EXERCISES FOR CHAPTER 6: Taylor and Maclaurin Series
1. Find the first 4 terms of the Taylor series for the following functions:
(a) ln x centered at a=1, (b)
1
centered at a=1, (c) sin x centered at a = .
x
4
Solution
1
1
2
6
, f (2) (x) = 2 , f (3) (x) = 3 , f (4 ) (x) = 4 and so
x
x
x
x
2
3
4
(x 1)
(x 1)
(x 1)
ln x = ln1 + (x 1) 1 +
(1) +
(2) +
(6) + …
2!
3!
4!
(x 1)2 (x 1)3 (x 1)4
= (x 1) +
+…
2
3
4
(a) f (x) = ln x . So f (1) (x) =
1
1
2
6
. So f (1) (x) = 2 , f (2) (x) = 3 , f (3) (x) = 4 and so
x
x
x
x
2
3
(x 1)
(x 1)
1
= 1 + (x 1) (1) +
(2) +
(6) + 2!
3!
x
= 1 (x 1) + (x 1)2 (x 1)3 (b) f (x) =
(c) f (x) = sin x . So f (1) (x) = cos x , f (2) (x) = sin x , f (3) (x) = cos x and so
(x )2
(x )3
2
2
2
4 (
4 ( 2 ) + sin x =
+ (x ) ( ) +
)+
2
4
2
2!
2
3!
2
2
3
)
)
(x
(x
2
4
4
=
1 + (x ) + 2
4
2
6
2. Find the first 3 terms of the Taylor series for the function sin x centered at a=0.5.
Use your answer to find an approximate value to sin( + )
2 10
Solution
f (x) = sin x . So
f (1) (x) = cos x ,
f (3) (x) = 4 sin x and so
f (2) (x) = 2 sin x ,
1 2
1 4
)
)
(x
(x
2
2
2
sin x = sin +
( ) +
( 4 ) + …
2
2!
4!
1
1
(x )2
(x )4
2 +4
2 +…
= 1 2
2!
4!
f (3) (x) = 3 cos x ,
70
K. A. Tsokos: Series and D.E.
1
1
( )2
( )4
1 1
sin ( + ) = 1 2 10 + 4 10 + …
2 10
2!
4!
= 1 0.0493 + 0.0004
= 0.9511
(To 4 decimal places the GDC gives the same answer.)
3. Find the Taylor series for the function x 4 + x 2 centered at a=1.
Solution
f (x) = x 4 + x 2 . f (1) (x) = 4x 3 + 1 , f (2) (x) = 12x 2 , f (3) (x) = 24x , f (4 ) (x) = 24 and all
other derivatives are zero. Thus
(x 1)2
(x 1)3
(x 1)4
12 +
24 +
24
2!
3!
4!
= 5(x 1) + 6(x 1)2 + 4(x 1)3 + (x 1)4
x 4 + x 2 = 0 + (x 1) 5 +
4. Find the first 4 terms in the Taylor series for (x 1)e x near x=1.
Solution
Either find the Taylor series for e x and then multiply by (x 1) :
f (x) = e x f (1) = e
f (1) (x) = e x f (1) (1) = e
f (2) (x) = e x f (2) (1) = e
f (3) (x) = e x f (3) (1) = e
so that
(x 1)2
(x 1)3 (x 1)e x (x 1) e + (x 1)e +
e+
e
2!
3!
(x 1)e + (x 1)2 e +
(x 1)3
(x 1)4
e+
e
2
6
or with a bit more work,
f (x) = (x 1)e x f (1) = 0
f (1) (x) = (x 1)e x + e x f (1) (1) = e
f (2) (x) = (x 1)e x + 2e x f (2) (1) = 2e
f (3) (x) = (x 1)e x + 3e x f (3) (1) = 3e
f (4 ) (x) = (x 1)e x + 4e x f (4 ) (1) = 4e
so that
(x 1)2
(x 1)3
(x 1)4
(x 1)e x 0 + (x 1)e +
2e +
3e +
4e
2!
3!
4!
(x 1)3
(x 1)4
(x 1)e + (x 1)2 e +
e+
e
2
6
71
K. A. Tsokos: Series and D.E.
5. Find the first 3 terms in the Maclaurin series for (a) sin 2 x , (b)
x
.
(d)
1 + x2
Solution
(a)
f (x) = sin 2 x f (0) = 0
f (1) (x) = 2 sin x cos x = sin 2x f (1) (0) = 0
f (2) (x) = 2 cos 2x f (2) (0) = 2
f (3) (x) = 4 sin 2x f (3) (0) = 0
f (4 ) (x) = 8 cos 2x f (4 ) (0) = 8
f (5) (x) = 16 sin 2x f (5) (0) = 0
f (6) (x) = 32 cos 2x f (6) (0) = 32
and so
x2
x4
x6
2
sin x 0 + 0 +
2+0
8+0+
32
2!
4!
6!
x 4 2x 6
x2 +
3
45
In later exercises you will see more efficient ways to do this expansion.
