10.E: POWER SERIES (EXERCISES)fliacob/An1/2018-2019/Resurse...16) Suppose that an annuity has a present value million dollars. What interest rate r would allow for perpetual annual

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10.E: POWER SERIES (EXERCISES)

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10.1: POWER SERIES AND FUNCTIONSIn the following exercises, state whether each statement is true, or give an example to show that it is false.

1) If converges, then as

Solution:True. If a series converges then its terms tend to zero.

2) converges at for any real numbers .

3) Given any sequence , there is always some , possibly very small, such that converges on .

Solution: False. It would imply that for . If , then does not tend to zero for any .

4) If has radius of convergence and if for all , then the radius of convergence of is greater than

or equal to .

5) Suppose that converges at . At which of the following points must the series also converge? Use the fact that

if converges at , then it converges at any point closer to than .

a.

b.

c.

d.

e.

f.

Solution: It must converge on and hence at: a. ; b. ; c. ; d. ; e. ; and f. .

6) Suppose that converges at . At which of the following points must the series also converge? Use the fact

that if converges at , then it converges at any point closer to than .

a.

b.

c.

d.

e.

f.

∑n=1

∞

anxn → 0anx

n n → ∞.

∑n=1

∞

anxn x = 0 an

an R > 0 ∑n=1

∞

anxn (−R,R)

→ 0anxn |x| < R =an n

n = (nxanxn )n x ≠ 0

∑n=1

∞

anxn R > 0 | | ≤ | |bn an n ∑

n=1

∞

bnxn

R

(x − 3∑n=0

∞

an )n x = 6

∑ (x − can )n x c x

x = 1

x = 2

x = 3

x = 0

x = 5.99

x = 0.000001

(0, 6] x = 1 x = 2 x = 3 x = 0 x = 5.99 x = 0.000001

(x + 1∑n=0

∞

an )n x = −2

∑ (x − can )n x c x

x = 2

x = −1

x = −3

x = 0

x = 0.99

x = 0.000001

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In the following exercises, suppose that as Find the radius of convergence for each series.

7)

Solution: so

8)

9)

Solution: so

10)

11)

Solution: so

12)

In the following exercises, find the radius of convergence and interval of convergence for with the given coefficients .

13)

Solution: so . so . When the series is harmonic and diverges. When the series is

alternating harmonic and converges. The interval of convergence is .

14)

15)

Solution: so so . When the series diverges by the divergence test. The interval of convergence is

.

16)

17)

Soluton: so . When the series diverges by the divergence test. The interval of convergence is

∣→ 1∣∣∣an+1

ann → ∞.

∑n=0

∞

an2nxn

∣= 2|x| ∣ ∣→ 2|x|∣

∣∣an+12n+1xn+1

an2nxn

an+1

anR =

1

2

∑n=0

∞anx

n

2n

∑n=0

∞anπ

nxn

en

∣= ∣ ∣→∣

∣∣

(an+1πe

)n+1xn+1

(anπe )nxn

π|x|

e

an+1

an

π|x|

eR =

e

π

sum∞n=0

(−1an )nxn

10n

(−1∑n=0

∞

an )nx2n

∣=∣ ∣∣ ∣→∣∣

∣∣

(−1an+1 )n+1x2n+2

(−1an )nx2nx2 an+1

anx2

∣

∣∣ R = 1

(−4∑n=0

∞

an )nx2n

R ∑anxn an

∑n=1

∞ (2x)n

n

=an2n

n→ 2x

xan+1

anR =

1

2x =

1

2x = −

1

2

I = [− , )1

2

1

2

(−1∑n=1

∞

)nxn

n−−√

∑n=1

∞nxn

2n

=ann

2n→

xan+1

an

x

2R = 2 x = ±2

I = (−2, 2)

∑n=1

∞nxn

en

∑n=1

∞n2xn

2n

=ann2

2nR = 2 x = ±2 I = (−2, 2).



18)

19)

Solution: so . When the series is an absolutely convergent p-series. The interval of convergence is

20)

21)

Solution: so the series converges for all by the ratio test and .

22)

In the following exercises, find the radius of convergence of each series.

23)

Solution: so so

24)

25)

Solution: so so

26)

27) where

Solution: so so

28)

In the following exercises, use the ratio test to determine the radius of convergence of each series.

29)

∑k=1

∞kexk

ek

∑k=1

∞πkxk

kπ

=akπk

kπR =

1

πx = ±

1

π

I = [− , ].1

π

1

π

∑n=1

∞xn

n!

∑n=1

∞ 10nxn

n!

= , = → 0 < 1an10n

n!

xan+1

an

10x

n + 1x I = (−∞, ∞)

(−1∑n=1

∞

)nxn

ln(2n)

∑k=1

∞ (k!)2xk

(2k)!

=ak(k!)2

(2k)!= →

ak+1

ak

(k + 1)2

(2k + 2)(2k + 1)

1

4R = 4

∑n=1

∞ (2n)!xn

n2n

∑k=1

∞k!

1 ⋅ 3 ⋅ 5 ⋯ (2k − 1)xk

=akk!

1 ⋅ 3 ⋅ 5 ⋯ (2k − 1)= →

ak+1

ak

k + 1

2k + 1

1

2R = 2

∑k=1

∞ 2 ⋅ 4 ⋅ 6 ⋯ 2k

(2k)!xk

∑n=1

∞xn

)(2nn

) =(nkn!

k!(n − k)!

=an1

)(2nn

= = →an+1

an

((n + 1)!)2

(2n + 2)!

2n!

(n!)2

(n + 1)2

(2n + 2)(2n + 1)

1

4R = 4

si n∑n=1

∞

n2 xn

∑n=1

∞ (n!)3

(3n)!xn



Solution: so

30)

31)

Solution: so so

32)

In the following exercises, given that with convergence in , find the power series for each function with the

given center a, and identify its interval of convergence.

33) (Hint:

Solution: on

34)

35)

Solution: on

36)

37)

Solution: on

38)

39)

Solution: on

40)

41)

Solution: on

= →an+1

an

(n + 1)3

(3n + 3)(3n + 2)(3n + 1)

1

27R = 27

∑n=1

∞ (n!23n )3

(3n)!xn

∑n=1

∞n!

nnxn

=ann!

nn= = ( →

an+1

an

(n + 1)!

n!

nn

(n + 1)n+1

n

n + 1)n

1

eR = e

∑n=1

∞ (2n)!

n2nxn

=1

1 − x∑n=0

∞

xn (−1, 1)

f(x) = ;a = 11

x= )

1

x

1

1 − (1 − x)

f(x) = (1 − x∑n=0

∞

)n I = (0, 2)

f(x) = ;a = 01

1 − x2

f(x) = ;a = 0x

1 − x2

∑n=0

∞

x2n+1 I = (−1, 1)

f(x) = ;a = 01

1 + x2

f(x) = ;a = 0x2

1 + x2

(−1∑n=0

∞

)nx2n+2I = (−1, 1)

f(x) = ;a = 11

2 − x

f(x) = ;a = 0.1

1 − 2x

∑n=0

∞

2nxn (− , )1

2

1

2

f(x) = ;a = 01

1 − 4x2

f(x) = ;a = 0x2

1 − 4x2

∑n=0

∞

4nx2n+2 (− , )1

2

1

2



42)

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

43) Explain why, if then whenever and, therefore, the radius of convergence of

is .

Solution: as and when . Therefore, converges when by the

nth root test.

44)

45)

Solution: so so

46)

47)

Solution: so so

48) Suppose that such that if is even. Explain why

49) Suppose that such that if is odd. Explain why

Solution: We can rewrite and since .

