{"id":958,"date":"2025-06-20T17:25:01","date_gmt":"2025-06-20T17:25:01","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=958"},"modified":"2025-09-12T14:34:57","modified_gmt":"2025-09-12T14:34:57","slug":"applications-of-series-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-series-learn-it-2\/","title":{"raw":"Applications of Series: Learn It 2","rendered":"Applications of Series: Learn It 2"},"content":{"raw":"

Common Functions Expressed as Taylor Series<\/h2>\r\nAt this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form [latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex]. The table below summarizes these important series.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Maclaurin Series for Common Functions<\/span><\/caption>\r\n
Function<\/th>\r\nMaclaurin Series<\/th>\r\nInterval of Convergence<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
[latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex]<\/td>\r\n[latex]-1 < x < 1[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]f\\left(x\\right)={e}^{x}[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/td>\r\n[latex]\\text{-}\\infty < x < \\infty [\/latex]<\/td>\r\n<\/tr>\r\n
[latex]f\\left(x\\right)=\\sin{x}[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/td>\r\n[latex]\\text{-}\\infty < x < \\infty [\/latex]<\/td>\r\n<\/tr>\r\n
[latex]f\\left(x\\right)=\\cos{x}[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/td>\r\n[latex]\\text{-}\\infty < x < \\infty [\/latex]<\/td>\r\n<\/tr>\r\n
[latex]f\\left(x\\right)=\\text{ln}\\left(1+x\\right)[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{n}[\/latex]<\/td>\r\n[latex]-1 < x\\le 1[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{2n+1}}{2n+1}[\/latex]<\/td>\r\n[latex]-1 < x\\le 1[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]f\\left(x\\right)={\\left(1+x\\right)}^{r}[\/latex]<\/td>\r\n[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}r\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]<\/td>\r\n[latex]-1 < x < 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
The convergence of the Maclaurin series for [latex]f\\left(x\\right)=\\ln\\left(1+x\\right)[\/latex] at [latex]x=1[\/latex] and for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex] at [latex]x=\\pm 1[\/latex] relies on Abel's theorem, which is beyond our current scope.<\/section>\r\n

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series above, to create Maclaurin series for other functions.<\/p>\r\n\r\n

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Find the Maclaurin series of each of the following functions by using one of the series listed in the table.<\/p>\r\n\r\n

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  1. [latex]f\\left(x\\right)=\\cos\\sqrt{x}[\/latex]<\/li>\r\n \t
  2. [latex]f\\left(x\\right)=\\text{sinh}x[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n
    \r\n
      \r\n \t
    1. Using the Maclaurin series for [latex]\\cos{x}[\/latex] we find that the Maclaurin series for [latex]\\cos\\sqrt{x}[\/latex] is given by\r\n<\/span>\r\n
      [latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}{\\left(\\sqrt{x}\\right)}^{2n}}{\\left(2n\\right)\\text{!}}& ={\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{\\left(-1\\right)}^{n}{x}^{n}}{\\left(2n\\right)\\text{!}}\\hfill \\\\ & =1-\\frac{x}{2\\text{!}}+\\frac{{x}^{2}}{4\\text{!}}-\\frac{{x}^{3}}{6\\text{!}}+\\frac{{x}^{4}}{8\\text{!}}-\\cdots .\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThis series converges to [latex]\\cos\\sqrt{x}[\/latex] for all [latex]x[\/latex] in the domain of [latex]\\cos\\sqrt{x}[\/latex]; that is, for all [latex]x\\ge 0[\/latex].<\/li>\r\n \t
    2. To find the Maclaurin series for [latex]\\text{sinh}x[\/latex], we use the fact that\r\n<\/span>\r\n
      [latex]\\text{sinh}x=\\frac{{e}^{x}-{e}^{\\text{-}x}}{2}[\/latex].<\/div>\r\n\r\n<\/span>\r\nUsing the Maclaurin series for [latex]{e}^{x}[\/latex], we see that the [latex]n\\text{th}[\/latex] term in the Maclaurin series for [latex]\\text{sinh}x[\/latex] is given by\r\n<\/span>\r\n
      [latex]\\frac{{x}^{n}}{n\\text{!}}-\\frac{{\\left(\\text{-}x\\right)}^{n}}{n\\text{!}}[\/latex].<\/div>\r\n\r\n<\/span>\r\nFor [latex]n[\/latex] even, this term is zero. For [latex]n[\/latex] odd, this term is [latex]\\frac{2{x}^{n}}{n\\text{!}}[\/latex]. Therefore, the Maclaurin series for [latex]\\text{sinh}x[\/latex] has only odd-order terms and is given by\r\n<\/span>\r\n
      [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}=x+\\frac{{x}^{3}}{3\\text{!}}+\\frac{{x}^{5}}{5\\text{!}}+\\cdots [\/latex].<\/div>\r\n <\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
      Watch the following video to see the worked solution to the example above.