{"id":131,"date":"2021-03-25T02:21:10","date_gmt":"2021-03-25T02:21:10","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/working-with-taylor-series-2\/"},"modified":"2022-01-03T18:46:31","modified_gmt":"2022-01-03T18:46:31","slug":"working-with-taylor-series-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/working-with-taylor-series-2\/","title":{"raw":"Problem Set: Working with Taylor Series","rendered":"Problem Set: Working with Taylor Series"},"content":{"raw":"

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.<\/p>\r\n\r\n

\r\n
\r\n
1.\u00a0<\/strong>[latex]{\\left(1-x\\right)}^{\\frac{1}{3}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

2.\u00a0<\/strong>[latex]{\\left(1+{x}^{2}\\right)}^{\\frac{-1}{3}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"523625\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"523625\"][latex]{\\left(1+{x}^{2}\\right)}^{\\frac{-1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}-\\frac{1}{3}\\hfill \\\\ \\hfill n\\hfill \\end{array}\\right){x}^{2n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
3.\u00a0<\/strong>[latex]{\\left(1-x\\right)}^{1.01}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

4.\u00a0<\/strong>[latex]{\\left(1 - 2x\\right)}^{\\frac{2}{3}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"850002\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"850002\"][latex]{\\left(1 - 2x\\right)}^{\\frac{2}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{2}^{n}\\left(\\begin{array}{c}\\frac{2}{3}\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex][\/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 substitution [latex]{\\left(b+x\\right)}^{r}={\\left(b+a\\right)}^{r}{\\left(1+\\frac{x-a}{b+a}\\right)}^{r}[\/latex] in the binomial expansion to find the Taylor series of each function with the given center.<\/p>\r\n\r\n

\r\n
\r\n
5.\u00a0<\/strong>[latex]\\sqrt{x+2}[\/latex] at [latex]a=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

6.\u00a0<\/strong>[latex]\\sqrt{{x}^{2}+2}[\/latex] at [latex]a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"182012\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"182012\"][latex]\\sqrt{2+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{\\left(\\frac{1}{2}\\right)-n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{2n};\\left(|{x}^{2}|<2\\right)[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
7.\u00a0<\/strong>[latex]\\sqrt{x+2}[\/latex] at [latex]a=1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

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

\r\n

[reveal-answer q=\"50062\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"50062\"][latex]\\sqrt{2x-{x}^{2}}=\\sqrt{1-{\\left(x - 1\\right)}^{2}}[\/latex] so [latex]\\sqrt{2x-{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){\\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

\r\n
\r\n
9.\u00a0<\/strong>[latex]{\\left(x - 8\\right)}^{\\frac{1}{3}}[\/latex] at [latex]a=9[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

10.\u00a0<\/strong>[latex]\\sqrt{x}[\/latex] at [latex]a=4[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"800620\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"800620\"][latex]\\sqrt{x}=2\\sqrt{1+\\frac{x - 4}{4}}[\/latex] so [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{1 - 2n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){\\left(x - 4\\right)}^{n}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
11.\u00a0<\/strong>[latex]{x}^{\\frac{1}{3}}[\/latex] at [latex]a=27[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

12.\u00a0<\/strong>[latex]\\sqrt{x}[\/latex] at [latex]x=9[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"486821\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"486821\"][latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{3}^{1 - 3n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){\\left(x - 9\\right)}^{n}[\/latex][\/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 binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most [latex]\\frac{1}{1000}[\/latex].<\/p>\r\n\r\n

\r\n
\r\n
13. [T]<\/strong> [latex]{\\left(15\\right)}^{\\frac{1}{4}}[\/latex] using [latex]{\\left(16-x\\right)}^{\\frac{1}{4}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

14. [T]<\/strong> [latex]{\\left(1001\\right)}^{\\frac{1}{3}}[\/latex] using [latex]{\\left(1000+x\\right)}^{\\frac{1}{3}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"598884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"598884\"][latex]10{\\left(1+\\frac{x}{1000}\\right)}^{\\frac{1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{10}^{1 - 3n}\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]. Using, for example, a fourth-degree estimate at [latex]x=1[\/latex] gives [latex]\\begin{array}{cc}\\hfill {\\left(1001\\right)}^{\\frac{1}{3}}& \\approx 10\\left(1+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 1\\hfill \\end{array}\\right){10}^{-3}+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 2\\hfill \\end{array}\\right){10}^{-6}+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 3\\hfill \\end{array}\\right){10}^{-9}+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 4\\hfill \\end{array}\\right){10}^{-12}\\right)\\hfill \\\\ & =10\\left(1+\\frac{1}{{3.10}^{3}}-\\frac{1}{{9.10}^{6}}+\\frac{5}{{81.10}^{9}}-\\frac{10}{{243.10}^{12}}\\right)=10.00333222...\\hfill \\end{array}[\/latex] whereas [latex]{\\left(1001\\right)}^{\\frac{1}{3}}=10.00332222839093...[\/latex]. Two terms would suffice for three-digit accuracy.[\/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 binomial approximation [latex]\\sqrt{1-x}\\approx 1-\\frac{x}{2}-\\frac{{x}^{2}}{8}-\\frac{{x}^{3}}{16}-\\frac{5{x}^{4}}{128}-\\frac{7{x}^{5}}{256}[\/latex] for [latex]|x|<1[\/latex] to approximate each number. Compare this value to the value given by a scientific calculator.<\/p>\r\n\r\n

\r\n
\r\n
15. [T]<\/strong> [latex]\\frac{1}{\\sqrt{2}}[\/latex] using [latex]x=\\frac{1}{2}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

16. [T]<\/strong> [latex]\\sqrt{5}=5\\times \\frac{1}{\\sqrt{5}}[\/latex] using [latex]x=\\frac{4}{5}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"580992\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"580992\"]The approximation is [latex]2.3152[\/latex]; the CAS value is [latex]2.23\\ldots[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
17. [T]<\/strong> [latex]\\sqrt{3}=\\frac{3}{\\sqrt{3}}[\/latex] using [latex]x=\\frac{2}{3}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

18. [T]<\/strong> [latex]\\sqrt{6}[\/latex] using [latex]x=\\frac{5}{6}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"21959\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"21959\"]The approximation is [latex]2.583\\ldots[\/latex]; the CAS value is [latex]2.449\\ldots[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
19.\u00a0<\/strong>Integrate the binomial approximation of [latex]\\sqrt{1-x}[\/latex] to find an approximation of [latex]{\\displaystyle\\int }_{0}^{x}\\sqrt{1-t}dt[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

20. [T]<\/strong> Recall that the graph of [latex]\\sqrt{1-{x}^{2}}[\/latex] is an upper semicircle of radius [latex]1[\/latex]. Integrate the binomial approximation of [latex]\\sqrt{1-{x}^{2}}[\/latex] up to order [latex]8[\/latex] from [latex]x=-1[\/latex] to [latex]x=1[\/latex] to estimate [latex]\\frac{\\pi }{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"438518\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"438518\"][latex]\\sqrt{1-{x}^{2}}=1-\\frac{{x}^{2}}{2}-\\frac{{x}^{4}}{8}-\\frac{{x}^{6}}{16}-\\frac{5{x}^{8}}{128}+\\cdots[\/latex]. Thus [latex]{\\displaystyle\\int }_{-1}^{1}\\sqrt{1-{x}^{2}}dx=x-\\frac{{x}^{3}}{6}-\\frac{{x}^{5}}{40}-\\frac{{x}^{7}}{7\\cdot 16}-\\frac{5{x}^{9}}{9\\cdot 128}+\\cdots{|}_{-1}^{1}\\approx 2-\\frac{1}{3}-\\frac{1}{20}-\\frac{1}{56}-\\frac{10}{9\\cdot 128}+\\text{error}=1.590..[\/latex]. whereas [latex]\\frac{\\pi }{2}=1.570..[\/latex].[\/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 expansion [latex]{\\left(1+x\\right)}^{\\frac{1}{3}}=1+\\frac{1}{3}x-\\frac{1}{9}{x}^{2}+\\frac{5}{81}{x}^{3}-\\frac{10}{243}{x}^{4}+\\cdots[\/latex] to write the first five terms (not necessarily a quartic polynomial) of each expression.<\/p>\r\n\r\n

\r\n
\r\n
21.\u00a0<\/strong>[latex]{\\left(1+4x\\right)}^{\\frac{1}{3}};a=0[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

22.\u00a0<\/strong>[latex]{\\left(1+4x\\right)}^{\\frac{4}{3}};a=0[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"991536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"991536\"][latex]{\\left(1+4x\\right)}^{\\frac{4}{3}}=\\left(1+4x\\right)\\left(1+4x\\right)^{\\frac{1}{3}}=\\left(1+4x\\right)\\left(1+\\frac{4x}{3}-\\frac{16x^{3}}{9}+\\frac{320x^{3}}{81}-\\frac{2560x^{4}}{243}\\right)=1+\\frac{16}{3}x+\\frac{32}{9}x^{2}-\\frac{256}{81}x^{3}+\\frac{1280}{243}x^{4}-\\frac{10240}{243}x^{5}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
23.\u00a0<\/strong>[latex]{\\left(3+2x\\right)}^{\\frac{1}{3}};a=-1[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

24.\u00a0<\/strong>[latex]{\\left({x}^{2}+6x+10\\right)}^{\\frac{1}{3}};a=-3[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"491111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"491111\"][latex]\\left(1+{\\left(x+3\\right)}^{2}\\right)^{\\frac{1}{3}} =1+\\frac{1}{3}{\\left(x+3\\right)}^{2}-\\frac{1}{9}{\\left(x+3\\right)}^{4}+\\frac{5}{81}{\\left(x+3\\right)}^{6}-\\frac{10}{243}{\\left(x+3\\right)}^{8}+\\cdots [\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
25.\u00a0<\/strong>Use [latex]{\\left(1+x\\right)}^{\\frac{1}{3}}=1+\\frac{1}{3}x-\\frac{1}{9}{x}^{2}+\\frac{5}{81}{x}^{3}-\\frac{10}{243}{x}^{4}+\\cdots[\/latex] with [latex]x=1[\/latex] to approximate [latex]{2}^{\\frac{1}{3}}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

