MATH 152: Calculus 2, SET8SOLUTIONS [Belmonte, 2018]11 Infinite Sequences & Series

11.10 Taylor & Maclaurin Series1. [771/8] Find the first four nonzero terms of the Taylor series

for f (x) = lnx centered at a = 1 using the definition.

The Taylor series for f (x) at x = 1 is∞

∑n=0

f (n) (1)n!

(x−1)n.

n 0 1 2 3 4f (n) (x) lnx x−1 −x−2 2x−3 −6x−4

f (n) (1) 0 1 −1 2 −6Hence the sum of first 4 nonzero terms of the Taylor series is

T4 (x) =∞

∑n=0

f (n) (1)n!

(x−1)n

=4

∑n=1

(−1)n+1 (n−1)!n!

(x−1)n

=4

∑n=1

(−1)n+1

n(x−1)n

= (x−1)− 12(x−1)2 +

13(x−1)3− 1

4(x−1)4 .

2. [771/16] Find the Maclaurin series for f (x) = xcosx and itsradius of convergence.

The Maclaurin series for cosx is∞

∑n=0

(−1)n

(2n)!x2n with radius of

convergence R = ∞. Accordingly, that for f (x) = xcosx is∞

∑n=0

(−1)n

(2n)!x2n+1 with R = ∞.

3. [771/24] Find the Taylor series for f (x) = cosx centered ata = π/2 and its radius of convergence.

Recall from trigonometry that sin(x− π

2

)=−cosx whence

cosx =−sin(x− π

2

). Recall the Maclaurin series for sinz is

∞

∑n=0

(−1)n

(2n+1)!z2n+1 with R = ∞. So set z = x− π

2 and negate

to get cosx =∞

∑n=0

(−1)n+1

(2n+1)!

(x− π

2

)2n+1with R = ∞.

4. [771/40] Find the Maclaurin series for f (x) = x2 ln(1+ x3

).

The Maclaurin series for ln(1+ z) is∞

∑n=1

(−1)n−1 zn

nwith

R = 1. Hence the Maclaurin series for f (x) = x2 ln(1+ x3

)is x2

∞

∑n=1

(−1)n−1(x3)n

nor

∞

∑n=1

(−1)n−1 x3n+2

nwith R = 1

since∣∣x3∣∣< 1 implies |x|< 1.

5. [771/44] Same as #4 for f (x) =

x− sinx

x3 if x 6= 016 if x = 0.

For x 6= 0 the Maclaurin sine series gives

f (x) =x−∑

∞k=0

(−1)k

(2k+1)! x2k+1

x3

=∞

∑k=1

(−1)k+1

(2k+1)!x2k−2

or∞

∑n=0

(−1)n

(2n+3)!x2n

with R = ∞ and f (0) = 16 to boot!

6. [772/48] Find the Maclaurin series for f (x) = tan−1(x3)

andits radius of convergence. Then graph both f and its first fewTaylor polynomials on the same plot. Comment.

The Maclaurin series for tan1 z is∞

∑n=0

(−1)n

2n+1z2n+1 with

R = 1. Set z = x3 to obtain∞

∑n=0

(−1)n

2n+1x6n+3 with R = 1 since∣∣x3

∣∣< 1 implies |x|< 1. We see from the graph below that asn increases the Taylor polynomials Tn (x) get closer andcloser to the graph of tan−1

(x3)

on (−1,1).

-1 -0.5 0 0.5 1

x

-1

-0.5

0

0.5

1

y

SET8, 772/48

7. [772/56] Evaluate the indefinite integral∫

arctan(x2)

dxas an infinite series.

Proceed in a manner similar to #6.∫tan−1 (x2) dx =

∫ ∞

∑n=0

(−1)n

2n+1(x2)2n+1

dx

=∫ ∞

∑n=0

(−1)n

2n+1x4n+2 dx

= C+∞

∑n=0

(−1)n

(2n+1)(4n+3)x4n+3

with R = 1.

1

8. [772/58] Use series to approximate the definite integral∫ 1

0sin(x4) dx to four decimal places.

Use the Maclaurin sine series.∫ 1

0sin(x4) dx =

∫ 1

0

∞

∑n=0

(−1)n

(2n+1)!(x4)2n+1

dx

=∫ 1

0

∞

∑n=0

(−1)n

(2n+1)!x8n+4 dx

=

(∞

∑n=0

(−1)n

(2n+1)!(8n+5)x8n+5

)∣∣∣10

=∞

∑n=0

(−1)n

(2n+1)!(8n+5)

=15− 1

78+

12520

− 1146160

+ · · ·

≈ 15− 1

78+

12520

≈ 0.1876

which is accurate to 4 decimal places by the ASET since1

146160 ≈ 6.84×10−6.

9. [772/62] Use series to evaluate the limit limx→0

1− cosx1+ x− ex .

Use Maclaurin series. As x→ 0, we have

1− cosx1+ x− ex =

1−(

1− x2

2 + x4

24 −·· ·)

1+ x−(

1+ x+ x2

2 + x3

6 + x4

24 + · · ·)

=x2

2 −x4

24 + · · ·− x2

2 −x3

6 −x4

24 −·· ·

=12 −

x2

24 + · · ·− 1

2 −x6 −

x2

24 −·· ·

→ 1/2−1/2

=−1.

10. [772/78] Find the sum 1− ln2+(ln2)2

2!− (ln2)3

3!+ · · · .

The Maclaurin series for ex is 1+ x+ x2

2! +x3

3! + · · · . So thesum of the series is ex evaluated at x =− ln2. In other words,e− ln2 =

(eln2)−1

= 2−1 = 12 .

2