CHAPTER 23 MACLAURIN’S SERIES
EXERCISE 97 Page 209
1. Determine the first four terms of the power series for sin 2x using Maclaurin’s series.
Let f(x) = sin 2x
f(0) = sin 0 = 0
f′(x) = 2 cos 2x
f ′(0) = 2 cos 0 = 2
f′′(x) = –4 sin 2x
f ′′(0) = –4 sin 0 = 0
f′′′(x) = –8 cos 2x
f ′′′(0) = –8 cos 0 = –8
f iv ( x) = 16 sin 2x
f iv (0) = 16 sin 0 = 0
f v ( x) = 32 cos 2x
f v (0) = 32 cos 0 = 32
f vi ( x) = –64 sin 2x
f
vii
( x) = –128 cos 2x
f vi (0) = –64 sin 0 = 0
f
vii
(0) = –128 cos 0 = –128
Maclaurin’s series states: f(x) = f(0) + xf ′(0) +
= 0 + x(2) +
x2
x3
f ′′(0) +
f ′′′(0) + …
3!
2!
x2
x3
x4
x5
x6
x7
(0) + (−8) + (0) + (32) + (0) + (−128)
2!
3!
4!
5!
6!
7!
8 x3 32 x 5 128 x 7
+
−
6
120 5040
i.e.
f(x) = 2 x −
i.e.
sin 2x = 2 x −
4 3 4 5
8 7
x + x −
x
3
15
315
2. Use Maclaurin’s series to produce a power series for cosh 3x as far as the term in x 6
Let f(x) = cosh 3x
f(0) = cosh 0 = 1
f′(x) = 3 sinh 3x
f ′(0) = 3 sinh 0 = 0
f′′(x) = 9 cosh 3x
f ′′(0) = 9 cosh 0 = 9
f′′′(x) = 27 sinh 3x
f ′′′(0) = 27 sinh 0 = 0
f iv ( x) = 81 cosh 3x
f iv (0) = 81 cosh 0 = 81
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© 2014, John Bird
f v ( x) = 243 sinh 3x
f v (0) = 243 sinh 0 = 0
f vi ( x) = 729 cosh 3x
f vi (0) = 729 cosh 0 = 729
x2
x3
f ′′(0) +
f ′′′(0) + …
2!
3!
Maclaurin’s series states: f(x) = f(0) + xf ′(0) +
= 1 + x(0) +
x2
x3
x4
x5
x6
(9) + (0) + (81) + (0) + (729) + ...
2!
3!
4!
5!
6!
i.e.
f(x) = 1 +
9 x 2 81x 4 729 x 6
+
+
2
24
720
i.e.
cosh 3x = 1 +
9 2 27 4 81 6
x +
x + x
2
8
80
3. Use Maclaurin's theorem to determine the first three terms of the power series for ln(1 + e x ).
Let f(x) = ln (1 + e x )
f ′(x) =
f ′′(x) =
f(0) = ln (1 + e0 ) = ln 2
ex
1 + ex
f ′(0) =
(1 + e x ) e x − e x ( e x )
2
(1 + e x )
e0
1
=
0
1+ e
2
f ′′(0) =
e0 )
(1 + e0 ) e0 − e0 (=
2
(1 + e0 )
Maclaurin’s series states: f(x) = f(0) + xf ′(0) +
2 −1 1
=
22
4
x2
x3
f ′′(0) +
f ′′′(0) + …
2!
3!
 1  x2  1 
= ln 2 + x   +   + ...
 2  2!  4 
ln (1 + e x ) = ln 2 +
i.e.
x x2
+
2 8
4. Determine the power series for cos 4t as far as the term in t6
Let f(t) = cos 4t
f ′(t) = –4 sin 4t
f ′′(t) = –16 cos 4t
f ′′′(t) = 64 sin 4t
f(0) = cos 0 = 1
f ′(0) = –4 sin 0 = 0
f ′′(0) = –16 cos 0 = –16
