{"id":184,"date":"2026-01-12T16:00:59","date_gmt":"2026-01-12T16:00:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/?post_type=chapter&p=184"},"modified":"2026-01-12T16:00:59","modified_gmt":"2026-01-12T16:00:59","slug":"power-series-and-applications-get-stronger-answer-key","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/chapter\/power-series-and-applications-get-stronger-answer-key\/","title":{"raw":"Power Series and Applications: Get Stronger Answer Key","rendered":"Power Series and Applications: Get Stronger Answer Key"},"content":{"raw":"

Introduction to Power Series<\/span><\/h2>\r\n
    \r\n \t
  1. True. If a series converges then its terms tend to zero.<\/li>\r\n \t
  2. False. It would imply that [latex]{a}{n}{x}^{n}\\to 0[\/latex] for [latex]|x|<R[\/latex]. If [latex]{a}{n}={n}^{n}[\/latex], then [latex]{a}_{n}{x}^{n}={\\left(nx\\right)}^{n}[\/latex] does not tend to zero for any [latex]x\\ne 0[\/latex].<\/li>\r\n \t
  3. It must converge on [latex]\\left(0,6\\right][\/latex] and hence at: a. [latex]x=1[\/latex]; b. [latex]x=2[\/latex]; c. [latex]x=3[\/latex]; d. [latex]x=0[\/latex]; e. [latex]x=5.99[\/latex]; and f. [latex]x=0.000001[\/latex].<\/li>\r\n \t
  4. [latex]|\\dfrac{{a}{n+1}{2}^{n+1}{x}^{n+1}}{{a}{n}{2}^{n}{x}^{n}}|=2|x||\\dfrac{{a}{n+1}}{{a}{n}}|\\to 2|x|[\/latex] so [latex]R=\\dfrac{1}{2}[\/latex]<\/li>\r\n \t
  5. [latex]|\\dfrac{{a}{n+1}{\\left(\\dfrac{\\pi }{e}\\right)}^{n+1}{x}^{n+1}}{{a}{n}{\\left(\\dfrac{\\pi }{e}\\right)}^{n}{x}^{n}}|=\\dfrac{\\pi |x|}{e}|\\dfrac{{a}{n+1}}{{a}{n}}|\\to \\dfrac{\\pi |x|}{e}[\/latex] so [latex]R=\\dfrac{e}{\\pi }[\/latex]<\/li>\r\n \t
  6. [latex]|\\dfrac{{a}{n+1}{\\left(-1\\right)}^{n+1}{x}^{2n+2}}{{a}{n}{\\left(-1\\right)}^{n}{x}^{2n}}|=|{x}^{2}||\\dfrac{{a}{n+1}}{{a}{n}}|\\to |{x}^{2}|[\/latex] so [latex]R=1[\/latex]<\/li>\r\n \t
  7. [latex]{a}{n}=\\dfrac{{2}^{n}}{n}[\/latex] so [latex]\\dfrac{{a}{n+1}x}{{a}_{n}}\\to 2x[\/latex]. so [latex]R=\\dfrac{1}{2}[\/latex]. When [latex]x=\\dfrac{1}{2}[\/latex] the series is harmonic and diverges. When [latex]x=-\\dfrac{1}{2}[\/latex] the series is alternating harmonic and converges. The interval of convergence is [latex]I=\\left[-\\dfrac{1}{2},\\dfrac{1}{2}\\right)[\/latex].<\/li>\r\n \t
  8. [latex]{a}{n}=\\dfrac{n}{{2}^{n}}[\/latex] so [latex]\\dfrac{{a}{n+1}x}{{a}_{n}}\\to \\dfrac{x}{2}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm2[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\right)[\/latex].<\/li>\r\n \t
  9. [latex]{a}_{n}=\\dfrac{{n}^{2}}{{2}^{n}}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\right)[\/latex].<\/li>\r\n \t
  10. [latex]{a}_{k}=\\dfrac{{\\pi }^{k}}{{k}^{\\pi }}[\/latex] so [latex]R=\\dfrac{1}{\\pi }[\/latex]. When [latex]x=\\pm\\dfrac{1}{\\pi }[\/latex] the series is an absolutely convergent p-series. The interval of convergence is [latex]I=\\left[-\\dfrac{1}{\\pi },\\dfrac{1}{\\pi }\\right][\/latex].<\/li>\r\n \t
  11. [latex]{a}{n}=\\dfrac{{10}^{n}}{n\\text{!}},\\dfrac{{a}{n+1}x}{{a}_{n}}=\\dfrac{10x}{n+1}\\to 0<1[\/latex] so the series converges for all x by the ratio test and [latex]I=\\left(\\text{-}\\infty ,\\infty \\right)[\/latex].<\/li>\r\n \t
  12. [latex]{a}{k}=\\dfrac{{\\left(k\\text{!}\\right)}^{2}}{\\left(2k\\right)\\text{!}}[\/latex] so [latex]\\dfrac{{a}{k+1}}{{a}_{k}}=\\dfrac{{\\left(k+1\\right)}^{2}}{\\left(2k+2\\right)\\left(2k+1\\right)}\\to \\dfrac{1}{4}[\/latex] so [latex]R=4[\/latex]<\/li>\r\n \t
  13. [latex]{a}{k}=\\dfrac{k\\text{!}}{1\\cdot 3\\cdot 5\\cdots\\left(2k - 1\\right)}[\/latex] so [latex]\\dfrac{{a}{k+1}}{{a}_{k}}=\\dfrac{k+1}{2k+1}\\to \\dfrac{1}{2}[\/latex] so [latex]R=2[\/latex]<\/li>\r\n \t
  14. [latex]{a}{n}=\\dfrac{1}{\\left(\\begin{array}{c}2n\\ n\\end{array}\\right)}[\/latex] so [latex]\\dfrac{{a}{n+1}}{{a}_{n}}=\\dfrac{{\\left(\\left(n+1\\right)\\text{!}\\right)}^{2}}{\\left(2n+2\\right)\\text{!}}\\dfrac{2n\\text{!}}{{\\left(n\\text{!}\\right)}^{2}}=\\dfrac{{\\left(n+1\\right)}^{2}}{\\left(2n+2\\right)\\left(2n+1\\right)}\\to \\dfrac{1}{4}[\/latex] so [latex]R=4[\/latex]<\/li>\r\n \t
  15. [latex]\\dfrac{{a}{n+1}}{{a}{n}}=\\dfrac{{\\left(n+1\\right)}^{3}}{\\left(3n+3\\right)\\left(3n+2\\right)\\left(3n+1\\right)}\\to \\dfrac{1}{27}[\/latex] so [latex]R=27[\/latex]<\/li>\r\n \t
  16. [latex]{a}{n}=\\dfrac{n\\text{!}}{{n}^{n}}[\/latex] so [latex]\\dfrac{{a}{n+1}}{{a}_{n}}=\\dfrac{\\left(n+1\\right)\\text{!}}{n\\text{!}}\\dfrac{{n}^{n}}{{\\left(n+1\\right)}^{n+1}}={\\left(\\dfrac{n}{n+1}\\right)}^{n}\\to \\dfrac{1}{e}[\/latex] so [latex]R=e[\/latex]<\/li>\r\n \t
  17. [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(1-x\\right)}^{n}[\/latex] on [latex]I=\\left(0,2\\right)[\/latex]<\/li>\r\n \t
  18. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n+1}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex]<\/li>\r\n \t
  19. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n+2}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex]<\/li>\r\n \t
  20. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}[\/latex] on [latex]\\left(-\\dfrac{1}{2},\\dfrac{1}{2}\\right)[\/latex]<\/li>\r\n \t
  21. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{4}^{n}{x}^{2n+2}[\/latex] on [latex]\\left(-\\dfrac{1}{2},\\dfrac{1}{2}\\right)[\/latex]<\/li>\r\n \t
  22. Converges on [latex]\\left(-1,1\\right)[\/latex] by the ratio test<\/li>\r\n \t
  23. Consider the series [latex]\\displaystyle\\sum {b}{k}{x}^{k}[\/latex] where [latex]{b}{k}={a}{k}[\/latex] if [latex]k={n}^{2}[\/latex] and [latex]{b}{k}=0[\/latex] otherwise. Then [latex]{b}{k}\\le {a}{k}[\/latex] and so the series converges on [latex]\\left(-1,1\\right)[\/latex] by the comparison test.<\/li>\r\n \t
  24. \"This\r\nThe approximation is more accurate near [latex]x=-1[\/latex]. The partial sums follow [latex]\\dfrac{1}{1-x}[\/latex] more closely as N increases but are never accurate near [latex]x=1[\/latex] since the series diverges there.<\/li>\r\n \t
  25. \"This\r\nThe approximation appears to stabilize quickly near both [latex]x=\\pm 1[\/latex].<\/li>\r\n<\/ol>\r\n

    Operations with Power Series<\/span><\/h2>\r\n
      \r\n \t
    1. [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/li>\r\n \t
    2. [latex]\\dfrac{4}{\\left(x - 3\\right)\\left(x+1\\right)}=\\dfrac{1}{x - 3}-\\dfrac{1}{x+1}=-\\dfrac{1}{3\\left(1-\\dfrac{x}{3}\\right)}-\\dfrac{1}{1-\\left(\\text{-}x\\right)}=-\\dfrac{1}{3}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\dfrac{x}{3}\\right)}^{n}-\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left({\\left(-1\\right)}^{n+1}-\\dfrac{1}{{3}^{n+1}}\\right){x}^{n}[\/latex]<\/li>\r\n \t
    3. [latex]\\dfrac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}=\\dfrac{1}{{x}^{2}-1}-\\dfrac{1}{4}\\dfrac{1}{1+{\\left(\\dfrac{x}{2}\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n}-\\dfrac{1}{4}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(\\dfrac{x}{2}\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\left(-1\\right)+{\\left(-1\\right)}^{n+1}\\dfrac{1}{{2n+2}}\\right){x}^{2n}[\/latex]<\/li>\r\n \t
