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Mathematics LibreTexts

10.4E: Exercises for Section 10.4

  • Page ID
    72457
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    In exercises 1 - 4, use appropriate substitutions to write down the Maclaurin series for the given binomial.

    1) \((1−x)^{1/3}\)

    2) \((1+x^2)^{−1/3}\)

    Answer:
    \(\displaystyle (1+x^2)^{−1/3}=\sum_{n=0}^∞\left(n^{−\frac{1}{3}}\right)x^{2n}\)

    3) \((1−x)^{1.01}\)

    4) \((1−2x)^{2/3}\)

    Answer:
    \(\displaystyle (1−2x)^{2/3}=\sum_{n=0}^∞(−1)^n2^n\left(n^{\frac{2}{3}}\right)x^n\)

     

    In exercises 5 - 12, use the substitution \((b+x)^r=(b+a)^r\left(1+\dfrac{x−a}{b+a}\right)^r\) in the binomial expansion to find the Taylor series of each function with the given center.

    5) \(\sqrt{x+2}\) at \(a=0\)

    6) \(\sqrt{x^2+2}\) at \(a=0\)

    Answer:
    \(\displaystyle \sqrt{2+x^2}=\sum_{n=0}^∞2^{(1/2)−n}\left(n^{\frac{1}{2}}\right)x^{2n};(|x^2|<2)\)

    7) \(\sqrt{x+2}\) at \(a=1\)

    8) \(\sqrt{2x−x^2}\) at \(a=1\) (Hint: \(2x−x^2=1−(x−1)^2\))

    Answer:
    \(\sqrt{2x−x^2}=\sqrt{1−(x−1)^2}\) so \(\displaystyle \sqrt{2x−x^2}=\sum_{n=0}^∞(−1)^n\left(n^{\frac{1}{2}}\right)(x−1)^{2n}\)

    9) \((x−8)^{1/3}\) at \(a=9\)

    10) \(\sqrt{x}\) at \(a=4\)

    Answer:
    \(\sqrt{x}=2\sqrt{1+\frac{x−4}{4}}\) so \(\displaystyle \sqrt{x}=\sum_{n=0}^∞2^{1−2n}\left(n^{\frac{1}{2}}\right)(x−4)^n\)

    11) \(x^{1/3}\) at \(a=27\)

    12) \(\sqrt{x}\) at \(x=9\)

    Answer:
    \(\displaystyle \sqrt{x}=\sum_{n=0}^∞3^{1−3n}\left(n^{\frac{1}{2}}\right)(x−9)^n\)

     

    In exercises 13 - 14, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most \(1/1000.\)

    13) [T] \((15)^{1/4}\) using \((16−x)^{1/4}\)

    14) [T] \((1001)^{1/3}\) using \((1000+x)^{1/3}\)

    Answer:
    \(\displaystyle 10(1+\frac{x}{1000})^{1/3}=\sum_{n=0}^∞10^{1−3n}(^{\frac{1}{3}}_n)x^n\). Using, for example, a fourth-degree estimate at \(x=1\) gives \((1001)^{1/3}≈10\left(1+\left(1^{\frac{1}{3}}\right)10^{−3}+\left(2^{\frac{1}{3}}\right)10^{−6}+\left(3^{\frac{1}{3}}\right)10^{−9}+\left(3^{\frac{1}{3}}\right)10^{−12}\right)=10\left(1+\frac{1}{3.10^3}−\frac{1}{9.10^6}+\frac{5}{81.10^9}−\frac{10}{243.10^{12}}\right)=10.00333222...\) whereas \((1001)^{1/3}=10.00332222839093....\) Two terms would suffice for three-digit accuracy.

     

    In exercises 15 - 18, use the binomial approximation \(\sqrt{1−x}≈1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{128}−\frac{7x^5}{256}\) for \(|x|<1\) to approximate each number. Compare this value to the value given by a scientific calculator.

