{"id":969,"date":"2025-06-20T17:25:36","date_gmt":"2025-06-20T17:25:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=969"},"modified":"2025-08-14T17:16:14","modified_gmt":"2025-08-14T17:16:14","slug":"power-series-and-applications-get-stronger","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/power-series-and-applications-get-stronger\/","title":{"raw":"Power Series and Applications: Get Stronger","rendered":"Power Series and Applications: Get Stronger"},"content":{"raw":"<h2><span data-sheets-root=\"1\">Introduction to Power Series<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (1-3), state whether each statement is true, or give an example to show that it is false.<\/strong><\/p>\r\n\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{x}^{n}[\/latex] converges, then [latex]{a}_{n}{x}^{n}\\to 0[\/latex] as [latex]n\\to \\infty [\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Given any sequence [latex]{a}_{n}[\/latex], there is always some [latex]R&gt;0[\/latex], possibly very small, such that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{x}^{n}[\/latex] converges on [latex]\\left(\\text{-}R,R\\right)[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that [latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{\\left(x - 3\\right)}^{n}[\/latex] converges at [latex]x=6[\/latex]. At which of the following points must the series also converge? Use the fact that if [latex]\\displaystyle\\sum {a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges at [latex]x[\/latex], then it converges at any point closer to [latex]c[\/latex] than [latex]x[\/latex].\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=5.99[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x=0.000001[\/latex]<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (4-6), suppose that [latex]|\\dfrac{{a}{n+1}}{{a}{n}}|\\to 1[\/latex] as [latex]n\\to \\infty [\/latex]. Find the radius of convergence for each series.<\/strong><\/p>\r\n\r\n<ol start=\"4\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{2}^{n}{x}^{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }\\dfrac{{a}{n}{\\pi }^{n}{x}^{n}}{{e}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{\\left(-1\\right)}^{n}{x}^{2n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (7-11), find the radius of convergence R and interval of convergence for [latex]\\displaystyle\\sum {a}{n}{x}^{n}[\/latex] with the given coefficients [latex]{a}{n}[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(2x\\right)}^{n}}{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n{x}^{n}}{{2}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{n}^{2}{x}^{n}}{{2}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{\\pi }^{k}{x}^{k}}{{k}^{\\pi }}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{10}^{n}{x}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (12-14), find the radius of convergence of each series.<\/strong><\/p>\r\n\r\n<ol start=\"12\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{\\left(k\\text{!}\\right)}^{2}{x}^{k}}{\\left(2k\\right)\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{k\\text{!}}{1\\cdot 3\\cdot 5\\text{...}\\left(2k - 1\\right)}{x}^{k}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{x}^{n}}{\\left(\\begin{array}{c}2n\\ n\\end{array}\\right)}[\/latex] where [latex]\\left(\\begin{array}{c}n\\ k\\end{array}\\right)=\\dfrac{n\\text{!}}{k\\text{!}\\left(n-k\\right)\\text{!}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (15-16), use the ratio test to determine the radius of convergence of each series.<\/strong><\/p>\r\n\r\n<ol start=\"15\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n\\text{!}\\right)}^{3}}{\\left(3n\\right)\\text{!}}{x}^{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n\\text{!}}{{n}^{n}}{x}^{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (17-21), given that [latex]\\dfrac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex] with convergence in [latex]\\left(-1,1\\right)[\/latex], find the power series for each function with the given center [latex]a[\/latex], and identify its interval of convergence.<\/strong><\/p>\r\n\r\n<ol start=\"17\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{x};a=1[\/latex] (Hint: [latex]\\dfrac{1}{x}=\\dfrac{1}{1-\\left(1-x\\right)}[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{x}{1-{x}^{2}};a=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{{x}^{2}}{1+{x}^{2}};a=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{1 - 2x};a=0[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{{x}^{2}}{1 - 4{x}^{2}};a=0[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (22-23), suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{x}^{n}[\/latex] satisfies [latex]\\underset{n\\to \\infty }{\\text{lim}}\\dfrac{{a}{n+1}}{{a}{n}}=1[\/latex] where [latex]{a}_{n}\\ge 0[\/latex] for each [latex]n[\/latex]. State whether each series converges on the full interval [latex]\\left(-1,1\\right)[\/latex], or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.<\/strong><\/p>\r\n\r\n<ol start=\"22\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{2n}{x}^{2n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{{n}^{2}}{x}^{{n}^{2}}[\/latex] (Hint: Let [latex]{b}{k}={a}{k}[\/latex] if [latex]k={n}^{2}[\/latex] for some n, otherwise [latex]{b}_{k}=0.[\/latex])<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (24-25), solve each problem.<\/strong>\r\n<ol start=\"24\">\r\n \t<li class=\"whitespace-normal break-words\">Plot the graphs of [latex]\\dfrac{1}{1-x}[\/latex] and of the partial sums [latex]{S}_{N}=\\displaystyle\\sum {n=0}^{N}{x}^{n}[\/latex] for [latex]n=10,20,30[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the approximation of [latex]\\dfrac{1}{1-x}[\/latex] by [latex]{S}{N}[\/latex] near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N increases.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Plot the graphs of the partial sums [latex]{S}_{n}=\\displaystyle\\sum _{n=1}^{N}\\dfrac{{x}^{n}}{{n}^{2}}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N increases.<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Operations with Power Series<\/span><\/h2>\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{n}}{n\\text{!