    4. [latex]\\dfrac{1}{x}\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{1}{{x}^{n}}=\\dfrac{1}{x}\\dfrac{1}{1-\\dfrac{1}{x}}=\\dfrac{1}{x - 1}[\/latex]<\/li>\r\n \t
    5. [latex]\\dfrac{1}{x - 3}\\dfrac{1}{1-\\dfrac{1}{{\\left(x - 3\\right)}^{2}}}=\\dfrac{x - 3}{{\\left(x - 3\\right)}^{2}-1}[\/latex]<\/li>\r\n \t
    6. [latex]P={P}{1}+\\cdots+{P}{20}[\/latex] where [latex]{P}_{k}=10,000\\dfrac{1}{{\\left(1+r\\right)}^{k}}[\/latex]. Then [latex]P=10,000\\displaystyle\\sum _{k=1}^{20}\\dfrac{1}{{\\left(1+r\\right)}^{k}}=10,000\\dfrac{1-{\\left(1+r\\right)}^{-20}}{r}[\/latex]. When [latex]r=0.03,P\\approx 10,000\\times 14.8775=148,775[\/latex]. When [latex]r=0.05,P\\approx 10,000\\times 12.4622=124,622[\/latex]. When [latex]r=0.07,P\\approx 105,940[\/latex].<\/li>\r\n \t
    7. In general, [latex]P=\\dfrac{C\\left(1-{\\left(1+r\\right)}^{\\text{-}N}\\right)}{r}[\/latex] for N years of payouts, or [latex]C=\\dfrac{Pr}{1-{\\left(1+r\\right)}^{\\text{-}N}}[\/latex]. For [latex]N=20[\/latex] and [latex]P=100,000[\/latex], one has [latex]C=6721.57[\/latex] when [latex]r=0.03;C=8024.26[\/latex] when [latex]r=0.05[\/latex]; and [latex]C\\approx 9439.29[\/latex] when [latex]r=0.07[\/latex].<\/li>\r\n \t
    8. In general, [latex]P=\\dfrac{C}{r}[\/latex]. Thus, [latex]r=\\dfrac{C}{P}=5\\times \\dfrac{{10}^{4}}{{10}^{6}}=0.05[\/latex].<\/li>\r\n \t
    9. [latex]\\left(x+{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\dfrac{x+{x}^{2}-{x}^{3}}{1-{x}^{3}}[\/latex]<\/li>\r\n \t
    10. [latex]\\left(x-{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\dfrac{x-{x}^{2}-{x}^{3}}{1-{x}^{3}}[\/latex]<\/li>\r\n \t
    11. [latex]{a}{n}=2,{b}{n}=n[\/latex] so [latex]{c}{n}=\\displaystyle\\sum {k=0}^{n}{b}{k}{a}{n-k}=2\\displaystyle\\sum _{k=0}^{n}k=\\left(n\\right)\\left(n+1\\right)[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n\\left(n+1\\right){x}^{n}[\/latex]<\/li>\r\n \t
    12. [latex]{a}{n}={b}{n}={2}^{\\text{-}n}[\/latex] so [latex]{c}{n}=\\displaystyle\\sum {k=1}^{n}{b}{k}{a}{n-k}={2}^{\\text{-}n}\\displaystyle\\sum _{k=1}^{n}1=\\dfrac{n}{{2}^{n}}[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n{\\left(\\dfrac{x}{2}\\right)}^{n}[\/latex]<\/li>\r\n \t
    13. The derivative of [latex]f[\/latex] is [latex]-\\dfrac{1}{{\\left(1+x\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex].<\/li>\r\n \t
    14. The indefinite integral of [latex]f[\/latex] is [latex]\\dfrac{1}{1+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}[\/latex].<\/li>\r\n \t
    15. [latex]f\\left(x\\right)=\\displaystyle\\sum {n=0}^{\\infty }{x}^{n}=\\dfrac{1}{1-x};{f}^{\\prime }\\left(\\dfrac{1}{2}\\right)=\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{n}{{2}^{n - 1}}={\\dfrac{d}{dx}{\\left(1-x\\right)}^{-1}|}{x=\\dfrac{1}{2}}={\\dfrac{1}{{\\left(1-x\\right)}^{2}}|}{x=\\dfrac{1}{2}}=4[\/latex] so [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{{2}^{n}}=2[\/latex].<\/li>\r\n \t
    16. [latex]f\\left(x\\right)=\\displaystyle\\sum {n=0}^{\\infty }{x}^{n}=\\dfrac{1}{1-x};f\\text{''}\\left(\\dfrac{1}{2}\\right)=\\displaystyle\\sum {n=2}^{\\infty }\\dfrac{n\\left(n - 1\\right)}{{2}^{n - 2}}={\\dfrac{{d}^{2}}{d{x}^{2}}{\\left(1-x\\right)}^{-1}|}{x=\\dfrac{1}{2}}={\\dfrac{2}{{\\left(1-x\\right)}^{3}}|}{x=\\dfrac{1}{2}}=16[\/latex] so [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{n\\left(n - 1\\right)}{{2}^{n}}=4[\/latex].<\/li>\r\n \t
    17. [latex]\\displaystyle\\int \\displaystyle\\sum {\\left(1-x\\right)}^{n}dx=\\displaystyle\\int \\displaystyle\\sum {\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}dx=\\displaystyle\\sum \\dfrac{{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n+1}}{n+1}[\/latex]<\/li>\r\n \t
    18. [latex]\\text{-}{\\displaystyle\\int }_{t=0}^{{x}^{2}}\\dfrac{1}{1-t}dt=\\text{-}\\displaystyle\\sum {n=0}^{\\infty }{\\displaystyle\\int }{0}^{{x}^{2}}{t}^{n}dx-\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{2\\left(n+1\\right)}}{n+1}=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{x}^{2n}}{n}[\/latex]<\/li>\r\n \t
    19. [latex]{\\displaystyle\\int }_{0}^{{x}^{2}}\\dfrac{dt}{1+{t}^{2}}=\\displaystyle\\sum {n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\displaystyle\\int }{0}^{{x}^{2}}{t}^{2n}dt={\\displaystyle\\sum {n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{t}^{2n+1}}{2n+1}|}{t=0}^{{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{4n+2}}{2n+1}[\/latex]<\/li>\r\n \t
    20. Term-by-term integration gives [latex]{\\displaystyle\\int }_{0}^{x}\\text{ln}tdt=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\dfrac{{\\left(x - 1\\right)}^{n+1}}{n\\left(n+1\\right)}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\left(\\dfrac{1}{n}-\\dfrac{1}{n+1}\\right){\\left(x - 1\\right)}^{n+1}=\\left(x - 1\\right)\\text{ln}x+\\displaystyle\\sum _{n=2}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{\\left(x - 1\\right)}^{n}}{n}= x\\text{ln}x-x[\/latex].<\/li>\r\n \t
    21. [latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{k}}{k}=\\text{-}\\text{ln}\\left(1-x\\right)[\/latex] so [latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{3k}}{6k}=-\\dfrac{1}{6}\\text{ln}\\left(1-{x}^{3}\\right)[\/latex]. The radius of convergence is equal to 1 by the ratio test.<\/li>\r\n \t
    22. If [latex]y={2}^{\\text{-}x}[\/latex], then [latex]\\displaystyle\\sum {k=1}^{\\infty }{y}^{k}=\\dfrac{y}{1-y}=\\dfrac{{2}^{\\text{-}x}}{1-{2}^{\\text{-}x}}=\\dfrac{1}{{2}^{x}-1}[\/latex]. If [latex]{a}{k}={2}^{\\text{-}kx}[\/latex], then [latex]\\dfrac{{a}{k+1}}{{a}{k}}={2}^{\\text{-}x}<1[\/latex] when [latex]x>0[\/latex]. So the series converges for all [latex]x>0[\/latex].<\/li>\r\n \t
    23. Answers will vary.<\/li>\r\n \t
    24. \"This\r\nThe solid curve is [latex]{S}{5}[\/latex]. The dashed curve is [latex]{S}{2}[\/latex], dotted is [latex]{S}{3}[\/latex], and dash-dotted is [latex]{S}{4}[\/latex]<\/li>\r\n \t
    25. When [latex]x=-\\dfrac{1}{2},\\text{-}\\text{ln}\\left(2\\right)=\\text{ln}\\left(\\dfrac{1}{2}\\right)=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n{2}^{n}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=11}^{\\infty }\\dfrac{1}{n{2}^{n}}<\\displaystyle\\sum _{n=11}^{\\infty }\\dfrac{1}{{2}^{n}}=\\dfrac{1}{{2}^{10}}[\/latex], one has [latex]\\displaystyle\\sum _{n=1}^{10}\\dfrac{1}{n{2}^{n}}=0.69306\\ldots[\/latex] whereas [latex]\\text{ln}\\left(2\\right)=0.69314\\ldots[\/latex]; therefore, [latex]N=10[\/latex].<\/li>\r\n \t
    26. [latex]6{S}{N}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=2\\sqrt{3}\\displaystyle\\sum {n=0}^{N}{\\left(-1\\right)}^{n}\\dfrac{1}{{3}^{n}\\left(2n+1\\right)}[\/latex]. One has [latex]\\pi -6{S}{4}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=0.00101\\ldots[\/latex] and [latex]\\pi -6{S}{5}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=0.00028\\ldots[\/latex] so [latex]N=5[\/latex] is the smallest partial sum with accuracy to within 0.001. Also, [latex]\\pi -6{S}{7}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=0.00002\\ldots[\/latex] while [latex]\\pi -6{S}{8}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=-0.000007\\ldots[\/latex] so [latex]N=8[\/latex] is the smallest N to give accuracy to within 0.00001.<\/li>\r\n<\/ol>\r\n

