5.4E: Exercises for Section 5.4
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( \newcommand{\kernel}{\mathrm{null}\,}\)
In exercises 1 - 4, use appropriate substitutions to write down the Maclaurin series for the given binomial.
1) (1−x)1/3
2) (1+x2)−1/3
- Answer
- (1+x2)−1/3=∞∑n=0(n−13)x2n
3) (1−x)1.01
4) (1−2x)2/3
- Answer
- (1−2x)2/3=∞∑n=0(−1)n2n(n23)xn
In exercises 5 - 12, use the substitution (b+x)r=(b+a)r(1+x−ab+a)r in the binomial expansion to find the Taylor series of each function with the given center.
5) √x+2 at a=0
6) √x2+2 at a=0
- Answer
- √2+x2=∞∑n=02(1/2)−n(n12)x2n;(|x2|<2)
7) √x+2 at a=1
8) √2x−x2 at a=1 (Hint: 2x−x2=1−(x−1)2)
- Answer
- √2x−x2=√1−(x−1)2 so √2x−x2=∞∑n=0(−1)n(n12)(x−1)2n
9) (x−8)1/3 at a=9
10) √x at a=4
- Answer
- √x=2√1+x−44 so √x=∞∑n=021−2n(n12)(x−4)n
11) x1/3 at a=27
12) √x at x=9
- Answer
- √x=∞∑n=031−3n(n12)(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
- 10(1+x1000)1/3=∞∑n=0101−3n(13n)xn. Using, for example, a fourth-degree estimate at x=1 gives (1001)1/3≈10(1+(113)10−3+(213)10−6+(313)10−9+(313)10−12)=10(1+13.103−19.106+581.109−10243.1012)=10.00333222... whereas (1001)1/3=10.00332222839093.... Two terms would suffice for three-digit accuracy.
In exercises 15 - 18, use the binomial approximation √1−x≈1−x2−x28−x316−5x4128−7x5256 for |x|<1 to approximate each number. Compare this value to the value given by a scientific calculator.
15) [T] 1√2 using x=12 in (1−x)1/2
16) [T] √5=5×1√5 using x=45 in (1−x)1/2
- Answer
- The approximation is 2.3152; the CAS value is 2.23….
17) [T] √3=3√3 using x=23 in (1−x)1/2
18) [T] √6 using x=56 in (1−x)1/2
- Answer
- The approximation is 2.583…; the CAS value is 2.449….
19) Integrate the binomial approximation of √1−x to find an approximation of ∫x0√1−tdt.
20) [T] Recall that the graph of √1−x2 is an upper semicircle of radius 1. Integrate the binomial approximation of √1−x2 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.
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)!}.
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.
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.