}}[\/latex] and [latex]g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{n}}{n\\text{!}}[\/latex], find the power series of [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)[\/latex] and of [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (2-3), use partial fractions to find the power series of each function.<\/strong><\/p>\r\n\r\n<ol start=\"2\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{4}{\\left(x - 3\\right)\\left(x+1\\right)}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (4-6), express each series as a rational function.<\/strong><\/p>\r\n\r\n<ol start=\"4\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{x}^{n}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{\\left(x - 3\\right)}^{2n - 1}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>The following exercises (6-8) explore applications of annuities.<\/strong><\/p>\r\n\r\n<ol start=\"6\">\r\n \t<li class=\"whitespace-normal break-words\">Calculate the present values [latex]P[\/latex] of an annuity in which [latex]$10,000[\/latex] is to be paid out annually for a period of [latex]20[\/latex] years, assuming interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate the annual payouts [latex]C[\/latex] to be given for [latex]20[\/latex] years on annuities having present value [latex]$100,000[\/latex] assuming respective interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that an annuity has a present value [latex]P=1\\text{million dollars}[\/latex]. What interest rate [latex]r[\/latex] would allow for perpetual annual payouts of [latex]$50,000[\/latex]?<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (9-10), express the sum of each power series in terms of geometric series, and then express the sum as a rational function.<\/strong><\/p>\r\n\r\n<ol start=\"9\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\\cdots[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]x-{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}-{x}^{6}+{x}^{7}-\\cdots[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (11-12), find the power series of [latex]f\\left(x\\right)g\\left(x\\right)[\/latex] given [latex]f[\/latex] and [latex]g[\/latex] as defined.<\/strong><\/p>\r\n\r\n<ol start=\"11\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=2\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n},g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }n{x}^{n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\dfrac{x}{2}\\right)}^{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, differentiate the given series expansion of [latex]f[\/latex] term-by-term to obtain the corresponding series expansion for the derivative of [latex]f[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"13\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{1+x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, integrate the given series expansion of [latex]f[\/latex] term-by-term from zero to [latex]x[\/latex] to obtain the corresponding series expansion for the indefinite integral of [latex]f[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"14\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{2x}{{\\left(1+{x}^{2}\\right)}^{2}}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\left(2n\\right){x}^{2n - 1}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (15-16), evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.<\/strong><\/p>\r\n\r\n<ol start=\"15\">\r\n \t<li class=\"whitespace-normal break-words\">Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{{2}^{n}}[\/latex] as [latex]{f}^{\\prime }\\left(\\dfrac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{n\\left(n - 1\\right)}{{2}^{n}}[\/latex] as [latex]f''\\left(\\dfrac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (17-20), given that [latex]\\dfrac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex], use term-by-term differentiation or integration to find power series for each function centered at the given point.<\/strong><\/p>\r\n\r\n<ol start=\"17\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\text{ln}x[\/latex] centered at [latex]x=1[\/latex] (Hint: [latex]x=1-\\left(1-x\\right)[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left(1-{x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\tan}^{-1}\\left({x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\text{ln}tdt[\/latex] where [latex]\\text{ln}\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\dfrac{{\\left(x - 1\\right)}^{n}}{n}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (21-22), using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.<\/strong><\/p>\r\n\r\n<ol start=\"21\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{3k}}{6k}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }{2}^{\\text{-}kx}[\/latex] using [latex]y={2}^{\\text{-}x}[\/latex]<\/li>\r\n<\/ol>\r\n<strong>For the following exercises (23-26), solve each problem.<\/strong>\r\n<ol start=\"23\">\r\n \t<li class=\"whitespace-normal break-words\">Differentiate the series [latex]E\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{n}}{n\\text{!}}[\/latex] term-by-term to show that [latex]E\\left(x\\right)[\/latex] is equal to its derivative.<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Suppose that the coefficients an of the series [latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{x}^{n}[\/latex] are defined by the recurrence relation [latex]{a}{n}=\\dfrac{{a}{n - 1}}{n}+\\dfrac{{a}{n - 2}}{n\\left(n - 1\\right)}[\/latex]. For [latex]{a}{0}=0[\/latex] and [latex]{a}{1}=1[\/latex], compute and plot the sums [latex]{S}{N}=\\displaystyle\\sum {n=0}^{N}{a}{n}{x}^{n}[\/latex] for [latex]N=2,3,4,5[\/latex] on [latex]\\left[-1,1\\right][\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Given the power series expansion [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\dfrac{{x}^{n}}{n}[\/latex], determine how many terms N of the sum evaluated at [latex]x=-\\dfrac{1}{2}[\/latex] are needed to approximate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within [latex]\\dfrac{1}{1000}[\/latex]. Evaluate the corresponding partial sum [latex]\\displaystyle\\sum _{n=1}^{N}{\\left(-1\\right)}^{n - 1}\\dfrac{{x}^{n}}{n}[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Recall that [latex]{\\tan}^{-1}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=\\dfrac{\\pi }{6}[\/latex]. Assuming an exact value of [latex]\\left(\\dfrac{1}{\\sqrt{3}}\\right)[\/latex], estimate [latex]\\dfrac{\\pi }{6}[\/latex] by evaluating partial sums [latex]{S}_{N}\\left(\\dfrac{1}{\\sqrt{3}}\\right)[\/latex] of the power series expansion [latex]{\\tan}^{-1}\\left(x\\right)=\\displaystyle\\sum {k=0}^{\\infty }{\\left(-1\\right)}^{k}\\dfrac{{x}^{2k+1}}{2k+1}[\/latex] at [latex]x=\\dfrac{1}{\\sqrt{3}}[\/latex]. What is the smallest number N such that [latex]6{S}{N}\\left(\\dfrac{1}{\\sqrt{3}}\\right)[\/latex] approximates \u03c0 accurately to within [latex]0.001[\/latex]? How many terms are needed for accuracy to within [latex]0.00001[\/latex]?