26.\u00a0<\/strong>Use the approximation [latex]{\\left(1-x\\right)}^{\\frac{2}{3}}=1-\\frac{2x}{3}-\\frac{{x}^{2}}{9}-\\frac{4{x}^{3}}{81}-\\frac{7{x}^{4}}{243}-\\frac{14{x}^{5}}{729}+\\cdots [\/latex] for [latex]|x|<1[\/latex] to approximate [latex]{2}^{\\frac{1}{3}}={2.2}^{\\frac{-2}{3}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"94833\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"94833\"]Twice the approximation is [latex]1.260\\ldots[\/latex] whereas [latex]{2}^{\\frac{1}{3}}=1.2599...[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
27.\u00a0<\/strong>Find the [latex]25\\text{th}[\/latex] derivative of [latex]f\\left(x\\right)={\\left(1+{x}^{2}\\right)}^{13}[\/latex] at [latex]x=0[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

28.\u00a0<\/strong>Find the [latex]99[\/latex] th derivative of [latex]f\\left(x\\right)={\\left(1+{x}^{4}\\right)}^{25}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"878181\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"878181\"][latex]{f}^{\\left(99\\right)}\\left(0\\right)=0[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find the Maclaurin series of each function.<\/p>\r\n\r\n

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

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

\r\n

[reveal-answer q=\"68512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68512\"][latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(\\text{ln}\\left(2\\right)x\\right)}^{n}}{n\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
31.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{\\sin{x}}{x}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

32.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{\\sin\\left(\\sqrt{x}\\right)}{\\sqrt{x}},\\left(x>0\\right)[\/latex],<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"544672\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"544672\"]For [latex]x>0,\\sin\\left(\\sqrt{x}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{\\frac{\\left(2n+1\\right)}{2}}}{\\sqrt{x}\\left(2n+1\\right)\\text{!}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{\\left(2n+1\\right)\\text{!}}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
33.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin\\left({x}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

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

\r\n

[reveal-answer q=\"916736\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"916736\"][latex]{e}^{{x}^{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{3n}}{n\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
35.\u00a0<\/strong>[latex]f\\left(x\\right)={\\cos}^{2}x[\/latex] using the identity [latex]{\\cos}^{2}x=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

36.\u00a0<\/strong>[latex]f\\left(x\\right)={\\sin}^{2}x[\/latex] using the identity [latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"617738\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"617738\"][latex]{\\sin}^{2}x=\\text{-}\\displaystyle\\sum _{k=1}^{\\infty }\\frac{{\\left(-1\\right)}^{k}{2}^{2k - 1}{x}^{2k}}{\\left(2k\\right)\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find the Maclaurin series of [latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}f\\left(t\\right)dt[\/latex] by integrating the Maclaurin series of [latex]f[\/latex] term by term. If [latex]f[\/latex] is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.<\/p>\r\n\r\n

\r\n
\r\n
37.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}{e}^{\\text{-}{t}^{2}}dt;f\\left(t\\right)={e}^{\\text{-}{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n}}{n\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

38.\u00a0<\/strong>[latex]F\\left(x\\right)={\\tan}^{-1}x;f\\left(t\\right)=\\frac{1}{1+{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{t}^{2n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"724107\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"724107\"][latex]{\\tan}^{-1}x=\\displaystyle\\sum _{k=0}^{\\infty }\\frac{{\\left(-1\\right)}^{k}{x}^{2k+1}}{2k+1}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
39.\u00a0<\/strong>[latex]F\\left(x\\right)={\\text{tanh}}^{-1}x;f\\left(t\\right)=\\frac{1}{1-{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{t}^{2n}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n
\r\n

40.\u00a0<\/strong>[latex]F\\left(x\\right)={\\sin}^{-1}x;f\\left(t\\right)=\\frac{1}{\\sqrt{1-{t}^{2}}}=\\displaystyle\\sum _{k=0}^{\\infty }\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ k\\hfill \\end{array}\\right)\\frac{{t}^{2k}}{k\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"446593\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"446593\"][latex]{\\sin}^{-1}x=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right)\\frac{{x}^{2n+1}}{\\left(2n+1\\right)n\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

41.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\frac{\\sin{t}}{t}dt;f\\left(t\\right)=\\frac{\\sin{t}}{t}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

42.\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{{x}^{n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"456078\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"456078\"][latex]F\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n+1}}{\\left(n+1\\right)\\left(2n\\right)\\text{!}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
43.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\frac{1-\\cos{t}}{{t}^{2}}dt;f\\left(t\\right)=\\frac{1-\\cos{t}}{{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n}}{\\left(2n+2\\right)\\text{!}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n
\r\n

44.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\frac{\\text{ln}\\left(1+t\\right)}{t}dt;f\\left(t\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{n}}{n+1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"562961\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"562961\"][latex]F\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\frac{{x}^{n}}{{n}^{2}}[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of [latex]f[\/latex].<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
45.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin\\left(x+\\frac{\\pi }{4}\\right)=\\sin{x}\\cos\\left(\\frac{\\pi }{4}\\right)+\\cos{x}\\sin\\left(\\frac{\\pi }{4}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

46.\u00a0<\/strong>[latex]f\\left(x\\right)=\\tan{x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"724595\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"724595\"][latex]x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}+\\cdots[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
47.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}\\left(\\cos{x}\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

48.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{x}\\cos{x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"919509\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"919509\"][latex]1+x-\\frac{{x}^{3}}{3}-\\frac{{x}^{4}}{6}+\\cdots[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
49.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{\\sin{x}}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

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

\r\n

[reveal-answer q=\"942787\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"942787\"][latex]1+{x}^{2}+\\frac{2{x}^{4}}{3}+\\frac{17{x}^{6}}{45}+\\cdots[\/latex][\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
51.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{tanh}x[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

52.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{\\tan\\sqrt{x}}{\\sqrt{x}}[\/latex] (see expansion for [latex]\\tan{x}[\/latex])<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"226951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"226951\"]Using the expansion for [latex]\\tan{x}[\/latex] gives [latex]1+\\frac{x}{3}+\\frac{2{x}^{2}}{15}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, find the radius of convergence of the Maclaurin series of each function.<\/p>\r\n\r\n

\r\n
\r\n
53.\u00a0<\/strong>[latex]\\text{ln}\\left(1+x\\right)[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

54.\u00a0<\/strong>[latex]\\frac{1}{1+{x}^{2}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"568068\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568068\"][latex]\\frac{1}{1+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}[\/latex] so [latex]R=1[\/latex] by the ratio test.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
55.\u00a0<\/strong>[latex]{\\tan}^{-1}x[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

56.\u00a0<\/strong>[latex]\\text{ln}\\left(1+{x}^{2}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"745194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"745194\"][latex]\\text{ln}\\left(1+{x}^{2}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n - 1}}{n}{x}^{2n}[\/latex] so [latex]R=1[\/latex] by the ratio test.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
57.\u00a0<\/strong>Find the Maclaurin series of [latex]\\text{sinh}x=\\frac{{e}^{x}-{e}^{\\text{-}x}}{2}[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

58.\u00a0<\/strong>Find the Maclaurin series of [latex]\\text{cosh}x=\\frac{{e}^{x}+{e}^{\\text{-}x}}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"734049\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734049\"]Add series of [latex]{e}^{x}[\/latex] and [latex]{e}^{\\text{-}x}[\/latex] term by term. Odd terms cancel and [latex]\\text{cosh}x=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{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

\r\n
\r\n
59.\u00a0<\/strong>Differentiate term by term the Maclaurin series of [latex]\\text{sinh}x[\/latex] and compare the result with the Maclaurin series of [latex]\\text{cosh}x[\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

60. [T]<\/strong> Let [latex]{S}_{n}\\left(x\\right)=\\displaystyle\\sum _{k=0}^{n}{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{\\left(2k+1\\right)\\text{!}}[\/latex] and [latex]{C}_{n}\\left(x\\right)=\\displaystyle\\sum _{n=0}^{n}{\\left(-1\\right)}^{k}\\frac{{x}^{2k}}{\\left(2k\\right)\\text{!}}[\/latex] denote the respective Maclaurin polynomials of degree [latex]2n+1[\/latex] of [latex]\\sin{x}[\/latex] and degree [latex]2n[\/latex] of [latex]\\cos{x}[\/latex]. Plot the errors [latex]\\frac{{S}_{n}\\left(x\\right)}{{C}_{n}\\left(x\\right)}-\\tan{x}[\/latex] for [latex]n=1,..,5[\/latex] and compare them to [latex]x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}+\\frac{17{x}^{7}}{315}-\\tan{x}[\/latex] on [latex]\\left(-\\frac{\\pi }{4},\\frac{\\pi }{4}\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"966220\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966220\"]\"This<\/p>\r\nThe ratio [latex]\\frac{{S}_{n}\\left(x\\right)}{{C}_{n}\\left(x\\right)}[\/latex] approximates [latex]\\tan{x}[\/latex] better than does [latex]{p}_{7}\\left(x\\right)=x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}+\\frac{17{x}^{7}}{315}[\/latex] for [latex]N\\ge 3[\/latex]. The dashed curves are [latex]\\frac{{S}_{n}}{{C}_{n}}-\\tan[\/latex] for [latex]n=1,2[\/latex]. The dotted curve corresponds to [latex]n=3[\/latex], and the dash-dotted curve corresponds to [latex]n=4[\/latex]. The solid curve is [latex]{p}_{7}-\\tan{x}[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
61.\u00a0<\/strong>Use the identity [latex]2\\sin{x}\\cos{x}=\\sin\\left(2x\\right)[\/latex] to find the power series expansion of [latex]{\\sin}^{2}x[\/latex] at [latex]x=0[\/latex]. (Hint:<\/em> Integrate the Maclaurin series of [latex]\\sin\\left(2x\\right)[\/latex] term by term.)<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