      Taylor and Maclaurin Series<\/span><\/h2>\r\n
        \r\n \t
      1. [latex]f\\left(-1\\right)=1;{f}^{\\prime }\\left(-1\\right)=-1;f\\text{''}\\left(-1\\right)=2;f\\left(x\\right)=1-\\left(x+1\\right)+{\\left(x+1\\right)}^{2}[\/latex]<\/li>\r\n \t
      2. [latex]{f}^{\\prime }\\left(x\\right)=2\\cos\\left(2x\\right);f\\text{''}\\left(x\\right)=-4\\sin\\left(2x\\right);{p}_{2}\\left(x\\right)=-2\\left(x-\\dfrac{\\pi }{2}\\right)[\/latex]<\/li>\r\n \t
      3. [latex]{f}^{\\prime }\\left(x\\right)=\\dfrac{1}{x};f\\text{''}\\left(x\\right)=-\\dfrac{1}{{x}^{2}};{p}_{2}\\left(x\\right)=0+\\left(x - 1\\right)-\\dfrac{1}{2}{\\left(x - 1\\right)}^{2}[\/latex]<\/li>\r\n \t
      4. [latex]{p}_{2}\\left(x\\right)=e+e\\left(x - 1\\right)+\\dfrac{e}{2}{\\left(x - 1\\right)}^{2}[\/latex]<\/li>\r\n \t
      5. [latex]\\dfrac{{d}^{2}}{d{x}^{2}}{x}^{\\dfrac{1}{3}}=-\\dfrac{2}{9{x}^{\\dfrac{5}{3}}}\\ge -0.00092\\ldots[\/latex] when [latex]x\\ge 28[\/latex] so the remainder estimate applies to the linear approximation [latex]{x}^{\\dfrac{1}{3}}\\approx {p}_{1}\\left(27\\right)=3+\\dfrac{x - 27}{27}[\/latex], which gives [latex]{\\left(28\\right)}^{\\dfrac{1}{3}}\\approx 3+\\dfrac{1}{27}=3.\\overline{037}[\/latex], while [latex]{\\left(28\\right)}^{\\dfrac{1}{3}}\\approx 3.03658[\/latex].<\/li>\r\n \t
      6. Using the estimate [latex]\\dfrac{{2}^{10}}{10\\text{!}}<0.000283[\/latex] we can use the Taylor expansion of order 9 to estimate [latex]e^{x}[\/latex] at [latex]x=2[\/latex]. as [latex]{e}^{2}\\approx {p}_{9}\\left(2\\right)=1+2+\\dfrac{{2}^{2}}{2}+\\dfrac{{2}^{3}}{6}+\\cdots+\\dfrac{{2}^{9}}{9\\text{!}}=7.3887\\ldots[\/latex] whereas [latex]{e}^{2}\\approx 7.3891[\/latex].<\/li>\r\n \t
      7. Since [latex]\\dfrac{{d}^{n}}{d{x}^{n}}\\left(\\text{ln}x\\right)={\\left(-1\\right)}^{n - 1}\\dfrac{\\left(n - 1\\right)\\text{!}}{{x}^{n}},{R}{1000}\\approx \\dfrac{1}{1001}[\/latex]. One has [latex]{p}{1000}\\left(1\\right)=\\displaystyle\\sum _{n=1}^{1000}\\dfrac{{\\left(-1\\right)}^{n - 1}}{n}\\approx 0.6936[\/latex] whereas [latex]\\text{ln}\\left(2\\right)\\approx 0.6931\\cdots[\/latex].<\/li>\r\n \t
      8. [latex]{\\displaystyle\\int }{0}^{1}\\left(1-{x}^{2}+\\dfrac{{x}^{4}}{2}-\\dfrac{{x}^{6}}{6}+\\dfrac{{x}^{8}}{24}-\\dfrac{{x}^{10}}{120}+\\dfrac{{x}^{12}}{720}\\right)dx[\/latex] [latex]=1-\\dfrac{{1}^{3}}{3}+\\dfrac{{1}^{5}}{10}-\\dfrac{{1}^{7}}{42}+\\dfrac{{1}^{9}}{9\\cdot 24}-\\dfrac{{1}^{11}}{120\\cdot 11}+\\dfrac{{1}^{13}}{720\\cdot 13}\\approx 0.74683[\/latex] whereas [latex]{\\displaystyle\\int }{0}^{1}{e}^{\\text{-}{x}^{2}}dx\\approx 0.74682[\/latex].<\/li>\r\n \t
      9. Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)[\/latex] is [latex]\\sin{z}[\/latex] or [latex]\\cos{z}[\/latex], we have [latex]M=1[\/latex]. Since [latex]|x - 0|\\le \\dfrac{\\pi }{2}[\/latex], we seek the smallest n such that [latex]\\dfrac{{\\pi }^{n+1}}{{2}^{n+1}\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=7[\/latex]. The remainder estimate is [latex]{R}_{7}\\le 0.00092[\/latex].<\/li>\r\n \t
      10. Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)=\\pm{e}^{\\text{-}z}[\/latex] one has [latex]M={e}^{3}[\/latex]. Since [latex]|x - 0|\\le 3[\/latex], one seeks the smallest n such that [latex]\\dfrac{{3}^{n+1}{e}^{3}}{\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=14[\/latex]. The remainder estimate is [latex]{R}_{14}\\le 0.000220[\/latex].<\/li>\r\n \t
      11. \"This\r\nSince [latex]\\sin{x}[\/latex] is increasing for small x and since [latex]\\mathrm{si}n\\text{''}x=\\text{-}\\sin{x}[\/latex], the estimate applies whenever [latex]{R}^{2}\\sin\\left(R\\right)\\le 0.2[\/latex], which applies up to [latex]R=0.596[\/latex].<\/li>\r\n \t
      12. \"This\r\nSince the second derivative of [latex]\\cos{x}[\/latex] is [latex]\\text{-}\\cos{x}[\/latex] and since [latex]\\cos{x}[\/latex] is decreasing away from [latex]x=0[\/latex], the estimate applies when [latex]{R}^{2}\\cos{R}\\le 0.2[\/latex] or [latex]R\\le 0.447[\/latex].<\/li>\r\n \t
      13. [latex]{\\left(x+1\\right)}^{3}-2{\\left(x+1\\right)}^{2}+2\\left(x+1\\right)[\/latex]<\/li>\r\n \t
      14. Values of derivatives are the same as for [latex]x=0[\/latex] so [latex]\\cos{x}={\\displaystyle\\sum _{n=0}^{\\infty }\\left(-1\\right)}^{n}\\dfrac{{\\left(x - 2\\pi \\right)}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/li>\r\n \t
      15. [latex]\\cos\\left(\\dfrac{\\pi }{2}\\right)=0,\\text{-}\\sin\\left(\\dfrac{\\pi }{2}\\right)=-1[\/latex] so [latex]\\cos{x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{\\left(x-\\dfrac{\\pi }{2}\\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex], which is also [latex]\\text{-}\\cos\\left(x-\\dfrac{\\pi }{2}\\right)[\/latex].<\/li>\r\n \t
      16. The derivatives are [latex]{f}^{\\left(n\\right)}\\left(1\\right)=e[\/latex] so [latex]{e}^{x}=e\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex].<\/li>\r\n \t
      17. [latex]\\dfrac{1}{{\\left(x - 1\\right)}^{3}}=\\text{-}\\left(\\dfrac{1}{2}\\right)\\dfrac{{d}^{2}}{d{x}^{2}}\\dfrac{1}{1-x}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\dfrac{\\left(n+2\\right)\\left(n+1\\right){x}^{n}}{2}\\right)[\/latex]<\/li>\r\n \t
      18. [latex]2-x=1-\\left(x - 1\\right)[\/latex]<\/li>\r\n \t
      19. [latex]{\\left(\\left(x - 1\\right)-1\\right)}^{2}={\\left(x - 1\\right)}^{2}-2\\left(x - 1\\right)+1[\/latex]<\/li>\r\n \t
      20. [latex]\\dfrac{1}{1-\\left(1-x\\right)}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex]<\/li>\r\n \t
      21. [latex]x\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(1-x\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n+1}+\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n}[\/latex]<\/li>\r\n \t
      22. [latex]{e}^{2x}={e}^{2\\left(x - 1\\right)+2}={e}^{2}\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{2}^{n}{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t
      23. [latex]x={e}^{2};{S}_{10}=\\dfrac{34,913}{4725}\\approx 7.3889947[\/latex]<\/li>\r\n \t
      24. [latex]\\sin\\left(2\\pi \\right)=0;{S}_{10}=8.27\\times {10}^{-5}[\/latex]<\/li>\r\n \t
      25. \"This\r\nThe difference is small on the interior of the interval but approaches [latex]1[\/latex] near the endpoints. The remainder estimate is [latex]|{R}_{4}|=\\dfrac{{\\pi }^{5}}{120}\\approx 2.552[\/latex].<\/li>\r\n \t
      26. \"This\r\nThe difference is on the order of [latex]{10}^{-4}[\/latex] on [latex]\\left[-1,1\\right][\/latex] while the Taylor approximation error is around [latex]0.1[\/latex] near [latex]\\pm 1[\/latex]. The top curve is a plot of [latex]{\\tan}^{2}x-{\\left(\\dfrac{{S}{5}\\left(x\\right)}{{C}{4}\\left(x\\right)}\\right)}^{2}[\/latex] and the lower dashed plot shows [latex]{t}^{2}-{\\left(\\dfrac{{S}{5}}{{C}{4}}\\right)}^{2}[\/latex].<\/li>\r\n \t
