4.4E: Exercises

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

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

3) \((1−x)^{1.01}\)

4) \((1−2x)^{2/3}\)

Answer

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

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

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

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 \(\displaystyle \sqrt{2x−x^2}=\sum_{n=0}^ \infty (−1)^n\left(n^{\frac{1}{2}}\right)(x−1)^{2n}\)

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 \(\displaystyle \sqrt{x}=\sum_{n=0}^ \infty 2^{1−2n}\left(n^{\frac{1}{2}}\right)(x−4)^n\)

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

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

Answer

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

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

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

Answer

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

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

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

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

Answer

The approximation is \(2.3152\); the CAS value is \(2.23 \ldots .\)

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

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

Answer

The approximation is \(2.583 \ldots \); the CAS value is \(2.449 \ldots .\)

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

20) [Technology Required] 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 \(\frac{ \pi }{2}\).

Answer

\(\sqrt{1−x^2}=1−\frac{x^2}{2}−\frac{x^4}{8}−\frac{x^6}{16}−\frac{5x^8}{128}+ \cdots .\) Thus \(\displaystyle \int ^1_{−1}\sqrt{1−x^2}\,dx=\left[x−\frac{x^3}{6}−\frac{x^5}{40}−\frac{x^7}{7 \cdot 16}−\frac{5x^9}{9 \cdot 128}+ \cdots \right]\Big|^1_{−1} \approx 2−\frac{1}{3}−\frac{1}{20}−\frac{1}{56}−\frac{10}{9 \cdot 128}+error=1.590...\) whereas \(\frac{ \pi }{2}=1.570...\)

In exercises 21 - 24, use the expansion \((1+x)^{1/3}=1+\frac{1}{3}x−\frac{1}{9}x^2+\frac{5}{81}x^3−\frac{10}{243}x^4+ \cdots \) 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

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

23) \((3+2x)^{1/3};\;a=−1\)

24) \((x^2+6x+10)^{1/3};\;a=−3\)

Answer

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

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

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

Answer

Twice the approximation is \(1.260 \ldots \) 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

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

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

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

Answer

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

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

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

Answer

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

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

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

In exercises 37 - 44, find the Maclaurin series of \(\displaystyle F(x)= \int ^x_0f(t)\,dt\) 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) \(\displaystyle F(x)= \int ^x_0e^{−t^2}\,dt;\; f(t)=e^{−t^2}=\sum_{n=0}^ \infty (−1)^n\frac{t^{2n}}{n!}\)

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

Answer

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

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

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

Answer

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

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

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

Answer

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

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

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

Answer

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

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

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

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

Answer

\(x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+ \cdots \)

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

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

Answer

\(1+x−\dfrac{x^3}{3}−\dfrac{x^4}{6}+ \cdots \)

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

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

Answer

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

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

\(\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^ \infty (−1)^nx^{2n}\) so \(R=1\) by the Ratio Test.

55) \(\tan^{−1}x\)

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

Answer

\(\displaystyle \ln(1+x^2)=\sum_{n=1}^ \infty \frac{(−1)^{n−1}}{n}x^{2n}\) 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 \(\displaystyle \cosh x=\sum_{n=0}^ \infty \frac{x^{2n}}{(2n)!}\).

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

60) [Technology Required] 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 \(\left(−\frac{ \pi }{4},\frac{ \pi }{4}\right)\).

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 \(N \geq 3\). 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 \(p_7−\tan x\).

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

Answer

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

63) [Technology Required] Suppose that a set of standardized test scores is normally distributed with mean \( \mu =100\) and standard deviation \( \Sigma =10\). 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 \(\frac{1}{\sqrt{2 \pi }}e^{−x^2/2}\) to estimate this probability.

Answer

The probability is \(\displaystyle p=\frac{1}{\sqrt{2 \pi }} \int ^{(b− \mu )/ \Sigma }_{(a− \mu )/ \Sigma }e^{−x^2/2}\,dx\) where \(a=90\) and \(b=100\), that is, \(\displaystyle p=\frac{1}{\sqrt{2 \pi }} \int ^1_{−1}e^{−x^2/2}\,dx=\frac{1}{\sqrt{2 \pi }} \int ^1_{−1}\sum_{n=0}^5(−1)^n\frac{x^{2n}}{2^nn!}\,dx=\frac{2}{\sqrt{2 \pi }}\sum_{n=0}^5(−1)^n\frac{1}{(2n+1)2^nn!} \approx 0.6827.\)

64) [Technology Required] Suppose that a set of standardized test scores is normally distributed with mean \( \mu =100\) and standard deviation \( \Sigma =10\). 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 \(\frac{1}{\sqrt{2 \pi }}e^{−x^2/2}\) to estimate this probability.