(b)
f (x) =
x
1 x2
f (0) = 0
1 x2 x
f (1) (x) =
2x
1
2 1 x2 =
f (0) = 1
2
(1 x 2 )3/2
1 x
3
f (2) (x) = (1 x 2 )5 /2 (2x) f (0) = 0
2
3
35
f (4 ) (x) = (1 x 2 )5 /2 (2) +
(1 x 2 )5 /2 (2x)(2x) f (0) = 3
2
22
(c)
f (x) = xe x f (0) = 0
f (1) (x) = e x xe x f (1) (0) = 1
f (2) (x) = e x e x + xe x f (2) (0) = 2
f (3) (x) = 2e x + e x xe x f (3) (0) = 3
so that
x2
x3
+3
2!
3!
3
x
x x2 +
2
xe x x 2
x
1 x
2
, (c) xe x ,
K. A. Tsokos: Series and D.E.
(d)
f (x) =
72
x
1
. Notice that since
= 1 x 2 + x 4 x 6 + , it follows that
2
1+ x
1 + x2
x
= x x3 + x5 x7 + .
2
1+ x
Of course by evaluating derivatives we also have:
f (0) (0) = 0
(1 + x 2 ) x(2x)
1 x2
f (1) (x) =
=
f (1) (0) = 1
(1 + x 2 )2
(1 + x 2 )2
2x(1 + x 2 )2 (1 x 2 )2(1 + x 2 )2x 2x(x 2 3)
(2)
f (x) =
=
f (2) (0) = 0
2 4
2 3
(1 + x )
(1 + x )
2
4
6(1 6x + x )
f (3) (x) = f (3) (0) = 6
(1 + x 2 )4
24x(5 10x 2 + x 4 )
(4 )
f (x) =
f (4 ) (0) = 0
2 5
(1 + x )
120(1 + 15x 2 15x 4 + x 6 )
f (5) (x) = f (5) (0) = 120
(1 + x 2 )6
so that the first three terms are:
x2
x3
x4
x5
x
=
0
+
x
1
+
0
+
(6)
+
0
+
120
2
6
24
120
1 + x2
= x x3 + x5
as before.
1+ x
.
6. Find the Maclaurin series for ln(1+ x) and hence that for ln 1 x Solution
The Maclaurin series for ln(1+ x) is standard:
x2 x3
x2 x3
ln(1 + x) = x + and so ln(1 x) = x .
2
3
2
3
3
5
x
x
x 2n 1
1+ x
Hence ln .
= ln(1 + x) ln(1 x) = 2(x + +
+ ) = 2
1 x 3
5
n =1 2n 1
7. The function ln(1+ x) is to be approximated by the first three terms of its Maclaurin
x2 x3
series, i.e. ln(1 + x) = x + . Estimate the maximum value of x for which the
2
3
approximation agrees with the exact value to 3 decimal places.
Solution
x4
6
x4
=
(1 + c)4 4! 4(1 + c)4
where c is between 0 and x. The maximum value of the remainder term is obtained
x4
x4
when c = 0 and so equals
. We must then have
< 5 10 4 and so x < 0.211 .
4
4
The absolute value of the remainder term in Lagrange form is
73
K. A. Tsokos: Series and D.E.
8. By using a suitable Maclaurin series given in the text find the sum to infinity of the
e2 e4 e6
3 5 7
following infinite series: (a) +
+ … , (b) 1 + + …
3! 5! 7!
2! 4! 6!
Solution
(a) sin = 0
(b) cos e
1
1
.
9. Using
1 x + x 2 x 3 + … find the Maclaurin series for the function
1+ x
2+x
Solution
1 1 1
Since
we have that
=
2 + x 2 1+ x 2
1 1 1
=
2 + x 2 1+ x 2
1
x x2 x3
(1 +
+ …)
2
2 4
8
2
3
x
1 x x
+…
= +
2 4 8 16
=
6
10. (a) Evaluate the integral
sin
2
x dx by first finding the Maclaurin approximation to
0
the integrand with 3 terms. (b) Evaluate the integral exactly and compare. (c) The
x8
2
fourth term in the Maclaurin expansion of sin x is . Confirm that your
315
estimate of the integral is consistent with the alternating series estimation theorem.
Solution
(a) The Maclaurin series for was shown in Exercise 5 to be sin 2 x x 2 Hence
/6
x 4 2x 6
x 3 x 5 2x 7
0 sin x dx 0 (x 3 + 45 )dx = ( 3 15 + 315 ) 0.0452941 .
0
(b) The exact integral is
/6
/6
/6 1 cos 2x
x sin 2x
sin( / 3)
2
sin
x
dx
=
dx
=
(
)
=
0
0
2
2
4
12
4
0
/6
/6
2
=
2
sin( / 3) 2 3 3
=
0.045293
12
24
4
x 4 2x 6
.
+
3
45
74
K. A. Tsokos: Series and D.E.
(c) The error is expected to have the same sign and less (in absolute value) as the first
term neglected i.e.
/6
0
x8
x9
dx = 315
9 315
0.045293 0.0452941 = 1.033 10
theorem.
