
# 10.4E: Exercises for Section 10.4


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

1) $$(1−x)^{1/3}$$

2) $$(1+x^2)^{−1/3}$$

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

3) $$(1−x)^{1.01}$$

4) $$(1−2x)^{2/3}$$

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

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 $$a=0$$

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

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

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

$$\sqrt{2x−x^2}=\sqrt{1−(x−1)^2}$$ 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 $$a=4$$

$$\sqrt{x}=2\sqrt{1+\frac{x−4}{4}}$$ 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 $$x=9$$

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

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 $$(1000+x)^{1/3}$$

$$\displaystyle 10(1+\frac{x}{1000})^{1/3}=\sum_{n=0}^∞10^{1−3n}(^{\frac{1}{3}}_n)x^n$$. 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 $$(1−x)^{1/2}$$

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 $$(1−x)^{1/2}$$

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 $$\frac{π}{2}$$.

$$\sqrt{1−x^2}=1−\frac{x^2}{2}−\frac{x^4}{8}−\frac{x^6}{16}−\frac{5x^8}{128}+⋯.$$ 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) $$(1+4x)^{4/3};\;a=0$$

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

23) $$(3+2x)^{1/3};\;a=−1$$

24) $$(x^2+6x+10)^{1/3};\;a=−3$$

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

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 $$2^{1/3}=2.2^{−2/3}$$.

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 $$f(x)=(1+x^4)^{25}$$.

$$f^{(99)}(0)=0$$

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

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

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

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

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

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

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) $$f(x)=e^{x^3}$$

$$\displaystyle e^{x^3}=\sum_{n=0}^∞\frac{x^{3n}}{n!}$$

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 $$\sin^2x=\frac{1}{2}−\frac{1}{2}\cos(2x)$$

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

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

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

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

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

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

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

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

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

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

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) $$f(x)=\tan x$$

$$x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+⋯$$

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

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

$$1+x−\dfrac{x^3}{3}−\dfrac{x^4}{6}+⋯$$

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

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

$$1+x^2+\dfrac{2x^4}{3}+\dfrac{17x^6}{45}+⋯$$

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

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

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) $$\dfrac{1}{1+x^2}$$

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

55) $$\tan^{−1}x$$

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

$$\displaystyle \ln(1+x^2)=\sum_{n=1}^∞\frac{(−1)^{n−1}}{n}x^{2n}$$ 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 $$\cosh x=\dfrac{e^x+e^{−x}}{2}$$.

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 $$\left(−\frac{π}{4},\frac{π}{4}\right)$$.

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 $$p_7−\tan x$$.

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 $$x^2y''$$.

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.

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 $$[−5,5].$$

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 $$a_{2n}=\dfrac{(−1)^n}{(2n)!}$$.

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 $$a_0,...,a_5$$.

$$\displaystyle y''=\sum_{n=0}^∞(n+2)(n+1)a_{n+2}x^n$$ 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 $$0.1$$.)

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 $$[0,2π]$$.

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.

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 $$\displaystyle 1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{2128}−\frac{7x^5}{256}$$.

$$\displaystyle ∫^{1/4}_0\sqrt{x}\left(1−\frac{x}{2}−\frac{x^2}{8}−\frac{x^3}{16}−\frac{5x^4}{128}−\frac{7x^5}{256}\right)\,dx =\frac{2}{3}2^{−3}−\frac{1}{2}\frac{2}{5}2^{−5}−\frac{1}{8}\frac{2}{7}2^{−7}−\frac{1}{16}\frac{2}{9}2^{−9}−\frac{5}{128}\frac{2}{11}2^{−11}−\frac{7}{256}\frac{2}{13}2^{−13}=0.0767732...$$ 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 $$T≈2π\sqrt{\frac{L}{g}}$$.

$$T≈2π\sqrt{\frac{10}{9.8}}\left(1+\frac{\sin^2(θ/12)}{4}\right)≈6.453$$ 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 $$T$$.

$$\displaystyle ∫^{π/2}_0\sin^4θ\, dθ=\frac{3π}{16}.$$ 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.