      27. [latex]\\dfrac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}\\to \\text{-}1[\/latex]<\/li>\r\n \t
      28. [latex]\\dfrac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}\\approx \\dfrac{\\left(1-\\dfrac{x}{2}+\\dfrac{{x}^{2}}{4\\text{!}}-\\cdots\\right)-1}{2x}\\to -\\dfrac{1}{4}[\/latex]<\/li>\r\n<\/ol>\r\n

        Applications of Series<\/span><\/h2>\r\n
          \r\n \t
        1. [latex]{\\left(1+{x}^{2}\\right)}^{\\dfrac{-1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}-\\dfrac{1}{3}\\hfill \\ \\hfill n\\hfill \\end{array}\\right){x}^{2n}[\/latex]<\/li>\r\n \t
        2. [latex]{\\left(1 - 2x\\right)}^{\\dfrac{2}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{2}^{n}\\left(\\begin{array}{c}\\dfrac{2}{3}\\hfill \\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]<\/li>\r\n \t
        3. [latex]\\sqrt{2+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{\\left(\\dfrac{1}{2}\\right)-n}\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){x}^{2n};\\left(|{x}^{2}|<2\\right)[\/latex]<\/li>\r\n \t
        4. [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}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){\\left(x - 1\\right)}^{2n}[\/latex]<\/li>\r\n \t
        5. [latex]\\sqrt{x}=2\\sqrt{1+\\dfrac{x - 4}{4}}[\/latex] so [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{1 - 2n}\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){\\left(x - 4\\right)}^{n}[\/latex]<\/li>\r\n \t
        6. [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{3}^{1 - 3n}\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){\\left(x - 9\\right)}^{n}[\/latex]<\/li>\r\n \t
        7. [latex]10{\\left(1+\\dfrac{x}{1000}\\right)}^{\\dfrac{1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{10}^{1 - 3n}\\left(\\begin{array}{c}\\dfrac{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)}^{\\dfrac{1}{3}}& \\approx 10\\left(1+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 1\\hfill \\end{array}\\right){10}^{-3}+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 2\\hfill \\end{array}\\right){10}^{-6}+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 3\\hfill \\end{array}\\right){10}^{-9}+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 4\\hfill \\end{array}\\right){10}^{-12}\\right)\\hfill \\ & =10\\left(1+\\dfrac{1}{{3.10}^{3}}-\\dfrac{1}{{9.10}^{6}}+\\dfrac{5}{{81.10}^{9}}-\\dfrac{10}{{243.10}^{12}}\\right)=10.00333222...\\hfill \\end{array}[\/latex] whereas [latex]{\\left(1001\\right)}^{\\dfrac{1}{3}}=10.00332222839093...[\/latex]. Two terms would suffice for three-digit accuracy.<\/li>\r\n \t
        8. The approximation is [latex]2.3152[\/latex]; the CAS value is [latex]2.23\\ldots[\/latex].<\/li>\r\n \t
        9. The approximation is [latex]2.583\\ldots[\/latex]; the CAS value is [latex]2.449\\ldots[\/latex].<\/li>\r\n \t
        10. [latex]\\sqrt{1-{x}^{2}}=1-\\dfrac{{x}^{2}}{2}-\\dfrac{{x}^{4}}{8}-\\dfrac{{x}^{6}}{16}-\\dfrac{5{x}^{8}}{128}+\\cdots[\/latex]. Thus [latex]{\\displaystyle\\int }{-1}^{1}\\sqrt{1-{x}^{2}}dx=x-\\dfrac{{x}^{3}}{6}-\\dfrac{{x}^{5}}{40}-\\dfrac{{x}^{7}}{7\\cdot 16}-\\dfrac{5{x}^{9}}{9\\cdot 128}+\\cdots{|}{-1}^{1}\\approx 2-\\dfrac{1}{3}-\\dfrac{1}{20}-\\dfrac{1}{56}-\\dfrac{10}{9\\cdot 128}+\\text{error}=1.590..[\/latex]. whereas [latex]\\dfrac{\\pi }{2}=1.570..[\/latex].<\/li>\r\n \t
        11. [latex]{\\left(1+4x\\right)}^{\\dfrac{4}{3}}=\\left(1+4x\\right)\\left(1+4x\\right)^{\\dfrac{1}{3}}=\\left(1+4x\\right)\\left(1+\\dfrac{4x}{3}-\\dfrac{16x^{3}}{9}+\\dfrac{320x^{3}}{81}-\\dfrac{2560x^{4}}{243}\\right)=1+\\dfrac{16}{3}x+\\dfrac{32}{9}x^{2}-\\dfrac{256}{81}x^{3}+\\dfrac{1280}{243}x^{4}-\\dfrac{10240}{243}x^{5}[\/latex]<\/li>\r\n \t
        12. [latex]\\left(1+{\\left(x+3\\right)}^{2}\\right)^{\\dfrac{1}{3}} =1+\\dfrac{1}{3}{\\left(x+3\\right)}^{2}-\\dfrac{1}{9}{\\left(x+3\\right)}^{4}+\\dfrac{5}{81}{\\left(x+3\\right)}^{6}-\\dfrac{10}{243}{\\left(x+3\\right)}^{8}+\\cdots [\/latex]<\/li>\r\n \t
        13. Twice the approximation is [latex]1.260\\ldots[\/latex] whereas [latex]{2}^{\\dfrac{1}{3}}=1.2599...[\/latex].<\/li>\r\n \t
        14. [latex]{f}^{\\left(99\\right)}\\left(0\\right)=0[\/latex]<\/li>\r\n \t
        15. [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{\\left(\\text{ln}\\left(2\\right)x\\right)}^{n}}{n\\text{!}}[\/latex]<\/li>\r\n \t
        16. For [latex]x>0,\\sin\\left(\\sqrt{x}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{\\dfrac{\\left(2n+1\\right)}{2}}}{\\sqrt{x}\\left(2n+1\\right)\\text{!}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{n}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/li>\r\n \t
        17. [latex]{e}^{{x}^{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{3n}}{n\\text{!}}[\/latex]<\/li>\r\n \t
        18. [latex]{\\sin}^{2}x=\\text{-}\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{k}{2}^{2k - 1}{x}^{2k}}{\\left(2k\\right)\\text{!}}[\/latex]<\/li>\r\n \t
        19. [latex]{\\tan}^{-1}x=\\displaystyle\\sum _{k=0}^{\\infty }\\dfrac{{\\left(-1\\right)}^{k}{x}^{2k+1}}{2k+1}[\/latex]<\/li>\r\n \t
        20. [latex]{\\sin}^{-1}x=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right)\\dfrac{{x}^{2n+1}}{\\left(2n+1\\right)n\\text{!}}[\/latex]<\/li>\r\n \t
        21. [latex]F\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{n+1}}{\\left(n+1\\right)\\left(2n\\right)\\text{!}}[\/latex]<\/li>\r\n \t
        22. [latex]F\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{x}^{n}}{{n}^{2}}[\/latex]<\/li>\r\n \t
        23. [latex]x+\\dfrac{{x}^{3}}{3}+\\dfrac{2{x}^{5}}{15}+\\cdots[\/latex]<\/li>\r\n \t
        24. [latex]1+x-\\dfrac{{x}^{3}}{3}-\\dfrac{{x}^{4}}{6}+\\cdots[\/latex]<\/li>\r\n \t
        25. [latex]1+{x}^{2}+\\dfrac{2{x}^{4}}{3}+\\dfrac{17{x}^{6}}{45}+\\cdots[\/latex]<\/li>\r\n \t
        26. Using the expansion for [latex]\\tan{x}[\/latex] gives [latex]1+\\dfrac{x}{3}+\\dfrac{2{x}^{2}}{15}[\/latex].<\/li>\r\n \t
        27. [latex]\\dfrac{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.<\/li>\r\n \t
        28. [latex]\\text{ln}\\left(1+{x}^{2}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n - 1}}{n}{x}^{2n}[\/latex] so [latex]R=1[\/latex] by the ratio test.<\/li>\r\n<\/ol>","rendered":"

          Introduction to Power Series<\/span><\/h2>\n
            \n
          1. True. If a series converges then its terms tend to zero.<\/li>\n
          2. False. It would imply that [latex]{a}{n}{x}^{n}\\to 0[\/latex] for [latex]|x|\n
          3. It must converge on [latex]\\left(0,6\\right][\/latex] and hence at: a. [latex]x=1[\/latex]; b. [latex]x=2[\/latex]; c. [latex]x=3[\/latex]; d. [latex]x=0[\/latex]; e. [latex]x=5.99[\/latex]; and f. [latex]x=0.000001[\/latex].<\/li>\n
          4. [latex]|\\dfrac{{a}{n+1}{2}^{n+1}{x}^{n+1}}{{a}{n}{2}^{n}{x}^{n}}|=2|x||\\dfrac{{a}{n+1}}{{a}{n}}|\\to 2|x|[\/latex] so [latex]R=\\dfrac{1}{2}[\/latex]<\/li>\n