62.\u00a0<\/strong>If [latex]y=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex], find the power series expansions of [latex]x{y}^{\\prime }[\/latex] and [latex]{x}^{2}y\\text{''}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"415665\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"415665\"]By the term-by-term differentiation theorem, [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=1}^{\\infty }n{a}_{n}{x}^{n - 1}[\/latex] so [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=1}^{\\infty }n{a}_{n}{x}^{n - 1}x{y}^{\\prime }=\\displaystyle\\sum _{n=1}^{\\infty }n{a}_{n}{x}^{n}[\/latex], whereas [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=2}^{\\infty }n\\left(n - 1\\right){a}_{n}{x}^{n - 2}[\/latex] so [latex]xy\\text{''}=\\displaystyle\\sum _{n=2}^{\\infty }n\\left(n - 1\\right){a}_{n}{x}^{n}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
63. [T]<\/strong> Suppose that [latex]y=\\displaystyle\\sum _{k=0}^{\\infty }{a}_{k}{x}^{k}[\/latex] satisfies [latex]{y}^{\\prime }=-2xy[\/latex] and [latex]y\\left(0\\right)=0[\/latex]. Show that [latex]{a}_{2k+1}=0[\/latex] for all [latex]k[\/latex] and that [latex]{a}_{2k+2}=\\frac{\\text{-}{a}_{2k}}{k+1}[\/latex]. Plot the partial sum [latex]{S}_{20}[\/latex] of [latex]y[\/latex] on the interval [latex]\\left[-4,4\\right][\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

64. [T]<\/strong> Suppose that a set of standardized test scores is normally distributed with mean [latex]\\mu =100[\/latex] and standard deviation [latex]\\sigma =10[\/latex]. Set up an integral that represents the probability that a test score will be between [latex]90[\/latex] and [latex]110[\/latex] and use the integral of the degree [latex]10[\/latex] Maclaurin polynomial of [latex]\\frac{1}{\\sqrt{2\\pi }}{e}^{\\frac{\\text{-}{x}^{2}}{2}}[\/latex] to estimate this probability.<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"82480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"82480\"]The probability is [latex]p=\\frac{1}{\\sqrt{2\\pi}}{\\displaystyle\\int }_{\\frac{(a-\\mu}{\\sigma }}^\\frac{(b-\\mu)}{\\sigma}{e}^{\\frac{-{x}^{2}}{2}}dx[\/latex] where [latex]a=90[\/latex] and [latex]b=100[\/latex], that is, [latex]p=\\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int }_{-1}^{1}{e}^{\\frac{\\text{-}{x}^{2}}{2}}dx=\\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int }_{-1}^{1}\\displaystyle\\sum _{n=0}^{5}{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{{2}^{n}n\\text{!}}dx=\\frac{2}{\\sqrt{2\\pi }}\\displaystyle\\sum _{n=0}^{5}{\\left(-1\\right)}^{n}\\frac{1}{\\left(2n+1\\right){2}^{n}n\\text{!}}\\approx 0.6827[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
65. [T]<\/strong> Suppose that a set of standardized test scores is normally distributed with mean [latex]\\mu =100[\/latex] and standard deviation [latex]\\sigma =10[\/latex]. Set up an integral that represents the probability that a test score will be between [latex]70[\/latex] and [latex]130[\/latex] and use the integral of the degree [latex]50[\/latex] Maclaurin polynomial of [latex]\\frac{1}{\\sqrt{2\\pi }}{e}^{\\frac{\\text{-}{x}^{2}}{2}}[\/latex] to estimate this probability.<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n

66. [T]<\/strong> Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]f\\left(x\\right)[\/latex] such that [latex]f\\left(0\\right)=1,{f}^{\\prime }\\left(0\\right)=0[\/latex], and [latex]f\\text{''}\\left(x\\right)=\\text{-}f\\left(x\\right)[\/latex]. Find a formula for [latex]{a}_{n}[\/latex] and plot the partial sum [latex]{S}_{N}[\/latex] for [latex]N=20[\/latex] on [latex]\\left[-5,5\\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"597939\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"597939\"]\"This<\/p>\r\nAs in the previous problem one obtains [latex]{a}_{n}=0[\/latex] if [latex]n[\/latex] is odd and [latex]{a}_{n}=\\text{-}\\left(n+2\\right)\\left(n+1\\right){a}_{n+2}[\/latex] if [latex]n[\/latex] is even, so [latex]{a}_{0}=1[\/latex] leads to [latex]{a}_{2n}=\\frac{{\\left(-1\\right)}^{n}}{\\left(2n\\right)\\text{!}}[\/latex].[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n
\r\n
67. [T]<\/strong> Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]f\\left(x\\right)[\/latex] such that [latex]f\\left(0\\right)=0,{f}^{\\prime }\\left(0\\right)=1[\/latex], and [latex]f\\text{''}\\left(x\\right)=\\text{-}f\\left(x\\right)[\/latex]. Find a formula for [latex]{a}_{n}[\/latex] and plot the partial sum [latex]{S}_{N}[\/latex] for [latex]N=10[\/latex] on [latex]\\left[-5,5\\right][\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n
\r\n
\r\n
\r\n
\r\n

68.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]y[\/latex] such that [latex]y\\text{''}-{y}^{\\prime }+y=0[\/latex] where [latex]y\\left(0\\right)=1[\/latex] and [latex]y^{\\prime} \\left(0\\right)=0[\/latex]. Find a formula that relates [latex]{a}_{n+2},{a}_{n+1}[\/latex], and [latex]{a}_{n}[\/latex] and compute [latex]{a}_{0},...,{a}_{5}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

[reveal-answer q=\"710623\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"710623\"][latex]y\\text{''}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+2\\right)\\left(n+1\\right){a}_{n+2}{x}^{n}[\/latex] and [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){a}_{n+1}{x}^{n}[\/latex] so [latex]y\\text{''}-{y}^{\\prime }+y=0[\/latex] implies that [latex]\\left(n+2\\right)\\left(n+1\\right){a}_{n+2}-\\left(n+1\\right){a}_{n+1}+{a}_{n}=0[\/latex] or [latex]{a}_{n}=\\frac{{a}_{n - 1}}{n}-\\frac{{a}_{n - 2}}{n\\left(n - 1\\right)}[\/latex] for all [latex]n\\cdot y\\left(0\\right)={a}_{0}=1[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)={a}_{1}=0[\/latex], so [latex]{a}_{2}=\\frac{1}{2},{a}_{3}=\\frac{1}{6},{a}_{4}=0[\/latex], and [latex]{a}_{5}=-\\frac{1}{120}[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

69.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]y[\/latex] such that [latex]y\\text{''}-{y}^{\\prime }+y=0[\/latex] where [latex]y\\left(0\\right)=0[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)=1[\/latex]. Find a formula that relates [latex]{a}_{n+2},{a}_{n+1}[\/latex], and [latex]{a}_{n}[\/latex] and compute [latex]{a}_{1},...,{a}_{5}[\/latex].<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n

The error in approximating the integral [latex]{\\displaystyle\\int }_{a}^{b}f\\left(t\\right)dt[\/latex] by that of a Taylor approximation [latex]{\\displaystyle\\int }_{a}^{b}{P}_{n}\\left(t\\right)dt[\/latex] is at most [latex]{\\displaystyle\\int }_{a}^{b}{R}_{n}\\left(t\\right)dt[\/latex]. In the following exercises, the Taylor remainder estimate [latex]{R}_{n}\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex] guarantees that the integral of the Taylor polynomial of the given order approximates the integral of [latex]f[\/latex] with an error less than [latex]\\frac{1}{10}[\/latex].<\/p>\r\n\r\n

    \r\n \t
  1. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than [latex]\\frac{1}{100}[\/latex].<\/li>\r\n \t
  2. Compare the accuracy of the polynomial integral estimate with the remainder estimate.<\/li>\r\n<\/ol>\r\n
    \r\n
    \r\n
    \r\n
    \r\n

    70. [T]<\/strong> [latex]{\\displaystyle\\int }_{0}^{\\pi }\\frac{\\sin{t}}{t}dt;{P}_{s}=1-\\frac{{x}^{2}}{3\\text{!}}+\\frac{{x}^{4}}{5\\text{!}}-\\frac{{x}^{6}}{7\\text{!}}+\\frac{{x}^{8}}{9\\text{!}}[\/latex] (You may assume that the absolute value of the ninth derivative of [latex]\\frac{\\sin{t}}{t}[\/latex] is bounded by [latex]0.1.[\/latex])<\/p>\r\n[reveal-answer q=\"711848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"711848\"]a. (Proof) b. We have [latex]{R}_{s}\\le \\frac{0.1}{\\left(9\\right)\\text{!}}{\\pi }^{9}\\approx 0.0082<0.01[\/latex]. We have [latex]{\\displaystyle\\int }_{0}^{\\pi }\\left(1-\\frac{{x}^{2}}{3\\text{!}}+\\frac{{x}^{4}}{5\\text{!}}-\\frac{{x}^{6}}{7\\text{!}}+\\frac{{x}^{8}}{9\\text{!}}\\right)dx=\\pi -\\frac{{\\pi }^{3}}{3\\cdot 3\\text{!}}+\\frac{{\\pi }^{5}}{5\\cdot 5\\text{!}}-\\frac{{\\pi }^{7}}{7\\cdot 7\\text{!}}+\\frac{{\\pi }^{9}}{9\\cdot 9\\text{!}}=1.852...[\/latex], whereas [latex]{\\displaystyle\\int }_{0}^{\\pi }\\frac{\\sin{t}}{t}dt=1.85194...[\/latex], so the actual error is approximately [latex]0.00006[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n