6
/6
1.04336 10 6 . The actual error is
0
consistently with the alternating series estimation
11. Find the Maclaurin series for x sin x .
Solution
x3 x5
x4 x6
2
Since sin x = x + , it follows that x sin x = x +
.
3! 5!
3! 5!
12. (a) Find the Maclaurin series for (1 + x)m where m is not necessarily an integer and
hence show that the formula for the binomial series works for non-integral
1
exponents as well. (b) Use your answer to find the expansion of
up to the
1 x2
term in x 6 .
Solution
f (x) = (1 + x)m f (0) = 1
f (1) (x) = m(1 + x)m 1 f (1) (0) = m
f (2) (x) = m(m 1)(1 + x)m 2 f (2) (0) = m(m 1)
f (3) (x) = m(m 1)(m 2)(1 + x)m 3 f (3) (0) = m(m 1)(m 2)
f (k ) (x) = m(m 1)(m 2)(m k + 1)(1 + x)m k f (k ) (0) = m(m 1)(m 2)(m k + 1)
and so
m(m 1) 2 m(m 1)(m 2) 3
x +
x +
2!
3!
m(m 1)(m 2)(m k + 1) k
x +
+
k!
This is an infinite series. If m is a positive integer the series will stop when k = m and
will agree with the standard binomial expansion.
(b)
1
= (1 x 2 )1/2
2
1 x
1 3
1 3 5 2 2
1
2
2
2
2
2 2
= 1 + (x ) +
(x ) +
(x 2 )3 + 2
2!
3!
(1 + x)m = 1 + mx +
1 2 3x 4 5x 6
x +
+
+
2
8
16
13. For Physics students. The kinetic energy of a relativistic particle is given by
1
. Here m is the constant mass of the particle, v its
K = ( 1)mc 2 where =
v2
1 2
c
= 1+
75
K. A. Tsokos: Series and D.E.
speed and c is the constant speed of light. Use the result of the previous problem to
1
show that for v c , K mv 2 .
2
Solution
v2 =
= 1 2 c v2 1 2
c
1
1/2
2
1 v2 3 v2 1+
+
+,
2 c 2 8 c 2 2
1 v2 3 v2 1+
+
+
2 c 2 8 c 2 2
1 v2 3 v4
1
3
v
and
so
This
is
K (1 +
+
+ 1)mc 2 = mv 2 + mv 2 + .
2
4
c
2c
8c
2
8
1
approximately K mv 2 if v c since the neglected terms are “small”.
2
14. Find the Maclaurin series for sin 2 x using the series for cos2x . Hence find
sin 2 x x 2
.
lim
x0
x4
Solution
(2x)2 (2x)4 (2x)6
1 cos 2x
Since sin 2 x =
and cos 2x = 1 +
+ it follows that
2!
4!
6!
2
(2x)2 (2x)4 (2x)6
+
+ 1 1 2!
4!
6!
sin 2 x =
2
4
6
2x
64x
+
+
2x 2 3
720
=
2
4
x
2x 6
= x2 +
+
3
45
x 4 2x 6
2
x
+
+ x2
2
2
sin x x
3
45
lim
= lim
x0
x0
x4
x4
1 2x 2
= lim( +
+ )
x0
3 45
1
=
3
15. Find the first 3 terms in the Maclaurin series for cos(sin x) . Hence or otherwise find
1 cos(sin x)
.
lim
x0
x2
Solution
76
K. A. Tsokos: Series and D.E.
From cos x = 1 x2 x4 x6
x3 x5
+
+ and sin x = x +
+ we have that
2! 4! 6!
3! 5!
x3 x5
x3 x5
+
+ )2 (x +
+ )4
3!
5!
3!
5!
cos(sin x) = 1 +
2!
4!
x 2 2x 4 x 4
+
+
+
= 1
2! 3!2! 4!
x 2 5x 4
= 1
+
+
2
24
Thus
x 2 5x 4
1
(1
+
+ )
1 cos(sin x)
2
24
lim
=
lim
x0
x0
x2
x2
x 2 5x 4
( + )
24
= lim 2
x0
x2
1 5x 2 = lim x0 2
24 (x =
1
2
16. Find the first 3 terms in the Maclaurin series for sin(sin x) . Hence or otherwise find
x sin(sin x)
lim
x0
x3
Solution
We know that sin x = x x3 x5
+
+ so
3! 5!
3
x3 x5
sin(sin x) = x +
+ 3! 5!
x 3 x 5 x 3 3x 5 x 5
+
+
+
+
3! 5! 3! 3!3! 5!
x3 x5
+
= x +
3 10
= x
Thus
5
x3 x5
x3 x5
+
+
+
+ x
x
3! 5!
3! 5!
+
+
3!
5!
77
lim
x0
K. A. Tsokos: Series and D.E.
x sin(sin x)
= lim
x0
x3
x (x x3 x5
+
+ )
3 10
x3
x3 x5
= lim 3 103
x0
x
1 x2 = lim x0 3
10 1
3
17. The equation e2 x = 3x 2 has a root near x = 0 . By finding a suitable polynomial
approximation to e2 x find an approximation to this root.