<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Taylor and Maclaurin Series<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (1-4), find the Taylor polynomials of degree two approximating the given function centered at the given point.<\/strong><\/p>\r\n\r\n<ol>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=1+x+{x}^{2}[\/latex] at [latex]a=-1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\sin\\left(2x\\right)[\/latex] at [latex]a=\\dfrac{\\pi }{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]a=1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{x}[\/latex] at [latex]a=1[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (5-7), verify that the given choice of n in the remainder estimate [latex]|{R}{n}|\\le \\dfrac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between a and the indicated point, yields [latex]|{R}{n}|\\le \\dfrac{1}{1000}[\/latex]. Find the value of the Taylor polynomial [latex]p_n[\/latex] of [latex]f[\/latex] at the indicated point.<\/strong><\/p>\r\n\r\n<ol start=\"5\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\left(28\\right)}^{\\dfrac{1}{3}};a=27,n=1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]e^2[\/latex]; [latex]a=0,n=9[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left(2\\right);a=1,n=1000[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate the approximation [latex]{e}^{x}\\approx 1+x+\\dfrac{{x}^{2}}{2}+\\cdots+\\dfrac{{x}^{6}}{720}[\/latex] evaluated at \u2212x2 to approximate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (9-10), find the smallest value of n such that the remainder estimate [latex]|{R}{n}|\\le \\dfrac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between a and the indicated point, yields [latex]|{R}{n}|\\le \\dfrac{1}{1000}[\/latex] on the indicated interval.<\/strong><\/p>\r\n\r\n<ol start=\"9\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\cos{x}[\/latex] on [latex]\\left[-\\dfrac{\\pi }{2},\\dfrac{\\pi }{2}\\right],a=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex] on [latex]\\left[-3,3\\right],a=0[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (11-12), the maximum of the right-hand side of the remainder estimate [latex]|{R}_{1}|\\le \\dfrac{\\text{max}|f\\text{''}\\left(z\\right)|}{2}{R}^{2}[\/latex] on [latex]\\left[a-R,a+R\\right][\/latex] occurs at a or [latex]a\\pm R[\/latex]. Estimate the maximum value of [latex]R[\/latex] such that [latex]\\dfrac{\\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 [latex]R[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"11\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sin{x}[\/latex] approximated by x, [latex]a=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos{x}[\/latex] approximated by [latex]1,a=0[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (13-17), find the Taylor series of the given function centered at the indicated point.<\/strong><\/p>\r\n\r\n<ol start=\"13\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]1+x+{x}^{2}+{x}^{3}[\/latex] at [latex]a=-1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos{x}[\/latex] at [latex]a=2\\pi [\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\cos{x}[\/latex] at [latex]x=\\dfrac{\\pi }{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{e}^{x}[\/latex] at [latex]a=1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{{\\left(x - 1\\right)}^{3}}[\/latex] at [latex]a=0[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (18-22), compute the Taylor series of each function around [latex]x=1[\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"18\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=2-x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\left(x - 2\\right)}^{2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{x}{4x - 2{x}^{2}-1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{2x}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (23-24), identify the value of [latex]x[\/latex] 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].<\/strong><\/p>\r\n\r\n<ol start=\"23\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{2}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n}{\\left(2\\pi \\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>The following exercises (25-26) make use of the functions [latex]{S}{5}\\left(x\\right)=x-\\dfrac{{x}^{3}}{6}+\\dfrac{{x}^{5}}{120}[\/latex] and [latex]{C}{4}\\left(x\\right)=1-\\dfrac{{x}^{2}}{2}+\\dfrac{{x}^{4}}{24}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex].<\/strong><\/p>\r\n\r\n<ol start=\"25\">\r\n \t<li class=\"whitespace-normal break-words\">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].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Compare [latex]\\dfrac{{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+\\dfrac{{x}^{3}}{3}+\\dfrac{2{x}^{5}}{15}[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (27-28), 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.<\/strong><\/p>\r\n\r\n<ol start=\"27\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to 0}{\\text{lim}}\\dfrac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to {0}^{+}}{\\text{lim}}\\dfrac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}[\/latex]<\/li>\r\n<\/ol>\r\n<h2><span data-sheets-root=\"1\">Applications of Series<\/span><\/h2>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (1-2), use appropriate substitutions to write down the Maclaurin series for the given binomial.