50) Suppose that converges on . Find the interval of convergence of .

51) Suppose that converges on . Find the interval of convergence of .

Solution: If then so converges.

In the following exercises, suppose that satisfies where for each . State whether each series

converges on the full interval , or if there is not enough information to draw a conclusion. Use the comparison test whenappropriate.

52)

f(x) = ;a = 2x2

5 − 4x + x2

| → r > 0,an|1/n | → |x|r < 1anxn|1/n |x| <

1

r

∑n=1

∞

anxn R =

1

r

| = | |x| → |x|ranxn|1/n

an|1/nn → ∞ |x|r < 1 |x| <

1

r∑n=1

∞

anxn |x| <

1

r

∑n=1

∞xn

nn

(∑k=1

∞k − 1

2k + 3)kxk

= (akk − 1

2k + 3)k ( → < 1ak)1/k 1

2R = 2

(∑k=1

∞ 2 − 1k2

+ 3k2)kxk

= ( − 1∑n=1

∞

an n1/n )nxn

= ( − 1an n1/n )n ( → 0an)1/n R = ∞

p(x) =∑n=0

∞

anxn = 0an n p(x) = p(−x).

p(x) =∑n=0

∞

anxn = 0an n p(x) = −p(−x).

p(x) =∑n=0

∞

a2n+1x2n+1 p(x) = p(−x) = −(−xx2n+1 )2n+1

p(x) =∑n=0

∞

anxn (−1, 1] p(Ax)

p(x) =∑n=0

∞

anxn (−1, 1] p(2x − 1)

x ∈ [0, 1], y = 2x − 1 ∈ [−1, 1] p(2x − 1) = p(y) =∑n=0

∞

anyn

p(x) =∑n=0

∞

anxn = 1lim

n→∞

an+1

an≥ 0an n

(−1, 1)

∑n=0

∞

anx2n



53)

Solution: Converges on by the ratio test

54) (Hint: )

55) (Hint: Let if for some , otherwise .)

Solution: Consider the series where if and otherwise. Then and so the series converges on by the comparison test.

56) Suppose that is a polynomial of degree . Find the radius and interval of convergence of .

57) [T] Plot the graphs of and of the partial sums for on the interval . Comment on

the approximation of by near and near as increases.

Solution: The approximation is more accurate near . The partial sums follow more closely as increases but are never

accurate near since the series diverges there.

58) [T] Plot the graphs of and of the partial sums for on the interval .

Comment on the behavior of the sums near and near as increases.

59) [T] Plot the graphs of the partial sums for on the interval . Comment on the behavior of

the sums near and near as increases.

Solution: The approximation appears to stabilize quickly near both .

∑n=0

∞

a2nx2n

(−1, 1)

∑n=0

∞

a2nxn x = ± x

2−−√

∑n=0

∞

an2xn2

=bk ak k = n2 n = 0bk

∑ bkxk =bk ak k = n2 = 0bk ≤bk ak

(−1, 1)

p(x) N p(n)∑n=1

∞

xn

1

1 − x=SN ∑

n=0

N

xn n = 10, 20, 30 [−0.99, 0.99]

1

1 − xSN x = −1 x = 1 N

x = −11

1 − xN

x = 1

−ln(1 − x) =SN ∑n=1

Nxn

nn = 10, 50, 100 [−0.99, 0.99]

x = −1 x = 1 N

=Sn ∑n=1

Nxn

n2n = 10, 50, 100 [−0.99, 0.99]

x = −1 x = 1 N

x = ±1



60) [T] Plot the graphs of the partial sums for on the interval . Comment on the

behavior of the sums near and near as increases.

61) [T] Plot the graphs of the partial sums for on the interval . Comment on how

these plots approximate as increases.

Solution: The polynomial curves have roots close to those of up to their degree and then the polynomials diverge from .

62) [T] Plot the graphs of the partial sums for on the interval . Comment on how these

plots approximate as increases.

10.2: PROPERTIES OF POWER SERIES

1) If and , find the power series of and of .

Solution: and .

2) If and , find the power series of and of .

= sinnSN ∑n=1

N

xn n = 10, 50, 100 [−0.99, 0.99]

x = −1 x = 1 N

= (−1SN ∑n=0

N

)nx2n+1

(2n + 1)!n = 3, 5, 10 [−2π, 2π]

sinx N

sinx sinx

= (−1SN ∑n=0

N

)nx2n

(2n)!n = 3, 5, 10 [−2π, 2π]

cosx N

f(x) =∑n=0

∞xn

n!g(x) = (−1∑

n=0

∞

)nxn

n!(f(x) + g(x))

1

2(f(x) − g(x))

1

2

(f(x) + g(x)) =1

2∑n=0

∞x2n

(2n)!(f(x) − g(x)) =

1

2∑n=0

∞x2n+1

(2n + 1)!

C(x) =∑n=0

∞x2n

(2n)!S(x) =∑

n=0

∞x2n+1

(2n + 1)!C(x) + S(x) C(x) − S(x)

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In the following exercises, use partial fractions to find the power series of each function.

3)

Solution:

4)

5)

Soution:

6)

In the following exercises, express each series as a rational function.

7)

Solution:

8)

9)

Solution:

10)

The following exercises 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, assuminginterest rates of , and .

Solution: where . Then . When

When When .

12) Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of and .

13) Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interestrates of and

4

(x − 3)(x + 1)

= − = − − = − ( − (−1 = ((−1 − )4

(x − 3)(x + 1)

1

x − 3

1

x + 1

1

3(1 − )x

3

1

1 − (−x)

1

3∑n=0

∞x

3)n ∑

n=0

∞

)nxn ∑n=0

∞

)n+1 1

3n + 1xn

3

(x + 2)(x − 1)

5

( + 4)( − 1)x2 x2

= − = − − (−1 ( = ((−1) + (−1 )5

( + 4)( − 1)x2 x2

1

− 1x2

1

4

1

1 + ( x

2)2

∑n=0

∞

x2n 1

4∑n=0

∞

)nx

2)n ∑

n=0

∞

)n+1 1

2n+2x2n

30

( + 1)( − 9)x2 x2

∑n=1

∞ 1

xn

= =1

x∑n=0

∞ 1

xn

1

x

1

1 − 1x

1

x − 1

∑n=1

∞ 1

x2n

∑n=1

∞ 1

(x − 3)2n−1

=1

x − 3

1

1 − 1

(x−3)2

x − 3

(x − 3 − 1)2

( − )∑n=1

∞ 1

(x − 3)2n−1

1

(x − 2)2n−1

r = 0.03, r = 0.05 r = 0.07

P = + ⋯ +P1 P20 = 10, 000Pk1

(1 + r)kP = 10, 000 = 10, 000∑

k=1

20 1

(1 + r)k1 − (1 + r)−20

r

r = 0.03,P ≈ 10, 000 × 14.8775 = 148, 775. r = 0.05,P ≈ 10, 000 × 12.4622 = 124, 622. r = 0.07,P ≈ 105, 940

r = 0.03, r = 0.05 r = 0.07

r = 0.03, r = 0.05, r = 0.07.



Solution: In general, for N years of payouts, or . For and , one has

when when ; and when .

14) Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interestrates of and .

15) Suppose that an annuity has a present value million dollars. What interest rate r would allow for perpetual annualpayouts of $50,000?

Solution: In general, Thus,

16) Suppose that an annuity has a present value million dollars. What interest rate r would allow for perpetual annualpayouts of $100,000?

In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rationalfunction.

17) (Hint: Group powers and .)

Solution:

18) (Hint: Group powers etc.)