    15) [T] \(\frac{1}{\sqrt{2}}\) using \(x=\frac{1}{2}\) in \((1−x)^{1/2}\)

    16) [T] \(\sqrt{5}=5×\frac{1}{\sqrt{5}}\) using \(x=\frac{4}{5}\) in \((1−x)^{1/2}\)

    Answer:
    The approximation is \(2.3152\); the CAS value is \(2.23….\)

    17) [T] \(\sqrt{3}=\frac{3}{\sqrt{3}}\) using \(x=\frac{2}{3}\) in \((1−x)^{1/2}\)

    18) [T] \(\sqrt{6}\) using \(x=\frac{5}{6}\) in \((1−x)^{1/2}\)

    Answer:
    The approximation is \(2.583…\); the CAS value is \(2.449….\)

     

    19) Integrate the binomial approximation of \(\sqrt{1−x}\) to find an approximation of \(\displaystyle ∫^x_0\sqrt{1−t}\,dt\).

    20) [T] Recall that the graph of \(\sqrt{1−x^2}\) is an upper semicircle of radius \(1\). Integrate the binomial approximation of \(\sqrt{1−x^2}\) up to order \(8\) from \(x=−1\) to \(x=1\) to estimate \(\frac{π}{2}\).

    Answer:
    \(\sqrt{1−x^2}=1−\frac{x^2}{2}−\frac{x^4}{8}−\frac{x^6}{16}−\frac{5x^8}{128}+⋯.\) Thus \(\displaystyle ∫^1_{−1}\sqrt{1−x^2}\,dx=\left[x−\frac{x^3}{6}−\frac{x^5}{40}−\frac{x^7}{7⋅16}−\frac{5x^9}{9⋅128}+⋯\right]\Big|^1_{−1}≈2−\frac{1}{3}−\frac{1}{20}−\frac{1}{56}−\frac{10}{9⋅128}+error=1.590...\) whereas \(\frac{π}{2}=1.570...\)

     

    In exercises 21 - 24, use the expansion \((1+x)^{1/3}=1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯\) to write the first five terms (not necessarily a quartic polynomial) of each expression.

    21) \((1+4x)^{1/3};\;a=0\)

    22) \((1+4x)^{4/3};\;a=0\)

    Answer:
    \((1+x)^{4/3}=(1+x)(1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯)=1+\frac{4x}{3}+\frac{2x^2}{9}−\frac{4x^3}{81}+\frac{5x^4}{243}+⋯\)

    23) \((3+2x)^{1/3};\;a=−1\)

    24) \((x^2+6x+10)^{1/3};\;a=−3\)

    Answer:
    \((1+(x+3)^2)^{1/3}=1+\frac{1}{3}(x+3)^2−\frac{1}{9}(x+3)^4+\frac{5}{81}(x+3)^6−\frac{10}{243}(x+3)^8+⋯\)

     

    25) Use \((1+x)^{1/3}=1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+⋯\) with \(x=1\) to approximate \(2^{1/3}\).

    26) Use the approximation \((1−x)^{2/3}=1−\frac{2x}{3}−\frac{x^2}{9}−\frac{4x^3}{81}−\frac{7x^4}{243}−\frac{14x^5}{729}+⋯\) for \(|x|<1\) to approximate \(2^{1/3}=2.2^{−2/3}\).

    Answer:
    Twice the approximation is \(1.260…\) whereas \(2^{1/3}=1.2599....\)

    27) Find the \(25^{\text{th}}\) derivative of \(f(x)=(1+x^2)^{13}\) at \(x=0\).

    28) Find the \(99^{\text{th}}\) derivative of \(f(x)=(1+x^4)^{25}\).

    Answer:
    \(f^{(99)}(0)=0\)

     

    In exercises 29 - 36, find the Maclaurin series of each function.

    29) \(f(x)=xe^{2x}\)

    30) \(f(x)=2^x\)

    Answer:
    \(\displaystyle \sum_{n=0}^∞\frac{(\ln(2)x)^n}{n!}\)

    31) \(f(x)=\dfrac{\sin x}{x}\)

    32) \(f(x)=\dfrac{\sin(\sqrt{x})}{\sqrt{x}},(x>0),\)

    Answer:
    For \(\displaystyle x>0,\, \sin(\sqrt{x})=\sum_{n=0}^∞(−1)^n\frac{x^{(2n+1)/2}}{\sqrt{x}(2n+1)!}=\sum_{n=0}^∞(−1)^n\frac{x^n}{(2n+1)!}\).

