120 Basic Engineering Mathematics
From equation (1),ex= 1 +x+x^2
2!+x^3
3!+x^4
4!+···Hence, e^0.^5 = 1 + 0. 5 +( 0. 5 )^2
( 2 )( 1 )+( 0. 5 )^3
( 3 )( 2 )( 1 )+( 0. 5 )^4
( 4 )( 3 )( 2 )( 1 )+( 0. 5 )^5
( 5 )( 4 )( 3 )( 2 )( 1 )+( 0. 5 )^6
( 6 )( 5 )( 4 )( 3 )( 2 )( 1 )
= 1 + 0. 5 + 0. 125 + 0. 020833
+ 0. 0026042 + 0. 0002604
+ 0. 0000217i.e. e^0.^5 = 1. 64872 ,correct to 6 significant
figures
Hence, 5 e^0.^5 = 5 ( 1. 64872 )=8.2436,correct to 5
significant figures.Problem 5. Determine the value of 3e−^1 , correct
to 4 decimal places, using the power series forexSubstitutingx=−1inthepowerseriesex= 1 +x+x^2
2!+x^3
3!+x^4
4!+···gives e−^1 = 1 +(− 1 )+(− 1 )^2
2!+(− 1 )^3
3!+(− 1 )^4
4!+···= 1 − 1 + 0. 5 − 0. 166667 + 0. 041667
− 0. 008333 + 0. 001389
− 0. 000198 +···
= 0 .367858 correct to 6 decimal placesHence, 3 e−^1 =( 3 )( 0. 367858 )=1.1036, correct to 4
decimal places.Problem 6. Expandex(x^2 − 1 )as far as the term
inx^5The power series forexisex= 1 +x+x^2
2!+x^3
3!+x^4
4!+x^5
5!+···Hence,ex(x^2 − 1 )=(
1 +x+x^2
2!+x^3
3!+x^4
4!+x^5
5!+···)
(x^2 − 1 )=(
x^2 +x^3 +x^4
2!+x^5
3!+···)−(
1 +x+x^2
2!+x^3
3!+x^4
4!+x^5
5!+···)Grouping like terms givesex(x^2 − 1 )=− 1 −x+(
x^2 −x^2
2!)
+(
x^3 −x^3
3!)+(
x^4
2!−x^4
4!)
+(
x^5
3!−x^5
5!)
+···=− 1 −x+1
2x^2 +5
6x^3 +11
24x^4 +19
120x^5when expanded as far as the term inx^5.Now try the following Practice ExercisePracticeExercise 63 Power series forex
(answers on page 347)- Evaluate 5. 6 e−^1 , correct to 4 decimal places,
using the power series forex. - Use the power series forextodetermine, cor-
rect to 4 significant figures, (a)e^2 (b)e−^0.^3
and check your results using a calculator. - Expand (1− 2 x)e^2 xas far as the term inx^4.
- Expand (2ex
2
)(x^1 /^2 ) to six terms.16.3 Graphs of exponential functions
Values ofexande−xobtained from a calculator, correct
to 2 decimal places, over a rangex=−3tox=3, are
shown in Table 16.1.