Exponential functions 29
where a and k are constants. In the series of equation (1),
let x be replaced by kx. Then,
a e kx = a
1 + (kx) +
(kx) 2
2!
+
(kx) 3
3!
+ · · ·
Thus 5 e 2x = 5
1 + (2x) +
(2x) 2
2!
+
(2x) 3
3!
+ · · ·
= 5
1 + 2x +
4x 2
2
+
8x 3
6
+ · · ·
i.e. 5 e 2x = 5
1 + 2x + 2x 2 +
4
3
x 3 + · · ·
Problem 4. Determine the value of 5 e 0.5 , correct
to 5 significant figures by using the power series
for e x .
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. 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 exercise
Exercise 15 Further problems on the power
series for e x
1. Evaluate 5.6 e
−1 , correct to 4 decimal places,
using the power series for e x .
[2.0601]
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 result by using a calculator.
[(a) 7.389 (b) 0.7408]
3. Expand (1 − 2x) e 2x as far as the term in x 4 .
1 − 2x 2 −
8x 3
3
− 2x 4
4. Expand
2 e x 2
x
1
2
to six terms.
⎡
⎢
⎢
⎣
2x
1
2 + 2x
5
2 + x
9
2 +
1
3
x
13
2
+
1
12
x
17
2 +
1
60
x
21
2
⎤
⎥
⎥
⎦
4.3 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 the following table.
where a and k are constants. In the series of equation (1),
let x be replaced by kx. Then,
a e kx = a
1 + (kx) +
(kx) 2
2!
+
(kx) 3
3!
+ · · ·
Thus 5 e 2x = 5
1 + (2x) +
(2x) 2
2!
+
(2x) 3
3!
+ · · ·
= 5
1 + 2x +
4x 2
2
+
8x 3
6
+ · · ·
i.e. 5 e 2x = 5
1 + 2x + 2x 2 +
4
3
x 3 + · · ·
Problem 4. Determine the value of 5 e 0.5 , correct
to 5 significant figures by using the power series
for e x .
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. 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 exercise
Exercise 15 Further problems on the power
series for e x
1. Evaluate 5.6 e
−1 , correct to 4 decimal places,
using the power series for e x .
[2.0601]
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 result by using a calculator.
[(a) 7.389 (b) 0.7408]
3. Expand (1 − 2x) e 2x as far as the term in x 4 .
1 − 2x 2 −
8x 3
3
− 2x 4
4. Expand
2 e x 2
x
1
2
to six terms.
⎡
⎢
⎢
⎣
2x
1
2 + 2x
5
2 + x
9
2 +
1
3
x
13
2
+
1
12
x
17
2 +
1
60
x
21
2
⎤
⎥
⎥
⎦
4.3 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 the following table.