=
Solution
We use the Maclaurin series for the exponential.
(2x)2
e2 x 1 + (2x) +
= 1 2x + 2x 2 . Hence
2!
1 2x + 2x 2 = 3x 2
x 2 + 2x 1 = 0
x=
2 ± 8 0.414
=
2
1.41
We want the positive root. The actual root is 0.390.
18. (a) Find the Maclaurin series for the function f (x) = ln(1 + x) and hence that for
ln(1 + x)
ln(1 + x)
. (b) By keeping the first four terms in the Maclaurin series for
x
x
ln(1 + x)
integrate the function
from x = 0 to x = 1 . (c) Estimate the magnitude and
x
sign of the error made in evaluating the integral in this way. (d) Confirm that your
results are consistent with the alternating series estimation theorem by evaluating
the integral on the GDC.
Solution
(a) With f (x) = ln(1 + x) we have f (0) = ln(1) = 0 .
1
f (1) (x) =
f (1) (0) = 1
1+ x
f (2) (x) = (1 + x)2 f (2) (0) = 1
f (3) (x) = (1)(2)(1 + x)3 f (3) (0) = (1)(2)
f (k ) (x) = (1)k 1 (k 1)!(1 + x) k f (k ) (0) = (1)k 1 (k 1)!
Hence
78
K. A. Tsokos: Series and D.E.
ln(1 + x) = x x2 x3 x4
x x2 x3
ln(1 + x)
+ + and so
= 1 +
+.
2 3
2
3
4
4
x
(b)
1
x2 x3 x4
ln(1 + x)
dx
=
x
+ +
0 x
2 2 32 4 2
0
1
1
1
1
+ 2 2 +
2
2
3
4
= 0.7986
= 1
(c) The error is less in magnitude than the first term neglected i.e.
1
= 0.0400 and has
52
the same sign (i.e. positive) so the estimate is an underestimate.
(d)The value of the integral is 0.8225. The error is thus 0.8225-0.7986=+0.0239.
19. Find the first 3 terms in the Maclaurin series for 1 x + x 2 .
Solution
Work first with the function f (x) = 1 x and then substitute x x 2 for x in the
answer.
f (0) = 1
1
1 1
f (0) = f (x) =
2
2 1 x
1
1 1
f (x) = ( )(1 x)3/2 (1) f (0) = 4
2 2
So we have that:
1 x = 1
1
1
x x2 + 2
8
and so
1
1
1 x + x 2 = 1 (x x 2 ) (x x 2 )2 + 2
8
2
2
x x
x
= 1 +
+
2 2
8
x 3x 2
+
= 1 +
2
8
20. (a) Find the Maclaurin series for the function f (x) =
1
. (b) Hence by
1 + x2
x 2n 1
. (c) Show that this series
2n 1
1
1 1 1
converges for 1 x 1 and hence show that 1 + + = .
3 5 7
4
integrating each term, show that arctan x = (1)n +1
79
K. A. Tsokos: Series and D.E.
Solution
(a) The Maclaurin series for f (x) =
1
is 1 x + x 2 x 3 + so that the series for
1+ x
1
is 1 x 2 + x 4 x 6 + . Including the remainder term we have:
1 + x2
2n + 2
1
2
4
6
n 2n
n +1 x
=
1
x
+
x
x
+
+
(1)
x
+
(1)
1 + c2
1 + x2
(b) Integrating
x
1
1
2
4
6
n 2n
0 1 + t 2 dt = 0 (1 t + t t + + (1) t )dt
f (x) =
x3 x5 x7
x 2n +1
+
+ + (1)n +1
+ Rn
3
5
7
2n + 1
x 2n + 2
t
where Rn = dt .
2
1
+
c
0
arctan x = x x 2n 1
.
2n 1
1
(c) For the remainder term it is true that:
2n + 3
2n + 3
x
x 2n + 2
x
x
1
t
2n + 2
since
1 , and so lim Rn = lim
=0
Rn = dt t
dt =
n
n 2n + 3
1 + c2
1 + c2
2n + 3
0
0
provided 1 x 1.
Hence arctan x = (1)n +1
Thus the series converges for 1 x 1 . Setting x = 1 gives
arctan1 =
1 1 1
1
= (1)n +1
= 1 + + .
3 5 7
4
2n 1
1
21. Find the Maclaurin series for the function
1
2 by using partial fractions or
1 x
otherwise.
Solution
The answer is obtained at once using the known series for
1
and letting x x 2 .
1 x
With partial fractions,
1
A
B
1
1
1
. By the cover up method, A = , B = so that
=
=
+
2
(1 x)(1 + x) 1 x 1 + x
1 x
2
2
1 1
1 1
= +
. Since
2
2 1 x 1+ x
1 x
1
= 1 + x + x2 + x3 + x4 + 1 x
1
= 1 x + x2 x3 + x4 + 1+ x
it follows that
80
K. A. Tsokos: Series and D.E.
(
)
1
1
= 2 + 2x 2 + 2x 4 + 2
2
1 x
= 1 + x2 + x4 + 22. Find the first 4 terms in the Maclaurin series for the function
x +1
by first
x 5x + 6
2
finding the partial fraction decomposition of the function.