          5. [latex]|\\dfrac{{a}{n+1}{\\left(\\dfrac{\\pi }{e}\\right)}^{n+1}{x}^{n+1}}{{a}{n}{\\left(\\dfrac{\\pi }{e}\\right)}^{n}{x}^{n}}|=\\dfrac{\\pi |x|}{e}|\\dfrac{{a}{n+1}}{{a}{n}}|\\to \\dfrac{\\pi |x|}{e}[\/latex] so [latex]R=\\dfrac{e}{\\pi }[\/latex]<\/li>\n
          6. [latex]|\\dfrac{{a}{n+1}{\\left(-1\\right)}^{n+1}{x}^{2n+2}}{{a}{n}{\\left(-1\\right)}^{n}{x}^{2n}}|=|{x}^{2}||\\dfrac{{a}{n+1}}{{a}{n}}|\\to |{x}^{2}|[\/latex] so [latex]R=1[\/latex]<\/li>\n
          7. [latex]{a}{n}=\\dfrac{{2}^{n}}{n}[\/latex] so [latex]\\dfrac{{a}{n+1}x}{{a}_{n}}\\to 2x[\/latex]. so [latex]R=\\dfrac{1}{2}[\/latex]. When [latex]x=\\dfrac{1}{2}[\/latex] the series is harmonic and diverges. When [latex]x=-\\dfrac{1}{2}[\/latex] the series is alternating harmonic and converges. The interval of convergence is [latex]I=\\left[-\\dfrac{1}{2},\\dfrac{1}{2}\\right)[\/latex].<\/li>\n
          8. [latex]{a}{n}=\\dfrac{n}{{2}^{n}}[\/latex] so [latex]\\dfrac{{a}{n+1}x}{{a}_{n}}\\to \\dfrac{x}{2}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm2[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\right)[\/latex].<\/li>\n
          9. [latex]{a}_{n}=\\dfrac{{n}^{2}}{{2}^{n}}[\/latex] so [latex]R=2[\/latex]. When [latex]x=\\pm[\/latex] the series diverges by the divergence test. The interval of convergence is [latex]I=\\left(-2,2\\right)[\/latex].<\/li>\n
          10. [latex]{a}_{k}=\\dfrac{{\\pi }^{k}}{{k}^{\\pi }}[\/latex] so [latex]R=\\dfrac{1}{\\pi }[\/latex]. When [latex]x=\\pm\\dfrac{1}{\\pi }[\/latex] the series is an absolutely convergent p-series. The interval of convergence is [latex]I=\\left[-\\dfrac{1}{\\pi },\\dfrac{1}{\\pi }\\right][\/latex].<\/li>\n
          11. [latex]{a}{n}=\\dfrac{{10}^{n}}{n\\text{!}},\\dfrac{{a}{n+1}x}{{a}_{n}}=\\dfrac{10x}{n+1}\\to 0<1[\/latex] so the series converges for all x by the ratio test and [latex]I=\\left(\\text{-}\\infty ,\\infty \\right)[\/latex].<\/li>\n
          12. [latex]{a}{k}=\\dfrac{{\\left(k\\text{!}\\right)}^{2}}{\\left(2k\\right)\\text{!}}[\/latex] so [latex]\\dfrac{{a}{k+1}}{{a}_{k}}=\\dfrac{{\\left(k+1\\right)}^{2}}{\\left(2k+2\\right)\\left(2k+1\\right)}\\to \\dfrac{1}{4}[\/latex] so [latex]R=4[\/latex]<\/li>\n
          13. [latex]{a}{k}=\\dfrac{k\\text{!}}{1\\cdot 3\\cdot 5\\cdots\\left(2k - 1\\right)}[\/latex] so [latex]\\dfrac{{a}{k+1}}{{a}_{k}}=\\dfrac{k+1}{2k+1}\\to \\dfrac{1}{2}[\/latex] so [latex]R=2[\/latex]<\/li>\n
          14. [latex]{a}{n}=\\dfrac{1}{\\left(\\begin{array}{c}2n\\ n\\end{array}\\right)}[\/latex] so [latex]\\dfrac{{a}{n+1}}{{a}_{n}}=\\dfrac{{\\left(\\left(n+1\\right)\\text{!}\\right)}^{2}}{\\left(2n+2\\right)\\text{!}}\\dfrac{2n\\text{!}}{{\\left(n\\text{!}\\right)}^{2}}=\\dfrac{{\\left(n+1\\right)}^{2}}{\\left(2n+2\\right)\\left(2n+1\\right)}\\to \\dfrac{1}{4}[\/latex] so [latex]R=4[\/latex]<\/li>\n
          15. [latex]\\dfrac{{a}{n+1}}{{a}{n}}=\\dfrac{{\\left(n+1\\right)}^{3}}{\\left(3n+3\\right)\\left(3n+2\\right)\\left(3n+1\\right)}\\to \\dfrac{1}{27}[\/latex] so [latex]R=27[\/latex]<\/li>\n
          16. [latex]{a}{n}=\\dfrac{n\\text{!}}{{n}^{n}}[\/latex] so [latex]\\dfrac{{a}{n+1}}{{a}_{n}}=\\dfrac{\\left(n+1\\right)\\text{!}}{n\\text{!}}\\dfrac{{n}^{n}}{{\\left(n+1\\right)}^{n+1}}={\\left(\\dfrac{n}{n+1}\\right)}^{n}\\to \\dfrac{1}{e}[\/latex] so [latex]R=e[\/latex]<\/li>\n
          17. [latex]f\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(1-x\\right)}^{n}[\/latex] on [latex]I=\\left(0,2\\right)[\/latex]<\/li>\n
          18. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n+1}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex]<\/li>\n
          19. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n+2}[\/latex] on [latex]I=\\left(-1,1\\right)[\/latex]<\/li>\n
          20. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{x}^{n}[\/latex] on [latex]\\left(-\\dfrac{1}{2},\\dfrac{1}{2}\\right)[\/latex]<\/li>\n
          21. [latex]\\displaystyle\\sum _{n=0}^{\\infty }{4}^{n}{x}^{2n+2}[\/latex] on [latex]\\left(-\\dfrac{1}{2},\\dfrac{1}{2}\\right)[\/latex]<\/li>\n
          22. Converges on [latex]\\left(-1,1\\right)[\/latex] by the ratio test<\/li>\n
          23. Consider the series [latex]\\displaystyle\\sum {b}{k}{x}^{k}[\/latex] where [latex]{b}{k}={a}{k}[\/latex] if [latex]k={n}^{2}[\/latex] and [latex]{b}{k}=0[\/latex] otherwise. Then [latex]{b}{k}\\le {a}{k}[\/latex] and so the series converges on [latex]\\left(-1,1\\right)[\/latex] by the comparison test.<\/li>\n
          24. \"This
            \nThe approximation is more accurate near [latex]x=-1[\/latex]. The partial sums follow [latex]\\dfrac{1}{1-x}[\/latex] more closely as N increases but are never accurate near [latex]x=1[\/latex] since the series diverges there.<\/li>\n
          25. \"This
            \nThe approximation appears to stabilize quickly near both [latex]x=\\pm 1[\/latex].<\/li>\n<\/ol>\n

            Operations with Power Series<\/span><\/h2>\n
              \n
            1. [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)+g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex] and [latex]\\dfrac{1}{2}\\left(f\\left(x\\right)-g\\left(x\\right)\\right)=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/li>\n
            2. [latex]\\dfrac{4}{\\left(x - 3\\right)\\left(x+1\\right)}=\\dfrac{1}{x - 3}-\\dfrac{1}{x+1}=-\\dfrac{1}{3\\left(1-\\dfrac{x}{3}\\right)}-\\dfrac{1}{1-\\left(\\text{-}x\\right)}=-\\dfrac{1}{3}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(\\dfrac{x}{3}\\right)}^{n}-\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left({\\left(-1\\right)}^{n+1}-\\dfrac{1}{{3}^{n+1}}\\right){x}^{n}[\/latex]<\/li>\n
            3. [latex]\\dfrac{5}{\\left({x}^{2}+4\\right)\\left({x}^{2}-1\\right)}=\\dfrac{1}{{x}^{2}-1}-\\dfrac{1}{4}\\dfrac{1}{1+{\\left(\\dfrac{x}{2}\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{x}^{2n}-\\dfrac{1}{4}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(\\dfrac{x}{2}\\right)}^{n}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\left(-1\\right)+{\\left(-1\\right)}^{n+1}\\dfrac{1}{{2n+2}}\\right){x}^{2n}[\/latex]<\/li>\n
            4. [latex]\\dfrac{1}{x}\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{1}{{x}^{n}}=\\dfrac{1}{x}\\dfrac{1}{1-\\dfrac{1}{x}}=\\dfrac{1}{x - 1}[\/latex]<\/li>\n
            5. [latex]\\dfrac{1}{x - 3}\\dfrac{1}{1-\\dfrac{1}{{\\left(x - 3\\right)}^{2}}}=\\dfrac{x - 3}{{\\left(x - 3\\right)}^{2}-1}[\/latex]<\/li>\n
            6. [latex]P={P}{1}+\\cdots+{P}{20}[\/latex] where [latex]{P}_{k}=10,000\\dfrac{1}{{\\left(1+r\\right)}^{k}}[\/latex]. Then [latex]P=10,000\\displaystyle\\sum _{k=1}^{20}\\dfrac{1}{{\\left(1+r\\right)}^{k}}=10,000\\dfrac{1-{\\left(1+r\\right)}^{-20}}{r}[\/latex]. When [latex]r=0.03,P\\approx 10,000\\times 14.8775=148,775[\/latex]. When [latex]r=0.05,P\\approx 10,000\\times 12.4622=124,622[\/latex]. When [latex]r=0.07,P\\approx 105,940[\/latex].<\/li>\n
