{"id":119,"date":"2021-03-25T02:21:08","date_gmt":"2021-03-25T02:21:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-2\/"},"modified":"2022-01-03T18:38:11","modified_gmt":"2022-01-03T18:38:11","slug":"taylor-and-maclaurin-series-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/taylor-and-maclaurin-series-2\/","title":{"raw":"Problem Set: Taylor and Maclaurin Series","rendered":"Problem Set: Taylor and Maclaurin Series"},"content":{"raw":"

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]f\\left(x\\right)=1+x+{x}^{2}[\/latex] at [latex]a=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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2.\u00a0<\/strong>[latex]f\\left(x\\right)=1+x+{x}^{2}[\/latex] at [latex]a=-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"754388\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"754388\"][latex]f\\left(-1\\right)=1;{f}^{\\prime }\\left(-1\\right)=-1;f\\text{''}\\left(-1\\right)=2;f\\left(x\\right)=1-\\left(x+1\\right)+{\\left(x+1\\right)}^{2}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]f\\left(x\\right)=\\cos\\left(2x\\right)[\/latex] at [latex]a=\\pi [\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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4.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin\\left(2x\\right)[\/latex] at [latex]a=\\frac{\\pi }{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"15938\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"15938\"][latex]{f}^{\\prime }\\left(x\\right)=2\\cos\\left(2x\\right);f\\text{''}\\left(x\\right)=-4\\sin\\left(2x\\right);{p}_{2}\\left(x\\right)=-2\\left(x-\\frac{\\pi }{2}\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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5.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sqrt{x}[\/latex] at [latex]a=4[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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6.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]a=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"776806\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"776806\"][latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{x};f\\text{''}\\left(x\\right)=-\\frac{1}{{x}^{2}};{p}_{2}\\left(x\\right)=0+\\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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7.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] at [latex]a=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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8.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{x}[\/latex] at [latex]a=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"175844\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"175844\"][latex]{p}_{2}\\left(x\\right)=e+e\\left(x - 1\\right)+\\frac{e}{2}{\\left(x - 1\\right)}^{2}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, verify that the given choice of n<\/em> in the remainder estimate [latex]|{R}_{n}|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where M<\/em> is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between a<\/em> and the indicated point, yields [latex]|{R}_{n}|\\le \\frac{1}{1000}[\/latex]. Find the value of the Taylor polynomial pn<\/sub><\/em> of [latex]f[\/latex] at the indicated point.<\/p>\r\n\r\n

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9. [T]<\/strong> [latex]\\sqrt{10};a=9,n=3[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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10. [T]<\/strong> [latex]{\\left(28\\right)}^{\\frac{1}{3}};a=27,n=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"889787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"889787\"][latex]\\frac{{d}^{2}}{d{x}^{2}}{x}^{\\frac{1}{3}}=-\\frac{2}{9{x}^{\\frac{5}{3}}}\\ge -0.00092\\ldots[\/latex] when [latex]x\\ge 28[\/latex] so the remainder estimate applies to the linear approximation [latex]{x}^{\\frac{1}{3}}\\approx {p}_{1}\\left(27\\right)=3+\\frac{x - 27}{27}[\/latex], which gives [latex]{\\left(28\\right)}^{\\frac{1}{3}}\\approx 3+\\frac{1}{27}=3.\\overline{037}[\/latex], while [latex]{\\left(28\\right)}^{\\frac{1}{3}}\\approx 3.03658[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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11. [T]<\/strong> [latex]\\sin\\left(6\\right);a=2\\pi ,n=5[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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12. [T]<\/strong> e<\/em>2<\/sup>; [latex]a=0,n=9[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"725074\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"725074\"]Using the estimate [latex]\\frac{{2}^{10}}{10\\text{!}}<0.000283[\/latex] we can use the Taylor expansion of order 9 to estimate [latex]e^{x}[\/latex] at [latex]x=2[\/latex]. as [latex]{e}^{2}\\approx {p}_{9}\\left(2\\right)=1+2+\\frac{{2}^{2}}{2}+\\frac{{2}^{3}}{6}+\\cdots+\\frac{{2}^{9}}{9\\text{!}}=7.3887\\ldots[\/latex] whereas [latex]{e}^{2}\\approx 7.3891[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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13. [T]<\/strong> [latex]\\cos\\left(\\frac{\\pi }{5}\\right);a=0,n=4[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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14. [T]<\/strong> [latex]\\text{ln}\\left(2\\right);a=1,n=1000[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"887302\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"887302\"]Since [latex]\\frac{{d}^{n}}{d{x}^{n}}\\left(\\text{ln}x\\right)={\\left(-1\\right)}^{n - 1}\\frac{\\left(n - 1\\right)\\text{!}}{{x}^{n}},{R}_{1000}\\approx \\frac{1}{1001}[\/latex]. One has [latex]{p}_{1000}\\left(1\\right)=\\displaystyle\\sum _{n=1}^{1000}\\frac{{\\left(-1\\right)}^{n - 1}}{n}\\approx 0.6936[\/latex] whereas [latex]\\text{ln}\\left(2\\right)\\approx 0.6931\\cdots[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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15.\u00a0<\/strong>Integrate the approximation [latex]\\sin{t}\\approx t-\\frac{{t}^{3}}{6}+\\frac{{t}^{5}}{120}-\\frac{{t}^{7}}{5040}[\/latex] evaluated at \u03c0t<\/em> to approximate [latex]{\\displaystyle\\int }_{0}^{1}\\frac{\\sin\\pi t}{\\pi t}dt[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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16.\u00a0<\/strong>Integrate the approximation [latex]{e}^{x}\\approx 1+x+\\frac{{x}^{2}}{2}+\\cdots+\\frac{{x}^{6}}{720}[\/latex] evaluated at \u2212x<\/em>2<\/sup> to approximate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"104204\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"104204\"][latex]{\\displaystyle\\int }_{0}^{1}\\left(1-{x}^{2}+\\frac{{x}^{4}}{2}-\\frac{{x}^{6}}{6}+\\frac{{x}^{8}}{24}-\\frac{{x}^{10}}{120}+\\frac{{x}^{12}}{720}\\right)dx[\/latex]<\/p>\r\n[latex]=1-\\frac{{1}^{3}}{3}+\\frac{{1}^{5}}{10}-\\frac{{1}^{7}}{42}+\\frac{{1}^{9}}{9\\cdot 24}-\\frac{{1}^{11}}{120\\cdot 11}+\\frac{{1}^{13}}{720\\cdot 13}\\approx 0.74683[\/latex] whereas [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx\\approx 0.74682[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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In the following exercises, find the smallest value of <\/span>n<\/em> such that the remainder estimate [latex]|{R}_{n}|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where <\/span>M<\/em> is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between <\/span>a<\/em> and the indicated point, yields [latex]|{R}_{n}|\\le \\frac{1}{1000}[\/latex] on the indicated interval.<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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17.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin{x}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right],a=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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18.\u00a0<\/strong>[latex]f\\left(x\\right)=\\cos{x}[\/latex] on [latex]\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right],a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"486315\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"486315\"]Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)[\/latex] is [latex]\\sin{z}[\/latex] or [latex]\\cos{z}[\/latex], we have [latex]M=1[\/latex]. Since [latex]|x - 0|\\le \\frac{\\pi }{2}[\/latex], we seek the smallest n such that [latex]\\frac{{\\pi }^{n+1}}{{2}^{n+1}\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=7[\/latex]. The remainder estimate is [latex]{R}_{7}\\le 0.00092[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