f ′′′(0) = 64 sin 0 = 0
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© 2014, John Bird
f iv (t ) = 256 cos 4t
f iv (0) = 256 cos 0 = 256
f v (t ) = –1024 sin 4t
f v (0) = –1024 sin 0 = 0
f vi (t ) = –4096 cos 4t
f vi (0) = – 4096 cos 0 = –4096
Maclaurin’s series states: f(t) = f(0) + tf ′(0) +
= 1 + t (0) +
t3
t2
f ′′(0) +
f ′′′(0) + …
3!
2!
t2
t3
t4
t5
t6
(−16) + (0) + (256) + (0) + (−4096) + ...
2!
3!
4!
5!
6!
16t 2 256t 4 4096t 6
+
−
2
24
720
i.e.
f(x) = 1 −
i.e.
cos 4t = 1 − 8t 2 +
32 4 256 6
t −
t
3
45
5. Expand e (3/2) x in a power series as far as the term in x3
Let f(x) = e (3/2) x
f(0) = e0 = 1
f′(x) =
3 3x
e2
2
f ′(0) =
3 0 3
e =
2
2
f′′(x) =
9 3x
e2
4
f ′′(0) =
9 0 9
e =
4
4
f′′′(x) =
27 3 x
e2
8
f ′′′(0) =
27
27 0
e =
8
8
Maclaurin’s series states: f(x) = f(0) + xf ′(0) +
x3
x2
f ′′(0) +
f ′′′(0) + …
2!
3!
 3  x 2  9  x3  27 
= 1 + x   +   +   + ...
 2  2!  4  3!  8 
i.e.
f(x) = 1 +
i.e.
e (3/2) x = 1 +
3 x 9 x 2 27 x 3
+
+
2
8
48
3
9
9
x + x 2 + x3
2
8
16
6. Develop, as far as the term in x4, the power series for sec 2x.
Let f(x) = sec 2x
f(0) = sec 0 = 1
361
© 2014, John Bird
f′(x) = 2 sec 2x tan 2x
f ′(0) = 0
f′′(x) = ( 2sec 2 x )( 2sec 2 2 x ) + ( tan 2 x )( 4sec 2 x tan 2 x )
2 2x
= 4sec 2 x [sec 2 2 x + tan
=
] 4sec 2 x [sec2 2 x + sec2 2 x − 1]
= 4sec 2 x [ 2sec 2 2 x=
− 1] 8sec3 2 x − 4sec 2 x
f ′′(0) = 8 – 4 = 4
f′′′(x) = 24sec 2 2 x ( 2sec 2 x tan 2 x ) − 8sec 2 x tan 2 x
f ′′′(0) = 0
= 48sec3 2 x tan 2 x − 8sec 2 x tan 2 x
f iv ( x) = ( 48sec3 2 x )( 2sec 2 2 x ) + ( tan 2 x )(144sec 2 2 x(2sec 2 x tan 2 x) ) 
− ( 8sec 2 x )( 2sec 2 2 x ) + ( tan 2 x )(16sec 2 x tan 2 x ) 
f iv (0) = 96 + 0 – 16 – 0 = 80
x2
x3
f ′′(0) +
f ′′′(0) + …
2!
3!
Maclaurin’s series states: f(x) = f(0) + xf ′(0) +
= 1 + x(0) +
x2
x3
x4
(4) + (0) + (80)
2!
3!
4!
= 1 + 2 x2 +
80 4
x
24
sec 2x = 1 + 2 x 2 +
i.e.
10 4
x as far as the term in x 4
3
7. Expand e 2θ cos 3θ as far as the term in θ 2 using Maclaurin’s series.
Let f(θ) = e 2θ cos 3θ
f(0) = e0 cos 0 = 1
f ′(θ) = ( e 2θ )( −3sin 3θ ) + ( cos 3θ )( 2 e 2θ )
= e 2θ ( 2 cos 3θ − 3sin 3θ )
f ′(0) = e0 ( 2 cos 0 − 3sin 0 ) = 2
f ′′(θ) = ( e 2θ )( −6sin 3θ − 9 cos 3θ ) + ( 2 cos 3θ − 3sin 3θ )( 2 e 2θ )