            7. In general, [latex]P=\\dfrac{C\\left(1-{\\left(1+r\\right)}^{\\text{-}N}\\right)}{r}[\/latex] for N years of payouts, or [latex]C=\\dfrac{Pr}{1-{\\left(1+r\\right)}^{\\text{-}N}}[\/latex]. For [latex]N=20[\/latex] and [latex]P=100,000[\/latex], one has [latex]C=6721.57[\/latex] when [latex]r=0.03;C=8024.26[\/latex] when [latex]r=0.05[\/latex]; and [latex]C\\approx 9439.29[\/latex] when [latex]r=0.07[\/latex].<\/li>\n
            8. In general, [latex]P=\\dfrac{C}{r}[\/latex]. Thus, [latex]r=\\dfrac{C}{P}=5\\times \\dfrac{{10}^{4}}{{10}^{6}}=0.05[\/latex].<\/li>\n
            9. [latex]\\left(x+{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\dfrac{x+{x}^{2}-{x}^{3}}{1-{x}^{3}}[\/latex]<\/li>\n
            10. [latex]\\left(x-{x}^{2}-{x}^{3}\\right)\\left(1+{x}^{3}+{x}^{6}+\\cdots\\right)=\\dfrac{x-{x}^{2}-{x}^{3}}{1-{x}^{3}}[\/latex]<\/li>\n
            11. [latex]{a}{n}=2,{b}{n}=n[\/latex] so [latex]{c}{n}=\\displaystyle\\sum {k=0}^{n}{b}{k}{a}{n-k}=2\\displaystyle\\sum _{k=0}^{n}k=\\left(n\\right)\\left(n+1\\right)[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n\\left(n+1\\right){x}^{n}[\/latex]<\/li>\n
            12. [latex]{a}{n}={b}{n}={2}^{\\text{-}n}[\/latex] so [latex]{c}{n}=\\displaystyle\\sum {k=1}^{n}{b}{k}{a}{n-k}={2}^{\\text{-}n}\\displaystyle\\sum _{k=1}^{n}1=\\dfrac{n}{{2}^{n}}[\/latex] and [latex]f\\left(x\\right)g\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }n{\\left(\\dfrac{x}{2}\\right)}^{n}[\/latex]<\/li>\n
            13. The derivative of [latex]f[\/latex] is [latex]-\\dfrac{1}{{\\left(1+x\\right)}^{2}}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\left(n+1\\right){x}^{n}[\/latex].<\/li>\n
            14. The indefinite integral of [latex]f[\/latex] is [latex]\\dfrac{1}{1+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{x}^{2n}[\/latex].<\/li>\n
            15. [latex]f\\left(x\\right)=\\displaystyle\\sum {n=0}^{\\infty }{x}^{n}=\\dfrac{1}{1-x};{f}^{\\prime }\\left(\\dfrac{1}{2}\\right)=\\displaystyle\\sum {n=1}^{\\infty }\\dfrac{n}{{2}^{n - 1}}={\\dfrac{d}{dx}{\\left(1-x\\right)}^{-1}|}{x=\\dfrac{1}{2}}={\\dfrac{1}{{\\left(1-x\\right)}^{2}}|}{x=\\dfrac{1}{2}}=4[\/latex] so [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{n}{{2}^{n}}=2[\/latex].<\/li>\n
            16. [latex]f\\left(x\\right)=\\displaystyle\\sum {n=0}^{\\infty }{x}^{n}=\\dfrac{1}{1-x};f\\text{''}\\left(\\dfrac{1}{2}\\right)=\\displaystyle\\sum {n=2}^{\\infty }\\dfrac{n\\left(n - 1\\right)}{{2}^{n - 2}}={\\dfrac{{d}^{2}}{d{x}^{2}}{\\left(1-x\\right)}^{-1}|}{x=\\dfrac{1}{2}}={\\dfrac{2}{{\\left(1-x\\right)}^{3}}|}{x=\\dfrac{1}{2}}=16[\/latex] so [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\dfrac{n\\left(n - 1\\right)}{{2}^{n}}=4[\/latex].<\/li>\n
            17. [latex]\\displaystyle\\int \\displaystyle\\sum {\\left(1-x\\right)}^{n}dx=\\displaystyle\\int \\displaystyle\\sum {\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}dx=\\displaystyle\\sum \\dfrac{{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n+1}}{n+1}[\/latex]<\/li>\n
            18. [latex]\\text{-}{\\displaystyle\\int }_{t=0}^{{x}^{2}}\\dfrac{1}{1-t}dt=\\text{-}\\displaystyle\\sum {n=0}^{\\infty }{\\displaystyle\\int }{0}^{{x}^{2}}{t}^{n}dx-\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{2\\left(n+1\\right)}}{n+1}=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{x}^{2n}}{n}[\/latex]<\/li>\n
            19. [latex]{\\displaystyle\\int }_{0}^{{x}^{2}}\\dfrac{dt}{1+{t}^{2}}=\\displaystyle\\sum {n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\displaystyle\\int }{0}^{{x}^{2}}{t}^{2n}dt={\\displaystyle\\sum {n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{t}^{2n+1}}{2n+1}|}{t=0}^{{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{4n+2}}{2n+1}[\/latex]<\/li>\n
            20. Term-by-term integration gives [latex]{\\displaystyle\\int }_{0}^{x}\\text{ln}tdt=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\dfrac{{\\left(x - 1\\right)}^{n+1}}{n\\left(n+1\\right)}=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n - 1}\\left(\\dfrac{1}{n}-\\dfrac{1}{n+1}\\right){\\left(x - 1\\right)}^{n+1}=\\left(x - 1\\right)\\text{ln}x+\\displaystyle\\sum _{n=2}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{\\left(x - 1\\right)}^{n}}{n}= x\\text{ln}x-x[\/latex].<\/li>\n
            21. [latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{k}}{k}=\\text{-}\\text{ln}\\left(1-x\\right)[\/latex] so [latex]\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{x}^{3k}}{6k}=-\\dfrac{1}{6}\\text{ln}\\left(1-{x}^{3}\\right)[\/latex]. The radius of convergence is equal to 1 by the ratio test.<\/li>\n
            22. If [latex]y={2}^{\\text{-}x}[\/latex], then [latex]\\displaystyle\\sum {k=1}^{\\infty }{y}^{k}=\\dfrac{y}{1-y}=\\dfrac{{2}^{\\text{-}x}}{1-{2}^{\\text{-}x}}=\\dfrac{1}{{2}^{x}-1}[\/latex]. If [latex]{a}{k}={2}^{\\text{-}kx}[\/latex], then [latex]\\dfrac{{a}{k+1}}{{a}{k}}={2}^{\\text{-}x}<1[\/latex] when [latex]x>0[\/latex]. So the series converges for all [latex]x>0[\/latex].<\/li>\n
            23. Answers will vary.<\/li>\n
            24. \"This
              \nThe solid curve is [latex]{S}{5}[\/latex]. The dashed curve is [latex]{S}{2}[\/latex], dotted is [latex]{S}{3}[\/latex], and dash-dotted is [latex]{S}{4}[\/latex]<\/li>\n
            25. When [latex]x=-\\dfrac{1}{2},\\text{-}\\text{ln}\\left(2\\right)=\\text{ln}\\left(\\dfrac{1}{2}\\right)=\\text{-}\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{1}{n{2}^{n}}[\/latex]. Since [latex]\\displaystyle\\sum _{n=11}^{\\infty }\\dfrac{1}{n{2}^{n}}<\\displaystyle\\sum _{n=11}^{\\infty }\\dfrac{1}{{2}^{n}}=\\dfrac{1}{{2}^{10}}[\/latex], one has [latex]\\displaystyle\\sum _{n=1}^{10}\\dfrac{1}{n{2}^{n}}=0.69306\\ldots[\/latex] whereas [latex]\\text{ln}\\left(2\\right)=0.69314\\ldots[\/latex]; therefore, [latex]N=10[\/latex].<\/li>\n
            26. [latex]6{S}{N}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=2\\sqrt{3}\\displaystyle\\sum {n=0}^{N}{\\left(-1\\right)}^{n}\\dfrac{1}{{3}^{n}\\left(2n+1\\right)}[\/latex]. One has [latex]\\pi -6{S}{4}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=0.00101\\ldots[\/latex] and [latex]\\pi -6{S}{5}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=0.00028\\ldots[\/latex] so [latex]N=5[\/latex] is the smallest partial sum with accuracy to within 0.001. Also, [latex]\\pi -6{S}{7}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=0.00002\\ldots[\/latex] while [latex]\\pi -6{S}{8}\\left(\\dfrac{1}{\\sqrt{3}}\\right)=-0.000007\\ldots[\/latex] so [latex]N=8[\/latex] is the smallest N to give accuracy to within 0.00001.<\/li>\n<\/ol>\n

              Taylor and Maclaurin Series<\/span><\/h2>\n
                \n
              1. [latex]f\\left(-1\\right)=1;{f}^{\\prime }\\left(-1\\right)=-1;f\\text{''}\\left(-1\\right)=2;f\\left(x\\right)=1-\\left(x+1\\right)+{\\left(x+1\\right)}^{2}[\/latex]<\/li>\n
              2. [latex]{f}^{\\prime }\\left(x\\right)=2\\cos\\left(2x\\right);f\\text{''}\\left(x\\right)=-4\\sin\\left(2x\\right);{p}_{2}\\left(x\\right)=-2\\left(x-\\dfrac{\\pi }{2}\\right)[\/latex]<\/li>\n