65) [Technology Required] Suppose that \(\displaystyle \sum_{n=0}^ \infty a_nx^n\) converges to a function \(f(x)\) such that \(f(0)=1,\, f^{\prime}(0)=0\), 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 \(a_{2n}=\dfrac{(−1)^n}{(2n)!}\).

66) [Technology Required] Suppose that \(\displaystyle \sum_{n=0}^ \infty a_nx^n\) converges to a function \(f(x)\) such that \(f(0)=0,\; f^{\prime}(0)=1\), and \(f''(x)=−f(x)\). Find a formula for an and plot the partial sum \(S_N\) for \(N=10\) on \([−5,5]\).

The error in approximating the integral \(\displaystyle \int ^b_af(t)\, dt\) by that of a Taylor approximation \(\displaystyle \int ^b_aPn(t) \,dt\) is at most \(\displaystyle \int ^b_aR_n(t) \,dt\). In exercises 67 - 68, the Taylor remainder estimate \(R_n \leq \frac{M}{(n+1)!}|x−a|^{n+1}\) 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.

67) [Technology Required] \(\displaystyle \int ^ \pi _0\frac{\sin t}{t}\, dt;\quad P_s=1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!}\) (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 \(R_s \leq \frac{0.1}{(9)!} \pi ^9 \approx 0.0082<0.01.\) We have \(\displaystyle \int ^ \pi _0\left(1−\frac{x^2}{3!}+\frac{x^4}{5!}−\frac{x^6}{7!}+\frac{x^8}{9!}\right)\,dx= \pi −\frac{ \pi ^3}{3 \cdot 3!}+\frac{ \pi ^5}{5 \cdot 5!}−\frac{ \pi ^7}{7 \cdot 7!}+\frac{ \pi ^9}{9 \cdot 9!}=1.852...,\) whereas \(\displaystyle \int ^ \pi _0\frac{\sin t}{t}\,dt=1.85194...\), so the actual error is approximately \(0.00006.\)

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

The following exercises (69-70) deal with Fresnel integrals.

69) The Fresnel integrals are defined by \(\displaystyle C(x)= \int ^x_0\cos(t^2)\,dt\) and \(\displaystyle S(x)= \int ^x_0\sin(t^2)\,dt\). 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 \([0,2 \pi ]\).

Answer

Since \(\displaystyle \cos(t^2)=\sum_{n=0}^ \infty (−1)^n\frac{t^{4n}}{(2n)!}\) and \(\displaystyle \sin(t^2)=\sum_{n=0}^ \infty (−1)^n\frac{t^{4n+2}}{(2n+1)!}\), one has \(\displaystyle S(x)=\sum_{n=0}^ \infty (−1)^n\frac{x^{4n+3}}{(4n+3)(2n+1)!}\) and \(\displaystyle C(x)=\sum_{n=0}^ \infty (−1)^n\frac{x^{4n+1}}{(4n+1)(2n)!}\). 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.

70) [Technology Required] 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 \(0 \leq t \leq 2 \pi \), the coordinates of which were computed in the previous exercise.

71) Estimate \(\displaystyle \int ^{1/4}_0\sqrt{x−x^2}\,dx\) 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

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

72) [Technology Required] Use Newton’s approximation of the binomial \(\sqrt{1−x^2}\) to approximate \( \pi \) 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 \(\frac{ \pi }{24}\). 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 \(y=\sqrt{x}\sqrt{1−x}=\sqrt{x} \times (\text{binomial expansion of} \sqrt{1−x})\) and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate \( \pi \).

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

Answer

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

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

75) Evaluate \(\displaystyle \int ^{ \pi /2}_0\sin^4 \theta \,d \theta \) in the approximation \(\displaystyle T=4\sqrt{\frac{L}{g}} \int ^{ \pi /2}_0\left(1+\frac{1}{2}k^2\sin^2 \theta +\frac{3}{8}k^4\sin^4 \theta + \cdots \right)\,d \theta \) to obtain an improved estimate for \(T\).

Answer

\(\displaystyle \int ^{ \pi /2}_0\sin^4 \theta \, d \theta =\frac{3 \pi }{16}.\) Hence \(T \approx 2 \pi \sqrt{\frac{L}{g}}\left(1+\frac{k^2}{4}+\frac{9}{256}k^4\right).\)

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