19.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{-2x}[\/latex] on [latex]\\left[-1,1\\right],a=0[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
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20.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex] on [latex]\\left[-3,3\\right],a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"105130\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"105130\"]Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)=\\pm{e}^{\\text{-}z}[\/latex] one has [latex]M={e}^{3}[\/latex]. Since [latex]|x - 0|\\le 3[\/latex], one seeks the smallest n such that [latex]\\frac{{3}^{n+1}{e}^{3}}{\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=14[\/latex]. The remainder estimate is [latex]{R}_{14}\\le 0.000220[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, the maximum of the right-hand side of the remainder estimate [latex]|{R}_{1}|\\le \\frac{\\text{max}|f\\text{''}\\left(z\\right)|}{2}{R}^{2}[\/latex] on [latex]\\left[a-R,a+R\\right][\/latex] occurs at <\/span>a<\/em> or [latex]a\\pm R[\/latex]. Estimate the maximum value of <\/span>R<\/em> such that [latex]\\frac{\\text{max}|f\\text{''}\\left(z\\right)|}{2}{R}^{2}\\le 0.1[\/latex] on [latex]\\left[a-R,a+R\\right][\/latex] by plotting this maximum as a function of <\/span>R<\/em>.<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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21. [T]<\/strong> ex<\/sup><\/em> approximated by [latex]1+x,a=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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22. [T]<\/strong> [latex]\\sin{x}[\/latex] approximated by x<\/em>, [latex]a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"237736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"237736\"]<\/span>\"This<\/p>\r\nSince [latex]\\sin{x}[\/latex] is increasing for small x and since [latex]\\mathrm{si}n\\text{''}x=\\text{-}\\sin{x}[\/latex], the estimate applies whenever [latex]{R}^{2}\\sin\\left(R\\right)\\le 0.2[\/latex], which applies up to [latex]R=0.596[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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23. [T]<\/strong> [latex]\\text{ln}x[\/latex] approximated by [latex]x - 1,a=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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24. [T]<\/strong> [latex]\\cos{x}[\/latex] approximated by [latex]1,a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"26158\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"26158\"]\"This<\/p>\r\nSince the second derivative of [latex]\\cos{x}[\/latex] is [latex]\\text{-}\\cos{x}[\/latex] and since [latex]\\cos{x}[\/latex] is decreasing away from [latex]x=0[\/latex], the estimate applies when [latex]{R}^{2}\\cos{R}\\le 0.2[\/latex] or [latex]R\\le 0.447[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find the Taylor series of the given function centered at the indicated point.<\/p>\r\n\r\n