Solution
x +1
x +1
x 5x + 6 (x 2)(x 3)
A
B
+
x2 x3
By the cover up method, A = 3, B = 4 , so that
3
4
x +1
+
2
x 5x + 6 x 2 x 3
We know that
1
1
1 x + x 2 x 3 + and so
1 + x + x 2 + x 3 + so that
1+ x
1 x
2
3
3
=
x2 2x
3 1 = 2 1 x 2
3
x x2 x3
(1 + +
+ + )
2
2 4
8
2
3x 3
3 3x 3x
+
+
+
= +
8
16
2 4
=
and
4
4
=
3 x
x3
4 1 = 3 1 x 3
4
x x2 x 3
= (1 + +
+
+ )
3
3 9
27
4 4x 4x 2 4x 3
= 81
3 9 27
so that
3 3x 3x 2 3x 3
4 4x 4x 2 4x 3
x +1
= +
+
+
+ 8
16
3 9 27
81
x 2 5x + 6 2 4
1 11x 49x 2 179x 3
+
+
+
= +
1296
6 36 216
23. (a)Find the first three non-zero terms of the Maclaurin series for f (x) = e x sin x .
7
f (x) x
(b) Hence or otherwise show that lim
= .
3
x0
6
x
Solution
We use
2
81
K. A. Tsokos: Series and D.E.
2
x2 x3
x4 x6
+ + e x = 1 x 2 +
+
2! 3!
2! 3!
x3 x5
+
sin x = x +
3! 5!
ex = 1 + x +
so that
x4 x6
x3 x5
+ )(x +
+ )
2! 3!
3! 5!
x3 x5
x3 x5
x4
x3 x5
= x +
+ x 2 (x +
+ ) + (x +
+ )
3! 5!
3! 5!
2!
3! 5!
x3 x5
x5 x5
3
= x +
+ x +
+
+
3! 5!
3! 2!
7x 3 27x 5
= x
+
+
6
40
e x sin x = (1 x 2 +
2
Hence
7x 3 27x 5
+
+ x
f (x) x
6
40
lim
= lim
x0
x0
x3
x3
7x 3 27x 5
+
+
40
= lim 6
x0
x3
7 27x 2
= lim( +
+ )
x0
6
40
7
=
6
24. Find the Maclaurin series for the functions e x and sin x , and hence expand esin x up
x
1
to the term in x 4 . Hence integrate esin x dx .
0
Solution
x2 x3
+ +
2! 3!
x3 x5
+
sin x = x +
3! 5!
so that
ex = 1 + x +
82
K. A. Tsokos: Series and D.E.
e
sin x
sin 2 x sin 3 x sin 4 x
= 1 + sin x +
+
+
+
2!
3!
4!
2
x3
= 1 + x + 3!
3
4
x3
x3
x3
+
+
+ x
x
x
3!
3!
3!
+
+
+
+
2!
3!
4!
2x 4
+ x3 + x4 + x
3!
+
+
+
= 1+ x ++
3!
3!
4!
2!
x 3 x 2 2x 4 x 3 x 4
+ +
+
= 1+ x +
3! 2! 2!3! 3! 4!
x2 x4
+
= 1+ x +
2
8
Hence
x2 3
1
e
1
1
sin x
0
x2 x4
x2 x3 x5
394
dx = (1 + x +
+ ) dx = x +
+ =
1.6417 .
2
8
2
6 40 0 240
0
The
GDC
1
gives
e
sin x
dx = 1.63187 .
0
25. The Maclaurin series for e z converges for all z including the case when z is a
complex number. Using this fact, write the Maclaurin series for e i and hence
prove Euler’s formula e i = cos + isin . Hence deduce the extraordinary relation
i
e +1 = 0 .
Solution
(i )2
(i )n
e = 1 + (i ) +
++
+
2!
n!
2 4
3 5
= 1
+
+ + i( +
+ )
2! 4!
3! 5!
= cos + i sin Substituting = gives ei = cos + i sin = 1 ei + 1 = 0 .
1
. (b) By substituting
26. (a) Find the Maclaurin series for the function f (x) =
1+ x
1
x ex , find an expansion of
in powers of ex . (c) Use the series you got in
1 + e x
(1)n +1
e x
e x
(b) to show that x dx = . (d) Evaluate the integral dx
x
e
+
1
n
1
+
e
n
=1
0
0
i
directly and hence deduce the value of
Solution
1
= 1 x + x 2 x 3 + = (1)n x n
(a)
1+ x
0
(1) n +1
n .
n=1
83
K. A. Tsokos: Series and D.E.
(b)
1
x
2 x
3x
=
1
e
+
e
e
+
=
(1)n e nx
1 + e x
0
e x
x
2 x
3x
=
e
e
+
e
=
(1)n +1e nx
1 + e x
1
nx
e x
n +1
nx
n +1 e
(c) dx = (1) e dx = (1)
1 + e x
n
1
1
0
0
0
(1)n +1
1 1
=
= 1 + 2 3
n
1
(1)n +1
e
x .
