\/latex], then that power series must be the Taylor series for [latex]f[\/latex] at [latex]a[\/latex].\r\n\r\n<\/section>To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as <strong>Taylor polynomials<\/strong>.","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Learn how to find Taylor polynomials of a given order for a function<\/li>\n<li>Estimate the remainder when using a Taylor series to approximate a function<\/li>\n<li>Determine when a Taylor series converges to the original function<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Overview of Taylor\/Maclaurin Series<\/h2>\n<p class=\"whitespace-normal break-words\">In the previous sections, you learned to find power series representations for functions related to geometric series. Now we&#8217;ll tackle a broader question: <strong>Which functions can be represented by power series, and how do we find these representations?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">We&#8217;ll also address a crucial follow-up: If we find a power series for a function [latex]f[\/latex] that converges on some interval, how do we prove the series actually converges to [latex]f[\/latex]?<\/p>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding the Right Coefficients<\/h3>\n<p class=\"whitespace-normal break-words\">Consider a function [latex]f[\/latex] that has a power series representation at [latex]x = a[\/latex]. The series takes the form:<\/p>\n<div id=\"fs-id1167025118355\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}={c}_{0}+{c}_{1}\\left(x-a\\right)+{c}_{2}{\\left(x-a\\right)}^{2}+\\cdots[\/latex].<\/div>\n<p>The key question is: <strong>What should the coefficients be?<\/strong> For now, we&#8217;ll focus on determining the coefficients and address convergence later.<\/p>\n<p class=\"whitespace-normal break-words\">If this series represents [latex]f[\/latex] at [latex]x = a[\/latex], we want the series to equal [latex]f(a)[\/latex] when [latex]x = a[\/latex]. Evaluating the series at [latex]x = a[\/latex]:<\/p>\n<div id=\"fs-id1167024866801\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}& ={c}_{0}+{c}_{1}\\left(a-a\\right)+{c}_{2}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={c}_{0}.\\hfill \\end{array}[\/latex]<\/div>\n<p class=\"whitespace-normal break-words\">Therefore, for the series to equal [latex]f(a)[\/latex], we need [latex]c_0 = f(a)[\/latex].<\/p>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Matching Derivatives<\/h3>\n<p class=\"whitespace-normal break-words\">We also want the first derivative of our power series to equal [latex]f'(a)[\/latex] at [latex]x = a[\/latex]. Differentiating term-by-term:<\/p>\n<div id=\"fs-id1167024875132\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)={c}_{1}+2{c}_{2}\\left(x-a\\right)+3{c}_{3}{\\left(x-a\\right)}^{2}+\\cdots[\/latex].<\/div>\n<p>At [latex]x = a[\/latex], this becomes:<\/p>\n<div id=\"fs-id1167024999505\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ \\hfill \\frac{d}{dx}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)& ={c}_{1}+2{c}_{2}\\left(a-a\\right)+3{c}_{3}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & ={c}_{1}.\\hfill \\end{array}[\/latex]<\/div>\n<p class=\"whitespace-normal break-words\">So for the derivative to match, we need [latex]c_1 = f'(a)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Continuing this pattern, the second and third derivatives are:<\/p>\n<div id=\"fs-id1167024955379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=2{c}_{2}+3\\cdot 2{c}_{3}\\left(x-a\\right)+4\\cdot 3{c}_{4}{\\left(x-a\\right)}^{2}+\\cdots[\/latex]<\/div>\n<p id=\"fs-id1167024962562\">and<\/p>\n<div id=\"fs-id1167025089047\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)=3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(x-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(x-a\\right)}^{2}+\\cdots[\/latex].<\/div>\n<p>At [latex]x = a[\/latex], these derivatives evaluate to:<\/p>\n<div id=\"fs-id1167025091472\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{2}}{d{x}^{2}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)& =2{c}_{2}+3\\cdot 2{c}_{3}\\left(a-a\\right)+4\\cdot 3{c}_{4}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & =2{c}_{2}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1167025018088\">and<\/p>\n<div id=\"fs-id1167025099708\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\frac{{d}^{3}}{d{x}^{3}}\\left(\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{\\left(x-a\\right)}^{n}\\right)& =3\\cdot 2{c}_{3}+4\\cdot 3\\cdot 2{c}_{4}\\left(a-a\\right)+5\\cdot 4\\cdot 3{c}_{5}{\\left(a-a\\right)}^{2}+\\cdots \\hfill \\\\ & =3\\cdot 2{c}_{3}\\hfill \\end{array}[\/latex]<\/div>\n<p class=\"whitespace-normal break-words\">For these to equal [latex]f''(a)[\/latex] and [latex]f'''(a)[\/latex] respectively, we need:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]c_2 = \\frac{f''(a)}{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]c_3 = \\frac{f'''(a)}{3 \\cdot 2} = \\frac{f'''(a)}{3!}[\/latex]<\/li>\n<\/ul>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">The General Pattern<\/h3>\n<p class=\"whitespace-normal break-words\">Following this pattern, if [latex]f[\/latex] has a power series representation at [latex]x = a[\/latex], the coefficients must be:<\/p>\n<p id=\"fs-id1167025239876\" style=\"text-align: center;\">[latex]{c}_{n}=\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This gives us the complete series:<\/p>\n<div id=\"fs-id1167024875578\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}=f\\left(a\\right)+{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)+\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\frac{f^{\\prime\\prime\\prime}\\left(a\\right)}{3\\text{!}}{\\left(x-a\\right)}^{3}+\\cdots[\/latex].<\/div>\n<p id=\"fs-id1167025239902\">This power series for [latex]f[\/latex] is known as the <strong>Taylor series<\/strong> for [latex]f[\/latex] at [latex]a[\/latex]. If [latex]a=0[\/latex], then this series is known as the <strong>Maclaurin series<\/strong> for [latex]f[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Taylor and Maclaurin series<\/h3>\n<p id=\"fs-id1167025101458\">If [latex]f[\/latex] has derivatives of all orders at [latex]x=a[\/latex], then the <strong>Taylor series<\/strong> for the function [latex]f[\/latex] at [latex]a[\/latex] is<\/p>\n<div id=\"fs-id1167025228678\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}=f\\left(a\\right)+{f}^{\\prime }\\left(a\\right)\\left(x-a\\right)+\\frac{f^{\\prime\\prime}\\left(a\\right)}{2\\text{!}}{\\left(x-a\\right)}^{2}+\\cdots +\\frac{{f}^{\\left(n\\right)}\\left(a\\right)}{n\\text{!}}{\\left(x-a\\right)}^{n}+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1167024875404\">The Taylor series for [latex]f[\/latex] at [latex]0[\/latex] is known as the <strong>Maclaurin series<\/strong> for [latex]f[\/latex].<\/p>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Uniqueness of Taylor Series<\/h3>\n<p class=\"whitespace-normal break-words\">Power series representations are unique, which leads to an important result: if a function [latex]f[\/latex] has a power series at [latex]a[\/latex], then it must be the Taylor series for [latex]f[\/latex] at [latex]a[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>theorem: uniqueness of Taylor series<\/h3>\n<p>If a function [latex]f[\/latex] has a power series at [latex]a[\/latex] that converges to [latex]f[\/latex] on some open interval containing [latex]a[\/latex], then that power series must be the Taylor series for [latex]f[\/latex] at [latex]a[\/latex].<\/p>\n<\/section>\n<p>To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as <strong>Taylor polynomials<\/strong>.<\/p>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/945"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/945\/revisions"}],"predecessor-version":[{"id":1964,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/945\/revisions\/1964"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/945\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=945"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=945"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=945"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=945"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}