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25.\u00a0<\/strong>[latex]{x}^{4}[\/latex] at [latex]a=-1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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26.\u00a0<\/strong>[latex]1+x+{x}^{2}+{x}^{3}[\/latex] at [latex]a=-1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"177164\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"177164\"][latex]{\\left(x+1\\right)}^{3}-2{\\left(x+1\\right)}^{2}+2\\left(x+1\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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27.\u00a0<\/strong>[latex]\\sin{x}[\/latex] at [latex]a=\\pi [\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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28.\u00a0<\/strong>[latex]\\cos{x}[\/latex] at [latex]a=2\\pi [\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"170314\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"170314\"]Values of derivatives are the same as for [latex]x=0[\/latex] so [latex]\\cos{x}={\\displaystyle\\sum _{n=0}^{\\infty }\\left(-1\\right)}^{n}\\frac{{\\left(x - 2\\pi \\right)}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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29.\u00a0<\/strong>[latex]\\sin{x}[\/latex] at [latex]x=\\frac{\\pi }{2}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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30.\u00a0<\/strong>[latex]\\cos{x}[\/latex] at [latex]x=\\frac{\\pi }{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"860054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"860054\"][latex]\\cos\\left(\\frac{\\pi }{2}\\right)=0,\\text{-}\\sin\\left(\\frac{\\pi }{2}\\right)=-1[\/latex] so [latex]\\cos{x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{\\left(x-\\frac{\\pi }{2}\\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex], which is also [latex]\\text{-}\\cos\\left(x-\\frac{\\pi }{2}\\right)[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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31.\u00a0<\/strong>[latex]{e}^{x}[\/latex] at [latex]a=-1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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32.\u00a0<\/strong>[latex]{e}^{x}[\/latex] at [latex]a=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"789609\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"789609\"]The derivatives are [latex]{f}^{\\left(n\\right)}\\left(1\\right)=e[\/latex] so [latex]{e}^{x}=e\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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33.\u00a0<\/strong>[latex]\\frac{1}{{\\left(x - 1\\right)}^{2}}[\/latex] at [latex]a=0[\/latex] (Hint:<\/em> Differentiate [latex]\\frac{1}{1-x}.[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n
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34.\u00a0<\/strong>[latex]\\frac{1}{{\\left(x - 1\\right)}^{3}}[\/latex] at [latex]a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"130185\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"130185\"][latex]\\frac{1}{{\\left(x - 1\\right)}^{3}}=\\text{-}\\left(\\frac{1}{2}\\right)\\frac{{d}^{2}}{d{x}^{2}}\\frac{1}{1-x}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\frac{\\left(n+2\\right)\\left(n+1\\right){x}^{n}}{2}\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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35.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\cos\\left(\\sqrt{t}\\right)dt;f\\left(t\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{n}}{\\left(2n\\right)\\text{!}}[\/latex] at [latex]a=0[\/latex] (Note<\/em>: [latex]f[\/latex] is the Taylor series of [latex]\\cos\\left(\\sqrt{t}\\right).[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, compute the Taylor series of each function around [latex]x=1[\/latex].<\/p>\r\n\r\n

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36.\u00a0<\/strong>[latex]f\\left(x\\right)=2-x[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"929411\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"929411\"][latex]2-x=1-\\left(x - 1\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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37.\u00a0<\/strong>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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38.\u00a0<\/strong>[latex]f\\left(x\\right)={\\left(x - 2\\right)}^{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"850290\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"850290\"][latex]{\\left(\\left(x - 1\\right)-1\\right)}^{2}={\\left(x - 1\\right)}^{2}-2\\left(x - 1\\right)+1[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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39.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}x[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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40.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"519610\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"519610\"][latex]\\frac{1}{1-\\left(1-x\\right)}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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41.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{2x-{x}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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42.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{x}{4x - 2{x}^{2}-1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"821048\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"821048\"][latex]x\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(1-x\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n+1}+\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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43.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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44.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"803103\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"803103\"][latex]{e}^{2x}={e}^{2\\left(x - 1\\right)+2}={e}^{2}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{2}^{n}{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

[T]<\/strong> In the following exercises, identify the value of <\/span>x<\/em> such that the given series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}[\/latex] is the value of the Maclaurin series of [latex]f\\left(x\\right)[\/latex] at [latex]x[\/latex]. Approximate the value of [latex]f\\left(x\\right)[\/latex] using [latex]{S}_{10}=\\displaystyle\\sum _{n=0}^{10}{a}_{n}[\/latex].<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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45. <\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{n\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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46.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{2}^{n}}{n\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"795438\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"795438\"][latex]x={e}^{2};{S}_{10}=\\frac{34,913}{4725}\\approx 7.3889947[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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47.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{\\left(2\\pi \\right)}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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48.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{\\left(2\\pi \\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"153059\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"153059\"][latex]\\sin\\left(2\\pi \\right)=0;{S}_{10}=8.27\\times {10}^{-5}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

The following exercises make use of the functions [latex]{S}_{5}\\left(x\\right)=x-\\frac{{x}^{3}}{6}+\\frac{{x}^{5}}{120}[\/latex] and [latex]{C}_{4}\\left(x\\right)=1-\\frac{{x}^{2}}{2}+\\frac{{x}^{4}}{24}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex].<\/p>\r\n\r\n

