120 Basic Engineering Mathematics
From equation (1),
e
x
= 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
+ 0.0000217
i.e.
e
0.5
= 1.64872, correct to 6 significant
figures
Hence, 5e
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 for e x
Substituting x = −1 in the power series
e
x
= 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 places
Hence, 3e −1 = (3)(0.367858) = 1.1036, correct to 4
decimal places.
Problem 6. Expand e x (x 2 − 1) as far as the term
in x 5
The power series for e x is
e
x
= 1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+ · · ·
Hence,
e
x
(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 gives
e
x
(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
2
x
2 +
5
6
x
3 +
11
24
x
4 +
19
120
x
5
when expanded as far as the term in x 5 .
Now try the following Practice Exercise
Practice Exercise 63 Power series for e x
(answers on page 347)
1. Evaluate 5.6e −1 , correct to 4 decimal places,
using the power series for e x .
2. Use the power series for e x to determine, correct to 4 significant figures, (a) e 2 (b) e −0.3
and check your results using a calculator.
3. Expand (1 − 2x)e 2x as far as the term in x 4 .
4. Expand (2e x 2 )(x 1/2 ) to six terms.
16.3 Graphs of exponential functions
Values of e x and e −x obtained from a calculator, correct
to 2 decimal places, over a range x = −3 to x = 3, are
shown in Table 16.1.
From equation (1),
e
x
= 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
+ 0.0000217
i.e.
e
0.5
= 1.64872, correct to 6 significant
figures
Hence, 5e
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 for e x
Substituting x = −1 in the power series
e
x
= 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 places
Hence, 3e −1 = (3)(0.367858) = 1.1036, correct to 4
decimal places.
Problem 6. Expand e x (x 2 − 1) as far as the term
in x 5
The power series for e x is
e
x
= 1 + x +
x 2
2!
+
x 3
3!
+
x 4
4!
+
x 5
5!
+ · · ·
Hence,
e
x
(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 gives
e
x
(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
2
x
2 +
5
6
x
3 +
11
24
x
4 +
19
120
x
5
when expanded as far as the term in x 5 .
Now try the following Practice Exercise
Practice Exercise 63 Power series for e x
(answers on page 347)
1. Evaluate 5.6e −1 , correct to 4 decimal places,
using the power series for e x .
2. Use the power series for e x to determine, correct to 4 significant figures, (a) e 2 (b) e −0.3
and check your results using a calculator.
3. Expand (1 − 2x)e 2x as far as the term in x 4 .
4. Expand (2e x 2 )(x 1/2 ) to six terms.
16.3 Graphs of exponential functions
Values of e x and e −x obtained from a calculator, correct
to 2 decimal places, over a range x = −3 to x = 3, are
shown in Table 16.1.