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\left(1+{x}^{2}\\right)}^{\\dfrac{-1}{3}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\left(1 - 2x\\right)}^{\\dfrac{2}{3}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (3-6), use the substitution [latex]{\\left(b+x\\right)}^{r}={\\left(b+a\\right)}^{r}{\\left(1+\\dfrac{x-a}{b+a}\\right)}^{r}[\/latex] in the binomial expansion to find the Taylor series of each function with the given center.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"3\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{{x}^{2}+2}[\/latex] at [latex]a=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{2x-{x}^{2}}[\/latex] at [latex]a=1[\/latex] (Hint: [latex]2x-{x}^{2}=1-{\\left(x - 1\\right)}^{2}[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{x}[\/latex] at [latex]a=4[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{x}[\/latex] at [latex]x=9[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most [latex]\\dfrac{1}{1000}[\/latex].<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\left(1001\\right)}^{\\dfrac{1}{3}}[\/latex] using [latex]{\\left(1000+x\\right)}^{\\dfrac{1}{3}}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (8-10), use the binomial approximation [latex]\\sqrt{1-x}\\approx 1-\\dfrac{x}{2}-\\dfrac{{x}^{2}}{8}-\\dfrac{{x}^{3}}{16}-\\dfrac{5{x}^{4}}{128}-\\dfrac{7{x}^{5}}{256}[\/latex] for [latex]|x|&lt;1[\/latex] to approximate each number. Compare this value to the value given by a scientific calculator.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"8\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{5}=5\\times \\dfrac{1}{\\sqrt{5}}[\/latex] using [latex]x=\\dfrac{4}{5}[\/latex] in [latex]{\\left(1-x\\right)}^{\\dfrac{1}{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sqrt{6}[\/latex] using [latex]x=\\dfrac{5}{6}[\/latex] in [latex]{\\left(1-x\\right)}^{\\dfrac{1}{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">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]\\dfrac{\\pi }{2}[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (11-14), use the expansion [latex]{\\left(1+x\\right)}^{\\dfrac{1}{3}}=1+\\dfrac{1}{3}x-\\dfrac{1}{9}{x}^{2}+\\dfrac{5}{81}{x}^{3}-\\dfrac{10}{243}{x}^{4}+\\cdots[\/latex] to write the first five terms (not necessarily a quartic polynomial) of each expression.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"11\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\left(1+4x\\right)}^{\\dfrac{4}{3}};a=0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\left({x}^{2}+6x+10\\right)}^{\\dfrac{1}{3}};a=-3[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use the approximation [latex]{\\left(1-x\\right)}^{\\dfrac{2}{3}}=1-\\dfrac{2x}{3}-\\dfrac{{x}^{2}}{9}-\\dfrac{4{x}^{3}}{81}-\\dfrac{7{x}^{4}}{243}-\\dfrac{14{x}^{5}}{729}+\\cdots [\/latex] for [latex]|x|&lt;1[\/latex] to approximate [latex]{2}^{\\dfrac{1}{3}}={2.2}^{\\dfrac{-2}{3}}[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Find the [latex]99[\/latex] th derivative of [latex]f\\left(x\\right)={\\left(1+{x}^{4}\\right)}^{25}[\/latex].<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (15-18), find the Maclaurin series of each function.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"15\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{\\sin\\left(\\sqrt{x}\\right)}{\\sqrt{x}},\\left(x&gt;0\\right)[\/latex],<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{{x}^{3}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\sin}^{2}x[\/latex] using the identity [latex]{\\sin}^{2}x=\\dfrac{1}{2}-\\dfrac{1}{2}\\cos\\left(2x\\right)[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (19-22), 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.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"19\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]F\\left(x\\right)={\\tan}^{-1}x;f\\left(t\\right)=\\dfrac{1}{1+{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{t}^{2n}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]F\\left(x\\right)={\\sin}^{-1}x;f\\left(t\\right)=\\dfrac{1}{\\sqrt{1-{t}^{2}}}=\\displaystyle\\sum _{k=0}^{\\infty }\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ k\\hfill \\end{array}\\right)\\dfrac{{t}^{2k}}{k\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[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}\\dfrac{{x}^{n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\dfrac{\\text{ln}\\left(1+t\\right)}{t}dt;f\\left(t\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{t}^{n}}{n+1}[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (23-26), compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of [latex]f[\/latex].<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"23\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\tan{x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{x}\\cos{x}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\sec}^{2}x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{\\tan\\sqrt{x}}{\\sqrt{x}}[\/latex] (see expansion for [latex]\\tan{x}[\/latex])<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (27-28), find the radius of convergence of the Maclaurin series of each function.<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"27\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{1+{x}^{2}}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left(1+{x}^{2}\\right)[\/latex]<\/li>\r\n<\/ol>","rendered":"<h2><span data-sheets-root=\"1\">Introduction to Power Series<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (1-3), state whether each statement is true, or give an example to show that it is false.