    33) \(f(x)=\sin(x^2)\)

    34) \(f(x)=e^{x^3}\)

    Answer:
    \(\displaystyle e^{x^3}=\sum_{n=0}^∞\frac{x^{3n}}{n!}\)

    35) \(f(x)=\cos^2x\) using the identity \(\cos^2x=\frac{1}{2}+\frac{1}{2}\cos(2x)\)

    36) \(f(x)=\sin^2x\) using the identity \(\sin^2x=\frac{1}{2}−\frac{1}{2}\cos(2x)\)

    Answer:
    \(\displaystyle \sin^2x=−\sum_{k=1}^∞\frac{(−1)^k2^{2k−1}x^{2k}}{(2k)!}\)

     

    In exercises 37 - 44, find the Maclaurin series of \(\displaystyle F(x)=∫^x_0f(t)\,dt\) by integrating the Maclaurin series of \(f\) term by term. If \(f\) is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

    37) \(\displaystyle F(x)=∫^x_0e^{−t^2}\,dt;\; f(t)=e^{−t^2}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{n!}\)

    38) \(\displaystyle F(x)=\tan^{−1}x;\; f(t)=\frac{1}{1+t^2}=\sum_{n=0}^∞(−1)^nt^{2n}\)

    Answer:
    \(\displaystyle \tan^{−1}x=\sum_{k=0}^∞\frac{(−1)^kx^{2k+1}}{2k+1}\)

    39) \(\displaystyle F(x)=\tanh^{−1}x; \; f(t)=\frac{1}{1−t^2}=\sum_{n=0}^∞t^{2n}\)

    40) \(\displaystyle F(x)=\sin^{−1}x; \; f(t)=\frac{1}{\sqrt{1−t^2}}=\sum_{k=0}^∞\left(k^{\frac{1}{2}}\right)\frac{t^{2k}}{k!}\)

    Answer:
    \(\displaystyle \sin^{−1}x=\sum_{n=0}^∞\left(n^{\frac{1}{2}}\right)\frac{x^{2n+1}}{(2n+1)n!}\)

    41) \(\displaystyle F(x)=∫^x_0\frac{\sin t}{t}\,dt; \; f(t)=\frac{\sin t}{t}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{(2n+1)!}\)

    42) \(\displaystyle F(x)=∫^x_0\cos\left(\sqrt{t}\right)\,dt; \; f(t)=\sum_{n=0}^∞(−1)^n\frac{x^n}{(2n)!}\)

    Answer:
    \(\displaystyle F(x)=\sum_{n=0}^∞(−1)^n\frac{x^{n+1}}{(n+1)(2n)!}\)

    43) \(\displaystyle F(x)=∫^x_0\frac{1−\cos t}{t^2}\,dt; \; f(t)=\frac{1−\cos t}{t^2}=\sum_{n=0}^∞(−1)^n\frac{t^{2n}}{(2n+2)!}\)

    44) \(\displaystyle F(x)=∫^x_0\frac{\ln(1+t)}{t}\,dt; \; f(t)=\sum_{n=0}^∞(−1)^n\frac{t^n}{n+1}\)

    Answer:
    \(\displaystyle F(x)=\sum_{n=1}^∞(−1)^{n+1}\frac{x^n}{n^2}\)

     

    In exercises 45 - 52, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of \(f\).

    45) \(f(x)=\sin\left(x+\frac{π}{4}\right)=\sin x\cos\left(\frac{π}{4}\right)+\cos x\sin\left(\frac{π}{4}\right)\)

    46) \(f(x)=\tan x\)

    Answer:
    \(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+⋯\)

    47) \(f(x)=\ln(\cos x)\)

    48) \(f(x)=e^x\cos x\)

    Answer:
    \(1+x−\dfrac{x^3}{3}−\dfrac{x^4}{6}+⋯\)

    49) \(f(x)=e^{\sin x}\)

    50) \(f(x)=\sec^2x\)

    Answer:
    \(1+x^2+\dfrac{2x^4}{3}+\dfrac{17x^6}{45}+⋯\)

    51) \(f(x)=\tanh x\)

    52) \(f(x)=\dfrac{\tan\sqrt{x}}{\sqrt{x}}\) (see expansion for \(\tan x\))

    Answer:
    Using the expansion for \(\tan x\) gives \(1+\dfrac{x}{3}+\dfrac{2x^2}{15}\).