    <\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    71. [T]<\/strong> [latex]{\\displaystyle\\int }_{0}^{2}{e}^{\\text{-}{x}^{2}}dx;{p}_{11}=1-{x}^{2}+\\frac{{x}^{4}}{2}-\\frac{{x}^{6}}{3\\text{!}}+\\cdots-\\frac{{x}^{22}}{11\\text{!}}[\/latex] (You may assume that the absolute value of the [latex]23\\text{rd}[\/latex] derivative of [latex]{e}^{\\text{-}{x}^{2}}[\/latex] is less than [latex]2\\times {10}^{14}.[\/latex])<\/div>\r\n<\/div>\r\n<\/div>\r\n

    The following exercises deal with Fresnel integrals<\/span>.<\/p>\r\n\r\n

    \r\n
    \r\n
    \r\n
    \r\n
    \r\n

    72.\u00a0<\/strong>The Fresnel integrals are defined by [latex]C\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\cos\\left({t}^{2}\\right)dt[\/latex] and [latex]S\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\sin\\left({t}^{2}\\right)dt[\/latex]. Compute the power series of [latex]C\\left(x\\right)[\/latex] and [latex]S\\left(x\\right)[\/latex] and plot the sums [latex]{C}_{N}\\left(x\\right)[\/latex] and [latex]{S}_{N}\\left(x\\right)[\/latex] of the first [latex]N=50[\/latex] nonzero terms on [latex]\\left[0,2\\pi \\right][\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    [reveal-answer q=\"447909\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"447909\"]\"This<\/p>\r\nSince [latex]\\cos\\left({t}^{2}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{4n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]\\sin\\left({t}^{2}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{4n+2}}{\\left(2n+1\\right)\\text{!}}[\/latex], one has [latex]S\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{4n+3}}{\\left(4n+3\\right)\\left(2n+1\\right)\\text{!}}[\/latex] and [latex]C\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{4n+1}}{\\left(4n+1\\right)\\left(2n\\right)\\text{!}}[\/latex]. The sums of the first [latex]50[\/latex] nonzero terms are plotted below with [latex]{C}_{50}\\left(x\\right)[\/latex] the solid curve and [latex]{S}_{50}\\left(x\\right)[\/latex] the dashed curve.[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    73. [T]<\/strong> The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates [latex]\\left(C\\left(t\\right),S\\left(t\\right)\\right)[\/latex]. Plot the curve [latex]\\left({C}_{50},{S}_{50}\\right)[\/latex] for [latex]0\\le t\\le 2\\pi [\/latex], the coordinates of which were computed in the previous exercise.<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n
    \r\n

    74.\u00a0<\/strong>Estimate [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{4}}\\sqrt{x-{x}^{2}}dx[\/latex] by approximating [latex]\\sqrt{1-x}[\/latex] using the binomial approximation [latex]1-\\frac{x}{2}-\\frac{{x}^{2}}{8}-\\frac{{x}^{3}}{16}-\\frac{5{x}^{4}}{2128}-\\frac{7{x}^{5}}{256}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n\r\n[reveal-answer q=\"550647\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"550647\"][latex]{\\displaystyle\\int }_{0}^{\\frac{1}{4}}\\sqrt{x}\\left(1-\\frac{x}{2}-\\frac{{x}^{2}}{8}-\\frac{{x}^{3}}{16}-\\frac{5{x}^{4}}{128}-\\frac{7{x}^{5}}{256}\\right)dx[\/latex]\r\n

    [latex]=\\frac{2}{3}{2}^{-3}-\\frac{1}{2}\\frac{2}{5}{2}^{-5}-\\frac{1}{8}\\frac{2}{7}{2}^{-7}-\\frac{1}{16}\\frac{2}{9}{2}^{-9}-\\frac{5}{128}\\frac{2}{11}{2}^{-11}-\\frac{7}{256}\\frac{2}{13}{2}^{-13}=0.0767732..[\/latex].<\/p>\r\n

    whereas [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{4}}\\sqrt{x-{x}^{2}}dx=0.076773[\/latex].<\/p>\r\n

    [\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n
    75. [T]<\/strong> Use Newton\u2019s approximation of the binomial [latex]\\sqrt{1-{x}^{2}}[\/latex] to approximate [latex]\\pi [\/latex] as follows. The circle centered at [latex]\\left(\\frac{1}{2},0\\right)[\/latex] with radius [latex]\\frac{1}{2}[\/latex] has upper semicircle [latex]y=\\sqrt{x}\\sqrt{1-x}[\/latex]. The sector of this circle bounded by the [latex]x[\/latex] -axis between [latex]x=0[\/latex] and [latex]x=\\frac{1}{2}[\/latex] and by the line joining [latex]\\left(\\frac{1}{4},\\frac{\\sqrt{3}}{4}\\right)[\/latex] corresponds to [latex]\\frac{1}{6}[\/latex] of the circle and has area [latex]\\frac{\\pi }{24}[\/latex]. This sector is the union of a right triangle with height [latex]\\frac{\\sqrt{3}}{4}[\/latex] and base [latex]\\frac{1}{4}[\/latex] and the region below the graph between [latex]x=0[\/latex] and [latex]x=\\frac{1}{4}[\/latex]. To find the area of this region you can write [latex]y=\\sqrt{x}\\sqrt{1-x}=\\sqrt{x}\\times \\left(\\text{binomial expansion of}\\sqrt{1-x}\\right)[\/latex] and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate [latex]\\pi [\/latex].<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n
    \r\n

    76.\u00a0<\/strong>Use the approximation [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}\\right)[\/latex] to approximate the period of a pendulum having length [latex]10[\/latex] meters and maximum angle [latex]{\\theta }_{\\text{max}}=\\frac{\\pi }{6}[\/latex] where [latex]k=\\sin\\left(\\frac{{\\theta }_{\\text{max}}}{2}\\right)[\/latex]. Compare this with the small angle estimate [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    [reveal-answer q=\"641478\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"641478\"][latex]T\\approx 2\\pi \\sqrt{\\frac{10}{9.8}}\\left(1+\\frac{{\\sin}^{2}\\left(\\frac{\\theta}{12}\\right)}{4}\\right)\\approx 6.453[\/latex] seconds. The small angle estimate is [latex]T\\approx 2\\pi \\sqrt{\\frac{10}{9.8}\\approx 6.347}[\/latex]. The relative error is around [latex]2[\/latex] percent.[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n
    77.\u00a0<\/strong>Suppose that a pendulum is to have a period of [latex]2[\/latex] seconds and a maximum angle of [latex]{\\theta }_{\\text{max}}=\\frac{\\pi }{6}[\/latex]. Use [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}\\right)[\/latex] to approximate the desired length of the pendulum. What length is predicted by the small angle estimate [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}?[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    \r\n
    \r\n
    \r\n

    78.\u00a0<\/strong>Evaluate [latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}{\\sin}^{4}\\theta d\\theta [\/latex] in the approximation [latex]T=4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta +\\frac{3}{8}{k}^{4}{\\sin}^{4}\\theta +\\cdots\\right)d\\theta [\/latex] to obtain an improved estimate for [latex]T[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    [reveal-answer q=\"25080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"25080\"][latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}{\\sin}^{4}\\theta d\\theta =\\frac{3\\pi }{16}[\/latex]. Hence [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}+\\frac{9}{256}{k}^{4}\\right)[\/latex].[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n
    79. [T]<\/strong> An equivalent formula for the period of a pendulum with amplitude [latex]{\\theta }_{\\text{max}}[\/latex] is [latex]T\\left({\\theta }_{\\text{max}}\\right)=2\\sqrt{2}\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{{\\theta }_{\\text{max}}}\\frac{d\\theta }{\\sqrt{\\cos\\theta }-\\cos\\left({\\theta }_{\\text{max}}\\right)}[\/latex] where [latex]L[\/latex] is the pendulum length and [latex]g[\/latex] is the gravitational acceleration constant. When [latex]{\\theta }_{\\text{max}}=\\frac{\\pi }{3}[\/latex] we get [latex]\\frac{1}{\\sqrt{\\cos{t} - \\frac{1}{2}}}\\approx \\sqrt{2}\\left(1+\\frac{{t}^{2}}{2}+\\frac{{t}^{4}}{3}+\\frac{181{t}^{6}}{720}\\right)[\/latex]. Integrate this approximation to estimate [latex]T\\left(\\frac{\\pi }{3}\\right)[\/latex] in terms of [latex]L[\/latex] and [latex]g[\/latex]. Assuming [latex]g=9.806[\/latex] meters per second squared, find an approximate length [latex]L[\/latex] such that [latex]T\\left(\\frac{\\pi }{3}\\right)=2[\/latex] seconds.<\/div>\r\n<\/div>\r\n<\/div>","rendered":"

    In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.<\/p>\n

    \n
    \n
    1.\u00a0<\/strong>[latex]{\\left(1-x\\right)}^{\\frac{1}{3}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    2.\u00a0<\/strong>[latex]{\\left(1+{x}^{2}\\right)}^{\\frac{-1}{3}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{\\left(1+{x}^{2}\\right)}^{\\frac{-1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}-\\frac{1}{3}\\hfill \\\\ \\hfill n\\hfill \\end{array}\\right){x}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    3.\u00a0<\/strong>[latex]{\\left(1-x\\right)}^{1.01}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    4.\u00a0<\/strong>[latex]{\\left(1 - 2x\\right)}^{\\frac{2}{3}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{\\left(1 - 2x\\right)}^{\\frac{2}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{2}^{n}\\left(\\begin{array}{c}\\frac{2}{3}\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, use the substitution [latex]{\\left(b+x\\right)}^{r}={\\left(b+a\\right)}^{r}{\\left(1+\\frac{x-a}{b+a}\\right)}^{r}[\/latex] in the binomial expansion to find the Taylor series of each function with the given center.<\/p>\n