Hence
dx
==
ln(1
+
e
)
=
ln
2
n = ln 2 .
0 1 + e x
0
n =1
27. (a) Starting from cos 1 and integration over the range [0, x] establish that
x2
sin x x . By integrating this inequality again show that cos x 1 . By
2
x3
integrating once more establish that sin x x . (b) Use these results and the
6
sin x
squeeze theorem to prove that lim
= 1.
x 0
x
x
(d)
Solution
x
x
0
0
(a) cos 1 cos d 1 d
i.e.
x
x
sin 0 0 sin x x
(b) From sin we have that
x
x
0
0
sin d d
2
cos 0 2
x
x
0
x2
cos x + 1 2
x2
cos x .
2
(c) Similarly,
i.e. that 1 2
0 (1 2 ) d 0 cos d
x
x
x
3
x
sin 0
6 0
x
x3
i.e. that sin x x .
6
x3
sin x
6
84
K. A. Tsokos: Series and D.E.
(d) So we have that
x3
6
sin x
x2
1
1
x
6
x2
sin x
lim 1 lim
lim(1 )
x0
x0
x0
6
x
sin x
1 lim
1
x0
x
sin x
lim
=1
x0
x
x sin x x x2 x4
x2
+
cos x 1 28. Use the method of the previous problem to prove that 1 .
2 24
2
Solution
x2
x3
sin x and once
From the last problem, cos x 1 . Integrating once, gives x 2
6
more,
x
x
3
(
)
d
0
0 sin d
6
x
2 4
x
cos 0
2 24 0
x2 x4
cos x + 1
2 24
x2 x4
1 +
cos x
2 24
x2 x4
+
cos x
1
2 24
x2 x4
x2
+
cos x 1 so that 1 .
2 24
2
29. (a) Find the Maclaurin series for ln(1+ x) . (b) Use the alternating series estimation
x2
theorem to prove that x ln(1+ x) x . (c) Hence or otherwise prove that
2
1
1
1
ln(n + 1) ln n < . (d) Using your result in (c) show that 1 + + + > ln(n + 1) .
n
2
n
1
(e) Hence deduce that the harmonic series diverges.
k
1
Solution
85
K. A. Tsokos: Series and D.E.
x2 x3
xn
= x
+
…
n
2
3
n=1
(b) For any value of x we have an alternating series and so the sum of the series lies
x2
ln(1 + x) x .
between 2 consecutive partial sums and so x 2
1
1
1
n +1 1
1
(c) ln(1 + x) x . Set x = , then ln(1 + ) < ln(
) < ln(n + 1) ln n <
n
n
n
n
n
n
1
(Note x = 0 .)
n
(d)
1
> ln 2 ln1
1
1
> ln 3 ln 2
1 1
1
1 + + + + > ln(n + 1)
2
2 3
n
1
> ln(n + 1) ln n add side by side
n
(a) ln(1+ x) = (1)n+1
(e) We then have that lim ln(n + 1) = and so
n
1
n
diverges.
n =1
30. (a) Find the Maclaurin series for the function f (x) =
1
and hence that for
1+ x
1
. (b) By integrating both sides of the Maclaurin series for
1 + x2
1
, show that the Maclaurin series for the function f (x) = arctan x is
f (x) =
1 + x2
x 2n +1
x3 x5 x7
. (c) Using the Maclaurin series for
x +
+ … = (1) n +1
3
5
7
2n + 1
0
f (x) = arctan x up to and including the term with x13 , show that
f (x) =
1/ 3
arctan x dx 0.158459 .
(c) Show that the exact value of the integral is
0
1 1 4
ln . (d) Hence deduce that an approximate value of 3 6 2 3
0
is 3.14159 . (e) To how many decimal places is this approximation expected to
be accurate?
Solution
1
1
(a) The series for f (x) =
is standard,
= 1 x + x 2 x 3 + and so also
1+ x
1+ x
1
= 1 x2 + x4 x6 + .
2
1+ x
(b)
1/ 3
arctan x dx =
86
K. A. Tsokos: Series and D.E.
x
arctan x =
dt
1+ t
2
0
1/ 3
=
(1 t
2
)
+ t 4 dt
0
x
t3 t5
= t + 3 5
0
= x
x3 x5
x 2n +1
+
+ (1)n +1
+
3
5
2n + 1
(c)
1/ 3
1/ 3
arctan x dx 0
0
x 3 x 5 x 7 x 9 x11 x13 x 3 + 5 7 + 9 11 + 13 dx
1/ 3
x 2 x 4 x 6 x 8 x10 x12
x14 = +
+
+
2 12 30 56 90 132 182 0
0.158459
1/ 3
The error is the next term neglected i.e. 0
x15
x16
dx = 15
240
1/ 3
= 6.3 10 7 .