19) (Hint: Group powers , and .)

Solution:

20) (Hint: Group powers and .)

In the following exercises, find the power series of given f and g as defined.

21)

Solution: so and

22) . Express the coefficients of in terms of .

23)

Solution: so and

24)

In the following exercises, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion forthe derivative of f.

P =C(1 − (1 + r ))−N

rC =

Pr

1 − (1 + r)−NN = 20 P = 100, 000

C = 6721.57 r = 0.03;C = 8024.26 r = 0.05 C ≈ 9439.29 r = 0.07

r = 0.03, r = 0.05, r = 0.07

P = 1

P = .C

rr = = 5 × = 0.05.

C

P

104

106

P = 10

x + − + + − + ⋯x2 x3 x4 x5 x6 , ,x3k x3k−1 x3k−2

(x + − )(1 + + + ⋯) =x2 x3 x3 x6 x + −x2 x3

1 − x3

x + − − + + − − + ⋯x2 x3 x4 x5 x6 x7 x8 , ,x4k x4k−1

x − − + − − + − ⋯x2 x3 x4 x5 x6 x7 ,x3k x3k−1 x3k−2

(x − − )(1 + + + ⋯) =x2 x3 x3 x6 x − −x2 x3

1 − x3

+ − + + − + ⋯x

2

x2

4

x3

8

x4

16

x5

32

x6

64, ( ,

x

2)3k x

2)3k−1 x

2)3k−2

f(x)g(x)

f(x) = 2 , g(x) = n∑n=0

∞

xn ∑

n=0

∞

xn

= 2, = nan bn = = 2 k = (n)(n + 1)cn ∑k=0

n

bkan−k ∑k=0

n

f(x)g(x) = n(n + 1)∑n=1

∞

xn

f(x) = , g(x) =∑n=1

∞

xn ∑n=1

∞ 1

nxn f(x)g(x) =Hn ∑

k=1

n 1

k

f(x) = g(x) = (∑n=1

∞x

2)n

= =an bn 2−n = = 1 =cn ∑k=1

n

bkan−k 2−n∑k=1

nn

2nf(x)g(x) = n(∑

n=1

∞x

2)n

f(x) = g(x) = n∑n=1

∞

xn



25)

Solution: The derivative of is .

26)

In the following exercises, integrate the given series expansion of term-by-term from zero to x to obtain the corresponding seriesexpansion for the indefinite integral of .

27)

Solution: The indefinite integral of is .

28)

In the following exercises, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.

29) Evaluate as where .

Solution: so



Solution: so


In the following exercises, given that , use term-by-term differentiation or integration to find power series for each

function centered at the given point.

33) centered at (Hint: )

Solution:

34) at

35) at

f(x) = = (−11

1 + x∑n=0

∞

)nxn

f − = − (−1 (n + 1)1

(1 + x)2∑n=0

∞

)n xn

f(x) = =1

1 − x2∑n=0

∞

x2n

f

f

f(x) = = (−1 (2n)2x

(1 + x2)2∑n=1

∞

)n x2n−1

f = (−11

1 + x2∑n=0

∞

)nx2n

f(x) = = 2 (−12x

1 + x2∑n=0

∞

)nx2n+1

∑n=1

∞n

2nf'( )

1

2f(x) =∑

n=0

∞

xn

f(x) = = ; f'( ) = = (1 − x = = 4∑n=0

∞

xn 1

1 − x

1

2∑n=1

∞n

2n−1

d

dx)−1 ∣x=1/2

1

(1 − x)2∣x=1/2 = 2.∑

n=1

∞n

2n

∑n=1

∞n

3nf'( )

1

3f(x) = x6n∑

n=0

∞

∑n=2

∞ n(n − 1)

2n( )f ′′ 1

2f(x) =∑

n=0

∞

xn

f(x) = = ; ( ) = = (1 − x = = 16∑n=0

∞

xn 1

1 − xf ′′ 1

2∑n=2

∞ n(n − 1)

2n−2

d2

dx2)−1 ∣x=1/2

2

(1 − x)3∣x=1/2 n = 4.∑

n=2

∞ (n − 1)

2n

∑n=0

∞ (−1)n

n + 1f(t)dt∫

1

0f(x) = (−1 =∑

n=0

∞

)nx2n 1

1 + x2

=1

1 − x∑n=0

∞

xn

f(x) = lnx x = 1 x = 1 − (1 − x)

∫ ∑(1 − x dx = ∫ ∑(−1 (x − 1 dx =∑)n )n )n(−1 (x − 1)n )n+1

n + 1

ln(1 − x) x = 0

ln(1 − )x2

x = 0



Solution:

36) at

37) at

Solution:

38) at

39) where

Solution: Term-by-term integration gives

40) [T] Evaluate the power series expansion at to show that 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

accurate to within 0.001, and find such an approximation.

41) [T] Subtract the infinite series of from to get a power series for . Evaluate at . What is the

smallest N such that the Nth partial sum of this series approximates with an error less than 0.001?

Solution: We have so . Thus, .

When we obtain . We have , while

and therefore, .

In the following exercises, using a substitution if indicated, express each series in terms of elementary functions and find the radiusof convergence of the sum.

42)

43)

Solution: so . The radius of convergence is equal to 1 by the ratio test.

44) using

45) using

Solution: If , then . If , then when . So the series

converges for all .

− dt = − dx − = −∫x2

t=0

1

1 − t∑n=0

∞

∫x2

0tn ∑

n=0

∞x2(n+1)

n + 1∑n=1

∞x2n

n

f(x) =2x

(1 − x2)2x = 0

f(x) = ta ( )n−1

x2 x = 0

= (−1 dt = (−1 = (−1∫x2

0

dt

1 + t2∑n=0

∞

)n ∫x2

0

t2n ∑n=0

∞

)nt2n+1

2n + 1∣x

2

t=0 ∑n=0

∞

)nx4n+2

2n + 1

f(x) = ln(1 + )x2 x = 0

f(x) = lntdt∫x

0

ln(x) = (−1∑n=1

∞

)n−1 (x − 1)n

n

lntdt = (−1 = (−1 ( − )(x − 1 = (x − 1)lnx + (−1 = xlnx − x.∫x

0

∑n=1

∞

)n−1 (x − 1)n+1

n(n + 1)∑n=1

∞

)n−1 1

n

1

n + 1)n+1 ∑

n=2

∞

)n(x − 1)n

n

ln(1 + x) = (−1∑n=1

∞

)n−1 xn

nx = 1 ln(2)

ln(2)

ln(1 − x) ln(1 + x) ln( )1 + x

1 − xx =

1

3ln(2)

ln(1 − x) = −∑n=1

∞xn

nln(1 + x) = (−1∑

n=1

∞

)n−1 xn

nln( ) = (1 + (−1 ) = 2

1 + x

1 − x∑n=1

∞

)n−1 xn

n∑n=1

∞x2n−1

2n − 1

x =1

3ln(2) = 2∑

n=1

∞ 1

(2n − 1)32n−12 = 0.69300 …∑n=1

3 1

(2n − 1)32n−1

2 = 0.69313 …∑n=1

4 1

(2n − 1)32n−1ln(2) = 0.69314 … ; N = 4

( − )∑k=0

∞

xk

x2k+1

∑k=1

∞x3k

6k

= −ln(1 − x)∑k=1

∞xk

k6∞ = − ln(1 − )∑

k=1

x3k

6k

1

6x3

(1 +∑k=1

∞

x2)−k y =1

1 + x2

∑k=1

∞

2−kxy = 2−x

y = 2−x = = =∑k=1

∞

yky

1 − y

2−x

1 − 2−x

1

− 12x=ak 2−kx = < 1

ak+1

ak2−x x > 0

x > 0



46) Show that, up to powers and , satisfies .