    \n
    \n
    5.\u00a0<\/strong>[latex]\\sqrt{x+2}[\/latex] at [latex]a=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    6.\u00a0<\/strong>[latex]\\sqrt{{x}^{2}+2}[\/latex] at [latex]a=0[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\sqrt{2+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{\\left(\\frac{1}{2}\\right)-n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{2n};\\left(|{x}^{2}|<2\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    7.\u00a0<\/strong>[latex]\\sqrt{x+2}[\/latex] at [latex]a=1[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    8.\u00a0<\/strong>[latex]\\sqrt{2x-{x}^{2}}[\/latex] at [latex]a=1[\/latex] (Hint:<\/em> [latex]2x-{x}^{2}=1-{\\left(x - 1\\right)}^{2}[\/latex])<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\sqrt{2x-{x}^{2}}=\\sqrt{1-{\\left(x - 1\\right)}^{2}}[\/latex] so [latex]\\sqrt{2x-{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){\\left(x - 1\\right)}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    9.\u00a0<\/strong>[latex]{\\left(x - 8\\right)}^{\\frac{1}{3}}[\/latex] at [latex]a=9[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    10.\u00a0<\/strong>[latex]\\sqrt{x}[\/latex] at [latex]a=4[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\sqrt{x}=2\\sqrt{1+\\frac{x - 4}{4}}[\/latex] so [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{1 - 2n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){\\left(x - 4\\right)}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    11.\u00a0<\/strong>[latex]{x}^{\\frac{1}{3}}[\/latex] at [latex]a=27[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    12.\u00a0<\/strong>[latex]\\sqrt{x}[\/latex] at [latex]x=9[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{3}^{1 - 3n}\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right){\\left(x - 9\\right)}^{n}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most [latex]\\frac{1}{1000}[\/latex].<\/p>\n

    \n
    \n
    13. [T]<\/strong> [latex]{\\left(15\\right)}^{\\frac{1}{4}}[\/latex] using [latex]{\\left(16-x\\right)}^{\\frac{1}{4}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    14. [T]<\/strong> [latex]{\\left(1001\\right)}^{\\frac{1}{3}}[\/latex] using [latex]{\\left(1000+x\\right)}^{\\frac{1}{3}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]10{\\left(1+\\frac{x}{1000}\\right)}^{\\frac{1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{10}^{1 - 3n}\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]. Using, for example, a fourth-degree estimate at [latex]x=1[\/latex] gives [latex]\\begin{array}{cc}\\hfill {\\left(1001\\right)}^{\\frac{1}{3}}& \\approx 10\\left(1+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 1\\hfill \\end{array}\\right){10}^{-3}+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 2\\hfill \\end{array}\\right){10}^{-6}+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 3\\hfill \\end{array}\\right){10}^{-9}+\\left(\\begin{array}{c}\\frac{1}{3}\\hfill \\\\ 4\\hfill \\end{array}\\right){10}^{-12}\\right)\\hfill \\\\ & =10\\left(1+\\frac{1}{{3.10}^{3}}-\\frac{1}{{9.10}^{6}}+\\frac{5}{{81.10}^{9}}-\\frac{10}{{243.10}^{12}}\\right)=10.00333222...\\hfill \\end{array}[\/latex] whereas [latex]{\\left(1001\\right)}^{\\frac{1}{3}}=10.00332222839093...[\/latex]. Two terms would suffice for three-digit accuracy.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, use the binomial approximation [latex]\\sqrt{1-x}\\approx 1-\\frac{x}{2}-\\frac{{x}^{2}}{8}-\\frac{{x}^{3}}{16}-\\frac{5{x}^{4}}{128}-\\frac{7{x}^{5}}{256}[\/latex] for [latex]|x|<1[\/latex] to approximate each number. Compare this value to the value given by a scientific calculator.<\/p>\n

    \n
    \n
    15. [T]<\/strong> [latex]\\frac{1}{\\sqrt{2}}[\/latex] using [latex]x=\\frac{1}{2}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    16. [T]<\/strong> [latex]\\sqrt{5}=5\\times \\frac{1}{\\sqrt{5}}[\/latex] using [latex]x=\\frac{4}{5}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    The approximation is [latex]2.3152[\/latex]; the CAS value is [latex]2.23\\ldots[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    17. [T]<\/strong> [latex]\\sqrt{3}=\\frac{3}{\\sqrt{3}}[\/latex] using [latex]x=\\frac{2}{3}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    18. [T]<\/strong> [latex]\\sqrt{6}[\/latex] using [latex]x=\\frac{5}{6}[\/latex] in [latex]{\\left(1-x\\right)}^{\\frac{1}{2}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    The approximation is [latex]2.583\\ldots[\/latex]; the CAS value is [latex]2.449\\ldots[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    19.\u00a0<\/strong>Integrate the binomial approximation of [latex]\\sqrt{1-x}[\/latex] to find an approximation of [latex]{\\displaystyle\\int }_{0}^{x}\\sqrt{1-t}dt[\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    20. [T]<\/strong> Recall that the graph of [latex]\\sqrt{1-{x}^{2}}[\/latex] is an upper semicircle of radius [latex]1[\/latex]. Integrate the binomial approximation of [latex]\\sqrt{1-{x}^{2}}[\/latex] up to order [latex]8[\/latex] from [latex]x=-1[\/latex] to [latex]x=1[\/latex] to estimate [latex]\\frac{\\pi }{2}[\/latex].<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\sqrt{1-{x}^{2}}=1-\\frac{{x}^{2}}{2}-\\frac{{x}^{4}}{8}-\\frac{{x}^{6}}{16}-\\frac{5{x}^{8}}{128}+\\cdots[\/latex]. Thus [latex]{\\displaystyle\\int }_{-1}^{1}\\sqrt{1-{x}^{2}}dx=x-\\frac{{x}^{3}}{6}-\\frac{{x}^{5}}{40}-\\frac{{x}^{7}}{7\\cdot 16}-\\frac{5{x}^{9}}{9\\cdot 128}+\\cdots{|}_{-1}^{1}\\approx 2-\\frac{1}{3}-\\frac{1}{20}-\\frac{1}{56}-\\frac{10}{9\\cdot 128}+\\text{error}=1.590..[\/latex]. whereas [latex]\\frac{\\pi }{2}=1.570..[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, use the expansion [latex]{\\left(1+x\\right)}^{\\frac{1}{3}}=1+\\frac{1}{3}x-\\frac{1}{9}{x}^{2}+\\frac{5}{81}{x}^{3}-\\frac{10}{243}{x}^{4}+\\cdots[\/latex] to write the first five terms (not necessarily a quartic polynomial) of each expression.<\/p>\n

    \n
    \n
    21.\u00a0<\/strong>[latex]{\\left(1+4x\\right)}^{\\frac{1}{3}};a=0[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    22.\u00a0<\/strong>[latex]{\\left(1+4x\\right)}^{\\frac{4}{3}};a=0[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{\\left(1+4x\\right)}^{\\frac{4}{3}}=\\left(1+4x\\right)\\left(1+4x\\right)^{\\frac{1}{3}}=\\left(1+4x\\right)\\left(1+\\frac{4x}{3}-\\frac{16x^{3}}{9}+\\frac{320x^{3}}{81}-\\frac{2560x^{4}}{243}\\right)=1+\\frac{16}{3}x+\\frac{32}{9}x^{2}-\\frac{256}{81}x^{3}+\\frac{1280}{243}x^{4}-\\frac{10240}{243}x^{5}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    23.\u00a0<\/strong>[latex]{\\left(3+2x\\right)}^{\\frac{1}{3}};a=-1[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    24.\u00a0<\/strong>[latex]{\\left({x}^{2}+6x+10\\right)}^{\\frac{1}{3}};a=-3[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\left(1+{\\left(x+3\\right)}^{2}\\right)^{\\frac{1}{3}} =1+\\frac{1}{3}{\\left(x+3\\right)}^{2}-\\frac{1}{9}{\\left(x+3\\right)}^{4}+\\frac{5}{81}{\\left(x+3\\right)}^{6}-\\frac{10}{243}{\\left(x+3\\right)}^{8}+\\cdots[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    25.\u00a0<\/strong>Use [latex]{\\left(1+x\\right)}^{\\frac{1}{3}}=1+\\frac{1}{3}x-\\frac{1}{9}{x}^{2}+\\frac{5}{81}{x}^{3}-\\frac{10}{243}{x}^{4}+\\cdots[\/latex] with [latex]x=1[\/latex] to approximate [latex]{2}^{\\frac{1}{3}}[\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    26.\u00a0<\/strong>Use the approximation [latex]{\\left(1-x\\right)}^{\\frac{2}{3}}=1-\\frac{2x}{3}-\\frac{{x}^{2}}{9}-\\frac{4{x}^{3}}{81}-\\frac{7{x}^{4}}{243}-\\frac{14{x}^{5}}{729}+\\cdots[\/latex] for [latex]|x|<1[\/latex] to approximate [latex]{2}^{\\frac{1}{3}}={2.2}^{\\frac{-2}{3}}[\/latex].<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    Twice the approximation is [latex]1.260\\ldots[\/latex] whereas [latex]{2}^{\\frac{1}{3}}=1.2599...[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    27.\u00a0<\/strong>Find the [latex]25\\text{th}[\/latex] derivative of [latex]f\\left(x\\right)={\\left(1+{x}^{2}\\right)}^{13}[\/latex] at [latex]x=0[\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    28.\u00a0<\/strong>Find the [latex]99[\/latex] th derivative of [latex]f\\left(x\\right)={\\left(1+{x}^{4}\\right)}^{25}[\/latex].<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{f}^{\\left(99\\right)}\\left(0\\right)=0[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, find the Maclaurin series of each function.<\/p>\n

