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The Functions e^x and e^jθ: The Function e^x

Module by: Louis Scharf. E-mail the author

Note:

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.

Exercise 1

(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 fnefne 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)
Figure 1: The Function f(x)=axf(x)=ax
Figure one is a cartesian graph with horizontal axis labeled, x, and containing a labeled curve. The curve begins along the x-axis in the second quadrant to the left of the graph, and begins increasing at an increasing rate to point (0, 1), and further to point (1, a). The curve is labeled f(x) = a^x.

If there is a special value for a such that

limΔx0aΔx-1Δx=1,limΔx0aΔx-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-1Δx,Δxsmall eΔx-1Δx,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 nn in order to make Δx0Δx0. With this definition for ee, it is clear that the function e x e x is defined to be (e)x(e)x :

ex=limΔx0(1+Δx)x/Δx.ex=limΔx0(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 theorem1 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=01n!.e=n=01n!.
(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.”

Exercise 2

(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 SNeSNe 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%.

Figure 2: Block Diagram for Interest Computations; (a) Simple Annual Interest, and (b) Monthly Compounding
(a)
Figure two contains two parts, a, and b. Both will be described from left to right. Part a begins with a variable, V_0, followed by an arrow pointing right. At the end of the arrow is a rectangle containing the expression (1 + x). Another arrow pointing right follows, and at the end of this arrow is an equation, V_1 = (1 + x) V_0. Part b begins with an arrow pointing right, labeled above as V_0. At the end of the arrow is a rectangle containing the expression (1 + x/12). After this rectangle is another arrow pointing right, labeled above as V_1. The end of this arrow is followed by a rectangle containing the expression (1 + x/12), and is followed by another arrow pointing to the right. At the end of this arrow are three evenly spaced dots, and is followed by a fourth arrow labeled above as V_11. This arrow is followed by another rectangle containing the expression (1 + x/12). A final arrow then follows, pointing to the right, and labeled above as V_12. Below this chain of rectangle and arrows is an equation that extends the entire length of the diagram, and reads V_12 = (1 + x/12)V_11 = (1 + x/12)^2 V_10 . . . = (1 + x/12)^12 V_0.
(b)
Figure 2(b) (pic003.png)

Exercise 3

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

V=exV0.V=exV0.
(18)

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?

Exercise 4

(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.

Footnotes

  1. Taylor's theorem says that a function may be completely characterized by all of its derivatives (provided they all exist).

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