<\/strong><\/p>\n<ol>\n<li class=\"whitespace-normal break-words\">If [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{x}^{n}[\/latex] converges, then [latex]{a}_{n}{x}^{n}\\to 0[\/latex] as [latex]n\\to \\infty[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Given any sequence [latex]{a}_{n}[\/latex], there is always some [latex]R>0[\/latex], possibly very small, such that [latex]\\displaystyle\\sum {n=1}^{\\infty }{a}{n}{x}^{n}[\/latex] converges on [latex]\\left(\\text{-}R,R\\right)[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that [latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{\\left(x - 3\\right)}^{n}[\/latex] converges at [latex]x=6[\/latex]. At which of the following points must the series also converge? Use the fact that if [latex]\\displaystyle\\sum {a}_{n}{\\left(x-c\\right)}^{n}[\/latex] converges at [latex]x[\/latex], then it converges at any point closer to [latex]c[\/latex] than [latex]x[\/latex].\n<ol style=\"list-style-type: lower-alpha;\">\n<li class=\"whitespace-normal break-words\">[latex]x=1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=5.99[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x=0.000001[\/latex]<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (4-6), suppose that [latex]|\\dfrac{{a}{n+1}}{{a}{n}}|\\to 1[\/latex] as [latex]n\\to \\infty[\/latex]. Find the radius of convergence for each series.<\/strong><\/p>\n<ol start=\"4\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{2}^{n}{x}^{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }\\dfrac{{a}{n}{\\pi }^{n}{x}^{n}}{{e}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{\\left(-1\\right)}^{n}{x}^{2n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (7-11), find the radius of convergence R and interval of convergence for [latex]\\displaystyle\\sum {a}{n}{x}^{n}[\/latex] with the given coefficients [latex]{a}{n}[\/latex].<\/strong><\/p>\n<ol start=\"7\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(2x\\right)}^{n}}{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n{x}^{n}}{{2}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{n}^{2}{x}^{n}}{{2}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{\\pi }^{k}{x}^{k}}{{k}^{\\pi }}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{10}^{n}{x}^{n}}{n\\text{!}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (12-14), find the radius of convergence of each series.<\/strong><\/p>\n<ol start=\"12\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{\\left(k\\text{!}\\right)}^{2}{x}^{k}}{\\left(2k\\right)\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{k\\text{!}}{1\\cdot 3\\cdot 5\\text{...}\\left(2k - 1\\right)}{x}^{k}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{x}^{n}}{\\left(\\begin{array}{c}2n\\ n\\end{array}\\right)}[\/latex] where [latex]\\left(\\begin{array}{c}n\\ k\\end{array}\\right)=\\dfrac{n\\text{!}}{k\\text{!}\\left(n-k\\right)\\text{!}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (15-16), use the ratio test to determine the radius of convergence of each series.<\/strong><\/p>\n<ol start=\"15\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(n\\text{!}\\right)}^{3}}{\\left(3n\\right)\\text{!}}{x}^{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n\\text{!}}{{n}^{n}}{x}^{n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (17-21), given that [latex]\\dfrac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex] with convergence in [latex]\\left(-1,1\\right)[\/latex], find the power series for each function with the given center [latex]a[\/latex], and identify its interval of convergence.<\/strong><\/p>\n<ol start=\"17\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{x};a=1[\/latex] (Hint: [latex]\\dfrac{1}{x}=\\dfrac{1}{1-\\left(1-x\\right)}[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{x}{1-{x}^{2}};a=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{{x}^{2}}{1+{x}^{2}};a=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{1 - 2x};a=0[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{{x}^{2}}{1 - 4{x}^{2}};a=0[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (22-23), suppose that [latex]p\\left(x\\right)=\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{x}^{n}[\/latex] satisfies [latex]\\underset{n\\to \\infty }{\\text{lim}}\\dfrac{{a}{n+1}}{{a}{n}}=1[\/latex] where [latex]{a}_{n}\\ge 0[\/latex] for each [latex]n[\/latex]. State whether each series converges on the full interval [latex]\\left(-1,1\\right)[\/latex], or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.<\/strong><\/p>\n<ol start=\"22\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{2n}{x}^{2n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{{n}^{2}}{x}^{{n}^{2}}[\/latex] (Hint: Let [latex]{b}{k}={a}{k}[\/latex] if [latex]k={n}^{2}[\/latex] for some n, otherwise [latex]{b}_{k}=0.[\/latex])<\/li>\n<\/ol>\n<p><strong>For the following exercises (24-25), solve each problem.<\/strong><\/p>\n<ol start=\"24\">\n<li class=\"whitespace-normal break-words\">Plot the graphs of [latex]\\dfrac{1}{1-x}[\/latex] and of the partial sums [latex]{S}_{N}=\\displaystyle\\sum {n=0}^{N}{x}^{n}[\/latex] for [latex]n=10,20,30[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the approximation of [latex]\\dfrac{1}{1-x}[\/latex] by [latex]{S}{N}[\/latex] near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N increases.<\/li>\n<li class=\"whitespace-normal break-words\">Plot the graphs of the partial sums [latex]{S}_{n}=\\displaystyle\\sum _{n=1}^{N}\\dfrac{{x}^{n}}{{n}^{2}}[\/latex] for [latex]n=10,50,100[\/latex] on the interval [latex]\\left[-0.99,0.99\\right][\/latex]. Comment on the behavior of the sums near [latex]x=-1[\/latex] and near [latex]x=1[\/latex] as N increases.<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Operations with Power Series<\/span><\/h2>\n<ol>\n<li class=\"whitespace-normal break-words\">If [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{n}}{n\\text{!}}[\/latex] and [latex]g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{n}}{n\\text{!}}[\/latex], find the power series of [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)[\/latex] and of [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (2-3), use partial fractions to find the power series of each function.<\/strong><\/p>\n<ol start=\"2\">\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{4}{\\left(x - 3\\right)\\left(x+1\\right)}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (4-6), express each series as a rational function.<\/strong><\/p>\n<ol start=\"4\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{x}^{n}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{{\\left(x - 3\\right)}^{2n - 1}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>The following exercises (6-8) explore applications of annuities.