              3. [latex]{f}^{\\prime }\\left(x\\right)=\\dfrac{1}{x};f\\text{''}\\left(x\\right)=-\\dfrac{1}{{x}^{2}};{p}_{2}\\left(x\\right)=0+\\left(x - 1\\right)-\\dfrac{1}{2}{\\left(x - 1\\right)}^{2}[\/latex]<\/li>\n
              4. [latex]{p}_{2}\\left(x\\right)=e+e\\left(x - 1\\right)+\\dfrac{e}{2}{\\left(x - 1\\right)}^{2}[\/latex]<\/li>\n
              5. [latex]\\dfrac{{d}^{2}}{d{x}^{2}}{x}^{\\dfrac{1}{3}}=-\\dfrac{2}{9{x}^{\\dfrac{5}{3}}}\\ge -0.00092\\ldots[\/latex] when [latex]x\\ge 28[\/latex] so the remainder estimate applies to the linear approximation [latex]{x}^{\\dfrac{1}{3}}\\approx {p}_{1}\\left(27\\right)=3+\\dfrac{x - 27}{27}[\/latex], which gives [latex]{\\left(28\\right)}^{\\dfrac{1}{3}}\\approx 3+\\dfrac{1}{27}=3.\\overline{037}[\/latex], while [latex]{\\left(28\\right)}^{\\dfrac{1}{3}}\\approx 3.03658[\/latex].<\/li>\n
              6. Using the estimate [latex]\\dfrac{{2}^{10}}{10\\text{!}}<0.000283[\/latex] we can use the Taylor expansion of order 9 to estimate [latex]e^{x}[\/latex] at [latex]x=2[\/latex]. as [latex]{e}^{2}\\approx {p}_{9}\\left(2\\right)=1+2+\\dfrac{{2}^{2}}{2}+\\dfrac{{2}^{3}}{6}+\\cdots+\\dfrac{{2}^{9}}{9\\text{!}}=7.3887\\ldots[\/latex] whereas [latex]{e}^{2}\\approx 7.3891[\/latex].<\/li>\n
              7. Since [latex]\\dfrac{{d}^{n}}{d{x}^{n}}\\left(\\text{ln}x\\right)={\\left(-1\\right)}^{n - 1}\\dfrac{\\left(n - 1\\right)\\text{!}}{{x}^{n}},{R}{1000}\\approx \\dfrac{1}{1001}[\/latex]. One has [latex]{p}{1000}\\left(1\\right)=\\displaystyle\\sum _{n=1}^{1000}\\dfrac{{\\left(-1\\right)}^{n - 1}}{n}\\approx 0.6936[\/latex] whereas [latex]\\text{ln}\\left(2\\right)\\approx 0.6931\\cdots[\/latex].<\/li>\n
              8. [latex]{\\displaystyle\\int }{0}^{1}\\left(1-{x}^{2}+\\dfrac{{x}^{4}}{2}-\\dfrac{{x}^{6}}{6}+\\dfrac{{x}^{8}}{24}-\\dfrac{{x}^{10}}{120}+\\dfrac{{x}^{12}}{720}\\right)dx[\/latex] [latex]=1-\\dfrac{{1}^{3}}{3}+\\dfrac{{1}^{5}}{10}-\\dfrac{{1}^{7}}{42}+\\dfrac{{1}^{9}}{9\\cdot 24}-\\dfrac{{1}^{11}}{120\\cdot 11}+\\dfrac{{1}^{13}}{720\\cdot 13}\\approx 0.74683[\/latex] whereas [latex]{\\displaystyle\\int }{0}^{1}{e}^{\\text{-}{x}^{2}}dx\\approx 0.74682[\/latex].<\/li>\n
              9. Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)[\/latex] is [latex]\\sin{z}[\/latex] or [latex]\\cos{z}[\/latex], we have [latex]M=1[\/latex]. Since [latex]|x - 0|\\le \\dfrac{\\pi }{2}[\/latex], we seek the smallest n such that [latex]\\dfrac{{\\pi }^{n+1}}{{2}^{n+1}\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=7[\/latex]. The remainder estimate is [latex]{R}_{7}\\le 0.00092[\/latex].<\/li>\n
              10. Since [latex]{f}^{\\left(n+1\\right)}\\left(z\\right)=\\pm{e}^{\\text{-}z}[\/latex] one has [latex]M={e}^{3}[\/latex]. Since [latex]|x - 0|\\le 3[\/latex], one seeks the smallest n such that [latex]\\dfrac{{3}^{n+1}{e}^{3}}{\\left(n+1\\right)\\text{!}}\\le 0.001[\/latex]. The smallest such value is [latex]n=14[\/latex]. The remainder estimate is [latex]{R}_{14}\\le 0.000220[\/latex].<\/li>\n
              11. \"This
                \nSince [latex]\\sin{x}[\/latex] is increasing for small x and since [latex]\\mathrm{si}n\\text{''}x=\\text{-}\\sin{x}[\/latex], the estimate applies whenever [latex]{R}^{2}\\sin\\left(R\\right)\\le 0.2[\/latex], which applies up to [latex]R=0.596[\/latex].<\/li>\n
              12. \"This
                \nSince the second derivative of [latex]\\cos{x}[\/latex] is [latex]\\text{-}\\cos{x}[\/latex] and since [latex]\\cos{x}[\/latex] is decreasing away from [latex]x=0[\/latex], the estimate applies when [latex]{R}^{2}\\cos{R}\\le 0.2[\/latex] or [latex]R\\le 0.447[\/latex].<\/li>\n
              13. [latex]{\\left(x+1\\right)}^{3}-2{\\left(x+1\\right)}^{2}+2\\left(x+1\\right)[\/latex]<\/li>\n
              14. Values of derivatives are the same as for [latex]x=0[\/latex] so [latex]\\cos{x}={\\displaystyle\\sum _{n=0}^{\\infty }\\left(-1\\right)}^{n}\\dfrac{{\\left(x - 2\\pi \\right)}^{2n}}{\\left(2n\\right)\\text{!}}[\/latex]<\/li>\n
              15. [latex]\\cos\\left(\\dfrac{\\pi }{2}\\right)=0,\\text{-}\\sin\\left(\\dfrac{\\pi }{2}\\right)=-1[\/latex] so [latex]\\cos{x}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{\\left(x-\\dfrac{\\pi }{2}\\right)}^{2n+1}}{\\left(2n+1\\right)\\text{!}}[\/latex], which is also [latex]\\text{-}\\cos\\left(x-\\dfrac{\\pi }{2}\\right)[\/latex].<\/li>\n
              16. The derivatives are [latex]{f}^{\\left(n\\right)}\\left(1\\right)=e[\/latex] so [latex]{e}^{x}=e\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex].<\/li>\n
              17. [latex]\\dfrac{1}{{\\left(x - 1\\right)}^{3}}=\\text{-}\\left(\\dfrac{1}{2}\\right)\\dfrac{{d}^{2}}{d{x}^{2}}\\dfrac{1}{1-x}=\\text{-}\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\dfrac{\\left(n+2\\right)\\left(n+1\\right){x}^{n}}{2}\\right)[\/latex]<\/li>\n
              18. [latex]2-x=1-\\left(x - 1\\right)[\/latex]<\/li>\n
              19. [latex]{\\left(\\left(x - 1\\right)-1\\right)}^{2}={\\left(x - 1\\right)}^{2}-2\\left(x - 1\\right)+1[\/latex]<\/li>\n
              20. [latex]\\dfrac{1}{1-\\left(1-x\\right)}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{\\left(x - 1\\right)}^{n}[\/latex]<\/li>\n
              21. [latex]x\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(1-x\\right)}^{2n}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n+1}+\\displaystyle\\sum _{n=0}^{\\infty }{2}^{n}{\\left(x - 1\\right)}^{2n}[\/latex]<\/li>\n
              22. [latex]{e}^{2x}={e}^{2\\left(x - 1\\right)+2}={e}^{2}\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{2}^{n}{\\left(x - 1\\right)}^{n}}{n\\text{!}}[\/latex]<\/li>\n
              23. [latex]x={e}^{2};{S}_{10}=\\dfrac{34,913}{4725}\\approx 7.3889947[\/latex]<\/li>\n
              24. [latex]\\sin\\left(2\\pi \\right)=0;{S}_{10}=8.27\\times {10}^{-5}[\/latex]<\/li>\n
              25. \"This
                \nThe difference is small on the interior of the interval but approaches [latex]1[\/latex] near the endpoints. The remainder estimate is [latex]|{R}_{4}|=\\dfrac{{\\pi }^{5}}{120}\\approx 2.552[\/latex].<\/li>\n
              26. \"This
                \nThe difference is on the order of [latex]{10}^{-4}[\/latex] on [latex]\\left[-1,1\\right][\/latex] while the Taylor approximation error is around [latex]0.1[\/latex] near [latex]\\pm 1[\/latex]. The top curve is a plot of [latex]{\\tan}^{2}x-{\\left(\\dfrac{{S}{5}\\left(x\\right)}{{C}{4}\\left(x\\right)}\\right)}^{2}[\/latex] and the lower dashed plot shows [latex]{t}^{2}-{\\left(\\dfrac{{S}{5}}{{C}{4}}\\right)}^{2}[\/latex].<\/li>\n
              27. [latex]\\dfrac{\\text{ln}\\left(1-{x}^{2}\\right)}{{x}^{2}}\\to \\text{-}1[\/latex]<\/li>\n
              28. [latex]\\dfrac{\\cos\\left(\\sqrt{x}\\right)-1}{2x}\\approx \\dfrac{\\left(1-\\dfrac{x}{2}+\\dfrac{{x}^{2}}{4\\text{!}}-\\cdots\\right)-1}{2x}\\to -\\dfrac{1}{4}[\/latex]<\/li>\n<\/ol>\n

                Applications of Series<\/span><\/h2>\n