     

    In exercises 53 - 56, find the radius of convergence of the Maclaurin series of each function.

    53) \(\ln(1+x)\)

    54) \(\dfrac{1}{1+x^2}\)

    Answer:
    \(\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^∞(−1)^nx^{2n}\) so \(R=1\) by the ratio test.

    55) \(\tan^{−1}x\)

    56) \(\ln(1+x^2)\)

    Answer:
    \(\displaystyle \ln(1+x^2)=\sum_{n=1}^∞\frac{(−1)^{n−1}}{n}x^{2n}\) so \(R=1\) by the ratio test.

     

    57) Find the Maclaurin series of \(\sinh x=\dfrac{e^x−e^{−x}}{2}\).

    58) Find the Maclaurin series of \(\cosh x=\dfrac{e^x+e^{−x}}{2}\).

    Answer:
    Add series of \(e^x\) and \(e^{−x}\) term by term. Odd terms cancel and \(\displaystyle \cosh x=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}\).

    59) Differentiate term by term the Maclaurin series of \(\sinh x\) and compare the result with the Maclaurin series of \(\cosh x\).

    60) [T] Let \(\displaystyle S_n(x)=\sum_{k=0}^n(−1)^k\frac{x^{2k+1}}{(2k+1)!}\) and \(\displaystyle C_n(x)=\sum_{n=0}^n(−1)^k\frac{x^{2k}}{(2k)!}\) denote the respective Maclaurin polynomials of degree \(2n+1\) of \(\sin x\) and degree \(2n\) of \(\cos x\). Plot the errors \(\dfrac{S_n(x)}{C_n(x)}−\tan x\) for \(n=1,..,5\) and compare them to \(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}−\tan x\) on \(\left(−\frac{π}{4},\frac{π}{4}\right)\).

    Answer:

    The ratio \(\dfrac{S_n(x)}{C_n(x)}\) approximates \(\tan x\) better than does \(p_7(x)=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}\) for \(N≥3\). The dashed curves are \(\dfrac{S_n}{C_n}−\tan x\) for \(n=1,\, 2\). The dotted curve corresponds to \(n=3\), and the dash-dotted curve corresponds to \(n=4\). The solid curve is \(p_7−\tan x\).

    This graph has two curves. The first one is a decreasing function passing through the origin. The second is a broken line which is an increasing function passing through the origin. The two curves are very close around the origin.

    61) Use the identity \(2\sin x\cos x=\sin(2x)\) to find the power series expansion of \(\sin^2x\) at \(x=0\). (Hint: Integrate the Maclaurin series of \(\sin(2x)\) term by term.)

    62) If \(\displaystyle y=\sum_{n=0}^∞a_nx^n\), find the power series expansions of \(xy′\) and \(x^2y''\).

    Answer:
    By the term-by-term differentiation theorem, \(\displaystyle y′=\sum_{n=1}^∞na_nx^{n−1}\) so \(\displaystyle y′=\sum_{n=1}^∞na_nx^{n−1}xy′=\sum_{n=1}^∞na_nx^n\), whereas \(\displaystyle y′=\sum_{n=2}^∞n(n−1)a_nx^{n−2}\) so \(\displaystyle xy''=\sum_{n=2}^∞n(n−1)a_nx^n\).

    63) [T] Suppose that \(\displaystyle y=\sum_{k=0}^∞a^kx^k\) satisfies \(y′=−2xy\) and \(y(0)=0\). Show that \(a_{2k+1}=0\) for all \(k\) and that \(a_{2k+2}=\dfrac{−a_{2k}}{k+1}\). Plot the partial sum \(S_{20}\) of \(y\) on the interval \([−4,4]\).

