10.2E: Exercises for Section 10.2
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1) If f(x)=∞∑n=0xnn! and g(x)=∞∑n=0(−1)nxnn!, find the power series of 12(f(x)+g(x)) and of 12(f(x)−g(x)).
- Answer
- 12(f(x)+g(x))=∞∑n=0x2n(2n)! and 12(f(x)−g(x))=∞∑n=0x2n+1(2n+1)!.
2) If C(x)=∞∑n=0x2n(2n)! and S(x)=∞∑n=0x2n+1(2n+1)!, find the power series of C(x)+S(x) and of C(x)−S(x).
In exercises 3 - 6, use partial fractions to find the power series of each function.
3) 4(x−3)(x+1)
- Answer
- 4(x−3)(x+1)=1x−3−1x+1=−13(1−x3)−11−(−x)=−13∞∑n=0(x3)n−∞∑n=0(−1)nxn=∞∑n=0((−1)n+1−13n+1)xn
4) 3(x+2)(x−1)
5) 5(x2+4)(x2−1)
- Answer
- 5(x2+4)(x2−1)=1x2−1−1411+(x2)2=−∞∑n=0x2n−14∞∑n=0(−1)n(x2)n=∞∑n=0((−1)+(−1)n+112n+2)x2n
6) 30(x2+1)(x2−9)
In exercises 7 - 10, express each series as a rational function.
7) ∞∑n=11xn
- Answer
- 1x∞∑n=01xn=1x⋅11−1x=1x−1
8) ∞∑n=11x2n
9) ∞∑n=11(x−3)2n−1
- Answer
- 1x−3⋅11−1(x−3)2=x−3(x−3)2−1
10) ∞∑n=1(1(x−3)2n−1−1(x−2)2n−1)
Exercises 11 - 16 explore applications of annuities.
11) Calculate the present values P of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of r=0.03,r=0.05, and r=0.07.
- Answer
- P=P1+⋯+P20 where Pk=10,0001(1+r)k. Then P=10,00020∑k=11(1+r)k=10,0001−(1+r)−20r. When r=0.03,P≈10,000×14.8775=148,775. When r=0.05,P≈10,000×12.4622=124,622. When r=0.07,P≈105,940.
12) Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of r=0.03,r=0.05 and r=0.07.
13) Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of r=0.03,r=0.05, and r=0.07.
- Answer
- In general, P=C(1−(1+r)−N)r for N years of payouts, or C=Pr1−(1+r)−N. For N=20 and P=100,000, one has C=6721.57 when r=0.03;C=8024.26 when r=0.05; and C≈9439.29 when r=0.07.
14) Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of r=0.03,r=0.05, and r=0.07.
15) Suppose that an annuity has a present value P=1 million dollars. What interest rate r would allow for perpetual annual payouts of $50,000?
- Answer
- In general, P=Cr. Thus, r=CP=5×104106=0.05.
16) Suppose that an annuity has a present value P=10 million dollars. What interest rate r would allow for perpetual annual payouts of $100,000?
In exercises 17 - 20, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.
17) x+x2−x3+x4+x5−x6+⋯ (Hint: Group powers x3k,x3k−1, and x3k−2.)
- Answer
- (x+x2−x3)(1+x3+x6+⋯)=x+x2−x31−x3
18) x+x2−x3−x4+x5+x6−x7−x8+⋯ (Hint: Group powers x4k,x4k−1, etc.)
19) x−x2−x3+x4−x5−x6+x7−⋯ (Hint: Group powers x3k,x3k−1, and x3k−2.)
- Answer
- (x−x2−x3)(1+x3+x6+⋯)=x−x2−x31−x3
20) x2+x24−x38+x416+x532−x664+⋯ (Hint: Group powers (x2)3k,(x2)3k−1, and (x2)3k−2.)
In exercises 21 - 24, find the power series of f(x)g(x) given f and g as defined.
21) f(x)=2∞∑n=0xn,g(x)=∞∑n=0nxn
- Answer
- an=2,bn=n so cn=n∑k=0bkan−k=2n∑k=0k=(n)(n+1) and f(x)g(x)=∞∑n=1n(n+1)xn
22) f(x)=∞∑n=1xn,g(x)=∞∑n=11nxn. Express the coefficients of f(x)g(x) in terms of Hn=n∑k=11k.