47) Differentiate the series term-by-term to show that is equal to its derivative.

Solution: Answers will vary.

48) Show that if is a sum of even powers, that is, if is odd, then 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 are defined by the recurrence relation . For

and , compute and plot the sums for on

Solution: The solid curve is . The dashed curve is , dotted is , and dash-dotted is

50) [T] Suppose that the coefficients an of the series are defined by the recurrence relation . For

and , compute and plot the sums for on .

51) [T] Given the power series expansion , determine how many terms N of the sum evaluated at

are needed to approximate accurate to within 1/1000. Evaluate the corresponding partial sum .

Solution: When . Since one has whereas

therefore,

52) [T] Given the power series expansion , use the alternating series test to determine how many terms

N of the sum evaluated at are needed to approximate accurate to within 1/1000. Evaluate the corresponding

partial sum .

53) [T] Recall that Assuming an exact value of , estimate by evaluating partial sums of the power

series expansion at . What is the smallest number such that approximates

accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?

x3 y3 E(x) =∑n=0

∞xn

n!E(x + y) = E(x)E(y)

E(x) =∑n=0

∞xn

n!E(x)

f(x) =∑n=0

∞

anxn = 0an n F = f(t)dt∫

x

0

∑n=0

∞

anxn = +an

an−1

n

an−2

n(n − 1)

= 0a0 = 1a1 =SN ∑n=0

N

anxn N = 2, 3, 4, 5 [−1, 1].

S5 S2 S3 S4

∑n=0

∞

anxn = −an

an−1

n−−√

an−2

n(n − 1)− −−−−−−√

= 1a0 = 0a1 =SN ∑n=0

N

anxn N = 2, 3, 4, 5 [−1, 1]

ln(1 + x) = (−1∑n=1

∞

)n−1 xn

n

x = −1/2 ln(2) (−1∑n=1

N

)n−1 xn

n

x = − , −ln(2) = ln( ) = −1

2

1

2∑n=1

∞ 1

n2n< = ,∑

n=11

∞ 1

n2n∑n=11

∞ 1

2n1

210= 0.69306 …∑

n=1

10 1

n2n

ln(2) = 0.69314 … ; N = 10.

ta (x) = (−1n−1 ∑

k=0

∞

)kx2k+1

2k + 1

x = 1 ta (1) =n−1 π

4

(−1∑k=0

N

)kx2k+1

2k + 1

ta ( ) = .n−1 1

3–

√

π

6)

1

3–

√

π

6( )SN

1

3–

√

ta (x) = (−1n−1 ∑k=0

∞

)kx2k+1

2k + 1x =

1

3–

√N 6 ( )SN

1

3–

√π



Solution: One has and so is

the smallest partial sum with accuracy to within 0.001. Also, while so

is the smallest N to give accuracy to within 0.00001.

10.3: TAYLOR AND MACLAURIN SERIESIn the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

1) at

2) at

Solution:

3) at

4) at

Solution:

5) at

6) at

Solution:

7) at

8) at

Solution:

In the following exercises, verify that the given choice of n in the remainder estimate , where M is the

maximum value of on the interval between a and the indicated point, yields . Find the value of the Taylor

polynomial of at the indicated point.

9) [T]

10) [T]

Solution: when so the remainder estimate applies to the linear approximation

, which gives , while

11) [T]

12) [T]

Solution: Using the estimate we can use the Taylor expansion of order 9 to estimate at . as

… whereas

13) [T]

6 ( ) = 2 (−1SN

1

3–√3–√ ∑n=0

N

)n1

(2n + 1).3nπ − 6 ( ) = 0.00101 …S4

1

3–√π − 6 ( ) = 0.00028 …S5

1

3–√N = 5

π − 6 ( ) = 0.00002 …S71

3–√π − 6 ( ) = −0.000007 …S8

1

3–√N = 8

f(x) = 1 + x + x2 a = 1

f(x) = 1 + x + x2 a = −1

f(−1) = 1; f'(−1) = −1; (−1) = 2; f(x) = 1 − (x + 1) + (x + 1f ′′ )2

f(x) = cos(2x) a = π

f(x) = sin(2x) a =π

2

f'(x) = 2cos(2x); (x) = −4sin(2x); (x) = −2(x − )f ′′ p2π

2

f(x) = x−−

√ a = 4

f(x) = lnx a = 1

f'(x) = ; (x) = − ; (x) = 0 + (x − 1) − (x − 11

xf ′′ 1

x2p2

1

2)2

f(x) =1

xa = 1

f(x) = ex a = 1

(x) = e + e(x − 1) + (x − 1p2e

2)2

| | ≤ (x − aRnM

(n + 1)!)n+1

(z)∣∣f(n+1) ∣∣ | | ≤Rn

1

1000pn f

;a = 9,n = 310−−

√

(28 ;a = 27,n = 1)1/3

= − ≥ −0.00092 …d2

dx2x1/3 2

9x5/3x ≥ 28

≈ (27) = 3 +x1/3 p1x − 27

27(28 ≈ 3 + = 3.)1/3 1

27037¯ (28 ≈ 3.03658.)1/3

sin(6);a = 2π,n = 5

;a = 0,n = 9e2

< 0.000283210

10!ex x = 2

≈ (2) = 1 + 2 + + + ⋯ + = 7.3887e2 p922

2

23

6

29

9!≈ 7.3891.e2

cos( );a = 0,n = 4π

5

https://math.libretexts.org/TextMaps/Calculus/Book%3A_Calculus_(OpenStax)/10%3A_Power_Series/10.3%3A_Taylor_and_Maclaurin_Series



14) [T]

Solution: Since . One has whereas

15) Integrate the approximation evaluated at t to approximate .

16) Integrate the approximation evaluated at to approximate

Solution:

whereas

In the following exercises, find the smallest value of n such that the remainder estimate , where M is the

maximum value of on the interval between a and the indicated point, yields on the indicated interval.

17) on

18) on

Solution: Since is or , we have . Since , we seek the smallest n such that .

The smallest such value is . The remainder estimate is

19) on

20) on

Solution: Since one has . Since , one seeks the smallest n such that . The

smallest such value is . The remainder estimate is

In the following exercises, the maximum of the right-hand side of the remainder estimate on

occurs at a or . Estimate the maximum value of R such that on by plotting this maximum

as a function of I.

21) [T] approximated by


Solution: Since is increasing for small and since , the estimate applies whenever , whichapplies up to

ln(2);a = 1,n = 1000

(lnx) = (−1 , ≈dn

dxn)n−1 (n − 1)!

xnR1000

1

1001(1) = ≈ 0.6936p1000 ∑

n=1

1000 (−1)n−1

n

ln(2) ≈ 0.6931 ⋯ .

sint ≈ t − + −t3

6

t5

120

t7

5040π dt∫

1

0

sinπt

πt

≈ 1 + x + + ⋯ +ex x2

2

x6

720−x

2 dx.∫1

0e

−x2

(1 − + − + − + )dx = 1 − + − + − + ≈ 0.74683∫1

0x

2 x4

2

x6

6

x8

24

x10

120

x12

720

13

3

15

10

17

42

19

9 ⋅ 24

111

120 ⋅ 11

113

720 ⋅ 13

dx ≈ 0.74682.∫1

0e

−x2

| | ≤ (x − aRnM

(n + 1)!)n+1

(z)∣∣f(n+1) ∣∣ | | ≤Rn

1

1000

f(x) = sinx [−π,π],a = 0

f(x) = cosx [− , ],a = 0π

2

π

2

(z)f (n+1) sinz cosz M = 1 |x − 0| ≤π

2≤ 0.001

πn+1

(n + 1)!2n+1

n = 7 ≤ 0.00092.R7

f(x) = e−2x [−1, 1],a = 0

f(x) = e−x [−3, 3],a = 0

(z) = ±f(n+1)

e−z

M = e3 |x − 0| ≤ 3 ≤ 0.001

3n+1e3

(n + 1)!n = 14 ≤ 0.000220.R14

| | ≤R1max| (z)|f ′′

2R

2 [a − R,a + R]

a ± R ≤ 0.1max| (z)|f ′′

2R2 [a − R,a + R]

ex 1 + x,a = 0

sinx x,a = 0

sinx x si x = −sinxn′′

sin(R) ≤ 0.2R2

R = 0.596.