    \n
    \n
    29.\u00a0<\/strong>[latex]f\\left(x\\right)=x{e}^{2x}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

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

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{\\left(\\text{ln}\\left(2\\right)x\\right)}^{n}}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    31.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{\\sin{x}}{x}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    32.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{\\sin\\left(\\sqrt{x}\\right)}{\\sqrt{x}},\\left(x>0\\right)[\/latex],<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    For [latex]x>0,\\sin\\left(\\sqrt{x}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{\\frac{\\left(2n+1\\right)}{2}}}{\\sqrt{x}\\left(2n+1\\right)\\text{!}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    33.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin\\left({x}^{2}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    34.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{{x}^{3}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{e}^{{x}^{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{3n}}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    35.\u00a0<\/strong>[latex]f\\left(x\\right)={\\cos}^{2}x[\/latex] using the identity [latex]{\\cos}^{2}x=\\frac{1}{2}+\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    36.\u00a0<\/strong>[latex]f\\left(x\\right)={\\sin}^{2}x[\/latex] using the identity [latex]{\\sin}^{2}x=\\frac{1}{2}-\\frac{1}{2}\\cos\\left(2x\\right)[\/latex]<\/p>\n<\/div>\n

    \n

    \n

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

    In the following exercises, find the Maclaurin series of [latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}f\\left(t\\right)dt[\/latex] by integrating the Maclaurin series of [latex]f[\/latex] term by term. If [latex]f[\/latex] is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.<\/p>\n

    \n
    \n
    37.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}{e}^{\\text{-}{t}^{2}}dt;f\\left(t\\right)={e}^{\\text{-}{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n}}{n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    38.\u00a0<\/strong>[latex]F\\left(x\\right)={\\tan}^{-1}x;f\\left(t\\right)=\\frac{1}{1+{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{t}^{2n}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{\\tan}^{-1}x=\\displaystyle\\sum _{k=0}^{\\infty }\\frac{{\\left(-1\\right)}^{k}{x}^{2k+1}}{2k+1}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    39.\u00a0<\/strong>[latex]F\\left(x\\right)={\\text{tanh}}^{-1}x;f\\left(t\\right)=\\frac{1}{1-{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{t}^{2n}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n
    \n

    40.\u00a0<\/strong>[latex]F\\left(x\\right)={\\sin}^{-1}x;f\\left(t\\right)=\\frac{1}{\\sqrt{1-{t}^{2}}}=\\displaystyle\\sum _{k=0}^{\\infty }\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ k\\hfill \\end{array}\\right)\\frac{{t}^{2k}}{k\\text{!}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]{\\sin}^{-1}x=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}\\frac{1}{2}\\hfill \\\\ n\\hfill \\end{array}\\right)\\frac{{x}^{2n+1}}{\\left(2n+1\\right)n\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    41.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\frac{\\sin{t}}{t}dt;f\\left(t\\right)=\\frac{\\sin{t}}{t}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/span><\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    42.\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{{x}^{n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]F\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{n+1}}{\\left(n+1\\right)\\left(2n\\right)\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    43.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\frac{1-\\cos{t}}{{t}^{2}}dt;f\\left(t\\right)=\\frac{1-\\cos{t}}{{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{2n}}{\\left(2n+2\\right)\\text{!}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n
    \n

    44.\u00a0<\/strong>[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\frac{\\text{ln}\\left(1+t\\right)}{t}dt;f\\left(t\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{n}}{n+1}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

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

    In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of [latex]f[\/latex].<\/span><\/p>\n<\/div>\n<\/div>\n

    \n
    \n
    45.\u00a0<\/strong>[latex]f\\left(x\\right)=\\sin\\left(x+\\frac{\\pi }{4}\\right)=\\sin{x}\\cos\\left(\\frac{\\pi }{4}\\right)+\\cos{x}\\sin\\left(\\frac{\\pi }{4}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    46.\u00a0<\/strong>[latex]f\\left(x\\right)=\\tan{x}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}+\\cdots[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    47.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{ln}\\left(\\cos{x}\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    48.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{x}\\cos{x}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]1+x-\\frac{{x}^{3}}{3}-\\frac{{x}^{4}}{6}+\\cdots[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    49.\u00a0<\/strong>[latex]f\\left(x\\right)={e}^{\\sin{x}}[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    50.\u00a0<\/strong>[latex]f\\left(x\\right)={\\sec}^{2}x[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]1+{x}^{2}+\\frac{2{x}^{4}}{3}+\\frac{17{x}^{6}}{45}+\\cdots[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    51.\u00a0<\/strong>[latex]f\\left(x\\right)=\\text{tanh}x[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    52.\u00a0<\/strong>[latex]f\\left(x\\right)=\\frac{\\tan\\sqrt{x}}{\\sqrt{x}}[\/latex] (see expansion for [latex]\\tan{x}[\/latex])<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    Using the expansion for [latex]\\tan{x}[\/latex] gives [latex]1+\\frac{x}{3}+\\frac{2{x}^{2}}{15}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

    In the following exercises, find the radius of convergence of the Maclaurin series of each function.<\/p>\n

    \n
    \n
    53.\u00a0<\/strong>[latex]\\text{ln}\\left(1+x\\right)[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    54.\u00a0<\/strong>[latex]\\frac{1}{1+{x}^{2}}[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\frac{1}{1+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}[\/latex] so [latex]R=1[\/latex] by the ratio test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    55.\u00a0<\/strong>[latex]{\\tan}^{-1}x[\/latex]<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    56.\u00a0<\/strong>[latex]\\text{ln}\\left(1+{x}^{2}\\right)[\/latex]<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]\\text{ln}\\left(1+{x}^{2}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n - 1}}{n}{x}^{2n}[\/latex] so [latex]R=1[\/latex] by the ratio test.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    57.\u00a0<\/strong>Find the Maclaurin series of [latex]\\text{sinh}x=\\frac{{e}^{x}-{e}^{\\text{-}x}}{2}[\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    58.\u00a0<\/strong>Find the Maclaurin series of [latex]\\text{cosh}x=\\frac{{e}^{x}+{e}^{\\text{-}x}}{2}[\/latex].<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    Add series of [latex]{e}^{x}[\/latex] and [latex]{e}^{\\text{-}x}[\/latex] term by term. Odd terms cancel and [latex]\\text{cosh}x=\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    59.\u00a0<\/strong>Differentiate term by term the Maclaurin series of [latex]\\text{sinh}x[\/latex] and compare the result with the Maclaurin series of [latex]\\text{cosh}x[\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    60. [T]<\/strong> Let [latex]{S}_{n}\\left(x\\right)=\\displaystyle\\sum _{k=0}^{n}{\\left(-1\\right)}^{k}\\frac{{x}^{2k+1}}{\\left(2k+1\\right)\\text{!}}[\/latex] and [latex]{C}_{n}\\left(x\\right)=\\displaystyle\\sum _{n=0}^{n}{\\left(-1\\right)}^{k}\\frac{{x}^{2k}}{\\left(2k\\right)\\text{!}}[\/latex] denote the respective Maclaurin polynomials of degree [latex]2n+1[\/latex] of [latex]\\sin{x}[\/latex] and degree [latex]2n[\/latex] of [latex]\\cos{x}[\/latex]. Plot the errors [latex]\\frac{{S}_{n}\\left(x\\right)}{{C}_{n}\\left(x\\right)}-\\tan{x}[\/latex] for [latex]n=1,..,5[\/latex] and compare them to [latex]x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}+\\frac{17{x}^{7}}{315}-\\tan{x}[\/latex] on [latex]\\left(-\\frac{\\pi }{4},\\frac{\\pi }{4}\\right)[\/latex].<\/p>\n<\/div>\n

    \n

    \n

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

    The ratio [latex]\\frac{{S}_{n}\\left(x\\right)}{{C}_{n}\\left(x\\right)}[\/latex] approximates [latex]\\tan{x}[\/latex] better than does [latex]{p}_{7}\\left(x\\right)=x+\\frac{{x}^{3}}{3}+\\frac{2{x}^{5}}{15}+\\frac{17{x}^{7}}{315}[\/latex] for [latex]N\\ge 3[\/latex]. The dashed curves are [latex]\\frac{{S}_{n}}{{C}_{n}}-\\tan[\/latex] for [latex]n=1,2[\/latex]. The dotted curve corresponds to [latex]n=3[\/latex], and the dash-dotted curve corresponds to [latex]n=4[\/latex]. The solid curve is [latex]{p}_{7}-\\tan{x}[\/latex].<\/div>\n<\/div>\n