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49. [T]<\/strong> Plot [latex]{\\sin}^{2}x-{\\left({S}_{5}\\left(x\\right)\\right)}^{2}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex]. Compare the maximum difference with the square of the Taylor remainder estimate for [latex]\\sin{x}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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50. [T]<\/strong> Plot [latex]{\\cos}^{2}x-{\\left({C}_{4}\\left(x\\right)\\right)}^{2}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex]. Compare the maximum difference with the square of the Taylor remainder estimate for [latex]\\cos{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"791988\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"791988\"]<\/span>\"This<\/p>\r\nThe difference is small on the interior of the interval but approaches [latex]1[\/latex] near the endpoints. The remainder estimate is [latex]|{R}_{4}|=\\frac{{\\pi }^{5}}{120}\\approx 2.552[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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51. [T]<\/strong> Plot [latex]|2{S}_{5}\\left(x\\right){C}_{4}\\left(x\\right)-\\sin\\left(2x\\right)|[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
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52. [T]<\/strong> Compare [latex]\\frac{{S}_{5}\\left(x\\right)}{{C}_{4}\\left(x\\right)}[\/latex] on [latex]\\left[-1,1\\right][\/latex] to [latex]\\tan{x}[\/latex]. Compare this with the Taylor remainder estimate for the approximation of [latex]\\tan{x}[\/latex] by [latex]x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

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[reveal-answer q=\"124883\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"124883\"]<\/span>\"This<\/p>\r\nThe difference is on the order of [latex]{10}^{-4}[\/latex] on [latex]\\left[-1,1\\right][\/latex] while the Taylor approximation error is around [latex]0.1[\/latex] near [latex]\\pm 1[\/latex]. The top curve is a plot of [latex]{\\tan}^{2}x-{\\left(\\frac{{S}_{5}\\left(x\\right)}{{C}_{4}\\left(x\\right)}\\right)}^{2}[\/latex] and the lower dashed plot shows [latex]{t}^{2}-{\\left(\\frac{{S}_{5}}{{C}_{4}}\\right)}^{2}[\/latex].[\/hidden-answer]<\/span><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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53. [T]<\/strong> Plot [latex]{e}^{x}-{e}_{4}\\left(x\\right)[\/latex] where [latex]{e}_{4}\\left(x\\right)=1+x+\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{6}+\\frac{{x}^{4}}{24}[\/latex] on [latex]\\left[0,2\\right][\/latex]. Compare the maximum error with the Taylor remainder estimate.<\/div>\r\n<\/div>\r\n<\/div>\r\n
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54.\u00a0<\/strong>(Taylor approximations and root finding.) Recall that Newton\u2019s method [latex]{x}_{n+1}={x}_{n}-\\frac{f\\left({x}_{n}\\right)}{f\\prime \\left({x}_{n}\\right)}[\/latex] approximates solutions of [latex]f\\left(x\\right)=0[\/latex] near the input [latex]{x}_{0}[\/latex].<\/p>\r\n\r\n

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  1. If [latex]f[\/latex] and [latex]g[\/latex] are inverse functions, explain why a solution of [latex]g\\left(x\\right)=a[\/latex] is the value [latex]f\\left(a\\right)\\text{of}f[\/latex].<\/li>\r\n \t
  2. Let [latex]{p}_{N}\\left(x\\right)[\/latex] be the [latex]N\\text{th}[\/latex] degree Maclaurin polynomial of [latex]{e}^{x}[\/latex]. Use Newton\u2019s method to approximate solutions of [latex]{p}_{N}\\left(x\\right)-2=0[\/latex] for [latex]N=4,5,6[\/latex].<\/li>\r\n \t
  3. Explain why the approximate roots of [latex]{p}_{N}\\left(x\\right)-2=0[\/latex] are approximate values of [latex]\\text{ln}\\left(2\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n
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    [reveal-answer q=\"312868\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"312868\"]a. Answers will vary. b. The following are the [latex]{x}_{n}[\/latex] values after [latex]10[\/latex] iterations of Newton\u2019s method to approximation a root of [latex]{p}_{N}\\left(x\\right)-2=0\\text{:}[\/latex] for [latex]N=4,x=0.6939...[\/latex]; for [latex]N=5,x=0.6932...[\/latex]; for [latex]N=6,x=0.69315...[\/latex];. (Note: [latex]\\text{ln}\\left(2\\right)=0.69314...[\/latex]) c. Answers will vary.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    In the following exercises, use the fact that if [latex]q\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges in an interval containing [latex]c[\/latex], then [latex]\\underset{x\\to c}{\\text{lim}}q\\left(x\\right)={a}_{0}^{}[\/latex] to evaluate each limit using Taylor series.<\/p>\r\n\r\n

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    55.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\text{lim}}\\frac{\\cos{x} - 1}{{x}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    56.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\text{lim}}\\frac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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    [reveal-answer q=\"899438\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"899438\"][latex]\\frac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}\\to \\text{-}1[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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    57.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\text{lim}}\\frac{{e}^{{x}^{2}}-{x}^{2}-1}{{x}^{4}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
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    58.\u00a0<\/strong>[latex]\\underset{x\\to {0}^{+}}{\\text{lim}}\\frac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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    [reveal-answer q=\"113846\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"113846\"][latex]\\frac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}\\approx \\frac{\\left(1-\\frac{x}{2}+\\frac{{x}^{2}}{4\\text{!}}-\\cdots\\right)-1}{2x}\\to -\\frac{1}{4}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

