This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.
Many of you know the number e as the base of the natural logarithm,
which has the value 2.718281828459045. . . . What you may not know is that
this number is actually defined as the limit of a sequence of approximating
numbers. That is,
e
=
lim
n
→
∞
f
n
e
=
lim
n
→
∞
f
n
(1)
f
n
=
(
1
+
1
n
)
n
,
n
=
1
,
2
,
...
f
n
=
(
1
+
1
n
)
n
,
n
=
1
,
2
,
...
(2)This means, simply, that the sequence of numbers (1+1)1,(1+12)2,(1+13)3(1+1)1,(1+12)2,(1+13)3,
. . . , gets arbitrarily close to 2.718281828459045. . . . But why should such
a sequence of numbers be so important? In the next several paragraphs we
answer this question.
(MATLAB) Write a MATLAB program to evaluate the expression fn=(1+1n)nfn=(1+1n)n for n=1,2,4,8,16,32,64n=1,2,4,8,16,32,64 to show that fn≈efn≈e for large
nn.
Derivatives and the Number ee. The number fn=(1+1n)nfn=(1+1n)n arises
in the study of derivatives in the following way. Consider the function
f(x)=ax,a>1f(x)=ax,a>1
(3)and ask yourself when the derivative of f(x)f(x) equals f(x)f(x). The function f(x)f(x)
is plotted in Figure 1 for a>1a>1. The slope of the function at point
x
x is
d
f
(
x
)
d
x
=
lim
Δ
x
→
0
a
x
+
Δ
x
-
a
x
Δ
x
=
α
x
lim
Δ
x
→
0
α
Δ
x
-
1
Δ
x
.
d
f
(
x
)
d
x
=
lim
Δ
x
→
0
a
x
+
Δ
x
-
a
x
Δ
x
=
α
x
lim
Δ
x
→
0
α
Δ
x
-
1
Δ
x
.
(4)If there is a special value for a such that
limΔx→0aΔx-1Δx=1,limΔx→0aΔx-1Δx=1,
(5)then ddxf(x)ddxf(x) would equal f(x)f(x). We call this value of aa the special (or exceptional) number ee and write
f
(
x
)
=
e
x
d
d
x
f
(
x
)
=
e
x
.
f
(
x
)
=
e
x
d
d
x
f
(
x
)
=
e
x
.
(6)The number ee would then be e=f(1)e=f(1). Let's write our condition that aΔx-1ΔxaΔx-1Δx
converges to 1 as
eΔx-1≅Δx,Δxsmall
eΔx-1≅Δx,Δxsmall
(7)or as
e≅(1+Δx)1/Δx.e≅(1+Δx)1/Δx.
(8)Our definition of e=limn→∞(1+1n)1/ne=limn→∞(1+1n)1/n amounts to defining Δx=1nΔx=1n and
allowing n→∞n→∞ in order to make Δx→0Δx→0. With this definition for ee, it is
clear that the function
e
x
e
x
is defined to be (e)x(e)x :
ex=limΔx→0(1+Δx)x/Δx.ex=limΔx→0(1+Δx)x/Δx.
(9)By letting Δx=xnΔx=xn we can write this definition in the more familiar form
e
x
=
lim
n
→
∞
(
1
+
x
n
)
n
e
x
=
lim
n
→
∞
(
1
+
x
n
)
n
(10)This is our fundamental definition for the function
e
x
e
x
. When evaluated at
x=1x=1, it produces the definition of
e
e given in Equation 1.
The derivative of
e
x
e
x
is, of course,
d
d
x
e
x
=
lim
n
→
∞
n
(
1
+
x
n
)n-11n=ex.
d
d
x
e
x
=
lim
n
→
∞
n
(
1
+
x
n
)n-11n=ex.
(11)This means that Taylor's theorem may be used to find another characterization for
e
x
e
x
:
e
x
=
∑
n
=
0
∞
[
d
n
d
x
n
e
x
]
x
=
0
1
n
!
x
n
=
∑
n
=
0
∞
x
n
n
!
.
e
x
=
∑
n
=
0
∞
[
d
n
d
x
n
e
x
]
x
=
0
1
n
!
x
n
=
∑
n
=
0
∞
x
n
n
!
.
(12)When this series expansion for
e
x
e
x
is evaluated at x=1x=1, it produces the
following series for ee:
e=∑n=0∞1n!.e=∑n=0∞1n!.
(13)In this formula, nn! is the product n(n-1)(n-2)⋯(
2
)
1
n(n-1)(n-2)⋯(2)1. Read nn! as "
n
n
factorial.”
(MATLAB) Write a MATLAB program to evaluate the sum
S
N
=
∑
n
=
0
N
1
n
!
S
N
=
∑
n
=
0
N
1
n
!
(14)for N=1,2,4,8,16,32,64N=1,2,4,8,16,32,64 to show that SN≅eSN≅e for large
N
N. Compare
S
64
S
64
with
f
64
f
64
from Exercise 1. Which approximation do you prefer?
Compound Interest and the Function
e
x
e
x
. There is an example
from your everyday life that shows even more dramatically how the function
e
x
e
x
arises. Suppose you invest
V
0
V
0
dollars in a savings account that offers
100
x
100x%
annual interest. (When x=0.01x=0.01, this is 1%; when x=0.10x=0.10, this is 10%
interest.) If interest is compounded only once per year, you have the simple interest formula for
V
1
V
1
, the value of your savings account after 1 compound
(in this case, 1 year):
V1=(1+x)V0V1=(1+x)V0.
This result is illustrated in the block diagram of Figure 2.2(a). In this diagram,
your input fortune V0 is processed by the “interest block” to produce your
output fortune V1. If interest is compounded monthly, then the annual interest
is divided into 12 equal parts and applied 12 times. The compounding formula
for V12, the value of your savings after 12 compounds (also 1 year) is
V12=(1+x12)12V0.V12=(1+x12)12V0.
(15)This result is illustrated in Figure 2(b). Can you read the block diagram?
The general formula for the value of an account that is compounded n times
per year is
Vn=(1+xn)nV0.Vn=(1+xn)nV0.
(16)
V
n
V
n
is the value of your account after
n
n compounds in a year, when the annual
interest rate is 100x%.
Verify in Equation 16 that a recursion is at work that terminates at
V
n
V
n
. That is, show that
V
i
+
1
=(1+xn)V1
V
i
+
1
=(1+xn)V1 for i=0,1,...,n-1i=0,1,...,n-1
produces the result Vn=(1+xn)nV0Vn=(1+xn)nV0.
Bankers have discovered the (apparent) appeal of infinite, or continuous, compounding:
V∞=limn→∞(1+xn)nV0.V∞=limn→∞(1+xn)nV0.
(17)We know that this is just
So, when deciding between 100x1100x1 % interest compounded daily and 100x2%100x2%
interest compounded continuously, we need only compare
(1+x1365)365
versus
e
x
2
.
(1+x1365)365versus
e
x
2
.
(19)We suggest that daily compounding is about as good as continuous compounding. What do you think? How about monthly compounding?
(MATLAB) Write a MATLAB program to compute and plot
simple interest, monthly interest, daily interest, and continuous interest versus
interest rate 100x100x. Use the curves to develop a strategy for saving money.
(What does "Endorsed by" mean?)
"Reviewer's Comments: 'I recommend this book as a "required primary textbook." This text attempts to lower the barriers for students that take courses such as Principles of Electrical Engineering, […]"