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

    \n
    \n
    61.\u00a0<\/strong>Use the identity [latex]2\\sin{x}\\cos{x}=\\sin\\left(2x\\right)[\/latex] to find the power series expansion of [latex]{\\sin}^{2}x[\/latex] at [latex]x=0[\/latex]. (Hint:<\/em> Integrate the Maclaurin series of [latex]\\sin\\left(2x\\right)[\/latex] term by term.)<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    62.\u00a0<\/strong>If [latex]y=\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex], find the power series expansions of [latex]x{y}^{\\prime }[\/latex] and [latex]{x}^{2}y\\text{''}[\/latex].<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    By the term-by-term differentiation theorem, [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=1}^{\\infty }n{a}_{n}{x}^{n - 1}[\/latex] so [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=1}^{\\infty }n{a}_{n}{x}^{n - 1}x{y}^{\\prime }=\\displaystyle\\sum _{n=1}^{\\infty }n{a}_{n}{x}^{n}[\/latex], whereas [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=2}^{\\infty }n\\left(n - 1\\right){a}_{n}{x}^{n - 2}[\/latex] so [latex]xy\\text{''}=\\displaystyle\\sum _{n=2}^{\\infty }n\\left(n - 1\\right){a}_{n}{x}^{n}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    63. [T]<\/strong> Suppose that [latex]y=\\displaystyle\\sum _{k=0}^{\\infty }{a}_{k}{x}^{k}[\/latex] satisfies [latex]{y}^{\\prime }=-2xy[\/latex] and [latex]y\\left(0\\right)=0[\/latex]. Show that [latex]{a}_{2k+1}=0[\/latex] for all [latex]k[\/latex] and that [latex]{a}_{2k+2}=\\frac{\\text{-}{a}_{2k}}{k+1}[\/latex]. Plot the partial sum [latex]{S}_{20}[\/latex] of [latex]y[\/latex] on the interval [latex]\\left[-4,4\\right][\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    64. [T]<\/strong> Suppose that a set of standardized test scores is normally distributed with mean [latex]\\mu =100[\/latex] and standard deviation [latex]\\sigma =10[\/latex]. Set up an integral that represents the probability that a test score will be between [latex]90[\/latex] and [latex]110[\/latex] and use the integral of the degree [latex]10[\/latex] Maclaurin polynomial of [latex]\\frac{1}{\\sqrt{2\\pi }}{e}^{\\frac{\\text{-}{x}^{2}}{2}}[\/latex] to estimate this probability.<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    The probability is [latex]p=\\frac{1}{\\sqrt{2\\pi}}{\\displaystyle\\int }_{\\frac{(a-\\mu}{\\sigma }}^\\frac{(b-\\mu)}{\\sigma}{e}^{\\frac{-{x}^{2}}{2}}dx[\/latex] where [latex]a=90[\/latex] and [latex]b=100[\/latex], that is, [latex]p=\\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int }_{-1}^{1}{e}^{\\frac{\\text{-}{x}^{2}}{2}}dx=\\frac{1}{\\sqrt{2\\pi }}{\\displaystyle\\int }_{-1}^{1}\\displaystyle\\sum _{n=0}^{5}{\\left(-1\\right)}^{n}\\frac{{x}^{2n}}{{2}^{n}n\\text{!}}dx=\\frac{2}{\\sqrt{2\\pi }}\\displaystyle\\sum _{n=0}^{5}{\\left(-1\\right)}^{n}\\frac{1}{\\left(2n+1\\right){2}^{n}n\\text{!}}\\approx 0.6827[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    65. [T]<\/strong> Suppose that a set of standardized test scores is normally distributed with mean [latex]\\mu =100[\/latex] and standard deviation [latex]\\sigma =10[\/latex]. Set up an integral that represents the probability that a test score will be between [latex]70[\/latex] and [latex]130[\/latex] and use the integral of the degree [latex]50[\/latex] Maclaurin polynomial of [latex]\\frac{1}{\\sqrt{2\\pi }}{e}^{\\frac{\\text{-}{x}^{2}}{2}}[\/latex] to estimate this probability.<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n

    66. [T]<\/strong> Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]f\\left(x\\right)[\/latex] such that [latex]f\\left(0\\right)=1,{f}^{\\prime }\\left(0\\right)=0[\/latex], and [latex]f\\text{''}\\left(x\\right)=\\text{-}f\\left(x\\right)[\/latex]. Find a formula for [latex]{a}_{n}[\/latex] and plot the partial sum [latex]{S}_{N}[\/latex] for [latex]N=20[\/latex] on [latex]\\left[-5,5\\right][\/latex].<\/p>\n<\/div>\n

    \n

    \n

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

    As in the previous problem one obtains [latex]{a}_{n}=0[\/latex] if [latex]n[\/latex] is odd and [latex]{a}_{n}=\\text{-}\\left(n+2\\right)\\left(n+1\\right){a}_{n+2}[\/latex] if [latex]n[\/latex] is even, so [latex]{a}_{0}=1[\/latex] leads to [latex]{a}_{2n}=\\frac{{\\left(-1\\right)}^{n}}{\\left(2n\\right)\\text{!}}[\/latex].<\/div>\n<\/div>\n

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

    \n
    \n
    67. [T]<\/strong> Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]f\\left(x\\right)[\/latex] such that [latex]f\\left(0\\right)=0,{f}^{\\prime }\\left(0\\right)=1[\/latex], and [latex]f\\text{''}\\left(x\\right)=\\text{-}f\\left(x\\right)[\/latex]. Find a formula for [latex]{a}_{n}[\/latex] and plot the partial sum [latex]{S}_{N}[\/latex] for [latex]N=10[\/latex] on [latex]\\left[-5,5\\right][\/latex].<\/div>\n<\/div>\n<\/div>\n
    \n
    \n
    \n
    \n
    \n

    68.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]y[\/latex] such that [latex]y\\text{''}-{y}^{\\prime }+y=0[\/latex] where [latex]y\\left(0\\right)=1[\/latex] and [latex]y^{\\prime} \\left(0\\right)=0[\/latex]. Find a formula that relates [latex]{a}_{n+2},{a}_{n+1}[\/latex], and [latex]{a}_{n}[\/latex] and compute [latex]{a}_{0},...,{a}_{5}[\/latex].<\/p>\n<\/div>\n

    \n

    \n

    Show Solution<\/span><\/p>\n
    [latex]y\\text{''}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+2\\right)\\left(n+1\\right){a}_{n+2}{x}^{n}[\/latex] and [latex]{y}^{\\prime }=\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){a}_{n+1}{x}^{n}[\/latex] so [latex]y\\text{''}-{y}^{\\prime }+y=0[\/latex] implies that [latex]\\left(n+2\\right)\\left(n+1\\right){a}_{n+2}-\\left(n+1\\right){a}_{n+1}+{a}_{n}=0[\/latex] or [latex]{a}_{n}=\\frac{{a}_{n - 1}}{n}-\\frac{{a}_{n - 2}}{n\\left(n - 1\\right)}[\/latex] for all [latex]n\\cdot y\\left(0\\right)={a}_{0}=1[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)={a}_{1}=0[\/latex], so [latex]{a}_{2}=\\frac{1}{2},{a}_{3}=\\frac{1}{6},{a}_{4}=0[\/latex], and [latex]{a}_{5}=-\\frac{1}{120}[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
    69.\u00a0<\/strong>Suppose that [latex]\\displaystyle\\sum _{n=0}^{\\infty }{a}_{n}{x}^{n}[\/latex] converges to a function [latex]y[\/latex] such that [latex]y\\text{''}-{y}^{\\prime }+y=0[\/latex] where [latex]y\\left(0\\right)=0[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)=1[\/latex]. Find a formula that relates [latex]{a}_{n+2},{a}_{n+1}[\/latex], and [latex]{a}_{n}[\/latex] and compute [latex]{a}_{1},...,{a}_{5}[\/latex].<\/span><\/div>\n<\/div>\n<\/div>\n

    The error in approximating the integral [latex]{\\displaystyle\\int }_{a}^{b}f\\left(t\\right)dt[\/latex] by that of a Taylor approximation [latex]{\\displaystyle\\int }_{a}^{b}{P}_{n}\\left(t\\right)dt[\/latex] is at most [latex]{\\displaystyle\\int }_{a}^{b}{R}_{n}\\left(t\\right)dt[\/latex]. In the following exercises, the Taylor remainder estimate [latex]{R}_{n}\\le \\frac{M}{\\left(n+1\\right)\\text{!}}{|x-a|}^{n+1}[\/latex] guarantees that the integral of the Taylor polynomial of the given order approximates the integral of [latex]f[\/latex] with an error less than [latex]\\frac{1}{10}[\/latex].<\/p>\n

      \n
    1. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than [latex]\\frac{1}{100}[\/latex].<\/li>\n
    2. Compare the accuracy of the polynomial integral estimate with the remainder estimate.<\/li>\n<\/ol>\n
      \n
      \n
      \n
      \n

      70. [T]<\/strong> [latex]{\\displaystyle\\int }_{0}^{\\pi }\\frac{\\sin{t}}{t}dt;{P}_{s}=1-\\frac{{x}^{2}}{3\\text{!}}+\\frac{{x}^{4}}{5\\text{!}}-\\frac{{x}^{6}}{7\\text{!}}+\\frac{{x}^{8}}{9\\text{!}}[\/latex] (You may assume that the absolute value of the ninth derivative of [latex]\\frac{\\sin{t}}{t}[\/latex] is bounded by [latex]0.1.[\/latex])<\/p>\n

      Show Solution<\/span><\/p>\n
      a. (Proof) b. We have [latex]{R}_{s}\\le \\frac{0.1}{\\left(9\\right)\\text{!}}{\\pi }^{9}\\approx 0.0082<0.01[\/latex]. We have [latex]{\\displaystyle\\int }_{0}^{\\pi }\\left(1-\\frac{{x}^{2}}{3\\text{!}}+\\frac{{x}^{4}}{5\\text{!}}-\\frac{{x}^{6}}{7\\text{!}}+\\frac{{x}^{8}}{9\\text{!}}\\right)dx=\\pi -\\frac{{\\pi }^{3}}{3\\cdot 3\\text{!}}+\\frac{{\\pi }^{5}}{5\\cdot 5\\text{!}}-\\frac{{\\pi }^{7}}{7\\cdot 7\\text{!}}+\\frac{{\\pi }^{9}}{9\\cdot 9\\text{!}}=1.852...[\/latex], whereas [latex]{\\displaystyle\\int }_{0}^{\\pi }\\frac{\\sin{t}}{t}dt=1.85194...[\/latex], so the actual error is approximately [latex]0.00006[\/latex].<\/div>\n<\/div>\n<\/div>\n
      <\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      71. [T]<\/strong> [latex]{\\displaystyle\\int }_{0}^{2}{e}^{\\text{-}{x}^{2}}dx;{p}_{11}=1-{x}^{2}+\\frac{{x}^{4}}{2}-\\frac{{x}^{6}}{3\\text{!}}+\\cdots-\\frac{{x}^{22}}{11\\text{!}}[\/latex] (You may assume that the absolute value of the [latex]23\\text{rd}[\/latex] derivative of [latex]{e}^{\\text{-}{x}^{2}}[\/latex] is less than [latex]2\\times {10}^{14}.[\/latex])<\/div>\n<\/div>\n<\/div>\n