Maclaurin’s series states: f(θ) = f(0) + θ f ′(0) +
= 1 + θ (2) +
θ2
2!
θ2
2!
f ′′(0) +
θ3
3!
f ′′(0) = – 9 + 4 = –5
f ′′′(0) + …
(−5)
362
© 2014, John Bird
5
e 2θ cos 3θ =+
1 2θ − θ 2 as far as the term in θ 2
2
i.e.
8. Determine the first three terms of the series for sin2 x by applying Maclaurin’s theorem.
Let f(x) = sin 2 x
f(0) = sin 0 = 0
f′(x) = 2 sin x cos x
f ′(0) = 2 sin 0 cos 0 = 0
f′′(x) = (2 sin x)(–sin x) + (cos x)(2 cos x)
= −2sin 2 x + 2 cos 2 x= 2 ( cos 2 x − sin 2 x )= 2 cos 2 x
f′′′(x) = –4 sin 2x
f ′′(0) = 2 cos 0 = 2
f ′′′(0) = –4 sin 0 = 0
f iv ( x) = –8 cos 2x
f iv (0) = –8 cos 0 = –8
f v ( x) = 16 sin 2x
f v (0) = 16 sin 0 = 0
f vi ( x) = 32 cos 2x
f vi (0) = 32 cos 0 = 32
Maclaurin’s series states: f(x) = f(0) + x f ′(0) +
= 0 + x(0) +
x3
x2
f ′′(0) +
f ′′′(0) + …
3!
2!
x2
x3
x4
x5
x6
(2) + (0) + (−8) + (0) + (32) + ...
2!
3!
4!
5!
6!
1
2 6
sin 2 x =x 2 − x 4 +
x to three terms
3
45
i.e.
9. Use Maclaurin’s series to determine the expansion of (3 + 2t)4
Let f(t) = (3 + 2t)4
f(0) = 34 = 81
f ′(t) = 4(3 + 2t)3 (2) = 8(3 + 2t)3
f ′(0) = 8(3)3 = 216
f ′′(t) = 24(3 + 2t)2 (2) = 48(3 + 2t)2
f ′′(0) = 48(3)2 = 432
f ′′′(t) = 96(3 + 2t)(2) = 192(3 + 2t)
f ′′′(0) = 192(3) = 576
f iv (t ) = 192(2) = 384
f iv (0) = 384
t2
t3
Maclaurin’s series states: f(t) = f(0) + t f ′(0) +
f ′′(0) +
f ′′′(0) + …
2!
3!
= 81 + t (216) +
t2
t3
t4
(432) + (576) + (384) + ...
2!
3!
4!
363
© 2014, John Bird
i.e.
f(t) = (3 + 2t)4 = 81 + 216t + 216t 2 + 96t 3 + 16t 4
364
© 2014, John Bird
EXERCISE 98 Page 210
1. Evaluate
∫
0.6
0.2
3esin θ d θ , correct to 3 decimal places, using Maclaurin’s series.
Let f(θ) = 3esin θ
f(0) = 3esin 0 = 3
f ′(θ) = ( 3esin θ )( cos θ )
f ′(0) = ( 3esin 0 )( cos 0 ) = 3
f ′′(θ) = ( 3esin θ )( − sin θ ) + ( cos θ )( 3esin θ cos θ )
= 3esin θ ( cos 2 θ − sin θ )
f ′′(0) = 3esin 0 ( cos 2 0 − sin 0 ) = 3
f ′′′(θ) = ( 3esin θ )( −2 cos θ sin θ − cos θ ) + ( cos 2 θ − sin θ )( 3esin θ cos θ )
f ′′′(0) = ( 3esin 0 )( −2 cos 0sin 0 − cos 0 ) + ( cos 2 0 − sin 0 )( 3esin 0 cos 0 ) = (3)(–1) + (1)(3) = 0
Maclaurin’s series states: f(θ) = f(0) + θ f ′(0) +
θ2
2!
f ′′(0) +
θ3
3!
f ′′′(0) + …
3
= 3 + 3θ + θ 2 + 0
2
Hence,
∫
0.6
0.2
3e
sin θ
3 
3
θ3 