Solution: Since the second derivative of is and since is decreasing away from , the estimate applies when or .

In the following exercises, find the Taylor series of the given function centered at the indicated point.

25) at

26) at

Solution:

27) at

28) at

Solution: Values of derivatives are the same as for so

29) at

30) at

Solution: so , which is also .

31) at

32) at

lnx x − 1,a = 1

cosx 1,a = 0

cosx −cosx cosx x = 0cosR ≤ 0.2R2 R ≤ 0.447

x4 a = −1

1 + x + +x2 x3 a = −1

(x + 1 − 2(x + 1 + 2(x + 1))3 )2

sinx a = π

cosx a = 2π

x = 0 cosx = (−1∑n=0

∞

)n(x − 2π)2n

(2n)!

sinx x =π

2

cosx x =π

2

cos( ) = 0, −sin( ) = −1π

2

π

2cosx = (−1∑

n=0

∞

)n+1(x − π

2)2n+1

(2n + 1)!−cos(x − )

π

2

ex a = −1

ex a = 1



Solution: The derivatives are so

33) at (Hint: Differentiate .)

34) at

Solution:

35) at a=0 (Note: is the Taylor series of

In the following exercises, compute the Taylor series of each function around .

36)

Solution:

37)

38)

Solution:

39)

40)

Solution:

41)

42)

Solution:

43)

44)

Solution:

[T] In the following exercises, identify the value of x such that the given series is the value of the Maclaurin series of at .

Approximate the value of using .

45)

(1) = ef (n) = e .ex ∑n=0

∞ (x − 1)n

n!

1

(x − 1)2a = 0

1

1 − x

1

(x − 1)3a = 0

= −( ) = − ( )1

(x − 1)3

1

2

d2

dx2

1

1 − x∑n=0

∞ (n + 2)(n + 1)xn

2

F (x) = cos( )dt; f(t) = (−1∫x

0t√ ∑

n=0

∞

)ntn

(2n)!f cos( ). )t√

x = 1

f(x) = 2 − x

2 − x = 1 − (x − 1)

f(x) = x3

f(x) = (x − 2)2

((x − 1) − 1 = (x − 1 − 2(x − 1) + 1)2 )2

f(x) = lnx

f(x) =1

x

= (−1 (x − 11

1 − (1 − x)∑n=0

∞

)n )n

f(x) =1

2x − x2

f(x) =x

4x − 2 − 1x2

x (1 − x = (x − 1 + (x − 1∑n=0

∞

2n )2n ∑n=0

∞

2n )2n+1 ∑n=0

∞

2n )2n

f(x) = e−x

f(x) = e2x

= =e2x e2(x−1)+2 e2∑n=0

∞ (x − 12n )n

n!

∑n=0

∞

an f(x) x

f(x) =S10 ∑n=0

10

an

∑n=0

∞ 1

n!



46)

Solution:

47)

48)

Solution:

The following exercises make use of the functions and on .

49) [T] Plot on . Compare the maximum difference with the square of the Taylor remainder estimate for

50) [T] Plot on . Compare the maximum difference with the square of the Taylor remainder estimate for .

Solution: The difference is small on the interior of the interval but approaches near the endpoints. The remainder estimate is

51) [T] Plot on .

52) [T] Compare on to . Compare this with the Taylor remainder estimate for the approximation of by

.

Solution: The difference is on the order of on while the Taylor approximation error is around near . The top curve

is a plot of and the lower dashed plot shows .

sum∞n=0

2n

n!

x = ; = ≈ 7.3889947e2 S1034, 913

4725

∑n=0

∞ (−1 (2π)n )2n

(2n)!

∑n=0

∞ (−1 (2π)n )2n+1

(2n + 1)!

sin(2π) = 0; = 8.27 ×S10 10−5

(x) = x − +S5x3

6

x5

120(x) = 1 − +C4

x2

2

x4

24[−π,π]

si x − ( (x)n2

S5 )2 [−π,π]sinx.

co x − ( (x)s2 C4 )2 [−π,π]cosx

1

| | = ≈ 2.552.R4π5

120

|2 (x) (x) − sin(2x)|S5 C4 [−π,π]

(x)S5

(x)C4

[−1, 1] tanx tanx

x + +x3

3

2x5

15

10−4 [−1, 1] 0.1 ±1

ta x − (n2 (x)S5

(x)C4)2 − (t2 S5

C4)2



53) [T] Plot where on . Compare the maximum error with the Taylor remainder

estimate.

54) (Taylor approximations and root finding.) Recall that Newton’s method approximates solutions of

near the input .

a. If and are inverse functions, explain why a solution of is the value of .

b. Let be the degree Maclaurin polynomial of . Use Newton’s method to approximate solutions of for

c. Explain why the approximate roots of are approximate values of

Solution: a. Answers will vary. b. The following are the values after iterations of Newton’s method to approximation a root of : for for for (Note: ) c. Answers

will vary.

In the following exercises, use the fact that if converges in an interval containing , then to

evaluate each limit using Taylor series.

55)

56)

Solution:

57)

58)

Solution:

10.4: WORKING WITH TAYLOR SERIESIn the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

1)

2)

− (x)ex e4 (x) = 1 + x + + +e4x2

2

x3

6

x4

24[0, 2]

= −xn+1 xn

f( )xn

( )f ′ xn

f(x) = 0

x0

f g g(x) = a f(a) f

(x)pN Nth ex (x) − 2 = 0pNN = 4, 5, 6.

(x) − 2 = 0pN ln(2).

xn 10(x) − 2 = 0pN N = 4,x = 0.6939...; N = 5,x = 0.6932...; N = 6,x = 0.69315...; . ln(2) = 0.69314...

q(x) = (x − c∑n=1

∞

an )n c q(x) =limx→c

a0

limx→0

cosx − 1

x2

limx→0

ln(1 − )x2

x2

→ −1ln(1 − )x2

x2

limx→0

− − 1ex2

x2

x4

limx→0+

cos( ) − 1x−−√

2x

≈ → −cos( ) − 1x−−√

2x

(1 − + − ⋯) − 1x

2x2

4!

2x

1

4

(1 − x)1/3

(1 + x2)−1/3

https://math.libretexts.org/TextMaps/Calculus/Book%3A_Calculus_(OpenStax)/10%3A_Power_Series/10.4%3A_Working_with_Taylor_Series



Solution:

3)

4)

Solution:

In the following exercises, use the substitution in the binomial expansion to find the Taylor series

of each function with the given center.

5) (\sqrt{x+2}\) at

6) at

Solution:

7) at

8) at (Hint: )

Solution: so

9) at

10) at

Solution: so

11) at

12) at

Solution:

In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimateaccurate to an error of at most

13) [T] using

14) [T] using

Solution: . Using, for example, a fourth-degree estimate at gives

whereas Two terms would suffice for three-digit accuracy.