23) f(x)=g(x)=∞∑n=1(x2)n
- Answer
- an=bn=2−n so cn=n∑k=1bkan−k=2−nn∑k=11=n2n and f(x)g(x)=∞∑n=1n(x2)n
24) f(x)=g(x)=∞∑n=1nxn
In exercises 25 - 26, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f.
25) f(x)=11+x=∞∑n=0(−1)nxn
- Answer
- The derivative of f is −1(1+x)2=−∞∑n=0(−1)n(n+1)xn.
26) f(x)=11−x2=∞∑n=0x2n
In exercises 27 - 28, integrate the given series expansion of f term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of f.
27) f(x)=2x(1+x2)2=∞∑n=1(−1)n(2n)x2n−1
- Answer
- The indefinite integral of f is 11+x2=∞∑n=0(−1)nx2n.
28) f(x)=2x1+x2=2∞∑n=0(−1)nx2n+1
In exercises 29 - 32, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.
29) Evaluate ∞∑n=1n2n as f′(12) where f(x)=∞∑n=0xn.
- Answer
- f(x)=∞∑n=0xn=11−x;f′(12)=∞∑n=1n2n−1=ddx(1−x)−1|x=1/2=1(1−x)2|x=1/2=4 so ∞∑n=1n2n=2.
30) Evaluate ∞∑n=1n3n as f′(13) where f(x)=∞∑n=0xn.
31) Evaluate ∞∑n=2n(n−1)2n as f″(12) where f(x)=∞∑n=0xn.
- Answer
- f(x)=∞∑n=0xn=11−x;f″(12)=∞∑n=2n(n−1)2n−2=d2dx2(1−x)−1|x=1/2=2(1−x)3|x=1/2=16 so ∞∑n=2n(n−1)2n=4.
32) Evaluate ∞∑n=0(−1)nn+1 as ∫10f(t)dt where f(x)=∞∑n=0(−1)nx2n=11+x2.
In exercises 33 - 39, given that 11−x=∞∑n=0xn, use term-by-term differentiation or integration to find power series for each function centered at the given point.
33) f(x)=lnx centered at x=1 (Hint: x=1−(1−x))
- Answer
- ∫∑(1−x)ndx=∫∑(−1)n(x−1)ndx=∑(−1)n(x−1)n+1n+1
34) ln(1−x) at x=0
35) ln(1−x2) at x=0
- Answer
- −∫x2t=011−tdt=−∞∑n=0∫x20tndx−∞∑n=0x2(n+1)n+1=−∞∑n=1x2nn
36) f(x)=2x(1−x2)2 at x=0
37) f(x)=tan−1(x2) at x=0
- Answer
- ∫x20dt1+t2=∞∑n=0(−1)n∫x20t2ndt=∞∑n=0(−1)nt2n+12n+1∣x2t=0=∞∑n=0(−1)nx4n+22n+1
38) f(x)=ln(1+x2) at x=0
39) f(x)=∫x0lntdt where ln(x)=∞∑n=1(−1)n−1(x−1)nn
- Answer
- Term-by-term integration gives ∫x0lntdt=∞∑n=1(−1)n−1(x−1)n+1n(n+1)=∞∑n=1(−1)n−1(1n−1n+1)(x−1)n+1=(x−1)lnx+∞∑n=2(−1)n(x−1)nn=xlnx−x.
40) [T] Evaluate the power series expansion ln(1+x)=∞∑n=1(−1)n−1xnn at x=1 to show that ln(2) is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate ln(2) accurate to within 0.001, and find such an approximation.
41) [T] Subtract the infinite series of ln(1−x) from ln(1+x) to get a power series for ln(1+x1−x). Evaluate at x=13. What is the smallest N such that the Nth partial sum of this series approximates ln(2) with an error less than 0.001?
- Answer
- We have ln(1−x)=−∞∑n=1xnn so ln(1+x)=∞∑n=1(−1)n−1xnn. Thus, ln(1+x1−x)=∞∑n=1(1+(−1)n−1)xnn=2∞∑n=1x2n−12n−1. When x=13 we obtain ln(2)=2∞∑n=1132n−1(2n−1). We have 23∑n=1132n−1(2n−1)=0.69300…, while 24∑n=1132n−1(2n−1)=0.69313… and ln(2)=0.69314…; therefore, N=4.
