Exponential functions 119
In problems 3 and 4, evaluate correct to 5 decimal
places.- (a)
1
7
e^3.^4629 (b) 8. 52 e−^1.^2651(c)5 e^2.^6921
3 e^1.^1171- (a)^5.^6823
e−^2.^1347
(b)
e^2.^1127 −e−^2.^1127
2(c)4(
e−^1.^7295 − 1)e^3.^6817- The length of a bar,l, at a temperature,θ,
is given byl=l 0 eαθ,wherel 0 andαare
constants. Evaluatel, correct to 4 significant
figures, where l 0 = 2 .587,θ= 321 .7and
α= 1. 771 × 10 −^4. - 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)}EvaluateDwhenk=75m andL=180m.16.2 The power series forex
The value ofexcan 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!+··· (1)(where 3!= 3 × 2 ×1 and is called ‘factorial3’).
The series is valid for all values ofx.
The series is said toconverge; i.e., if all the terms are
added, an actual value forex(wherexis a real number)
is obtained. The more terms that are taken, the closer
will be the value ofexto its actual value. The value of
the exponente, 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 +···= 2. 71828i.e. e=2.7183,correct to 4 decimal places.The value ofe^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. 0000000026i.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 deter-
mine the value ofe^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 the series of equation (1),xis replaced by−x,
thene−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 ex may
be used to evaluate any exponential functionof the form
aekx,whereaandkare constants.
In the series of equation (1), letxbe replaced bykx.
Thenaekx=a{
1 +(kx)+(kx)^2
2!+(kx)^3
3!+···}Thus, 5 e^2 x= 5{
1 +( 2 x)+( 2 x)^2
2!+( 2 x)^3
3!+···}= 5{
1 + 2 x+4 x^2
2+8 x^3
6+···}i.e. 5 e^2 x= 5{
1 + 2 x+ 2 x^2 +4
3x^3 +···}Problem 4. Determine the value of 5e^0.^5 , correct
to 5 significant figures, by using the power series
forex