In the following exercises, use the binomial approximation for to approximate

each number. Compare this value to the value given by a scientific calculator.

(1 + = )x2)−1/3 ∑n=0

∞

(− 1

3n x2n

(1 − x)1.01

(1 − 2x)2/3

(1 − 2x = (−1 ))2/3 ∑n=0

∞

)n2n(23n xn

(b + x = (b + a (1 +)r )rx − a

b + a)r

a = 0

+ 2x2− −−−−√ a = 0

= ) ; (∣ ∣< 2)2 + x2− −−−−√ ∑

n=0

∞

2(1/2)−n(1

2n x2n x2

x + 2− −−−−√ a = 1

2x − x2− −−−−−√ a = 1 2x − = 1 − (x − 1x2 )2

=2x − x2− −−−−−√ 1 − (x − 1)2

− −−−−−−−−−√ = (−1 )(x − 12x − x2− −−−−−

√ ∑n=0

∞

)n(1

2n )2n

(x − 8)1/3a = 9

x−−√ a = 4

= 2x−−√ 1 +x − 4

4

− −−−−−−−√ = )(x − 4x−−√ ∑

n=0

∞

21−2n(1

2n )n

x1/3 a = 27

x−−√ x = 9

= )(x − 9x−−√ ∑n=0

∞

31−3n(1

2n )n

1/1000.

(15)1/4 (16 − x)1/4

(1001)1/3 (1000 + x)1/3

10(1 + = )x

1000)1/3 ∑

n=0

∞

101−3n(13n xn x = 1

(1001 ≈ 10(1 + ) + ) + ) + ) ) = 10(1 + − + − ) = 10.00333222...)1/3 (13

1 10−3 (13

2 10−6 (13

3 10−9 (13

4 10−12 1

3.103

1

9.106

5

81.109

10

243.1012

(1001 = 10.00332222839093....)1/3

≈ 1 − − − − −1 − x− −−−−√

x

2

x2

8

x3

16

5x4

128

7x5

256|x| < 1



15) [T] using in

16) [T] using in

Solution: The approximation is ; the CAS value is

17) [T] using in

18) [T] using in

Solution: The approximation is ; the CAS value is

19) Integrate the binomial approximation of to find an approximation of .

20) [T] Recall that the graph of is an upper semicircle of radius . Integrate the binomial approximation of up toorder from to to estimate .

Solution: Thus

whereas

In the following exercises, use the expansion to write the first five terms (not

necessarily a quartic polynomial) of each expression.

21)

22)

Solution:

23)

24)

Solution:

25) Use with to approximate .

26) Use the approximation for to approximate .

Solution: Twice the approximation is whereas

27) Find the derivative of at .

28) Find the th derivative of .

Solution:

In the following exercises, find the Maclaurin series of each function.

1

2–√

x =1

2(1 − x)1/2

= 5 ×5–√1

5–

√x =

4

5(1 − x)1/2

2.3152 2.23 … .

=3–√3

3–√x =

2

3(1 − x)1/2

6–√ x =5

6(1 − x)1/2

2.583 … 2.449 … .

1 − x− −−−−

√ dt∫x

01 − t− −−−

√

1 − x2− −−−−√ 1 1 − x2− −−−−√

8 x = −1 x = 1π

2

= 1 − − − − + ⋯ .1 − x2− −−−−√ x2

2

x4

8

x6

16

5x8

128

dx = x − − − − + ⋯ ≈ 2 − − − − + error = 1.590...∫1

−11 − x

2− −−−−√ x3

6

x5

40

x7

7 ⋅ 16

5x9

9 ⋅ 128∣1−1

1

3

1

20

1

56

10

9 ⋅ 128

= 1.570...π

2

(1 + x = 1 + x − + − + ⋯)1/3 1

3

1

9x

2 5

81x

3 10

243x

4

(1 + 4x ;a = 0)1/3

(1 + 4x ;a = 0)4/3

(1 + x = (1 + x)(1 + x − + − + ⋯) = 1 + + − + + ⋯)4/3 1

3

1

9x

2 5

81x

3 10

243x

4 4x

3

2x2

9

4x3

81

5x4

243

(3 + 2x ;a = −1)1/3

( + 6x + 10 ;a = −3x2 )1/3

(1 + (x + 3 = 1 + (x + 3 − (x + 3 + (x + 3 − (x + 3 + ⋯)2)1/3 1

3)2 1

9)4 5

81)6 10

243)8

(1 + x = 1 + x − + − + ⋯)1/3 1

3

1

9x2 5

81x3 10

243x4 x = 1 21/3

(1 − x = 1 − − − − − + ⋯)2/3 2x

3

x2

9

4x3

81

7x4

243

14x5

729|x| < 1 =21/3 2.2−2/3

1.260 … = 1.2599....21/3

25th f(x) = (1 + x2)13 x = 0

99 f(x) = (1 + x4)25

(0) = 0f (99)



29)

30)

Solution:

31)

32)

Solution: For .

33)

34)

Solution:

35) using the identity

36) using the identity

Solution:

In the following exercises, find the Maclaurin series of by integrating the Maclaurin series of term by term. If

is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

37)

38)

Solution:

39)

40)

Solution:

41)

42)

f(x) = xe2x

f(x) = 2x

∑n=0

∞ (ln(2)x)n

n!

f(x) =sinx

x

f(x) = , (x > 0),sin( )x−−√

x−−√

x > 0, sin( ) = (−1 = (−1x−−√ ∑n=0

∞

)nx(2n+1)/2

(2n + 1)!x−−√∑n=0

∞

)nxn

(2n + 1)!

f(x) = sin( )x2

f(x) = ex3

=ex3 ∑

n=0

∞x3n

n!

f(x) = co xs2 co x = + cos(2x)s2 1

2

1

2

f(x) = si xn2 si x = − cos(2x)n2 1

2

1

2

si x = −n2 ∑

k=1

∞ (−1)k22k−1x2k

(2k)!

F (x) = f(t)dt∫x

0f f

F (x) = dt; f(t) = = (−1∫x

0e

−t2e

−t2 ∑n=0

∞

)nt2n

n!

F (x) = ta x; f(t) = = (−1n−1 1

1 + t2∑n=0

∞

)nt2n

ta x =n−1 ∑

k=0

∞ (−1)kx2k+1

2k + 1

F (x) = tan x; f(t) = =h−1 1

1 − t2∑n=0

∞

t2n

F (x) = si x; f(t) = = )n−1 1

1 − t2− −−−−√

∑k=0

∞

(1

2k

t2k

k!

si x = )n−1 ∑n=0

∞

(1

2n

x2n+1

(2n + 1)n!

F (x) = dt; f(t) = = (−1∫x

0

sint

t

sint

t∑n=0

∞

)nt2n

(2n + 1)!

F (x) = cos( )dt; f(t) = (−1∫x

0t√ ∑

n=0

∞

)nxn

(2n)!



Solution:

43)

44)

Solution:

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurinseries of .

45)

46)

Solution:

47)

48)

Solution:

49)

50)

Solution:

51)

52) (see expansion for )

Solution: Using the expansion for gives .

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

53)

54)

Solution: so by the ratio test.

55)

56)

Solution: so by the ratio test.

F (x) = (−1∑n=0

∞

)nxn+1

(n + 1)(2n)!

F (x) = dt; f(t) = = (−1∫x

0

1 − cost

t2

1 − cost

t2∑n=0

∞

)nt2n

(2n + 2)!