    In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.<\/p>\n

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    1.\u00a0<\/strong>[latex]f\\left(x\\right)=1+x+{x}^{2}[\/latex] at [latex]a=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    2.\u00a0<\/strong>[latex]f\\left(x\\right)=1+x+{x}^{2}[\/latex] at [latex]a=-1[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]f\\left(-1\\right)=1;{f}^{\\prime }\\left(-1\\right)=-1;f\\text{''}\\left(-1\\right)=2;f\\left(x\\right)=1-\\left(x+1\\right)+{\\left(x+1\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    3.\u00a0<\/strong>[latex]f\\left(x\\right)=\\cos\\left(2x\\right)[\/latex] at [latex]a=\\pi[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    4.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin\\left(2x\\right)[\/latex] at [latex]a=\\frac{\\pi }{2}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{f}^{\\prime }\\left(x\\right)=2\\cos\\left(2x\\right);f\\text{''}\\left(x\\right)=-4\\sin\\left(2x\\right);{p}_{2}\\left(x\\right)=-2\\left(x-\\frac{\\pi }{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    5.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sqrt{x}[\/latex] at [latex]a=4[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    6.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]a=1[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{f}^{\\prime }\\left(x\\right)=\\frac{1}{x};f\\text{''}\\left(x\\right)=-\\frac{1}{{x}^{2}};{p}_{2}\\left(x\\right)=0+\\left(x - 1\\right)-\\frac{1}{2}{\\left(x - 1\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    7.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] at [latex]a=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    8.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{x}[\/latex] at [latex]a=1[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{p}_{2}\\left(x\\right)=e+e\\left(x - 1\\right)+\\frac{e}{2}{\\left(x - 1\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, verify that the given choice of n<\/em> in the remainder estimate [latex]|{R}_{n}|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where M<\/em> is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between a<\/em> and the indicated point, yields [latex]|{R}_{n}|\\le \\frac{1}{1000}[\/latex]. Find the value of the Taylor polynomial pn<\/sub><\/em> of [latex]f[\/latex] at the indicated point.<\/p>\n

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    9. [T]<\/strong> [latex]\\sqrt{10};a=9,n=3[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    10. [T]<\/strong> [latex]{\\left(28\\right)}^{\\frac{1}{3}};a=27,n=1[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\frac{{d}^{2}}{d{x}^{2}}{x}^{\\frac{1}{3}}=-\\frac{2}{9{x}^{\\frac{5}{3}}}\\ge -0.00092\\ldots[\/latex] when [latex]x\\ge 28[\/latex] so the remainder estimate applies to the linear approximation [latex]{x}^{\\frac{1}{3}}\\approx {p}_{1}\\left(27\\right)=3+\\frac{x - 27}{27}[\/latex], which gives [latex]{\\left(28\\right)}^{\\frac{1}{3}}\\approx 3+\\frac{1}{27}=3.\\overline{037}[\/latex], while [latex]{\\left(28\\right)}^{\\frac{1}{3}}\\approx 3.03658[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    11. [T]<\/strong> [latex]\\sin\\left(6\\right);a=2\\pi ,n=5[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    12. [T]<\/strong> e<\/em>2<\/sup>; [latex]a=0,n=9[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Using the estimate [latex]\\frac{{2}^{10}}{10\\text{!}}<0.000283[\/latex] we can use the Taylor expansion of order 9 to estimate [latex]e^{x}[\/latex] at [latex]x=2[\/latex]. as [latex]{e}^{2}\\approx {p}_{9}\\left(2\\right)=1+2+\\frac{{2}^{2}}{2}+\\frac{{2}^{3}}{6}+\\cdots+\\frac{{2}^{9}}{9\\text{!}}=7.3887\\ldots[\/latex] whereas [latex]{e}^{2}\\approx 7.3891[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    13. [T]<\/strong> [latex]\\cos\\left(\\frac{\\pi }{5}\\right);a=0,n=4[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    14. [T]<\/strong> [latex]\\text{ln}\\left(2\\right);a=1,n=1000[\/latex]<\/p>\n<\/div>\n

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    Since [latex]\\frac{{d}^{n}}{d{x}^{n}}\\left(\\text{ln}x\\right)={\\left(-1\\right)}^{n - 1}\\frac{\\left(n - 1\\right)\\text{!}}{{x}^{n}},{R}_{1000}\\approx \\frac{1}{1001}[\/latex]. One has [latex]{p}_{1000}\\left(1\\right)=\\displaystyle\\sum _{n=1}^{1000}\\frac{{\\left(-1\\right)}^{n - 1}}{n}\\approx 0.6936[\/latex] whereas [latex]\\text{ln}\\left(2\\right)\\approx 0.6931\\cdots[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    15.\u00a0<\/strong>Integrate the approximation [latex]\\sin{t}\\approx t-\\frac{{t}^{3}}{6}+\\frac{{t}^{5}}{120}-\\frac{{t}^{7}}{5040}[\/latex] evaluated at \u03c0t<\/em> to approximate [latex]{\\displaystyle\\int }_{0}^{1}\\frac{\\sin\\pi t}{\\pi t}dt[\/latex].<\/div>\n<\/div>\n<\/div>\n
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    16.\u00a0<\/strong>Integrate the approximation [latex]{e}^{x}\\approx 1+x+\\frac{{x}^{2}}{2}+\\cdots+\\frac{{x}^{6}}{720}[\/latex] evaluated at \u2212x<\/em>2<\/sup> to approximate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx[\/latex].<\/p>\n<\/div>\n