d θ = ∫  3 + 3θ + θ 2  d θ = 3θ + θ 2 + 
0.2
2 
2
2  0.2


0.6
0.6
3
3

0.6 )  
0.2 ) 
(
(
3
3
2
2
= 3(0.6) + ( 0.6 ) +
 − 3(0.2) + ( 0.2 ) +

2
2  
2
2 

 

= (1.8 + 0.54 + 0.108) – (0.6 + 0.06 + 0.004)
= 2.448 – 0.664 = 1.784, correct to 3 decimal places
2. Use Maclaurin’s theorem to expand cos 2θ and hence evaluate, correct to 2 decimal places,
∫
1
0
cos 2θ
1
θ3
dθ
Let f(θ) = cos 2θ
f ′(θ) = –2 sin 2θ
f ′′(θ) = –4 cos 2θ
f ′′′ (θ) = 8 sin 2θ
f(0) = cos 0 = 1
f ′(0) = –2 sin 0 = 0
f ′′(0) = –4 cos 0 = – 4
f ′′′(0) = 8 sin 0 = 0
365
© 2014, John Bird
f iv ( x) = 16 cos 2θ
f iv (0) = 16 cos 0 = 16
f v ( x) = –32 sin 2θ
f v (0) = –32 sin 0 = 0
f vi ( x) = –64 cos 2θ
f vi (0) = –64 cos 0 = – 64
θ2
Maclaurin’s series states: f(θ) = f(0) + θf ′(0) +
θ2
= 1 + θ (0) +
f(x) = 1 − 2θ 2 +
i.e.
2!
2!
θ3
(−4) +
3!
(0) +
θ3
3!
θ4
4!
f ′′′(0) + …
(16) +
θ5
5!
(0) +
θ6
(−64) + ...
6!
16 θ 4 64 θ 6
−
+ ...
24
720
2
4
cos 2θ = 1 − 2θ 2 + θ 4 − θ 6 + ...
3
45
i.e.
cos 2θ
Hence,
1
θ3
2
4
2 4 4 6
θ
θ
1 − 2θ 2 + θ 4 − θ 6 + ...
2
θ
1
2
3
45
3 − 45 + ...
=
=
−
+
1
1
1
1
1
θ3
θ3
=θ
Hence,
f ′′(0) +
∫
1
0
cos 2θ
1
θ3
d=
θ
∫
1
0
−
θ3
1
3
θ3
θ3
5
2 11 4 17
− 2θ 3 + θ 3 − θ 3
3
45
5
2 11 4 17 
 − 13
3 +
2
θ
θ
θ 3 − θ 3  dθ
−

3
45


1
 2
8
14
20 
θ 3 2θ 3  2  θ 3  4  θ 3 
=  −
+ 
−
2
8
14  15  20 
3




3
3
3 0
 3
1 1 

= 1.5 − 0.75 + −  − ( 0 ) = 0.88, correct to 2 decimal places
7 75 

3. Determine the value of
∫
1
0
θ cos θ d θ , correct to 2 significant figures, using Maclaurin’s series.
From Problem 1, page 206, cos θ = 1 −
θ2
2!
+
θ4
4!
−
θ6
6!
+ ...
5
Since
θ =θ
1
2
then
9
13
1
θ2 θ2 θ 2
 θ2 θ4 θ6

−
+ ...
θ cos θ = θ 1 − + − + ...  = θ 2 − +
2! 4! 6!
2 24 720


1
2
366
© 2014, John Bird
Thus,
∫
1
θ cos θ d=
θ
0
∫
1
0
5
9
13
 1

2
2
2
θ
θ
θ
θ 2 −
+
−
+ ...  d θ


2 24 720


1
 3

7
11
13
θ 2

θ2
θ2
θ2
=  −
+
−
+ ...
3
7
11
13


(2)
(24)
(720)
2
2
2
 2
0
1 
2 1 1
= − +
−
 − ( 0 ) = 0.53, correct to 2 significant places
 3 7 132 4680 
4. Use Maclaurin’s theorem to expand
decimal places,
From page 207,
Hence,
∫
0.5
0
∫
0.5
0
x ln(x + 1) as a power series. Hence evaluate, correct to 3
x ln( x + 1) d x .
ln( x + 1) =x −
x 2 x3 x 4
+ − + ...
2 3 4
1
x 2 x3 x 4 x5