F (x) = dt; f(t) = (−1∫x

0

ln(1 + t)

t∑n=0

∞

)ntn

n + 1

F (x) = (−1∑n=1

∞

)n+1 xn

n2

f

f(x) = sin(x + ) = sinxcos( ) + cosxsin( )π

4

π

4

π

4

f(x) = tanx

x + + + ⋯x3

3

2x5

15

f(x) = ln(cosx)

f(x) = cosxex

1 + x − − + ⋯x3

3

x4

6

f(x) = esinx

f(x) = se xc2

1 + + + + ⋯x2 2x4

3

17x6

45

f(x) = tanhx

f(x) =tan x−−√

x−−√tanx

tanx 1 + +x

3

2x2

15

ln(1 + x)

1

1 + x2

= (−11

1 + x2∑n=0

∞

)nx2n R = 1

ta xn−1

ln(1 + )x2

ln(1 + ) =x2 ∑n=1

∞ (−1)n−1

nx2n R = 1



57) Find the Maclaurin series of .

58) Find the Maclaurin series of .

Solution: Add series of and term by term. Odd terms cancel and .

59) Differentiate term by term the Maclaurin series of and compare the result with the Maclaurin series of .

60) [T] Let and denote the respective Maclaurin polynomials of degree

of and degree of . Plot the errors for and compare them to

on .

Solution: The ratio approximates better than does for . The dashed curves are

for . The dotted curve corresponds to , and the dash-dotted curve corresponds to . The solid curve is

.

61) Use the identity to find the power series expansion of at . (Hint: Integrate the Maclaurin seriesof term by term.)

62) If , find the power series expansions of and .

Solution: By the term-by-term differentiation theorem, so , whereas

so .

63) [T] Suppose that satisfies and . Show that for all and that . Plot the

partial sum of on the interval .

64) [T] Suppose that a set of standardized test scores is normally distributed with mean and standard deviation . Setup an integral that represents the probability that a test score will be between and and use the integral of the degree

Maclaurin polynomial of to estimate this probability.

sinhx =−ex e−x

2

coshx =+ex e−x

2

ex

e−x coshx =∑

n=0

∞x2n

(2n)!

sinhx coshx

(x) = (−1Sn ∑k=0

n

)kx2k+1

(2k + 1)!(x) = (−1Cn ∑

n=0

n

)kx2k

(2k)!2n + 1

sinx 2n cosx − tanx(x)Sn

(x)Cn

n = 1, . . , 5 x + + + − tanxx3

3

2x5

15

17x7

315

(− , )π

4

π

4

(x)Sn

(x)Cn

tanx (x) = x + + +p7x3

3

2x5

15

17x7

315N ≥ 3

− tanSn

Cn

n = 1, 2 n = 3 n = 4

− tanxp7

2sinxcosx = sin(2x) si xn2 x = 0sin(2x)

y =∑n=0

∞

anxn xy' x2y′′

y' = n∑n=1

∞

anxn−1 y' = n xy' = n∑

n=1

∞

anxn−1 ∑

n=1

∞

anxn

y' = n(n − 1)∑n=2

∞

anxn−2 x = n(n − 1)y′′ ∑

n=2

∞

anxn

y =∑k=0

∞

akxk y' = −2xy y(0) = 0 = 0a2k+1 k =a2k+2−a2k

k + 1

S20 y [−4, 4]

μ = 100 σ = 1090 110 10

1

2π−−√e

− /2x2



Solution: The probability is where and , that is,

65) [T] Suppose that a set of standardized test scores is normally distributed with mean and standard deviation . Setup an integral that represents the probability that a test score will be between and and use the integral of the degree

Maclaurin polynomial of to estimate this probability.

66) [T] Suppose that converges to a function such that , and . Find a formula for

and plot the partial sum for on

Solution: As in the previous problem one obtains if is odd and if is even, so leads to

.

67) [T] Suppose that converges to a function such that , and . Find a formula for an

and plot the partial sum for on .

68) Suppose that converges to a function such that where and Find a formula that

relates and an and compute .

Solution: and so implies that

or for all and so

, and .

69) Suppose that converges to a function such that where and . Find a formula that

relates , and an and compute .

The error in approximating the integral by that of a Taylor approximation is at most . In the

following exercises, the Taylor remainder estimate guarantees that the integral of the Taylor polynomial of

the given order approximates the integral of with an error less than .

a. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less

than .

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

p = dx1

2π−−√∫

(b−μ)/σ

(a−μ)/σe− /2x2

a = 90 b = 100

p = dx = (−1 dx = (−1 ≈ 0.6827.1

2π−−√∫

1

−1e− /2x2 1

2π−−√∫

1

−1∑n=0

5

)nx2n

n!2n2

2π−−√∑n=0

5

)n1

(2n + 1) n!2n

μ = 100 σ = 1070 130 50

1

2π−−√

e− /2x2

∑n=0

∞

anxn f(x) f(0) = 1, f'(0) = 0 (x) = −f(x)f ′′ an

SN N = 20 [−5, 5].

= 0an n = −(n + 2)(n + 1)an an+2 n = 1a0

=a2n(−1)n

(2n)!

∑n=0

∞

anxn

f(x) f(0) = 0, f'(0) = 1 (x) = −f(x)f′′

SN N = 10 [−5, 5]

∑n=0

∞

anxn

y − y' + y = 0y′′

y(0) = 1 (0) = 0.y′

, ,an+2 an+1 , . . . ,a0 a5

= (n + 2)(n + 1)y′′ ∑

n=0

∞

an+2xn

y' = (n + 1)∑n=0

∞

an+1xn − y' + y = 0y

′′

(n + 2)(n + 1) − (n + 1) + = 0an+2 an+1 an = −anan−1

n

an−2

n(n − 1)n ⋅ y(0) = = 1a0 y'(0) = = 0,a1

= , = , = 0a21

2a3

1

6a4 = −a5

1

120

∑n=0

∞

anxn y − y' + y = 0y′′ y(0) = 0 y'(0) = 1

,an+2 an+1 , . . . ,a1 a5

f(t)dt∫b

a

Pn(t)dt∫b

a

(t)dt∫b

a

Rn

≤ |x − aRn

M

(n + 1)!|n+1

f1

10

1

100



70) [T] (You may assume that the absolute value of the ninth derivative of is

bounded by .)

Solution: a. (Proof) b. We have We have

whereas , so the actual

error is approximately

71) [T] (You may assume that the absolute value of the derivative of is

less than .)

The following exercises deal with Fresnel integrals.

72) The Fresnel integrals are defined by and . Compute the power series of and

and plot the sums and of the first nonzero terms on .

Solution: Since and , one has and

. The sums of the first nonzero terms are plotted below with the solid curve and

the dashed curve.

73) [T] The Fresnel integrals are used in design applications for roadways and railways and other applications because of thecurvature properties of the curve with coordinates . Plot the curve for , the coordinates of whichwere computed in the previous exercise.

74) Estimate by approximating using the binomial approximation .

Solution:

whereas

dt; = 1 − + − +∫π

0

sint

tPs

x2

3!

x4

5!

x6

7!

x8

9!

sint

t

0.1

≤ ≈ 0.0082 < 0.01.Rs0.1

(9)!π9

(1 − + − + )dx = π − + − + = 1.852...,∫π

0

x2

3!

x4

5!

x6

7!

x8

9!

π3

3 ⋅ 3!

π5

5 ⋅ 5!

π7

7 ⋅ 7!