<\/strong><\/p>\n<ol start=\"6\">\n<li class=\"whitespace-normal break-words\">Calculate the present values [latex]P[\/latex] of an annuity in which [latex]$10,000[\/latex] is to be paid out annually for a period of [latex]20[\/latex] years, assuming interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Calculate the annual payouts [latex]C[\/latex] to be given for [latex]20[\/latex] years on annuities having present value [latex]$100,000[\/latex] assuming respective interest rates of [latex]r=0.03,r=0.05[\/latex], and [latex]r=0.07[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that an annuity has a present value [latex]P=1\\text{million dollars}[\/latex]. What interest rate [latex]r[\/latex] would allow for perpetual annual payouts of [latex]$50,000[\/latex]?<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (9-10), express the sum of each power series in terms of geometric series, and then express the sum as a rational function.<\/strong><\/p>\n<ol start=\"9\">\n<li class=\"whitespace-normal break-words\">[latex]x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\\cdots[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]x-{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}-{x}^{6}+{x}^{7}-\\cdots[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (11-12), find the power series of [latex]f\\left(x\\right)g\\left(x\\right)[\/latex] given [latex]f[\/latex] and [latex]g[\/latex] as defined.<\/strong><\/p>\n<ol start=\"11\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=2\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n},g\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }n{x}^{n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\dfrac{x}{2}\\right)}^{n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, differentiate the given series expansion of [latex]f[\/latex] term-by-term to obtain the corresponding series expansion for the derivative of [latex]f[\/latex].<\/strong><\/p>\n<ol start=\"13\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{1+x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, integrate the given series expansion of [latex]f[\/latex] term-by-term from zero to [latex]x[\/latex] to obtain the corresponding series expansion for the indefinite integral of [latex]f[\/latex].<\/strong><\/p>\n<ol start=\"14\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{2x}{{\\left(1+{x}^{2}\\right)}^{2}}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}\\left(2n\\right){x}^{2n - 1}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (15-16), evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.<\/strong><\/p>\n<ol start=\"15\">\n<li class=\"whitespace-normal break-words\">Evaluate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{{2}^{n}}[\/latex] as [latex]{f}^{\\prime }\\left(\\dfrac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{n\\left(n - 1\\right)}{{2}^{n}}[\/latex] as [latex]f''\\left(\\dfrac{1}{2}\\right)[\/latex] where [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (17-20), given that [latex]\\dfrac{1}{1-x}=\\displaystyle\\sum _{n=0}^{\\infty }{x}^{n}[\/latex], use term-by-term differentiation or integration to find power series for each function centered at the given point.<\/strong><\/p>\n<ol start=\"17\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\text{ln}x[\/latex] centered at [latex]x=1[\/latex] (Hint: [latex]x=1-\\left(1-x\\right)[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left(1-{x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\tan}^{-1}\\left({x}^{2}\\right)[\/latex] at [latex]x=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\text{ln}tdt[\/latex] where [latex]\\text{ln}\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\dfrac{{\\left(x - 1\\right)}^{n}}{n}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (21-22), using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.<\/strong><\/p>\n<ol start=\"21\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{3k}}{6k}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{k=1}^{\\infty }{2}^{\\text{-}kx}[\/latex] using [latex]y={2}^{\\text{-}x}[\/latex]<\/li>\n<\/ol>\n<p><strong>For the following exercises (23-26), solve each problem.<\/strong><\/p>\n<ol start=\"23\">\n<li class=\"whitespace-normal break-words\">Differentiate the series [latex]E\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{n}}{n\\text{!}}[\/latex] term-by-term to show that [latex]E\\left(x\\right)[\/latex] is equal to its derivative.<\/li>\n<li class=\"whitespace-normal break-words\">Suppose that the coefficients an of the series [latex]\\displaystyle\\sum {n=0}^{\\infty }{a}{n}{x}^{n}[\/latex] are defined by the recurrence relation [latex]{a}{n}=\\dfrac{{a}{n - 1}}{n}+\\dfrac{{a}{n - 2}}{n\\left(n - 1\\right)}[\/latex]. For [latex]{a}{0}=0[\/latex] and [latex]{a}{1}=1[\/latex], compute and plot the sums [latex]{S}{N}=\\displaystyle\\sum {n=0}^{N}{a}{n}{x}^{n}[\/latex] for [latex]N=2,3,4,5[\/latex] on [latex]\\left[-1,1\\right][\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Given the power series expansion [latex]\\text{ln}\\left(1+x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\dfrac{{x}^{n}}{n}[\/latex], determine how many terms N of the sum evaluated at [latex]x=-\\dfrac{1}{2}[\/latex] are needed to approximate [latex]\\text{ln}\\left(2\\right)[\/latex] accurate to within [latex]\\dfrac{1}{1000}[\/latex]. Evaluate the corresponding partial sum [latex]\\displaystyle\\sum _{n=1}^{N}{\\left(-1\\right)}^{n - 1}\\dfrac{{x}^{n}}{n}[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Recall that [latex]{\\tan}^{-1}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=\\dfrac{\\pi }{6}[\/latex]. Assuming an exact value of [latex]\\left(\\dfrac{1}{\\sqrt{3}}\\right)[\/latex], estimate [latex]\\dfrac{\\pi }{6}[\/latex] by evaluating partial sums [latex]{S}_{N}\\left(\\dfrac{1}{\\sqrt{3}}\\right)[\/latex] of the power series expansion [latex]{\\tan}^{-1}\\left(x\\right)=\\displaystyle\\sum {k=0}^{\\infty }{\\left(-1\\right)}^{k}\\dfrac{{x}^{2k+1}}{2k+1}[\/latex] at [latex]x=\\dfrac{1}{\\sqrt{3}}[\/latex]. What is the smallest number N such that [latex]6{S}{N}\\left(\\dfrac{1}{\\sqrt{3}}\\right)[\/latex] approximates \u03c0 accurately to within [latex]0.001[\/latex]? How many terms are needed for accuracy to within [latex]0.00001[\/latex]?