    64) [T] Suppose that a set of standardized test scores is normally distributed with mean \(μ=100\) and standard deviation \(σ=10\). Set up an integral that represents the probability that a test score will be between \(90\) and \(110\) and use the integral of the degree \(10\) Maclaurin polynomial of \(\frac{1}{\sqrt{2π}}e^{−x^2/2}\) to estimate this probability.

    Answer:
    The probability is \(\displaystyle p=\frac{1}{\sqrt{2π}}∫^{(b−μ)/σ}_{(a−μ)/σ}e^{−x^2/2}\,dx\) where \(a=90\) and \(b=100\), that is, \(\displaystyle p=\frac{1}{\sqrt{2π}}∫^1_{−1}e^{−x^2/2}\,dx=\frac{1}{\sqrt{2π}}∫^1_{−1}\sum_{n=0}^5(−1)^n\frac{x^{2n}}{2^nn!}\,dx=\frac{2}{\sqrt{2π}}\sum_{n=0}^5(−1)^n\frac{1}{(2n+1)2^nn!}≈0.6827.\)

    65) [T] Suppose that a set of standardized test scores is normally distributed with mean \(μ=100\) and standard deviation \(σ=10\). Set up an integral that represents the probability that a test score will be between \(70\) and \(130\) and use the integral of the degree \(50\) Maclaurin polynomial of \(\frac{1}{\sqrt{2π}}e^{−x^2/2}\) to estimate this probability.

    66) [T] Suppose that \(\displaystyle \sum_{n=0}^∞a_nx^n\) converges to a function \(f(x)\) such that \(f(0)=1,\, f′(0)=0\), and \(f''(x)=−f(x)\). Find a formula for \(a_n\) and plot the partial sum \(S_N\) for \(N=20\) on \([−5,5].\)

    Answer:

    As in the previous problem one obtains \(a_n=0\) if \(n\) is odd and \(a_n=−(n+2)(n+1)a_{n+2}\) if \(n\) is even, so \(a_0=1\) leads to \(a_{2n}=\dfrac{(−1)^n}{(2n)!}\).

    This graph is a wave curve symmetrical about the origin. It has a peak at y = 1 above the origin. It has lowest points at -3 and 3.

    67) [T] Suppose that \(\displaystyle \sum_{n=0}^∞a_nx^n\) converges to a function \(f(x)\) such that \(f(0)=0,\; f′(0)=1\), and \(f''(x)=−f(x)\). Find a formula for an and plot the partial sum \(S_N\) for \(N=10\) on \([−5,5]\).

    68) Suppose that \(\displaystyle \sum_{n=0}^∞a_nx^n\) converges to a function \(y\) such that \(y''−y′+y=0\) where \(y(0)=1\) and \(y'(0)=0.\) Find a formula that relates \(a_{n+2},\;a_{n+1},\) and an and compute \(a_0,...,a_5\).

    Answer:
    \(\displaystyle y''=\sum_{n=0}^∞(n+2)(n+1)a_{n+2}x^n\) and \(\displaystyle y′=\sum_{n=0}^∞(n+1)a_{n+1}x^n\) so \(y''−y′+y=0\) implies that \((n+2)(n+1)a_{n+2}−(n+1)a_{n+1}+a_n=0\) or \(a_n=\dfrac{a_{n−1}}{n}−\dfrac{a_{n−2}}{n(n−1)}\) for all \(n⋅y(0)=a_0=1\) and \(y′(0)=a_1=0,\) so \(a_2=\frac{1}{2},\;a_3=\frac{1}{6}\;,a_4=0\), and \(a_5=−\frac{1}{120}\).

    69) Suppose that \(\displaystyle \sum_{n=0}^∞a_nx^n\) converges to a function \(y\) such that \(y''−y′+y=0\) where \(y(0)=0\) and \(y′(0)=1\). Find a formula that relates \(a_{n+2},\;a_{n+1}\), and an and compute \(a_1,...,a_5\).