0
(d)
1/ 3
1/ 3
arctan x dx = x arctan x 0
1/ 3
0
0
x
dx
1 + x2
1
3 1 ln (1 + x 2 ) 1/
0
2
3
arctan
=
3
1
1
1
= 6 ln (1 + ) + ln (1)
3
2
3 2
1 1 4
=
ln 3 6 2 3
1 4
1 1 4
ln = 0.158459 = 6 3 0.158459 + ln 3.14159
2 3
3 6 2 3
(f) The error in (c) was estimated to be at most 6.3 10 7 in absolute value so we can
expect agreement with to 5 decimal places (due to rounding).
(e)
31. This problem is hard. Consider the function f (x) = sin(n arcsin x) for x < 1 . (a)
Find the first and second derivatives of the function f (x) . (b) Hence show that
(1 x 2 ) f (x) xf (x) + n 2 f (x) = 0 . (c) Use mathematical induction to prove that
for k = 0,1, 2, (1 x 2 ) f (k + 2) (x) (2k + 1)xf (k +1) (x) + (n 2 k 2 ) f (k ) (x) = 0 where
f (k ) (x) denotes the kth derivative of f (x) . (d) Hence derive the Maclaurin series
87
K. A. Tsokos: Series and D.E.
for f (x) . (e) Deduce that if n is odd, the function f (x) is a polynomial of degree n.
sin(n arcsin x)
(f) Using your answer in (d) or otherwise, find lim
.
x0
x
Solution
nx cos(n arcsin x) n 2 sin(n arcsin x)
n
(a) f (x) = cos(n arcsin x)
and f (x) =
.
(1 x 2 )3/2
1 x2
1 x2
(b)
(1 x 2 ) f (x) xf (x) + n 2 f (x) =
nx cos(n arcsin x) n 2 sin(n arcsin x) xn cos(n arcsin x)
+ n 2 sin(n arcsin x)
= (1 x 2 ) 2 3/2
2
2
(1 x )
1 x
1 x
=
xn cos(n arcsin x)
nx cos(n arcsin x)
n 2 sin(n arcsin x) + n 2 sin(n arcsin x)
2 1/2
2
(1 x )
1 x
=0
(c) The proposition was proven true for k = 0 . Assume that it is also true for some
k 0 i.e. that (1 x 2 ) f (k + 2) (x) (2k + 1)xf (k +1) (x) + (n 2 k 2 ) f (k ) (x) = 0 .
Then differentiating once we find
(1 x 2 ) f (k + 3) (x) 2xf (k + 2) (x) (2k + 1) f (k +1) (x) + (2k + 1)xf (k + 2) (x) + (n 2 k 2 ) f (k +1) (x) = 0
(1 x 2 ) f (k + 3) (x) + (2k + 3)xf (k + 2) (x) + (n 2 k 2 2k 1) f (k +1) (x) = 0
(1 x 2 ) f (k + 3) (x) + (2k + 3)xf (k + 2) (x) + (n 2 (k + 1)2 ) f (k +1) (x) = 0
which is precisely what needs to be proved. The statement is true for k=0, and when
assumed true for k it was proved true for k+1. So it is true for all k .
(d) f (0) = sin(n arcsin 0) = 0 , f (0) = cos(n arcsin 0)
n
= n and f (0) = 0 . Using
1 02
(1 x 2 ) f (k + 2) (x) (2k + 1)xf (k +1) (x) + (n 2 k 2 ) f (k ) (x) = 0 evaluated at x=0 we get
f (k + 2) (0) = (k 2 n 2 ) f (k ) (0) and since f (0) = 0 all even derivatives vanish. The odd
derivatives are:
f (3) (0) = (12 n 2 ) f (1) (0) = (12 n 2 )n
f (5) (0) = (32 n 2 ) f (3) (0) = (32 n 2 )(12 n 2 )n and so on.
Thus,
(12 n 2 )n 3 (32 n 2 )(12 n 2 )n 5
sin(n arcsin x) = nx +
x +
x +
3!
5!
(e) If n is odd, the series will stop and the function is thus a polynomial of degree n.
(12 n 2 )n 3
nx
+
x +
sin(n arcsin x)
3!
= lim
= n.
(f) lim
x0
x0
x
x
88
K. A. Tsokos: Series and D.E.
32. (a) Write down the Maclaurin series for the function f (x) = sin x and hence that for
sin x
. (b) Use this series to find an approximate value of the integral
f (x) =
x
1
sin x
0 x dx by keeping the first three terms in the Maclaurin series. (c) Estimate the
error made in evaluating this integral in this way. (d) Confirm your answer by
numerically evaluating the integral on your GDC.
Solution
x3 x5
x 2n +1
+
+ (1)n +1
+ and so
3! 5!
(2n + 1)!
x2 x4
sin x
x 2n
= 1
+
+ (1)n +1
+ . Hence
3! 5!