<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Taylor and Maclaurin Series<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (1-4), find the Taylor polynomials of degree two approximating the given function centered at the given point.<\/strong><\/p>\n<ol>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=1+x+{x}^{2}[\/latex] at [latex]a=-1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\sin\\left(2x\\right)[\/latex] at [latex]a=\\dfrac{\\pi }{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\text{ln}x[\/latex] at [latex]a=1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{x}[\/latex] at [latex]a=1[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (5-7), verify that the given choice of n in the remainder estimate [latex]|{R}{n}|\\le \\dfrac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between a and the indicated point, yields [latex]|{R}{n}|\\le \\dfrac{1}{1000}[\/latex]. Find the value of the Taylor polynomial [latex]p_n[\/latex] of [latex]f[\/latex] at the indicated point.<\/strong><\/p>\n<ol start=\"5\">\n<li class=\"whitespace-normal break-words\">[latex]{\\left(28\\right)}^{\\dfrac{1}{3}};a=27,n=1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]e^2[\/latex]; [latex]a=0,n=9[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left(2\\right);a=1,n=1000[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Integrate the approximation [latex]{e}^{x}\\approx 1+x+\\dfrac{{x}^{2}}{2}+\\cdots+\\dfrac{{x}^{6}}{720}[\/latex] evaluated at \u2212x2 to approximate [latex]{\\displaystyle\\int }_{0}^{1}{e}^{\\text{-}{x}^{2}}dx[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (9-10), find the smallest value of n such that the remainder estimate [latex]|{R}{n}|\\le \\dfrac{M}{\\left(n+1\\right)\\text{!}}{\\left(x-a\\right)}^{n+1}[\/latex], where [latex]M[\/latex] is the maximum value of [latex]|{f}^{\\left(n+1\\right)}\\left(z\\right)|[\/latex] on the interval between a and the indicated point, yields [latex]|{R}{n}|\\le \\dfrac{1}{1000}[\/latex] on the indicated interval.<\/strong><\/p>\n<ol start=\"9\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\cos{x}[\/latex] on [latex]\\left[-\\dfrac{\\pi }{2},\\dfrac{\\pi }{2}\\right],a=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{\\text{-}x}[\/latex] on [latex]\\left[-3,3\\right],a=0[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (11-12), the maximum of the right-hand side of the remainder estimate [latex]|{R}_{1}|\\le \\dfrac{\\text{max}|f\\text{''}\\left(z\\right)|}{2}{R}^{2}[\/latex] on [latex]\\left[a-R,a+R\\right][\/latex] occurs at a or [latex]a\\pm R[\/latex]. Estimate the maximum value of [latex]R[\/latex] such that [latex]\\dfrac{\\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 [latex]R[\/latex].<\/strong><\/p>\n<ol start=\"11\">\n<li class=\"whitespace-normal break-words\">[latex]\\sin{x}[\/latex] approximated by x, [latex]a=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cos{x}[\/latex] approximated by [latex]1,a=0[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (13-17), find the Taylor series of the given function centered at the indicated point.<\/strong><\/p>\n<ol start=\"13\">\n<li class=\"whitespace-normal break-words\">[latex]1+x+{x}^{2}+{x}^{3}[\/latex] at [latex]a=-1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cos{x}[\/latex] at [latex]a=2\\pi[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\cos{x}[\/latex] at [latex]x=\\dfrac{\\pi }{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{e}^{x}[\/latex] at [latex]a=1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{{\\left(x - 1\\right)}^{3}}[\/latex] at [latex]a=0[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (18-22), compute the Taylor series of each function around [latex]x=1[\/latex].<\/strong><\/p>\n<ol start=\"18\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=2-x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\left(x - 2\\right)}^{2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{1}{x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{x}{4x - 2{x}^{2}-1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{2x}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (23-24), identify the value of [latex]x[\/latex] 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].<\/strong><\/p>\n<ol start=\"23\">\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{2}^{n}}{n\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n}{\\left(2\\pi \\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>The following exercises (25-26) make use of the functions [latex]{S}{5}\\left(x\\right)=x-\\dfrac{{x}^{3}}{6}+\\dfrac{{x}^{5}}{120}[\/latex] and [latex]{C}{4}\\left(x\\right)=1-\\dfrac{{x}^{2}}{2}+\\dfrac{{x}^{4}}{24}[\/latex] on [latex]\\left[\\text{-}\\pi ,\\pi \\right][\/latex].<\/strong><\/p>\n<ol start=\"25\">\n<li class=\"whitespace-normal break-words\">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].<\/li>\n<li class=\"whitespace-normal break-words\">Compare [latex]\\dfrac{{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+\\dfrac{{x}^{3}}{3}+\\dfrac{2{x}^{5}}{15}[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (27-28), 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.<\/strong><\/p>\n<ol start=\"27\">\n<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to 0}{\\text{lim}}\\dfrac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\underset{x\\to {0}^{+}}{\\text{lim}}\\dfrac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}[\/latex]<\/li>\n<\/ol>\n<h2><span data-sheets-root=\"1\">Applications of Series<\/span><\/h2>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (1-2), use appropriate substitutions to write down the Maclaurin series for the given binomial.