     

    The error in approximating the integral \(\displaystyle ∫^b_af(t)\, dt\) by that of a Taylor approximation \(\displaystyle ∫^b_aPn(t) \,dt\) is at most \(\displaystyle ∫^b_aR_n(t) \,dt\). In exercises 70 - 71, the Taylor remainder estimate \(R_n≤\frac{M}{(n+1)!}|x−a|^{n+1}\) guarantees that the integral of the Taylor polynomial of the given order approximates the integral of \(f\) with an error less than \(\frac{1}{10}\).

    a. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than \(\frac{1}{100}\).

    b. Compare the accuracy of the polynomial integral estimate with the remainder estimate.

    70) [T] \(\displaystyle ∫^π_0\frac{\sin t}{t}\, dt;\quad P_s=1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!}\) (You may assume that the absolute value of the ninth derivative of \(\frac{\sin t}{t}\) is bounded by \(0.1\).)

    Answer:
    a. (Proof)
    b. We have \(R_s≤\frac{0.1}{(9)!}π^9≈0.0082<0.01.\) We have \(\displaystyle ∫^π_0\left(1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!}\right)\,dx=π−\frac{π^3}{3⋅3!}+\frac{π^5}{5⋅5!}−\frac{π^7}{7⋅7!}+\frac{π^9}{9⋅9!}=1.852...,\) whereas \(\displaystyle ∫^π_0\frac{\sin t}{t}\,dt=1.85194...\), so the actual error is approximately \(0.00006.\)

    71) [T] \(\displaystyle ∫^2_0e^{−x^2}\,dx;\; p_{11}=1−x^2+\frac{x^4}{2}−\frac{x^6}{3!}+⋯−\frac{x^{22}}{11!}\) (You may assume that the absolute value of the \(23^{\text{rd}}\) derivative of \(e^{−x^2}\) is less than \(2×10^{14}\).)

     

    The following exercises (72-73) deal with Fresnel integrals.

    72) The Fresnel integrals are defined by \(\displaystyle C(x)=∫^x_0\cos(t^2)\,dt\) and \(\displaystyle S(x)=∫^x_0\sin(t^2)\,dt\). Compute the power series of \(C(x)\) and \(S(x)\) and plot the sums \(C_N(x)\) and \(S_N(x)\) of the first \(N=50\) nonzero terms on \([0,2π]\).

    Answer:

    Since \(\displaystyle \cos(t^2)=\sum_{n=0}^∞(−1)^n\frac{t^{4n}}{(2n)!}\) and \(\displaystyle \sin(t^2)=\sum_{n=0}^∞(−1)^n\frac{t^{4n+2}}{(2n+1)!}\), one has \(\displaystyle S(x)=\sum_{n=0}^∞(−1)^n\frac{x^{4n+3}}{(4n+3)(2n+1)!}\) and \(\displaystyle C(x)=\sum_{n=0}^∞(−1)^n\frac{x^{4n+1}}{(4n+1)(2n)!}\). The sums of the first \(50\) nonzero terms are plotted below with \(C_{50}(x)\) the solid curve and \(S_{50}(x)\) the dashed curve.

    This graph has two curves. The first one is a solid curve labeled Csub50(x). It begins at the origin and is a wave that gradually decreases in amplitude. The highest it reaches is y = 1. The second curve is labeled Ssub50(x). It is a wave that gradually decreases in amplitude. The highest it reaches is 0.9. It is very close to the pattern of the first curve with a slight shift to the right.

    73) [T] The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates \((C(t),S(t))\). Plot the curve \((C_{50},S_{50})\) for \(0≤t≤2π\), the coordinates of which were computed in the previous exercise.

     

    74) Estimate \(\displaystyle ∫^{1/4}_0\sqrt{x−x^2}\,dx\) by approximating \(\sqrt{1−x}\) using the binomial approximation \(\displaystyle 1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{2128}−\frac{7x^5}{256}\).