x 2  x − + − + + ...  d x
0
2 3 4 5


5
7
9
11
13
 3

0.5
x2 x2 x2 x 2 x 2 

2
...
= ∫
x −
+
−
+
−
0 
2
3
4
5
6 


x ln(1 + x=
)d x
∫
0.5
0.5
 5

7
9
11
13
15
 x2

2
2
2
2
2
x
x
x
x
x
=  −
+
−
+
−
+ ...
 5 2  7  3  9  4  11  5  13  6  15 

 
 
 
 
 
 2

2
2
2
 2
 2
0
5
7
9
11
13
1
2
1
2
2

=  ( 0.5 ) 2 − ( 0.5 ) 2 + ( 0.5 ) 2 − (0.5) 2 + (0.5) 2 + ... − [ 0]
7
27
22
65
5

= 0.07071 – 0.0126 + 0.00327 – 0.00100 + 0.00034 – …
= 0.06072… = 0.061, correct to 3 decimal places
367
© 2014, John Bird
EXERCISE 99 Page 212
 x3 − 2 x + 1 
1. Determine: lim 

x →1 2 x 3 + 3 x − 5


1
 x3 − 2 x + 1 
 3x 2 − 2  3 − 2
= lim 
=
lim 
=

x →1 2 x 3 + 3 x − 5
x →1 6 x 2 + 3

 6+3 9


 sin x 
2. Determine: lim 

x →0
 x 
 sin x 
 cos x  cos 0 1
=1
lim  =
lim  =

 =
x →0
1
1
 x  x →0  1 
 ln(1 + x) 
3. Determine: lim 

x →0
 x 
 1  1
1+ x  1
ln(1
)
x
+


=1
lim =
lim
=



x →0
 x  x →0  1  1


 x 2 − sin 3 x 
4. Determine: lim 

x →0
 3x + x 2 
 x 2 − sin 3 x 
 2 x − 3cos 3 x  −3
lim 
= lim 
= –1
=

x →0
 3 x + x 2  x →0  3 + 2 x  3
 sin θ − θ cos θ 
5. Determine: lim 

θ →0
θ3


 cos θ − (θ )( − sin θ ) + cos θ  
 sin θ − θ cos θ 
θ sin θ 
= lim 
lim 
= lim 



3
2
θ →0
θ →0
θ
3θ

 θ →0 
 3θ 2 

θ cos θ + sin θ 
θ (− sin θ ) + cos θ (1) + cos θ  1 + 1 2 1
= lim 
=
=
lim 
=

 =
θ →0
6θ
6
6
6 3

 θ →0 

368
© 2014, John Bird
 ln t 
6. Determine: lim 

t →1 t 2 − 1


1 1
t  1
1
t
ln


=
=
lim =
lim




t →1 t 2 − 1
2

 t →1  2t  2
 
 sinh x − sin x 
7. Determine: lim 

x →0
x3


 cosh x − cos x 
 sinh x + sin x 
 cosh x + cos x 
 sinh x − sin x 
= lim
lim 
=
lim
=
lim 






x →0
x3
3x 2
6x
6

 x →0 
 x →0 
 x →0 

=
1+1 2 1
= =
3
6
6
 sin θ − 1 
8. Determine: lim 

π
θ →  ln sin θ 
2




cos θ
π

 sin θ − 1 
= lim 
sin θ } sin = 1
=
=
lim 
{
 lim

π
π
π
2
θ→  1 
θ →  ln sin θ 
2 
2
cos θ  θ → 2

  sin θ 

 sec t − 1 
9. Determine: lim 

t →0
 t sin t 
 (sec t )(sec 2 t ) + (tan t )(sec t tan t ) 
 sec t − 1 
 sec t tan t 
=
lim 
lim
=
lim 




t →0
t (− sin t ) + cos t + cos t
 t sin t  t →0  t cos t + sin t  t →0 

=
1+ 0
1
=
0 +1+1 2
369
© 2014, John Bird