(2n + 1)!
x
(b)
1
1
x2 x4
sin x
x 2n
n
0 x dx = 0 1 3! + 5! + (1) (2n + 1)! + dx
(a) sin x = x 1
x5
x3
x 2n +1
n
+
+ (1)
+
= x
3 3! 5 5!
(2n + 1)(2n + 1)!
0
1
1
+
3 3! 5 5!
0.946111
1
(c) The series is alternating so by the alternating series estimation theorem the error is
less than the absolute value of the fourth term in the series i.e.
1
= 0.0000283447 .
7 7!
(d) The value of the integral is 0.946083 . We have that S S3 = 0.0000280447 . The
error has the same sign as the first term neglected (the fourth) and is smaller (in absolute
value) in accordance with the alternating series estimation theorem.
1
1
33. Consider the improper integrals I = sin( ) dx and J = sin 2 ( ) dx . (a) By
x
x
1
1
1
1
1
sin u
sin 2 u
making the substitution u = , show that I = 2 du and J = 2 du . (b) By
x
u
u
0
0
using appropriate MacLaurin series, determine whether the integrals I and J exist.
Solution
0
1
1
1
sin u
sin u
(a) u =
and so du = 2 dx . Thus I = 2 du = 2 du and similarly for J.
x
x
u
u
1
0
(Note the change in the limits of integration.)
(b) We have that
89
K. A. Tsokos: Series and D.E.
1
1 u u3
sin u
u 2n 1
I = 2 du = + + (1)n
+ du . All terms are finite except
u
u 3! 5!
(2n + 1)!
0
0
1
1
for the first,
1
u du = ln u
1
0
0
2
For J use
= lim+ ln that diverges. Hence the integral does not exist.
0
sin u 1 cos 2u
=
so that
2u 2
u2
1
2n
1 u2 u4
n u
J = 2 1 (1 +
+ (1)
+ ) du
2u 2! 4!
(2n)!
0
2n
1 u2 u4
n u
+
(1)
+ ) du
0 2u 2 2! 4!
(2n)!
1
=
1
2n 2
1 1 u2
n u
= + (1)
+ ) du
2 0 2! 4!
(2n)!
The integral is therefore finite.
34. For Physics students. In Probability, the exponential probability distribution
dP
function is defined by
= e t where is a positive constant. A particular
dt
application of this distribution is to the decay of radioactive nuclei. In that case, the
probability that a particular nucleus will decay at some time T in the interval from
t = 0 to t = is given by P(0 T ) = e t dt . (a) Show that the p.d.f.
0
dP
= e t
dt
is well defined i.e. show that
e
t
dt = 1 . (b) Evaluate
0
P(0 T ) = e t dt . (c) The probability that a particular nucleus will decay
0
within a time interval called the half-life is . Deduce that the half-life is given by
ln 2
. (d) Using your answer to (b) in the case where 1 , explain why the
constant is called the probability of decay per unit time. (Hint: Do a MacLaurin
expansion of the answer in (b).)
Solution
(a)
e
0
t
dt = lim e
t
0
e t
dt = lim
= lim (e 1) = 1 .
0
e t
(b) P(0 T ) = e dt = = 1 e .
0
0
1
ln 2
1
(c) = 1 e T e T = T =
2
2
t
90
K. A. Tsokos: Series and D.E.
P(0 T ) = 1 e
(d)
=
2 2
1 (1 +
+ …) .
2!
Thus,
approximately,
P(0 T )
, is the probability of decay per unit time if 1 .
35. Consider the infinite series
(
3
n 3 + 1 n ) . (a) Find the first three terms of the
1
Maclaurin expansion of the function f (x) = 3 1 + x . (b) Use your result in (a) to
1
show that for large n the general term of the infinite series behaves as
. (c)
3n 2
Hence show that the infinite series
(
3
n 3 + 1 n ) converges.
1
Solution
(a) f (x) = (1 + x)1/ 3 . Hence f (0) = 1 and
1
1
= ,
f (0) = (1 + x)2 / 3
3
3
x=0
2
1 2
=
f (0) = ( )(1 + x)5 / 3
9
3 3
x=0
x 2x 2
+
Thus, f (x) = (1 + x) 1 + 3
9
1
1
(b) 3 n 3 + 1 = n 3 1 + 3 . When n is large, 3 is small and we may approximate
n
n
1
3 1+
by the first few terms of the MacLaurin series for f (x) = (1 + x)1/ 3 with
3
n
1
x = 3 . In other words,
n
1
1
2
3 1+
1+ 3 6 +.
3
n
3n
9n
1/ 3
Hence the general term
3
n 3 + 1 n behaves as
1
1
2
1
2
1
n n(1 + 3 6 + ) n = 2 5 2 when n is large.
3
n
3n
9n
3n
9n
3n
1
(c) Applying the limit comparison test with the series 2 we find
1 n
n3 1+
1
2
5
2
n3 + 1 n
9n = lim ( 1 2 ) = 1 .
lim
= lim 3n
n
n
n 3
1
1
9n 3
3
2
2
n
n
1
Hence the series converges since 2 does, being a p-series with p = 2 .
1 n
3