[Calculus] 6. Exponential and Logarithm

2024-02-24

Lemma 6.1 Let $m\in\N$. For every $a\in[0,\infty)$, there exists a unique $c\in[0,\infty)$, such that $c^m=a$.

Proof (Uniqueness)

Consider a map $f:[0,\infty)\to [0,\infty)(\sube\R)$, from $x$ to $x^m$. It is needed to show the map is injective.

Given two $x_1\ne x_2\in[0,\infty)$. Then $\frac{x_1}{x_2}\ne1$, i.e., $\left(\frac{x_1}{x_2}\right)^m\ne1$, thus $x_1^m\ne x_2^m$.

(Existence)

Case 1, $a=0$, then we can find $c=0$.

Case 2, $a>0$. Since the map $f:x\mapsto x^m$ is continuous. (Since $f(x)=x$ is continuous, the products by itself is also continous.)

$f(0) = 0 < a$. On the other hand, $f(a+1)=(a+1)^m > a$.

By the intermediate value theorem, $\exists c\in(0,a+1)[f(c)=c^m=a]$. □

Definition 6.2 For any $c\ge0$, we define $a^{\frac{1}{m}}$ to be such a number $c$. In other words, $c=a^{\frac{1}{m}} \Lrarr c^m=a\land c\ge0$.

More generally, for any $r\in\Q$, we define $a^r$ to be $(a^n)^m$ if $r=\frac{n}{m}$ for some $n\in\Z,m\in\N$.

Proposition 6.3 Let $a>0$ and $r_1,r_2\in\Q$.

$a^{r_1}a^{r_2}=a^{r_1+r_2}$, $(a^{r_1})^{r_2}=a^{r_1r_2}$. $r_1 < r_2, a > 1 \Rarr a^{r_1} < a^{r_2}$. If $a<1$, then $a^{r_1}>a^{r_2}$. $\lim_{n\to\infty}a^{\frac{1}{n}}=1$.

Proof Let $r_1=n_1/m_1, r_2=n_2/m_2$, where $n_1,n_2\in\Z, m_1,m_2\in\N$.

(1.1) By def, there exist two numbers $b_1,b_2$, such that $b_1^{m_1}=a^{n_1},b_2^{m_2}=a^{n_2}$.

Therefore,

\[ (b_1b_2)^{m_1m_2}= (b_1^{m_1})^{m_2}(b_2^{m_2})^{m_1}= a^{n_1m_2}a^{n_2m_1}= a^{n_1m_2+n_2m_1} \]

i.e.,

\[ b_1b_2=\left(a^{n_1m_2+n_2m_1}\right)^{\frac{1}{m_1m_2}} = a^{\frac{n_1}{m_1}+\frac{n_2}{m_2}} = a^{r_1+r_2} \]

And $b_1b_2$ is actually $a^{r_1}a^{r_2}$.

(1.2) By def, there exist two number $b_1,b_2$, such that $b_1^{m_1}=a^{n_1}$ and $b_2^{m_2}=b_1^{n_2}$.

Therefore,

\[ b_2^{m_1m_2}= (b_2^{m_2})^{m_1}= b_1^{n_2m_1}= a^{n_1n_2} \]

i.e., $b_2=a^{r_1r_2}$.

And, $b_2=b_1^{r_2}=(a^{r_1})^{r_2}$, then the proof is done.

(2)

\[ \frac{a^{r_1}}{a^{r_2}}= a^{r_1-r_2}= \left(\frac{1}{a}\right)^{r_2-r_1} \]

since $r_1< r _2$, then $r_2-r_1 > 0$. And

\[ \left(\frac{1}{a}\right)^{r_2-r_1} < 1^{r_2-r_1}=1 \]

Thus, $a^{r_1} < a^{r_2}$.

(3) Suppose $a>1$, then $a^{\frac{1}{n}}>1$.

Let $a=1+e$ for some positive $e\in\R$.

Then $a=(1+e)^n > C_n^0y^0 + C_n^1y^1=1+ny$, by binomial theorem.

Therefore, $0 < y < \frac{a-1}{n}$.

And $\lim_{n\to\infty}\frac{a-1}{n}=0$, by Squeeze Test, $\lim_{n\to\infty}y=0$.

Thus, $\lim_{n\to\infty}a^{\frac{1}{n}}=1$.

Similarly when $a<1$. □

Definition 6.4 Choose a sequence $r_n(n\in\N)$, such that $r_n\in\Q$ for all $n$, and $\lim_{n\to\infty}r_n=x$, and then define $a^x:=\lim_{n\to\infty}a^{r_n}$ if the limit exists.

