(last updated: 8:36:43 AM, October 10, 2020)
In this supplement, we develop the number \(e\approx 2.718\) and the exponential function \(e^x\) and the natural logarithm \(\ln x\).
Prove the following results by induction on \(n\).
Lemma 1. \(2^{n-1}\leq n!\) for \(n\geq1\)
Lemma 2. (finite arithmetic series) Consider a sequence that is a linear function of \(n\) given by \(a_n=mn+b\) for \(m,b\in\mathbb R\). Then \(a_{n+1}=a_n+b\) and \(a_1=m\). The summation of the first few terms is given by the average of the first and last terms multiplied by the number of terms: \[\sum_{k=1}^n a_k = \frac{n(a_1+a_n)}{2}\]
In particular, \(\sum_{k=1}^n k = \frac{n(n+1)}{2}\)
Lemma 3. (finite geometric series) Consider the sequence \(a_n=r^n\) for some \(r\in\mathbb R\). Then \[\sum_{k=0}^n r^k = \frac{1-r^n}{1-r}\]
Corollary 4. (summation of reciprocal powers of 2) \[\sum_{k=1}^n \frac1{2^k} = 1-\frac{1}{2^{n-1}}\]
The above Corollary is a direct application of the geometric summation formula, but you need to pay attention to the starting index carefully.
Lemma 5. (summation of reciprocal factorials upper bound) \[\sum_{k=0}^n \frac1{k!} < 3\]
To prove this last result, note the first Lemma in these notes, that \(2^{k-1}\leq k!\) for \(k\geq 1\). Note that \(0!=1\) by definition of factorial and that \(1!=1\), \(2!=2\cdot1=2\), \(3!=3\cdot2\cdot1=6\), etc.
Theorem 6. The sequence \(a_n=\left(1+\frac1n\right)^n\) converges to a real number \(A\) that satisfies \(2<A\leq 3\).
Prove that the above sequence is increasing and bounded above by \(3\). Here are the steps:
Then \[\frac{a_{n+1}}{a_n}=\frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1{n}\right)^{n}}\]
and then do some algebra to get \[ \frac{a_{n+1}}{a_n}= \frac{\left(\frac{n+2}{n+1}\right)^n}{\left(\frac{n+1}{n}\right)^n} \cdot \left(\frac{n+2}{n+1}\right) \] then \[\frac{a_{n+1}}{a_n}=\frac{\left(\frac{n+2}{n+1}\right)^{n}}{\left(\frac{n+1}{n}\right)^{n}}\cdot \left(\frac{n+2}{n+1}\right)\] then \[\frac{a_{n+1}}{a_n}=\left(1-\frac1{(n+1)^2}\right)^n\cdot \left(1+\frac{1}{n+1}\right)\] then show that \[\left(1-\frac1{(n+1)^2}\right)^n\geq 1-\frac{n}{(n+1)^2}\] Then simplify to get the result.
Here is the difficult work required to get the upper bound.
\(\qquad \qquad \qquad \qquad \displaystyle \left(1+\frac1{n}\right)^n=\sum_{k=0}^n{n\choose k} \frac1{n^k} \qquad \qquad \qquad \qquad (1)\)
Note that \[ \begin{aligned} {n\choose k} &=\frac{n!}{k!(n-k)!}\\ &=\frac{n(n-1)(n-2)(n-3)\cdots 3\cdot2\cdot1}{k!(n-k)!}\\ &=\frac{n(n-1)(n-2)(n-3)\cdots (n-(k-1))\cdot (n-k)!}{k!(n-k)!}\\ &=\frac{n(n-1)(n-2)(n-3)\cdots (n-(k-1))}{k!} \end{aligned} \]
Now see that in the end there are \(k\) terms being multiplied in the numerator. So we can break the \(n^k\) in the denominator of (1).
\[ \begin{aligned} {n\choose k}\frac1{n^k} &=\frac{n(n-1)(n-2)(n-3)\cdots (n-(k-1))}{k!}\frac1{n^k}\\ &=\frac{n(n-1)(n-2)(n-3)\cdots (n-(k-1))}{k!}\frac1{n \cdot n\cdot n\cdots n} \quad \text{ (with $k$ $n$'s in denominator)}\\ &=\frac{n}{n}\cdot \frac{(n-1)}{n}\cdot \frac{(n-2)}{n}\frac{(n-3)}{n} \cdots \frac{(n-(k-1))}{n} \cdot \frac1{k!}\\ &=1\cdot \left(1-\frac1n\right)\cdot \left(1-\frac2n\right)\cdot \left(1-\frac3n\right) \cdots \left(1-\frac{k-1}n\right) \cdot \frac1{k!}\\ &=\prod_{m=1}^{k-1} \left(1-\frac{m}n\right)\cdot \frac1{k!} \end{aligned} \]
Note that I have simplified the above using product notation \(\prod_{k=1}^n a_k=a_1\cdot a_2\cdot a_3\cdots a_n\). Also note that for the above calculation to be valid, we require \(k\geq2\).
Note that \(\prod_{m=1}^{k-1} \left(1-\frac{m}n\right)\cdot \frac1{k!} < \frac1{k!}\) since the multiplicative factors \(\left(1-\frac{m}n\right)<1\) for any \(m,n\in\mathbb N\). So we have that \[{n\choose k}\frac1{n^k}=\prod_{m=1}^{k-1} \left(1-\frac{m}n\right)\cdot \frac1{k!}<\frac1{k!}\] as long as \(k\geq 2\).