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    [latex]{\\displaystyle\\int }_{0}^{1}\\left(1-{x}^{2}+\\frac{{x}^{4}}{2}-\\frac{{x}^{6}}{6}+\\frac{{x}^{8}}{24}-\\frac{{x}^{10}}{120}+\\frac{{x}^{12}}{720}\\right)dx[\/latex]<\/p>\n

    [latex]=1-\\frac{{1}^{3}}{3}+\\frac{{1}^{5}}{10}-\\frac{{1}^{7}}{42}+\\frac{{1}^{9}}{9\\cdot 24}-\\frac{{1}^{11}}{120\\cdot 11}+\\frac{{1}^{13}}{720\\cdot 13}\\approx 0.74683[\/latex] whereas [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx\\approx 0.74682[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    In the following exercises, find the smallest value of <\/span>n<\/em> such that the remainder estimate [latex]|{R}_{n}|\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where <\/span>M<\/em> is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between <\/span>a<\/em> and the indicated point, yields [latex]|{R}_{n}|\\le \\frac{1}{1000}[\/latex] on the indicated interval.<\/span><\/p>\n<\/div>\n<\/div>\n

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    17.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin{x}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right],a=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    18.\u00a0<\/strong>[latex]f\\left(x\\right)=\\cos{x}[\/latex] on [latex]\\left[-\\frac{\\pi }{2},\\frac{\\pi }{2}\\right],a=0[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)[\/latex] is [latex]\\sin{z}[\/latex] or [latex]\\cos{z}[\/latex], we have [latex]M=1[\/latex]. Since [latex]|x - 0|\\le \\frac{\\pi }{2}[\/latex], we seek the smallest n such that [latex]\\frac{{\\pi }^{n+1}}{{2}^{n+1}\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=7[\/latex]. The remainder estimate is [latex]{R}_{7}\\le 0.00092[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    19.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{-2x}[\/latex] on [latex]\\left[-1,1\\right],a=0[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
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    20.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex] on [latex]\\left[-3,3\\right],a=0[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)=\\pm{e}^{\\text{-}z}[\/latex] one has [latex]M={e}^{3}[\/latex]. Since [latex]|x - 0|\\le 3[\/latex], one seeks the smallest n such that [latex]\\frac{{3}^{n+1}{e}^{3}}{\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=14[\/latex]. The remainder estimate is [latex]{R}_{14}\\le 0.000220[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, the maximum of the right-hand side of the remainder estimate [latex]|{R}_{1}|\\le \\frac{\\text{max}|f\\text{''}\\left(z\\right)|}{2}{R}^{2}[\/latex] on [latex]\\left[a-R,a+R\\right][\/latex] occurs at <\/span>a<\/em> or [latex]a\\pm R[\/latex]. Estimate the maximum value of <\/span>R<\/em> such that [latex]\\frac{\\text{max}|f\\text{''}\\left(z\\right)|}{2}{R}^{2}\\le 0.1[\/latex] on [latex]\\left[a-R,a+R\\right][\/latex] by plotting this maximum as a function of <\/span>R<\/em>.<\/span><\/p>\n<\/div>\n<\/div>\n

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    21. [T]<\/strong> ex<\/sup><\/em> approximated by [latex]1+x,a=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    22. [T]<\/strong> [latex]\\sin{x}[\/latex] approximated by x<\/em>, [latex]a=0[\/latex]<\/p>\n<\/div>\n

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    <\/p>\n

    Show Solution<\/span><\/p>\n
    <\/span>\"This<\/p>\n

    Since [latex]\\sin{x}[\/latex] is increasing for small x and since [latex]\\mathrm{si}n\\text{''}x=\\text{-}\\sin{x}[\/latex], the estimate applies whenever [latex]{R}^{2}\\sin\\left(R\\right)\\le 0.2[\/latex], which applies up to [latex]R=0.596[\/latex].<\/div>\n<\/div>\n

    <\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    23. [T]<\/strong> [latex]\\text{ln}x[\/latex] approximated by [latex]x - 1,a=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    24. [T]<\/strong> [latex]\\cos{x}[\/latex] approximated by [latex]1,a=0[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    \"This<\/p>\n

    Since the second derivative of [latex]\\cos{x}[\/latex] is [latex]\\text{-}\\cos{x}[\/latex] and since [latex]\\cos{x}[\/latex] is decreasing away from [latex]x=0[\/latex], the estimate applies when [latex]{R}^{2}\\cos{R}\\le 0.2[\/latex] or [latex]R\\le 0.447[\/latex].<\/div>\n<\/div>\n

    <\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, find the Taylor series of the given function centered at the indicated point.<\/p>\n

