Higher Engineering Mathematics, Sixth Edition

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 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]


6. 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,