So we have that \[\begin{aligned} \left(1+\frac1{n}\right)^n&=\sum_{k=0}^n{n\choose k} \frac1{n^k} \\ &=1+1+\sum_{k=2}^n{n\choose k} \frac1{n^k}\\ &=1+1+\sum_{k=2}^n\prod_{m=1}^{k-1} \left(1-\frac{m}n\right)\cdot \frac1{k!}\\ &<1+1+\sum_{k=2}^n\frac1{k!} \end{aligned} \]
By Lemma 6 we have that \(\sum_{k=2}^n\frac1{k!}<1\). Thus we can conclude that \(\left(1+\frac1{n}\right)^n<3\).
Since \(a_n=\left(1+\frac1{n}\right)^n\) is an increasing sequence and is bounded above by \(3\) we know it converges by the monotone convergence theorem, and that by comparison its limit must satisfy \(A\leq 3\). Note that we cannot yet exclude \(A=3\) as a possibility. To prove that \(A<3\), we would need to find a better upper bound.
Definition 7. (definition of the real number \(e\)) \[e \overset{def}{:=} \lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n\] Note that \(2<e\leq 3\). Plugging in \(n=10^6\) gives \(e\approx 2.718\).
Now that we know \(e\in\mathbb R\) and \(2<e\leq 3\) we can discuss raising \(e\) to exponents. We can easily understand raising it to natural number powers and roots, and by extension rational exponents. This is all covered by my Chapter 1 supplementary notes on exponentiation and roots. In fact since \(e>1\) we see that \(e^x\) is defined for all \(r\in\mathbb R\). For \(x\in\mathbb N\), we have that \(e^x=e\cdot e\cdot e\cdots e\) and \(e^{1/x}=b\) such that \(b^x=e\). Note that I didn’t prove the existence of arbitrary roots, but I proved the existence of \(\sqrt2\) and argued that the argument extends to \(\sqrt a\) for any \(a>0\). For \(x=\frac mn\in\mathbb Q\), we similarly argue that \(e^x=b\) such that \(e^m=b^n\). Of course we define \(e^0=1\). For \(x>0\) and irrational we define \[e^x=\sup\{e^q \mid q\in\mathbb Q, 0<q<x\}\] and for \(x<0\) and irrational we define \(e^x=\frac1{e^{|x|}}\).
So now we have the function \(f(x)=e^x\). Using the properties of exponents (again see my supplementary notes) we have that \(x_1<x_2\) implies \(e^{x_1}<e^{x_2}\) i.e. that \(e^x\) is strictly increasing. Of course \(e^x\) satisfies all exponential properties such as \(e^{-x}=\frac1{e^x}\) and \(e^{x_1} \cdot e^{x_2}=e^{x_1+x_2}\) (note that this last property is actually a bit tricky to prove), etc.
In this way we can see that \(f(x)=e^x\) is a strictly increasing function (\(x_1<x_2\) implies \(e^{x_1}<e^{x_2}\)), and it is one to one, i.e. \(e^x=e^y\) if an only if \(x=y\). Its domain is all of \(\mathbb R\), and its range is \((0,\infty)\). Since \(f\) is one to one, we can define its inverse function.
Definition 8. Given \(f(x)=e^x\) defined above. We define its inverse function as \[\ln x=f^{-1}(x)\] We call \(f^{-1}\) the natural logarithm.
Since \(f(x)=e^x\) is strictly increasing with domain \(\mathbb R\) and range \((0,\infty)\), then its inverse function \(f^{-1}(x)=\ln x\) is also strictly increasing and has domain \((0,\infty)\) and range \(\mathbb R\). See the textbook Section 1.2 for more on functions and relations (and my brief supplementary notes). See the textbook Section 1.5 for more on inverse functions (and my brief supplementary notes).
We have yet to cover infinite series, but it is also true that \(e=\sum_{k=0}^\infty \frac1{k!}\). We can understand this in terms of sequences though. Define the sequence \(s_n=\sum_{k=0}^n \frac1{k!}\). Then we can show that \(s_n\rightarrow e\).
Here is one way to do that:
We know that \(a_n=\left(1+\frac1n\right)^n\) converges to \(e\). So we just need to show that for \(s_n\) and \(a_n\) are close and that when \(n\) is large, they are extremely close. We have already seen that \(a_n\) and \(s_n\) are somewhat related in previous calculations in these supplementary notes. In fact it can be shown that \(a_n<s_n\) for all \(n\) and that \(0<s_n-a_n<\frac3{2n}\) for all \(n\). Getting this bound is a bit tricky, but it can be done!
Then choose a cutoff \(n^*\) so that \(|a_n-e|<\frac\epsilon2\) and \(|s_n-a_n|<\frac\epsilon2\) so that a triangle inequality argument gives \(|s_n-e|<\epsilon\).
There are a variety of ways to develop the number \(e\) and the natural logarithm and of course different proofs or mathematical arguments for what is presented here. This means there are other ways to develop general exponentiation as well. They all end up being equivalent, giving us the mathematics that we are familiar with though.
In particular, another popular path, which is probably easier computationally is to develop all the theory of derivatives and integrals and define \[L(x)\overset{def}{:=}\int_1^x \frac1t dt\] Then we can show this function satisfies all of the properties we expect of the natural logarithm. So we then define \[\ln x\overset{def}{:=}L(x).\] Then we can define \[e^x\overset{def}{:=}L^{-1}(x)\] to be its inverse function. The properties of \(e^x\) then follow. SOme issues still crop up with \(x\) is irrational though. If we follow this path though, then we sidestep the issue of defining irrational exponents by defining \[x^\alpha\overset{def}{:=}e^{\alpha \ln x}\] for arbitrary \(x>0\) and \(\alpha\in\mathbb R\). It can be quite tricky to truly understand the distinction between defining a function in this way and the fact that we have already encountered these functions many times in previous courses. It takes really careful thought.
\[ \diamond \S \diamond \]