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    25.\u00a0<\/strong>[latex]{x}^{4}[\/latex] at [latex]a=-1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    26.\u00a0<\/strong>[latex]1+x+{x}^{2}+{x}^{3}[\/latex] at [latex]a=-1[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{\\left(x+1\\right)}^{3}-2{\\left(x+1\\right)}^{2}+2\\left(x+1\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    27.\u00a0<\/strong>[latex]\\sin{x}[\/latex] at [latex]a=\\pi[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    28.\u00a0<\/strong>[latex]\\cos{x}[\/latex] at [latex]a=2\\pi[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    Values of derivatives are the same as for [latex]x=0[\/latex] so [latex]\\cos{x}={\\displaystyle\\sum _{n=0}^{\\infty }\\left(-1\\right)}^{n}\\frac{{\\left(x - 2\\pi \\right)}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    29.\u00a0<\/strong>[latex]\\sin{x}[\/latex] at [latex]x=\\frac{\\pi }{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    30.\u00a0<\/strong>[latex]\\cos{x}[\/latex] at [latex]x=\\frac{\\pi }{2}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\cos\\left(\\frac{\\pi }{2}\\right)=0,\\text{-}\\sin\\left(\\frac{\\pi }{2}\\right)=-1[\/latex] so [latex]\\cos{x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{\\left(x-\\frac{\\pi }{2}\\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex], which is also [latex]\\text{-}\\cos\\left(x-\\frac{\\pi }{2}\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    31.\u00a0<\/strong>[latex]{e}^{x}[\/latex] at [latex]a=-1[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    32.\u00a0<\/strong>[latex]{e}^{x}[\/latex] at [latex]a=1[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    The derivatives are [latex]{f}^{\\left(n\\right)}\\left(1\\right)=e[\/latex] so [latex]{e}^{x}=e\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    33.\u00a0<\/strong>[latex]\\frac{1}{{\\left(x - 1\\right)}^{2}}[\/latex] at [latex]a=0[\/latex] (Hint:<\/em> Differentiate [latex]\\frac{1}{1-x}.[\/latex])<\/div>\n<\/div>\n<\/div>\n
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    34.\u00a0<\/strong>[latex]\\frac{1}{{\\left(x - 1\\right)}^{3}}[\/latex] at [latex]a=0[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\frac{1}{{\\left(x - 1\\right)}^{3}}=\\text{-}\\left(\\frac{1}{2}\\right)\\frac{{d}^{2}}{d{x}^{2}}\\frac{1}{1-x}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\frac{\\left(n+2\\right)\\left(n+1\\right){x}^{n}}{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    35.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\cos\\left(\\sqrt{t}\\right)dt;f\\left(t\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{n}}{\\left(2n\\right)\\text{!}}[\/latex] at [latex]a=0[\/latex] (Note<\/em>: [latex]f[\/latex] is the Taylor series of [latex]\\cos\\left(\\sqrt{t}\\right).[\/latex])<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, compute the Taylor series of each function around [latex]x=1[\/latex].<\/p>\n

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    36.\u00a0<\/strong>[latex]f\\left(x\\right)=2-x[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]2-x=1-\\left(x - 1\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    37.\u00a0<\/strong>[latex]f\\left(x\\right)={x}^{3}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    38.\u00a0<\/strong>[latex]f\\left(x\\right)={\\left(x - 2\\right)}^{2}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{\\left(\\left(x - 1\\right)-1\\right)}^{2}={\\left(x - 1\\right)}^{2}-2\\left(x - 1\\right)+1[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    39.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}x[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    40.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{x}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\frac{1}{1-\\left(1-x\\right)}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    41.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{1}{2x-{x}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    42.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{x}{4x - 2{x}^{2}-1}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]x\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(1-x\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n+1}+\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    43.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    44.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{2x}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]{e}^{2x}={e}^{2\\left(x - 1\\right)+2}={e}^{2}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{2}^{n}{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    [T]<\/strong> In the following exercises, identify the value of <\/span>x<\/em> such that the given series [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}[\/latex] is the value of the Maclaurin series of [latex]f\\left(x\\right)[\/latex] at [latex]x[\/latex]. Approximate the value of [latex]f\\left(x\\right)[\/latex] using [latex]{S}_{10}=\\displaystyle\\sum _{n=0}^{10}{a}_{n}[\/latex].<\/span><\/p>\n<\/div>\n<\/div>\n

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    45. <\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{1}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    46.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{2}^{n}}{n\\text{!}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]x={e}^{2};{S}_{10}=\\frac{34,913}{4725}\\approx 7.3889947[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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    47.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{\\left(2\\pi \\right)}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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    48.\u00a0<\/strong>[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(-1\\right)}^{n}{\\left(2\\pi \\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    [latex]\\sin\\left(2\\pi \\right)=0;{S}_{10}=8.27\\times {10}^{-5}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    The following exercises make use of the functions [latex]{S}_{5}\\left(x\\right)=x-\\frac{{x}^{3}}{6}+\\frac{{x}^{5}}{120}[\/latex] and [latex]{C}_{4}\\left(x\\right)=1-\\frac{{x}^{2}}{2}+\\frac{{x}^{4}}{24}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex].<\/p>\n

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    49. [T]<\/strong> Plot [latex]{\\sin}^{2}x-{\\left({S}_{5}\\left(x\\right)\\right)}^{2}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex]. Compare the maximum difference with the square of the Taylor remainder estimate for [latex]\\sin{x}[\/latex].<\/div>\n<\/div>\n<\/div>\n
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    50. [T]<\/strong> Plot [latex]{\\cos}^{2}x-{\\left({C}_{4}\\left(x\\right)\\right)}^{2}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex]. Compare the maximum difference with the square of the Taylor remainder estimate for [latex]\\cos{x}[\/latex].<\/p>\n<\/div>\n

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    Show Solution<\/span><\/p>\n
    <\/span>\"This<\/p>\n