                  \n
                1. [latex]{\\left(1+{x}^{2}\\right)}^{\\dfrac{-1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}-\\dfrac{1}{3}\\hfill \\ \\hfill n\\hfill \\end{array}\\right){x}^{2n}[\/latex]<\/li>\n
                2. [latex]{\\left(1 - 2x\\right)}^{\\dfrac{2}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}{2}^{n}\\left(\\begin{array}{c}\\dfrac{2}{3}\\hfill \\ n\\hfill \\end{array}\\right){x}^{n}[\/latex]<\/li>\n
                3. [latex]\\sqrt{2+{x}^{2}}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{\\left(\\dfrac{1}{2}\\right)-n}\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){x}^{2n};\\left(|{x}^{2}|<2\\right)[\/latex]<\/li>\n
                4. [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}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){\\left(x - 1\\right)}^{2n}[\/latex]<\/li>\n
                5. [latex]\\sqrt{x}=2\\sqrt{1+\\dfrac{x - 4}{4}}[\/latex] so [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{2}^{1 - 2n}\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){\\left(x - 4\\right)}^{n}[\/latex]<\/li>\n
                6. [latex]\\sqrt{x}=\\displaystyle\\sum _{n=0}^{\\infty }{3}^{1 - 3n}\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right){\\left(x - 9\\right)}^{n}[\/latex]<\/li>\n
                7. [latex]10{\\left(1+\\dfrac{x}{1000}\\right)}^{\\dfrac{1}{3}}=\\displaystyle\\sum _{n=0}^{\\infty }{10}^{1 - 3n}\\left(\\begin{array}{c}\\dfrac{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)}^{\\dfrac{1}{3}}& \\approx 10\\left(1+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 1\\hfill \\end{array}\\right){10}^{-3}+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 2\\hfill \\end{array}\\right){10}^{-6}+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 3\\hfill \\end{array}\\right){10}^{-9}+\\left(\\begin{array}{c}\\dfrac{1}{3}\\hfill \\ 4\\hfill \\end{array}\\right){10}^{-12}\\right)\\hfill \\ & =10\\left(1+\\dfrac{1}{{3.10}^{3}}-\\dfrac{1}{{9.10}^{6}}+\\dfrac{5}{{81.10}^{9}}-\\dfrac{10}{{243.10}^{12}}\\right)=10.00333222...\\hfill \\end{array}[\/latex] whereas [latex]{\\left(1001\\right)}^{\\dfrac{1}{3}}=10.00332222839093...[\/latex]. Two terms would suffice for three-digit accuracy.<\/li>\n
                8. The approximation is [latex]2.3152[\/latex]; the CAS value is [latex]2.23\\ldots[\/latex].<\/li>\n
                9. The approximation is [latex]2.583\\ldots[\/latex]; the CAS value is [latex]2.449\\ldots[\/latex].<\/li>\n
                10. [latex]\\sqrt{1-{x}^{2}}=1-\\dfrac{{x}^{2}}{2}-\\dfrac{{x}^{4}}{8}-\\dfrac{{x}^{6}}{16}-\\dfrac{5{x}^{8}}{128}+\\cdots[\/latex]. Thus [latex]{\\displaystyle\\int }{-1}^{1}\\sqrt{1-{x}^{2}}dx=x-\\dfrac{{x}^{3}}{6}-\\dfrac{{x}^{5}}{40}-\\dfrac{{x}^{7}}{7\\cdot 16}-\\dfrac{5{x}^{9}}{9\\cdot 128}+\\cdots{|}{-1}^{1}\\approx 2-\\dfrac{1}{3}-\\dfrac{1}{20}-\\dfrac{1}{56}-\\dfrac{10}{9\\cdot 128}+\\text{error}=1.590..[\/latex]. whereas [latex]\\dfrac{\\pi }{2}=1.570..[\/latex].<\/li>\n
                11. [latex]{\\left(1+4x\\right)}^{\\dfrac{4}{3}}=\\left(1+4x\\right)\\left(1+4x\\right)^{\\dfrac{1}{3}}=\\left(1+4x\\right)\\left(1+\\dfrac{4x}{3}-\\dfrac{16x^{3}}{9}+\\dfrac{320x^{3}}{81}-\\dfrac{2560x^{4}}{243}\\right)=1+\\dfrac{16}{3}x+\\dfrac{32}{9}x^{2}-\\dfrac{256}{81}x^{3}+\\dfrac{1280}{243}x^{4}-\\dfrac{10240}{243}x^{5}[\/latex]<\/li>\n
                12. [latex]\\left(1+{\\left(x+3\\right)}^{2}\\right)^{\\dfrac{1}{3}} =1+\\dfrac{1}{3}{\\left(x+3\\right)}^{2}-\\dfrac{1}{9}{\\left(x+3\\right)}^{4}+\\dfrac{5}{81}{\\left(x+3\\right)}^{6}-\\dfrac{10}{243}{\\left(x+3\\right)}^{8}+\\cdots[\/latex]<\/li>\n
                13. Twice the approximation is [latex]1.260\\ldots[\/latex] whereas [latex]{2}^{\\dfrac{1}{3}}=1.2599...[\/latex].<\/li>\n
                14. [latex]{f}^{\\left(99\\right)}\\left(0\\right)=0[\/latex]<\/li>\n
                15. [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{\\left(\\text{ln}\\left(2\\right)x\\right)}^{n}}{n\\text{!}}[\/latex]<\/li>\n
                16. For [latex]x>0,\\sin\\left(\\sqrt{x}\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{\\dfrac{\\left(2n+1\\right)}{2}}}{\\sqrt{x}\\left(2n+1\\right)\\text{!}}=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{n}}{\\left(2n+1\\right)\\text{!}}[\/latex].<\/li>\n
                17. [latex]{e}^{{x}^{3}}=\\displaystyle\\sum _{n=0}^{\\infty }\\dfrac{{x}^{3n}}{n\\text{!}}[\/latex]<\/li>\n
                18. [latex]{\\sin}^{2}x=\\text{-}\\displaystyle\\sum _{k=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{k}{2}^{2k - 1}{x}^{2k}}{\\left(2k\\right)\\text{!}}[\/latex]<\/li>\n
                19. [latex]{\\tan}^{-1}x=\\displaystyle\\sum _{k=0}^{\\infty }\\dfrac{{\\left(-1\\right)}^{k}{x}^{2k+1}}{2k+1}[\/latex]<\/li>\n
                20. [latex]{\\sin}^{-1}x=\\displaystyle\\sum _{n=0}^{\\infty }\\left(\\begin{array}{c}\\dfrac{1}{2}\\hfill \\ n\\hfill \\end{array}\\right)\\dfrac{{x}^{2n+1}}{\\left(2n+1\\right)n\\text{!}}[\/latex]<\/li>\n
                21. [latex]F\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{\\left(-1\\right)}^{n}\\dfrac{{x}^{n+1}}{\\left(n+1\\right)\\left(2n\\right)\\text{!}}[\/latex]<\/li>\n
                22. [latex]F\\left(x\\right)=\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}\\dfrac{{x}^{n}}{{n}^{2}}[\/latex]<\/li>\n
                23. [latex]x+\\dfrac{{x}^{3}}{3}+\\dfrac{2{x}^{5}}{15}+\\cdots[\/latex]<\/li>\n
                24. [latex]1+x-\\dfrac{{x}^{3}}{3}-\\dfrac{{x}^{4}}{6}+\\cdots[\/latex]<\/li>\n
                25. [latex]1+{x}^{2}+\\dfrac{2{x}^{4}}{3}+\\dfrac{17{x}^{6}}{45}+\\cdots[\/latex]<\/li>\n
                26. Using the expansion for [latex]\\tan{x}[\/latex] gives [latex]1+\\dfrac{x}{3}+\\dfrac{2{x}^{2}}{15}[\/latex].<\/li>\n
                27. [latex]\\dfrac{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.<\/li>\n
                28. [latex]\\text{ln}\\left(1+{x}^{2}\\right)=\\displaystyle\\sum _{n=1}^{\\infty }\\dfrac{{\\left(-1\\right)}^{n - 1}}{n}{x}^{2n}[\/latex] so [latex]R=1[\/latex] by the ratio test.<\/li>\n<\/ol>\n","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":109,"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\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/184"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":1,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/184\/revisions"}],"predecessor-version":[{"id":195,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/184\/revisions\/195"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/parts\/109"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapters\/184\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/media?parent=184"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/pressbooks\/v2\/chapter-type?post=184"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/contributor?post=184"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/qrpracticepages\/wp-json\/wp\/v2\/license?post=184"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}