π9

9 ⋅ 9!dt = 1.85194...∫

π

0

sint

t0.00006.

dx; = 1 − + − + ⋯ −∫2

0e

−x2

p11 x2 x4

2

x6

3!

x22

11!23rd e

−x2

2 × 1014

C(x) = cos( )dt∫x

0

t2 S(x) = sin( )dt∫x

0

t2 C(x)

S(x) (x)CN (x)SN N = 50 [0, 2π]

cos( ) = (−1t2 ∑

n=0

∞

)nt4n

(2n)!sin( ) = (−1t

2 ∑n=0

∞

)nt4n+2

(2n + 1)!S(x) u (−1=s m∞

n=0 )nx4n+3

(4n + 3)(2n + 1)!

C(x) = (−1∑n=0

∞

)nx4n+1

(4n + 1)(2n)!50 (x)C50 (x)S50

(C(t),S(t)) ( , )C50 S50 0 ≤ t ≤ 2π

dx∫1/4

0x − x

2− −−−−√ 1 − x− −−−−√ 1 − − − − −

x

2

x2

8

x3

16

5x4

2128

7x5

256

(1 − − − − − )dx = − − − − − = 0.0767732...∫1/4

0

x−−√x

2

x2

8

x3

16

5x4

128

7x5

256

2

32−3 1

2

2

52−5 1

8

2

72−7 1

16

2

92−9 5

128

2

112−11 7

256

2

132−13

dx = 0.076773.∫1/4

0x − x2− −−−−

√



75) [T] Use Newton’s approximation of the binomial to approximate as follows. The circle centered at with radius

has upper semicircle . The sector of this circle bounded by the -axis between and and by the line

joining corresponds to of the circle and has area . This sector is the union of a right triangle with height and base

and the region below the graph between and . To find the area of this region you can write

and integrate term by term. Use this approach with the binomialapproximation from the previous exercise to estimate .

76) Use the approximation to approximate the period of a pendulum having length meters and maximum

angle where . Compare this with the small angle estimate .

Solution: seconds. The small angle estimate is . The relative error is

around percent.

77) Suppose that a pendulum is to have a period of seconds and a maximum angle of . Use to

approximate the desired length of the pendulum. What length is predicted by the small angle estimate ?

78) Evaluate in the approximation to obtain an improved

estimate for .

Solution: Hence

79) [T] An equivalent formula for the period of a pendulum with amplitude is

where is the pendulum length and is the gravitational acceleration constant. When we get

. Integrate this approximation to estimate in terms of and . Assuming

meters per second squared, find an approximate length such that seconds.

CHAPTER REVIEW EXERCISETrue or False? In the following exercises, justify your answer with a proof or a counterexample.

1) If the radius of convergence for a power series is , then the radius of convergence for the series is also .

Solution: True

2) Power series can be used to show that the derivative of is . (Hint: Recall that )

3) For small values of

Solution: True

4) The radius of convergence for the Maclaurin series of is .

1 − x2− −−−−√ π ( , 0)

1

21

2y = x

−−√ 1 − x

− −−−−√ x x = 0 x =1

2

( , )1

4

3–

√

4

1

6

π

24

3–

√

41

4x = 0 x =

1

4y = = × (binomial expansion of )x

−−√ 1 − x

− −−−−√ x−−

√ 1 − x− −−−−√

π

T ≈ 2π (1 + )L

g

−−

√k2

410

=θmaxπ

6k = sin( )

θmax

2T ≈ 2π

L

g

−−

√

T ≈ 2π (1 + ) ≈ 6.45310

9.8

− −−√ si (θ/12)n2

4T ≈ 2π ≈ 6.347

10

9.8

− −−−−−−−−−√

2

2 =θmaxπ

6T ≈ 2π (1 + )

L

g

−−

√k2

4

T ≈ 2πL

g

−−

√

si θdθ∫π/2

0n4 T = 4 (1 + si θ + si θ + ⋯)dθ

L

g

−−

√ ∫π/2

0

1

2k2 n2 3

8k4 n4

T

si θdθ = .∫π/2

0n4 3π

16T ≈ 2π (1 + + ).

L

g

−−

√k2

4

9

256k4

axθm T ( ) = 2θmax 2–

√L

g

−−

√ ∫θmax

0

dθ

− cos( )cosθ− −−−√ θmax

L g =θmax

π

3

≈ (1 + + + )1

cost − 1/2− −−−−−−−√2–

√t2

2

t4

3

181t6

720T ( )

π

3L g

g = 9.806 L T ( ) = 2π

3

∑n=0

∞

anxn 5 n∑

n=1

∞

anxn−1 5

ex ex = .ex ∑n=0

∞ 1

n!xn

x, sinx ≈ x.

f(x) = 3x 3



In the following exercises, find the radius of convergence and the interval of convergence for the given series.

5)

Solution: ROC: ; IOC:

6)

7)

Solution: ROC: IOC:

8)

In the following exercises, find the power series representation for the given function. Determine the radius of convergence and theinterval of convergence for that series.

9)

Solution: ROC: ; IOC:

10)

In the following exercises, find the power series for the given function using term-by-term differentiation or integration.

11)

Solution: integration:

12)

In the following exercises, evaluate the Taylor series expansion of degree four for the given function at the specified point. What isthe error in the approximation?

13)

Solution: exact

14)

In the following exercises, find the Maclaurin series for the given function.

15)

Solution:

16)

(x − 1∑n=0

∞

n2 )n

1 (0, 2)

∑n=0

∞xn

nn

∑n=0

∞ 3nxn

12n

12; (−16, 8)

(x − e∑n=0

∞ 2n

en)n

f(x) =x2

x + 3

;∑n=0

∞ (−1)n

3n+1xn 3 (−3, 3)

f(x) =8x + 2

2 − 3x + 1x2

f(x) = ta (2x)n−1

(2x∑n=0

∞ (−1)n

2n + 1)2n+1

f(x) =x

(2 + x2)2

f(x) = − 2 + 4,a = −3x3 x2

(x) = (x + 3 − 11(x + 3 + 39(x + 3) − 41;p4 )3 )2

f(x) = ,a = 4e1/(4x)

f(x) = cos(3x)

∑n=0

∞ (−1 (3x)n )2n

2n!

f(x) = ln(x + 1)



In the following exercises, find the Taylor series at the given value.

17)

Solution:

18)

In the following exercises, find the Maclaurin series for the given function.

19)

Solution:

20)

In the following exercises, find the Maclaurin series for by integrating the Maclaurin series of term by term.

21)

Solution:

22)

23) Use power series to prove Euler’s formula:

Solution: Answers may vary.

The following exercises consider problems of annuity payments.

24) For annuities with a present value of million, calculate the annual payouts given over years assuming interest rates of ,and

25) A lottery winner has an annuity that has a present value of million. What interest rate would they need to live on perpetualannual payments of ?

Solution:

26) Calculate the necessary present value of an annuity in order to support annual payouts of given over years assuminginterest rates of ,and

f(x) = sinx,a =π

2

(x −∑n=0

∞ (−1)n

(2n)!

π

2)2n

f(x) = ,a = 13

x

f(x) = − 1e−x2

∑n=1

∞ (−1)n

n!x2n

f(x) = cosx − xsinx

F (x) = f(t)dt∫x

0f(x)

f(x) =sinx

x

F (x) =∑n=0

∞ (−1)n

(2n + 1)(2n + 1)!x2n+1

f(x) = 1 − ex

= cosx + isinxeix

$1 25 110

$10$250, 000

2.5

$15, 000 251 10

Documents

10.E: POWER SERIES (EXERCISES)fliacob/An1/2018-2019/Resurse...16) Suppose that an annuity has a present value million dollars. What interest rate r would allow for perpetual annual