    The difference is small on the interior of the interval but approaches [latex]1[\/latex] near the endpoints. The remainder estimate is [latex]|{R}_{4}|=\\frac{{\\pi }^{5}}{120}\\approx 2.552[\/latex].<\/div>\n<\/div>\n

    <\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    51. [T]<\/strong> Plot [latex]|2{S}_{5}\\left(x\\right){C}_{4}\\left(x\\right)-\\sin\\left(2x\\right)|[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex].<\/div>\n<\/div>\n<\/div>\n
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    52. [T]<\/strong> Compare [latex]\\frac{{S}_{5}\\left(x\\right)}{{C}_{4}\\left(x\\right)}[\/latex] on [latex]\\left[-1,1\\right][\/latex] to [latex]\\tan{x}[\/latex]. Compare this with the Taylor remainder estimate for the approximation of [latex]\\tan{x}[\/latex] by [latex]x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}[\/latex].<\/p>\n<\/div>\n

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    <\/p>\n

    Show Solution<\/span><\/p>\n
    <\/span>\"This<\/p>\n

    The difference is on the order of [latex]{10}^{-4}[\/latex] on [latex]\\left[-1,1\\right][\/latex] while the Taylor approximation error is around [latex]0.1[\/latex] near [latex]\\pm 1[\/latex]. The top curve is a plot of [latex]{\\tan}^{2}x-{\\left(\\frac{{S}_{5}\\left(x\\right)}{{C}_{4}\\left(x\\right)}\\right)}^{2}[\/latex] and the lower dashed plot shows [latex]{t}^{2}-{\\left(\\frac{{S}_{5}}{{C}_{4}}\\right)}^{2}[\/latex].<\/div>\n<\/div>\n

    <\/span><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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    53. [T]<\/strong> Plot [latex]{e}^{x}-{e}_{4}\\left(x\\right)[\/latex] where [latex]{e}_{4}\\left(x\\right)=1+x+\\frac{{x}^{2}}{2}+\\frac{{x}^{3}}{6}+\\frac{{x}^{4}}{24}[\/latex] on [latex]\\left[0,2\\right][\/latex]. Compare the maximum error with the Taylor remainder estimate.<\/div>\n<\/div>\n<\/div>\n
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    54.\u00a0<\/strong>(Taylor approximations and root finding.) Recall that Newton\u2019s method [latex]{x}_{n+1}={x}_{n}-\\frac{f\\left({x}_{n}\\right)}{f\\prime \\left({x}_{n}\\right)}[\/latex] approximates solutions of [latex]f\\left(x\\right)=0[\/latex] near the input [latex]{x}_{0}[\/latex].<\/p>\n

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    1. If [latex]f[\/latex] and [latex]g[\/latex] are inverse functions, explain why a solution of [latex]g\\left(x\\right)=a[\/latex] is the value [latex]f\\left(a\\right)\\text{of}f[\/latex].<\/li>\n
    2. Let [latex]{p}_{N}\\left(x\\right)[\/latex] be the [latex]N\\text{th}[\/latex] degree Maclaurin polynomial of [latex]{e}^{x}[\/latex]. Use Newton\u2019s method to approximate solutions of [latex]{p}_{N}\\left(x\\right)-2=0[\/latex] for [latex]N=4,5,6[\/latex].<\/li>\n
    3. Explain why the approximate roots of [latex]{p}_{N}\\left(x\\right)-2=0[\/latex] are approximate values of [latex]\\text{ln}\\left(2\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n
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      Show Solution<\/span><\/p>\n
      a. Answers will vary. b. The following are the [latex]{x}_{n}[\/latex] values after [latex]10[\/latex] iterations of Newton\u2019s method to approximation a root of [latex]{p}_{N}\\left(x\\right)-2=0\\text{:}[\/latex] for [latex]N=4,x=0.6939...[\/latex]; for [latex]N=5,x=0.6932...[\/latex]; for [latex]N=6,x=0.69315...[\/latex];. (Note: [latex]\\text{ln}\\left(2\\right)=0.69314...[\/latex]) c. Answers will vary.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises, use the fact that if [latex]q\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges in an interval containing [latex]c[\/latex], then [latex]\\underset{x\\to c}{\\text{lim}}q\\left(x\\right)={a}_{0}^{}[\/latex] to evaluate each limit using Taylor series.<\/p>\n

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      55.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\text{lim}}\\frac{\\cos{x} - 1}{{x}^{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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      56.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\text{lim}}\\frac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}[\/latex]<\/p>\n<\/div>\n

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      Show Solution<\/span><\/p>\n
      [latex]\\frac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}\\to \\text{-}1[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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      57.\u00a0<\/strong>[latex]\\underset{x\\to 0}{\\text{lim}}\\frac{{e}^{{x}^{2}}-{x}^{2}-1}{{x}^{4}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
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      58.\u00a0<\/strong>[latex]\\underset{x\\to {0}^{+}}{\\text{lim}}\\frac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}[\/latex]<\/p>\n<\/div>\n

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      Show Solution<\/span><\/p>\n
      [latex]\\frac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}\\approx \\frac{\\left(1-\\frac{x}{2}+\\frac{{x}^{2}}{4\\text{!}}-\\cdots\\right)-1}{2x}\\to -\\frac{1}{4}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
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