In exercises 42 - 45, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.
42) ∞∑k=0(xk−x2k+1)
43) ∞∑k=1x3k6k
- Answer
- ∞∑k=1xkk=−ln(1−x) so ∞∑k=1x3k6k=−16ln(1−x3). The radius of convergence is equal to 1 by the ratio test.
44) ∞∑k=1(1+x2)−k using y=11+x2
45) ∞∑k=12−kx using y=2−x
- Answer
- If y=2−x, then ∞∑k=1yk=y1−y=2−x1−2−x=12x−1. If ak=2−kx, then ak+1ak=2−x<1 when x>0. So the series converges for all x>0.
46) Show that, up to powers x3 and y3, E(x)=∞∑n=0xnn! satisfies E(x+y)=E(x)E(y).
47) Differentiate the series E(x)=∞∑n=0xnn! term-by-term to show that E(x) is equal to its derivative.
- Answer
- Answers will vary.
48) Show that if f(x)=∞∑n=0anxn is a sum of even powers, that is, an=0 if n is odd, then F=∫x0f(t)dt is a sum of odd powers, while if I is a sum of odd powers, then F is a sum of even powers.
49) [T] Suppose that the coefficients an of the series ∞∑n=0anxn are defined by the recurrence relation an=an−1n+an−2n(n−1). For a0=0 and a1=1, compute and plot the sums SN=N∑n=0anxn for N=2,3,4,5 on [−1,1].
- Answer
-
The solid curve is S5. The dashed curve is S2, dotted is S3, and dash-dotted is S4
50) [T] Suppose that the coefficients an of the series ∞∑n=0anxn are defined by the recurrence relation an=an−1√n−an−2√n(n−1). For a0=1 and a1=0, compute and plot the sums SN=N∑n=0anxn for N=2,3,4,5 on [−1,1].
51) [T] Given the power series expansion ln(1+x)=∞∑n=1(−1)n−1xnn, determine how many terms N of the sum evaluated at x=−1/2 are needed to approximate ln(2) accurate to within 1/1000. Evaluate the corresponding partial sum N∑n=1(−1)n−1xnn.
- Answer
- When x=−12,−ln(2)=ln(12)=−∞∑n=11n2n. Since ∞∑n=111n2n<∞∑n=1112n=1210, one has 10∑n=11n2n=0.69306… whereas ln(2)=0.69314…; therefore, N=10.
52) [T] Given the power series expansion tan−1(x)=∞∑k=0(−1)kx2k+12k+1, use the alternating series test to determine how many terms N of the sum evaluated at x=1 are needed to approximate \tan^{−1}(1)=\frac{π}{4} accurate to within 1/1000. Evaluate the corresponding partial sum \displaystyle \sum_{k=0}^N(−1)^k\frac{x^{2k+1}}{2k+1}.
53) [T] Recall that \tan^{−1}\left(\frac{1}{\sqrt{3}}\right)=\frac{π}{6}. Assuming an exact value of \frac{1}{\sqrt{3}}), estimate \frac{π}{6} by evaluating partial sums S_N\left(\frac{1}{\sqrt{3}}\right) of the power series expansion \displaystyle \tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1} at x=\frac{1}{\sqrt{3}}. What is the smallest number N such that 6S_N\left(\frac{1}{\sqrt{3}}\right) approximates π accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?
- Answer
- \displaystyle 6S_N\left(\frac{1}{\sqrt{3}}\right)=2\sqrt{3}\sum_{n=0}^N(−1)^n\frac{1}{3^n(2n+1).} One has π−6S_4\left(\frac{1}{\sqrt{3}}\right)=0.00101… and π−6S_5\left(\frac{1}{\sqrt{3}}\right)=0.00028… so N=5 is the smallest partial sum with accuracy to within 0.001. Also, π−6S_7\left(\frac{1}{\sqrt{3}}\right)=0.00002… while π−6S_8\left(\frac{1}{\sqrt{3}}\right)=−0.000007… so N=8 is the smallest N to give accuracy to within 0.00001.
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.