2.2: The Function e^x
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,
\[e=\lim_{n→∞}f_n \nonumber \]
\[f_n=(1+\frac 1 n)^n\;,\;n=1,2,... \nonumber \]
This means, simply, that the sequence of numbers \((1+1)^1,(1+\frac 1 2)^2,(1+\frac 1 3)^3, . . . \), 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 \(f_n=(1+\frac 1 n)^n\) for \(n=1,2,4,8,16,32,64\) to show that \(f_n≈e\) for large \(n\).
Derivatives and the Number e
The number \(f_n=(1+\frac 1 n)^n\) arises in the study of derivatives in the following way. Consider the function
\[f(x)=a^x\;,\;a>1 \nonumber \]
and ask yourself when the derivative of \(f(x)\) equals \(f(x)\). The function \(f(x)\) is plotted in the Figure for \(a>1\). The slope of the function at point \(x\) is
\[df(x)dx=\lim_{Δx→0}{\frac {a^{x+Δx}−a^x} {Δx}} = α^x \lim_{Δx→0} {\frac {α^{Δx}−1} {Δx}} \nonumber \]
If there is a special value for \(a\) such that
\[lim_{Δx→0} \frac {a^{Δx}−1} {Δx}=1 \nonumber \]
then \(\frac d {dx} f(x)\) would equal \(f(x)\). We call this value of a the special (or exceptional) number \(e\) and write
\[f(x)=e^x \nonumber \]
\[\frac d {dx}f(x)=e^x \nonumber \]
The number \(e\) would then be \(e=f(1)\). Let's write our condition that \(\frac {a^{Δx}−1} {Δx}\) converges to 1 as
\[e^{Δx}−1≅Δx\;,\;Δx\;\mathrm{small} \nonumber \]
or as
\[e≅(1+Δx)^{1/Δx} \nonumber \]
Our definition of \(e=\lim_{n→∞}(1+1n)^{1/n}\) amounts to defining \(Δx=\frac 1 n\) and allowing \(n→∞\) in order to make \(Δx→0\). With this definition for \(e\), it is clear that the function \(e^x\) is defined to be \((e)^x\):
\[e^x=\lim_{Δx→0}(1+Δx)^{x/Δx} \nonumber \]
By letting \(Δx=\frac x n\) we can write this definition in the more familiar form
\[e^x=\lim_{n→∞}(1+\frac x n)^n \nonumber \]
This is our fundamental definition for the function \(e^x\). When evaluated at \(x=1\), it produces the definition of \(e\) given in Equation.
The derivative of \(e^x\) is, of course,
\[\frac d {dx} e^x = \lim_{n→∞}n(1+\frac x n)^{n−1}\frac 1 n = e^x \nonumber \]
This means that Taylor's theorem 1 may be used to find another characterization for \(e^x\):
\[e^x=∑^∞_{n=0}[\frac {d^n} {dx^n} e^x]_{x=0} \frac {x^n} {n!} = ∑^∞_{n=0}\frac {x^n} {n!} \nonumber \]
When this series expansion for \(e^x\) is evaluated at x=1, it produces the following series for e:
\[e=∑_{n=0}^∞\frac 1 {n!} \nonumber \]
In this formula, \(n!\) is the product \(n(n−1)(n−2)⋯(2)1\). Read \(n!\) as "n factorial.”
(MATLAB) Write a MATLAB program to evaluate the sum
\[S_N=∑_{n=0}^N\frac 1 {n!} \nonumber \]
for \(N=1,2,4,8,16,32,64\) to show that \(S_N≅e\) for large \(N\). Compare \(S_{64}\) with \(f_{64}\) from Exercise 1. Which approximation do you prefer?
Compound Interest and the Function e x
There is an example from your everyday life that shows even more dramatically how the function \(e^x\) arises. Suppose you invest \(V_0\) dollars in a savings account that offers 100x% annual interest. (When x=0.01, this is 1%; when x=0.10, this is 10% interest.) If interest is compounded only once per year, you have the simple interest formula for \(V_1\), the value of your savings account after 1 compound (in this case, 1 year):
\(V_1=(1+x)V_0\). This result is illustrated in the block diagram of the Figure. In this diagram, your input fortune \(V_0\) is processed by the “interest block” to produce your output fortune \(V_1\). If interest is compounded monthly, then the annual interest is divided into 12 equal parts and applied 12 times. The compounding formula for \(V_{12}\), the value of your savings after 12 compounds (also 1 year) is
\[V_{12}=(1+\frac x {12})^{12}V_0 \nonumber \]
This result is illustrated in Figure. Can you read the block diagram? The general formula for the value of an account that is compounded n times per year is
\[V_n=(1+\frac x n)^nV^0. \nonumber \]
\(V_n\) is the value of your account after n compounds in a year, when the annual interest rate is 100x%.
Verify in the Equation that a recursion is at work that terminates at \(V_n\). That is, show that \(V_i+1=(1+\frac x n)V_1\) for \(i=0,1,...,n−1\) produces the result \(V_n=(1+\frac x n)^nV_0\).
Bankers have discovered the (apparent) appeal of infinite, or continuous, compounding:
\[V_∞=\lim_{n→∞}(1+\frac x n)^nV_0 \nonumber \]
We know that this is just
\[V_∞=e^xV_0 \nonumber \]
So, when deciding between 100x 2 % interest compounded daily and 100x 2 % interest compounded continuously, we need only compare
\[(1+\frac {x_1} {365})^{365}\;\;versus\;\;e^{x_2} \nonumber \]
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 100x. Use the curves to develop a strategy for saving money.
Footnotes
Taylor's theorem says that a function may be completely characterized by all of its derivatives (provided they all exist)