    Answer:
    \(\displaystyle ∫^{1/4}_0\sqrt{x}\left(1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{128}−\frac{7x^5}{256}\right)\,dx =\frac{2}{3}2^{−3}−\frac{1}{2}\frac{2}{5}2^{−5}−\frac{1}{8}\frac{2}{7}2^{−7}−\frac{1}{16}\frac{2}{9}2^{−9}−\frac{5}{128}\frac{2}{11}2^{−11}−\frac{7}{256}\frac{2}{13}2^{−13}=0.0767732...\) whereas \(\displaystyle ∫^{1/4}_0\sqrt{x−x^2}\, dx=0.076773.\)

    75) [T] Use Newton’s approximation of the binomial \(\sqrt{1−x^2}\) to approximate \(π\) as follows. The circle centered at \((\frac{1}{2},0)\) with radius \(\frac{1}{2}\) has upper semicircle \(y=\sqrt{x}\sqrt{1−x}\). The sector of this circle bounded by the \(x\)-axis between \(x=0\) and \(x=\frac{1}{2}\) and by the line joining \((\frac{1}{4},\frac{\sqrt{3}}{4})\) corresponds to \(\frac{1}{6}\) of the circle and has area \(\frac{π}{24}\). This sector is the union of a right triangle with height \(\frac{\sqrt{3}}{4}\) and base \(\frac{1}{4}\) and the region below the graph between \(x=0\) and \(x=\frac{1}{4}\). To find the area of this region you can write \(y=\sqrt{x}\sqrt{1−x}=\sqrt{x}×(\text{binomial expansion of} \sqrt{1−x})\) and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate \(π\).

    76) Use the approximation \(T≈2π\sqrt{\frac{L}{g}}(1+\frac{k^2}{4})\) to approximate the period of a pendulum having length \(10\) meters and maximum angle \(θ_{max}=\frac{π}{6}\) where \(k=\sin\left(\frac{θ_{max}}{2}\right)\). Compare this with the small angle estimate \(T≈2π\sqrt{\frac{L}{g}}\).

    Answer:
    \(T≈2π\sqrt{\frac{10}{9.8}}\left(1+\frac{\sin^2(θ/12)}{4}\right)≈6.453\) seconds. The small angle estimate is \(T≈2π\sqrt{\frac{10}{9.8}≈6.347}\). The relative error is around \(2\) percent.

    77) Suppose that a pendulum is to have a period of \(2\) seconds and a maximum angle of \(θ_{max}=\frac{π}{6}\). Use \(T≈2π\sqrt{\dfrac{L}{g}}\left(1+\dfrac{k^2}{4}\right)\) to approximate the desired length of the pendulum. What length is predicted by the small angle estimate \(T≈2π\sqrt{\frac{L}{g}}\)?

    78) Evaluate \(\displaystyle ∫^{π/2}_0\sin^4θ\,dθ\) in the approximation \(\displaystyle T=4\sqrt{\frac{L}{g}}∫^{π/2}_0\left(1+\frac{1}{2}k^2\sin^2θ+\frac{3}{8}k^4\sin^4θ+⋯\right)\,dθ\) to obtain an improved estimate for \(T\).

    Answer:
    \(\displaystyle ∫^{π/2}_0\sin^4θ\, dθ=\frac{3π}{16}.\) Hence \(T≈2π\sqrt{\frac{L}{g}}\left(1+\frac{k^2}{4}+\frac{9}{256}k^4\right).\)

    79) [T] An equivalent formula for the period of a pendulum with amplitude \(\displaystyle θ_{max}\) is \(T(θ_{max})=2\sqrt{2}\sqrt{\frac{L}{g}}∫^{θ_{max}}_0\frac{dθ}{\sqrt{\cos θ}−\cos(θ_{max})}\) where \(L\) is the pendulum length and \(g\) is the gravitational acceleration constant. When \(θ_{max}=\frac{π}{3}\) we get \(\dfrac{1}{\sqrt{\cos t−1/2}}≈\sqrt{2}\left(1+\frac{t^2}{2}+\frac{t^4}{3}+\frac{181t^6}{720}\right)\). Integrate this approximation to estimate \(T(\frac{π}{3})\) in terms of \(L\) and \(g\). Assuming \(g=9.806\) meters per second squared, find an approximate length \(L\) such that \(T(\frac{π}{3})=2\) seconds.

     

    Contributors

    Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

     

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