However, we need to show the above definition is well-defined. There are some probloms needed to be fixed:

Does $a^{r_n}(n\in\N)$ converge? The result is relevant to the sequence we choose?

Proof

(1) The goal also means showing $a^{r_n}$ is a Cauchy sequence.

Consider this

\[ \vert a^{r_n}-a^{r_m} | = a^{r_m}|a^{r_n-r_m}-1| \]

Since $r_n$ is convergent, then it is bounded, i.e., $\exists M_1,M_2\in\Z\forall n\in\N[M_1 \le r_n \le M_2]$.

Thus, $a^{r_m}\le\max\{a^{M_1},a^{M_2}\}=: M(>0)$ for all $n\in\N$.

And we have $\lim_{n\to\infty}a^{\frac{1}{n}}=1$, i.e., $\forall\epsilon>0\exists N\in\N[k\ge N \Rarr \left|a^{\frac{1}{k}}-1\right|<\epsilon]$.

Choose the restriction to be $\frac{\epsilon}{M}$.

Since $\{r_n\}$ converges, it is a Cauchy sequence, i.e., when $n,m$ reach big enough, then $r_n-r_m$ will be small enough, so that $r_n-r_m<\frac{1}{k}$.

Then we have $a^{r_n-r_m} < a^{\frac{1}{k}} < 1+\frac{\epsilon}{M}$. Similarly, we can obtain $1-\frac{\epsilon}{M} < a^{\frac{1}{k}} < a^{r_n-r_m}$. (by Archimedean property?)

Therefore, we have:

\[ \vert a^{r_n}-a^{r_m}| = a^{r_m}|a^{r_n-r_m}-1| < M\cdot\frac{\epsilon}{M}=\epsilon \]

In other words, $\{a^{r_n}\}$ is Cauchy, then convergent.

(2) Goal: If there is another sequence $r_n'$ is also convergent to $x$, then $\lim_{n\to\infty}a^{r_n'}=\lim_{n\to\infty}a^{r_n}$.

\[ \lim{n\to\infty}a^{r_n-r_n'}=1 \] □

Proposition 6.5

Given $a>0$,

For $x_1,x_2\in\R$, then $a^{x_1}a^{x_2}=a^{x_1+x_2}$, and $(a^{x_1})^{x_2}=a^{x_1x_2}$.
If $x_1 < x_2$, then $a^{x_1} < a^{x_2}$ when $a>1$, vice versa for $a < 1$.
$f:x\mapsto a^x$ is a continuous map.
$\lim_{x\to0}a^x=1$.

Proof

Suppose there are two sequences $b_n,c_n(\in\Q)$, such that $\lim_{n\to\infty}b_n=x_1$ and $\lim_{n\to\infty}c_n=x_2$ where $x_1,x_2\in\R$.

(1.1) $\lim_{n\to\infty}(a_n+b_n)=x_1+x_2$.

Since $\lim_{n\to\infty}a^{b_n}=a^{x_1},\lim_{n\to\infty}a^{c_n}=a^{x_2}$, we have

\[ a^{x_1}a^{x_2}=\lim_{n\to\infty}(a^{b_n}a^{c_n})= \lim_{n\to\infty}a^{b_n+c_n}=a^{x_1+x_2} \]

(1.2) Choose $x_n,y_n\in\Q$, specially $x_n$ grows non-decreasingly to $x$ as $n\to\infty$, so as $y_n$.

Without loss of generality, supppose $a>1, x,y>0$.

$a^x \ge a^{x_n} > 1$, then $(a^x)^y \ge (a^{x_n})^{y_n}$.

By the law of exponentials of rationals, $(a^{x_n})^{y_n} = a^{x_ny_n}$ which converges to $a^{xy}$ non-decreasingly.

Therefore, $(a^x)^y \ge a^{xy}$, too.

Choose another $x_y,y_n$, which grows non-increasingly to $x,y$. After the same operations, we obtain $(a^x)^y \le a^{xy}$.

Thus, $(a^x)^y = a^{xy}$.

(2)

\[ \frac{a^{x_1}}{a^{x_2}}=a^{x_1-x_2} \]

Since $x_1 - x_2 < 0$ and $a>1$, $a^{x_1-x_2}<1$. Vice versa.

(3)

Our goal is to show that for any $x_0\in\R$, we have $\lim_{x\to x_0}a^x = a^{x_0}$.

We already have $\lim_{n\to\infty}a^{\frac{1}{n}}=1$, and also $\lim_{n\to\infty}a^{\frac{-1}{n}}=1$.

Which means that for any positive $\epsilon$, there exists an $N\in\N$, such that

\[ 1-\epsilon < a^{\frac{\pm1}{n}} < 1 + \epsilon \]

for all $n\ge N$. It specially holds when $n=N$.