      The following exercises deal with Fresnel integrals<\/span>.<\/p>\n

      \n
      \n
      \n
      \n
      \n

      72.\u00a0<\/strong>The Fresnel integrals are defined by [latex]C\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\cos\\left({t}^{2}\\right)dt[\/latex] and [latex]S\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\sin\\left({t}^{2}\\right)dt[\/latex]. Compute the power series of [latex]C\\left(x\\right)[\/latex] and [latex]S\\left(x\\right)[\/latex] and plot the sums [latex]{C}_{N}\\left(x\\right)[\/latex] and [latex]{S}_{N}\\left(x\\right)[\/latex] of the first [latex]N=50[\/latex] nonzero terms on [latex]\\left[0,2\\pi \\right][\/latex].<\/p>\n<\/div>\n

      \n

      \n

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

      Since [latex]\\cos\\left({t}^{2}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{4n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]\\sin\\left({t}^{2}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{t}^{4n+2}}{\\left(2n+1\\right)\\text{!}}[\/latex], one has [latex]S\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{4n+3}}{\\left(4n+3\\right)\\left(2n+1\\right)\\text{!}}[\/latex] and [latex]C\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\frac{{x}^{4n+1}}{\\left(4n+1\\right)\\left(2n\\right)\\text{!}}[\/latex]. The sums of the first [latex]50[\/latex] nonzero terms are plotted below with [latex]{C}_{50}\\left(x\\right)[\/latex] the solid curve and [latex]{S}_{50}\\left(x\\right)[\/latex] the dashed curve.<\/div>\n<\/div>\n

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

      73. [T]<\/strong> The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates [latex]\\left(C\\left(t\\right),S\\left(t\\right)\\right)[\/latex]. Plot the curve [latex]\\left({C}_{50},{S}_{50}\\right)[\/latex] for [latex]0\\le t\\le 2\\pi[\/latex], the coordinates of which were computed in the previous exercise.<\/span><\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      \n
      \n

      74.\u00a0<\/strong>Estimate [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{4}}\\sqrt{x-{x}^{2}}dx[\/latex] by approximating [latex]\\sqrt{1-x}[\/latex] using the binomial approximation [latex]1-\\frac{x}{2}-\\frac{{x}^{2}}{8}-\\frac{{x}^{3}}{16}-\\frac{5{x}^{4}}{2128}-\\frac{7{x}^{5}}{256}[\/latex].<\/p>\n<\/div>\n

      \n
      Show Solution<\/span><\/p>\n
      [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{4}}\\sqrt{x}\\left(1-\\frac{x}{2}-\\frac{{x}^{2}}{8}-\\frac{{x}^{3}}{16}-\\frac{5{x}^{4}}{128}-\\frac{7{x}^{5}}{256}\\right)dx[\/latex]<\/p>\n

      [latex]=\\frac{2}{3}{2}^{-3}-\\frac{1}{2}\\frac{2}{5}{2}^{-5}-\\frac{1}{8}\\frac{2}{7}{2}^{-7}-\\frac{1}{16}\\frac{2}{9}{2}^{-9}-\\frac{5}{128}\\frac{2}{11}{2}^{-11}-\\frac{7}{256}\\frac{2}{13}{2}^{-13}=0.0767732..[\/latex].<\/p>\n

      whereas [latex]{\\displaystyle\\int }_{0}^{\\frac{1}{4}}\\sqrt{x-{x}^{2}}dx=0.076773[\/latex].<\/p>\n

      <\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      \n
      \n
      75. [T]<\/strong> Use Newton\u2019s approximation of the binomial [latex]\\sqrt{1-{x}^{2}}[\/latex] to approximate [latex]\\pi[\/latex] as follows. The circle centered at [latex]\\left(\\frac{1}{2},0\\right)[\/latex] with radius [latex]\\frac{1}{2}[\/latex] has upper semicircle [latex]y=\\sqrt{x}\\sqrt{1-x}[\/latex]. The sector of this circle bounded by the [latex]x[\/latex] -axis between [latex]x=0[\/latex] and [latex]x=\\frac{1}{2}[\/latex] and by the line joining [latex]\\left(\\frac{1}{4},\\frac{\\sqrt{3}}{4}\\right)[\/latex] corresponds to [latex]\\frac{1}{6}[\/latex] of the circle and has area [latex]\\frac{\\pi }{24}[\/latex]. This sector is the union of a right triangle with height [latex]\\frac{\\sqrt{3}}{4}[\/latex] and base [latex]\\frac{1}{4}[\/latex] and the region below the graph between [latex]x=0[\/latex] and [latex]x=\\frac{1}{4}[\/latex]. To find the area of this region you can write [latex]y=\\sqrt{x}\\sqrt{1-x}=\\sqrt{x}\\times \\left(\\text{binomial expansion of}\\sqrt{1-x}\\right)[\/latex] and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate [latex]\\pi[\/latex].<\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      \n
      \n

      76.\u00a0<\/strong>Use the approximation [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}\\right)[\/latex] to approximate the period of a pendulum having length [latex]10[\/latex] meters and maximum angle [latex]{\\theta }_{\\text{max}}=\\frac{\\pi }{6}[\/latex] where [latex]k=\\sin\\left(\\frac{{\\theta }_{\\text{max}}}{2}\\right)[\/latex]. Compare this with the small angle estimate [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}[\/latex].<\/p>\n<\/div>\n

      \n

      \n

      Show Solution<\/span><\/p>\n
      [latex]T\\approx 2\\pi \\sqrt{\\frac{10}{9.8}}\\left(1+\\frac{{\\sin}^{2}\\left(\\frac{\\theta}{12}\\right)}{4}\\right)\\approx 6.453[\/latex] seconds. The small angle estimate is [latex]T\\approx 2\\pi \\sqrt{\\frac{10}{9.8}\\approx 6.347}[\/latex]. The relative error is around [latex]2[\/latex] percent.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      77.\u00a0<\/strong>Suppose that a pendulum is to have a period of [latex]2[\/latex] seconds and a maximum angle of [latex]{\\theta }_{\\text{max}}=\\frac{\\pi }{6}[\/latex]. Use [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}\\right)[\/latex] to approximate the desired length of the pendulum. What length is predicted by the small angle estimate [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}?[\/latex]<\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      \n
      \n

      78.\u00a0<\/strong>Evaluate [latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}{\\sin}^{4}\\theta d\\theta[\/latex] in the approximation [latex]T=4\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}\\left(1+\\frac{1}{2}{k}^{2}{\\sin}^{2}\\theta +\\frac{3}{8}{k}^{4}{\\sin}^{4}\\theta +\\cdots\\right)d\\theta[\/latex] to obtain an improved estimate for [latex]T[\/latex].<\/p>\n<\/div>\n

      \n

      \n

      Show Solution<\/span><\/p>\n
      [latex]{\\displaystyle\\int }_{0}^{\\frac{\\pi}{2}}{\\sin}^{4}\\theta d\\theta =\\frac{3\\pi }{16}[\/latex]. Hence [latex]T\\approx 2\\pi \\sqrt{\\frac{L}{g}}\\left(1+\\frac{{k}^{2}}{4}+\\frac{9}{256}{k}^{4}\\right)[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n
      \n
      79. [T]<\/strong> An equivalent formula for the period of a pendulum with amplitude [latex]{\\theta }_{\\text{max}}[\/latex] is [latex]T\\left({\\theta }_{\\text{max}}\\right)=2\\sqrt{2}\\sqrt{\\frac{L}{g}}{\\displaystyle\\int }_{0}^{{\\theta }_{\\text{max}}}\\frac{d\\theta }{\\sqrt{\\cos\\theta }-\\cos\\left({\\theta }_{\\text{max}}\\right)}[\/latex] where [latex]L[\/latex] is the pendulum length and [latex]g[\/latex] is the gravitational acceleration constant. When [latex]{\\theta }_{\\text{max}}=\\frac{\\pi }{3}[\/latex] we get [latex]\\frac{1}{\\sqrt{\\cos{t} - \\frac{1}{2}}}\\approx \\sqrt{2}\\left(1+\\frac{{t}^{2}}{2}+\\frac{{t}^{4}}{3}+\\frac{181{t}^{6}}{720}\\right)[\/latex]. Integrate this approximation to estimate [latex]T\\left(\\frac{\\pi }{3}\\right)[\/latex] in terms of [latex]L[\/latex] and [latex]g[\/latex]. Assuming [latex]g=9.806[\/latex] meters per second squared, find an approximate length [latex]L[\/latex] such that [latex]T\\left(\\frac{\\pi }{3}\\right)=2[\/latex] seconds.<\/div>\n<\/div>\n<\/div>\n\n\t\t\t
      \n\t\t\t

      Candela Citations<\/h3>\n\t\t\t\t\t
      \n\t\t\t\t\t\t
      \n\t\t\t\t\t\t\t
      CC licensed content, Shared previously<\/div>