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]{\\left(1+{x}^{2}\\right)}^{\\dfrac{-1}{3}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{\\left(1 - 2x\\right)}^{\\dfrac{2}{3}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (3-6), use the substitution [latex]{\\left(b+x\\right)}^{r}={\\left(b+a\\right)}^{r}{\\left(1+\\dfrac{x-a}{b+a}\\right)}^{r}[\/latex] in the binomial expansion to find the Taylor series of each function with the given center.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"3\">\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{{x}^{2}+2}[\/latex] at [latex]a=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{2x-{x}^{2}}[\/latex] at [latex]a=1[\/latex] (Hint: [latex]2x-{x}^{2}=1-{\\left(x - 1\\right)}^{2}[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{x}[\/latex] at [latex]a=4[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{x}[\/latex] at [latex]x=9[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercise, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most [latex]\\dfrac{1}{1000}[\/latex].<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"7\">\n<li class=\"whitespace-normal break-words\">[latex]{\\left(1001\\right)}^{\\dfrac{1}{3}}[\/latex] using [latex]{\\left(1000+x\\right)}^{\\dfrac{1}{3}}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (8-10), use the binomial approximation [latex]\\sqrt{1-x}\\approx 1-\\dfrac{x}{2}-\\dfrac{{x}^{2}}{8}-\\dfrac{{x}^{3}}{16}-\\dfrac{5{x}^{4}}{128}-\\dfrac{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.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"8\">\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{5}=5\\times \\dfrac{1}{\\sqrt{5}}[\/latex] using [latex]x=\\dfrac{4}{5}[\/latex] in [latex]{\\left(1-x\\right)}^{\\dfrac{1}{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sqrt{6}[\/latex] using [latex]x=\\dfrac{5}{6}[\/latex] in [latex]{\\left(1-x\\right)}^{\\dfrac{1}{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">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]\\dfrac{\\pi }{2}[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (11-14), use the expansion [latex]{\\left(1+x\\right)}^{\\dfrac{1}{3}}=1+\\dfrac{1}{3}x-\\dfrac{1}{9}{x}^{2}+\\dfrac{5}{81}{x}^{3}-\\dfrac{10}{243}{x}^{4}+\\cdots[\/latex] to write the first five terms (not necessarily a quartic polynomial) of each expression.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"11\">\n<li class=\"whitespace-normal break-words\">[latex]{\\left(1+4x\\right)}^{\\dfrac{4}{3}};a=0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{\\left({x}^{2}+6x+10\\right)}^{\\dfrac{1}{3}};a=-3[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Use the approximation [latex]{\\left(1-x\\right)}^{\\dfrac{2}{3}}=1-\\dfrac{2x}{3}-\\dfrac{{x}^{2}}{9}-\\dfrac{4{x}^{3}}{81}-\\dfrac{7{x}^{4}}{243}-\\dfrac{14{x}^{5}}{729}+\\cdots[\/latex] for [latex]|x|<1[\/latex] to approximate [latex]{2}^{\\dfrac{1}{3}}={2.2}^{\\dfrac{-2}{3}}[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\">Find the [latex]99[\/latex] th derivative of [latex]f\\left(x\\right)={\\left(1+{x}^{4}\\right)}^{25}[\/latex].<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (15-18), find the Maclaurin series of each function.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"15\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={2}^{x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{\\sin\\left(\\sqrt{x}\\right)}{\\sqrt{x}},\\left(x>0\\right)[\/latex],<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{{x}^{3}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\sin}^{2}x[\/latex] using the identity [latex]{\\sin}^{2}x=\\dfrac{1}{2}-\\dfrac{1}{2}\\cos\\left(2x\\right)[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (19-22), 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.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"19\">\n<li class=\"whitespace-normal break-words\">[latex]F\\left(x\\right)={\\tan}^{-1}x;f\\left(t\\right)=\\dfrac{1}{1+{t}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{t}^{2n}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]F\\left(x\\right)={\\sin}^{-1}x;f\\left(t\\right)=\\dfrac{1}{\\sqrt{1-{t}^{2}}}=\\displaystyle\\sum _{k=0}^{\\infty }\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ k\\hfill \\end{array}\\right)\\dfrac{{t}^{2k}}{k\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[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}\\dfrac{{x}^{n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]F\\left(x\\right)={\\displaystyle\\int }_{0}^{x}\\dfrac{\\text{ln}\\left(1+t\\right)}{t}dt;f\\left(t\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{t}^{n}}{n+1}[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (23-26), compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of [latex]f[\/latex].<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"23\">\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\tan{x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={e}^{x}\\cos{x}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)={\\sec}^{2}x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]f\\left(x\\right)=\\dfrac{\\tan\\sqrt{x}}{\\sqrt{x}}[\/latex] (see expansion for [latex]\\tan{x}[\/latex])<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>In the following exercises (27-28), find the radius of convergence of the Maclaurin series of each function.<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\" start=\"27\">\n<li class=\"whitespace-normal break-words\">[latex]\\dfrac{1}{1+{x}^{2}}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\text{ln}\\left(1+{x}^{2}\\right)[\/latex]<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":29,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":673,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/969"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/969\/revisions"}],"predecessor-version":[{"id":1812,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/969\/revisions\/1812"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/673"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/969\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=969"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=969"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=969"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=969"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}