28 Higher Engineering Mathematics
Using a calculator,
v = 300e
−0.1063829 ...
= 300(0.89908025 ...)
= 269.7 volts
Now try the following exercise
Exercise 14 Further problems on
evaluating exponential functions
1. Evaluate the following, correct to 4 significant
figures: (a) e −1.8 (b) e −0.78 (c) e 10
[(a) 0.1653 (b) 0.4584 (c) 22030]
2. Evaluate the following, correct to 5 significant
figures:
(a) e 1.629 (b) e −2.7483 (c) 0.62e 4.178
[(a) 5.0988 (b) 0.064037 (c) 40.446]
In Problems 3 and 4, evaluate correct to 5 decimal
places:
3. (a)
1
7
e 3.4629 (b) 8.52e −1.2651 (c)
5e 2.6921
3e 1.1171
[(a) 4.55848 (b) 2.40444 (c) 8.05124]
4. (a)
5.6823
e −2.1347
(b)
e 2.1127 − e −2.1127
2
(c)
4(e −1.7295 − 1)
e 3.6817
[(a) 48.04106 (b) 4.07482 (c) −0.08286]
5. The length of a bar, l, at a temperature θ
is given by l = l 0 e
αθ , where l 0 and α are
constants. Evaluate 1, correct to 4 significant figures, where l 0 = 2.587, θ = 321.7 and
α = 1.771 × 10 −4 .
[2.739]
6. When a chain of length 2L is suspended from
two points, 2D metres apart, on the same horizontal level: D = k
ln
L+
√
L 2 +k 2
k
. Evaluate D when k = 75 m and L = 180 m.
[120.7m]
4.2 The power series for e x
The value of e x can be calculated to any required degree
of accuracy since it is defined in terms of the following
power series:
e
x
= 1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+ · · ·
(where 3! = 3 ×2 × 1 and is called ‘factorial 3’)
The series is valid for all values of x.
The series is said to converge, i.e. if all the terms are
added, an actual value for e x (where x is a real number)
is obtained. The more terms that are taken, the closer
will be the value of e x to its actual value. The value of
the exponent e, correct to say 4 decimal places, may be
determined by substituting x = 1 in the power series of
equation (1). Thus,
e
1
= 1 + 1 +
(1) 2
2!
+
(1) 3
3!
+
(1) 4
4!
+
(1) 5
5!
+
(1) 6
6!
+
(1) 7
7!
+
(1) 8
8!
+ · · ·
= 1 + 1 + 0.5 + 0.16667 + 0.04167
+ 0.00833 + 0.00139 + 0.00020
+ 0.00002 + · · ·
i.e. e = 2.71828 = 2.7183,
correct to 4 decimal places
The value of e 0.05 , correct to say 8 significant figures,
is found by substituting x = 0.05 in the power series for
e x . Thus
e
0.05
= 1 + 0.05 +
(0.05) 2
2!
+
(0.05) 3
3!
+
(0.05)
4
4!
+
(0.05)
5
5!
+ · · ·
= 1 + 0.05 + 0.00125 + 0.000020833
+ 0.000000260 + 0.000000003
and by adding,
e
0.05
= 1.0512711, correct to 8 significant figures
In this example, successive terms in the series grow
smaller very rapidly and it is relatively easy to determine the value of e 0.05 to a high degree of accuracy.
However, when x is nearer to unity or larger than unity,
a very large number of terms are required for an accurate
result.
If in the series of equation (1), x is replaced by −x, then,
e −x = 1 + (−x) +
(−x) 2
2!
+
(−x) 3
3!
+ · · ·
i.e. e −x = 1 − x +
x
2
2!
−
x
3
3!
+ · · ·
In a similar manner the power series for e x may be used
to evaluate any exponential function of the form a e kx ,
Using a calculator,
v = 300e
−0.1063829 ...
= 300(0.89908025 ...)
= 269.7 volts
Now try the following exercise
Exercise 14 Further problems on
evaluating exponential functions
1. Evaluate the following, correct to 4 significant
figures: (a) e −1.8 (b) e −0.78 (c) e 10
[(a) 0.1653 (b) 0.4584 (c) 22030]
2. Evaluate the following, correct to 5 significant
figures:
(a) e 1.629 (b) e −2.7483 (c) 0.62e 4.178
[(a) 5.0988 (b) 0.064037 (c) 40.446]
In Problems 3 and 4, evaluate correct to 5 decimal
places:
3. (a)
1
7
e 3.4629 (b) 8.52e −1.2651 (c)
5e 2.6921
3e 1.1171
[(a) 4.55848 (b) 2.40444 (c) 8.05124]
4. (a)
5.6823
e −2.1347
(b)
e 2.1127 − e −2.1127
2
(c)
4(e −1.7295 − 1)
e 3.6817
[(a) 48.04106 (b) 4.07482 (c) −0.08286]
5. The length of a bar, l, at a temperature θ
is given by l = l 0 e
αθ , where l 0 and α are
constants. Evaluate 1, correct to 4 significant figures, where l 0 = 2.587, θ = 321.7 and
α = 1.771 × 10 −4 .
[2.739]
6. When a chain of length 2L is suspended from
two points, 2D metres apart, on the same horizontal level: D = k
ln
L+
√
L 2 +k 2
k
. Evaluate D when k = 75 m and L = 180 m.
[120.7m]
4.2 The power series for e x
The value of e x can be calculated to any required degree
of accuracy since it is defined in terms of the following
power series:
e
x
= 1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+ · · ·
(where 3! = 3 ×2 × 1 and is called ‘factorial 3’)
The series is valid for all values of x.
The series is said to converge, i.e. if all the terms are
added, an actual value for e x (where x is a real number)
is obtained. The more terms that are taken, the closer
will be the value of e x to its actual value. The value of
the exponent e, correct to say 4 decimal places, may be
determined by substituting x = 1 in the power series of
equation (1). Thus,
e
1
= 1 + 1 +
(1) 2
2!
+
(1) 3
3!
+
(1) 4
4!
+
(1) 5
5!
+
(1) 6
6!
+
(1) 7
7!
+
(1) 8
8!
+ · · ·
= 1 + 1 + 0.5 + 0.16667 + 0.04167
+ 0.00833 + 0.00139 + 0.00020
+ 0.00002 + · · ·
i.e. e = 2.71828 = 2.7183,
correct to 4 decimal places
The value of e 0.05 , correct to say 8 significant figures,
is found by substituting x = 0.05 in the power series for
e x . Thus
e
0.05
= 1 + 0.05 +
(0.05) 2
2!
+
(0.05) 3
3!
+
(0.05)
4
4!
+
(0.05)
5
5!
+ · · ·
= 1 + 0.05 + 0.00125 + 0.000020833
+ 0.000000260 + 0.000000003
and by adding,
e
0.05
= 1.0512711, correct to 8 significant figures
In this example, successive terms in the series grow
smaller very rapidly and it is relatively easy to determine the value of e 0.05 to a high degree of accuracy.
However, when x is nearer to unity or larger than unity,
a very large number of terms are required for an accurate
result.
If in the series of equation (1), x is replaced by −x, then,
e −x = 1 + (−x) +
(−x) 2
2!
+
(−x) 3
3!
+ · · ·
i.e. e −x = 1 − x +
x
2
2!
−
x
3
3!
+ · · ·
In a similar manner the power series for e x may be used
to evaluate any exponential function of the form a e kx ,