Exponential functions 119
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
4. (a)
5.6823
e −2.1347
(b)
e 2.1127 − e −2.1127
2
(c)
4
e −1.7295 − 1
e 3.6817
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 l, correct to 4 significant
figures, where l 0 = 2.587, θ = 321.7 and
α = 1.771 × 10 −4 .
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.
16.2 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!
+ · · ·
(1)
(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 + · · ·
= 2.71828
i.e.
e = 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 + 0.0000000026
i.e. 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
ae kx , where a and k are constants.
In the series of equation (1), let x be replaced by kx.
Then
ae
kx
= a
1 + (kx ) +
(kx )
2
2!
+
(kx )
3
3!
+ · · ·
Thus, 5e
2x
= 5
1 + (2x ) +
(2x)
2
2!
+
(2x)
3
3!
+ · · ·
= 5
1 + 2x +
4x 2
2
+
8x 3
6
+ · · ·
i.e.
5e
2x
= 5
1 + 2x + 2x
2
+
4
3
x
3
+ · · ·
Problem 4. Determine the value of 5e 0.5 , correct
to 5 significant figures, by using the power series
for e x
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
4. (a)
5.6823
e −2.1347
(b)
e 2.1127 − e −2.1127
2
(c)
4
e −1.7295 − 1
e 3.6817
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 l, correct to 4 significant
figures, where l 0 = 2.587, θ = 321.7 and
α = 1.771 × 10 −4 .
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.
16.2 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!
+ · · ·
(1)
(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 + · · ·
= 2.71828
i.e.
e = 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 + 0.0000000026
i.e. 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
ae kx , where a and k are constants.
In the series of equation (1), let x be replaced by kx.
Then
ae
kx
= a
1 + (kx ) +
(kx )
2
2!
+
(kx )
3
3!
+ · · ·
Thus, 5e
2x
= 5
1 + (2x ) +
(2x)
2
2!
+
(2x)
3
3!
+ · · ·
= 5
1 + 2x +
4x 2
2
+
8x 3
6
+ · · ·
i.e.
5e
2x
= 5
1 + 2x + 2x
2
+
4
3
x
3
+ · · ·
Problem 4. Determine the value of 5e 0.5 , correct
to 5 significant figures, by using the power series
for e x