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
- 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] - 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: - (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]
- (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]
- The length of a bar,l, at a temperatureθ
is given byl=l 0 eαθ,wherel 0 andαare
constants. Evaluate 1, correct to 4 signifi-
cant figures, wherel 0 = 2. 587 ,θ= 321 .7and
α= 1. 771 × 10 −^4. [2.739] - When a chain of length 2Lis suspended from
two points, 2Dmetres apart, on the same hor-
izontal level:D=k
{
ln
(
L+
√
L^2 +k^2
k
)}
.Eval-
uateDwhenk=75m andL=180m.
[120.7m]
4.2 The power series for ex
The value of excan be calculated to any required degree
of accuracy since it is defined in terms of the following
power series:
ex= 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 ofx.
The series is said toconverge, i.e. if all the terms are
added, an actual value for ex(wherexis a real number)
is obtained. The more terms that are taken, the closer
will be the value of exto its actual value. The value of
the exponent e, correct to say 4 decimal places, may be
determined by substitutingx=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 substitutingx= 0 .05 in the power series for
ex. 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 deter-
mine the value of e^0.^05 to a high degree of accuracy.
However, whenxis nearer to unity or larger than unity,
a very large number of terms are required for an accurate
result.
If in theseries of equation (1),xis 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 exmay be used
to evaluate any exponential function of the formaekx,