Let $\delta = \frac{1}{N}$. Then when $|x-x_0| < \delta$, i.e., $\frac{-1}{N} < x - x_0 < \frac{1}{N}$.

By monotonicity of exponentials, we have $1-\epsilon < a^{\frac{-1}{N}} < a^{x-x_0} < a^{\frac{1}{N}} < 1 + \epsilon$.

Then

\[ \vert a^x - a^{x_0}| = a^{x_0} | a^{x-x_0} - 1 | < a^{x_0} \epsilon \]

What we need to do is choose the first positive number to be $\frac{\epsilon}{a^{x_0}}$.

(4)

By (3), $a^x$ is continuous, then $\lim_{x\to0}a^x = a^0 = 1$. □

Proposition 6.6 $f(x)=a^x$ is a bijection if $a>0$ and $a\ne1$. The inverse map $\log_a:(0,\infty)\to\R$ of exponential functions is also continuous.

Proof

(1)

(Injective) Proved by the monotonicity of exponentials

(Surjective)

For any real number $r$, there must exist two $x_1,x_2\in\R$ such that $a^{x_1} < r < a^{x_2}$. It is easy to see the contradiction if no such kind of $x_1,x_2$.

Then by intermediate value theorem, there exists a $c\in(x_1,x_2)$, such that $a^c=r$. And hence the map is surjective.

(2)

For $x,x_0\in(0,\infty)$, let $y=\log_ax, y_0=\log_ax_0$ which are both in $\R$.

Considering $|y-y_0| < \epsilon$ for any given $\epsilon>0$, we have $y < y_0+\epsilon$, with assuming $a > 1$:

\[j a^y < a^{y_0+\epsilon}\\ a^y-a^{y_0} < a^{y_0}(a^\epsilon-1)\\ x-x_0 < \delta $$

for some $\delta >0$.

Basically, for any $\epsilon > 0$, there exists a $\delta > 0$, such that if $x - x_0 < \delta$ we have $|y - y_0 = \log_a x - \log_a x_0| < \epsilon$, i.e. $\lim_{x\to x_0} \log_a x = \log_a x_0$. □

Proposition 6.7 Let $a>0$, and for any $b,c>0$.

$\log_a(bc) = \log_ab + \log_ac$
$\log_ac = \log_ab \cdot \log_bc$ if $b\ne1$
If $a>1$, then $\log_ab>\log_ac$, vice versa for $0

Proof (1)

\[ a^{\log_a(bc)}=a^{\log_ab+\log_ac}\\ bc=a^{\log_ab}\cdot a^{\log_ac}=bc \]

(2) and (3) are both similar with (1) □

Proposition 6.8 $\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ where $x\in\R$ and $n\in\N$.

And we call this limit as a number $e$, Euler's number.

Proof We already know that $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n=e$, which means for any $\forall \epsilon > 0$, there exists an $N\in\N$, such that when the index $n$ is greater than $N$, we always have $e - \epsilon < \left(1+\frac{1}{n}\right)^n < e + \epsilon$.

Let $f(x) = \left(1+\frac{1}{x}\right)^x$, and $F(n) = \left(1+\frac{1}{n}\right)^n$.

Firstly, we restrict $x$ to be greater than $n$ for some $n\ge N$, i.e., $n < x$.

We claim that $F(n) < f(x)$.

Were this false, we have $\left(1+\frac{1}{n}\right)^n \ge \left(1+\frac{1}{x}\right)^x$.

We choose $x=2n$, then

\begin{align*} \left(1+\frac{1}{n}\right)^n $\ge \left(1+\frac{1}{x}\right)^x\\ &\ge \left(1+\frac{1}{2n}\right)^{2n}\\ \left(\frac{2n(n+1)}{n(2n+1)}\right)^n &\ge \left(1+\frac{1}{n}\right)^n \end{align*}

It is obvious that the numbers inside parenthesis are both greater than 1, then by monotonicity, we have

\begin{align*} \frac{2(n+1)}{2n+1} &\ge \frac{2n+1}{2n}\\ 0 &\ge 1 \end{align*}

This induces a contradiction. Therefore $F(n) < f(x)$, then $e - \epsilon < F(n) < f(x)$.

Secondly, by Archimedean Property, we can find a natural number $m$, such that $x

We claim that $f(x) < F(m)$.

Since $m$ is even, there exists a $k\in\N$ such that $2k=m$. We can let $m$ to be big enough so that $x$ can become $k$.

Were the claim false, we have

\begin{align*} \left(1+\frac{1}{x}\right)^x &\ge \left(1+\frac{1}{m}\right)^m \\ \left(1+\frac{1}{k}\right)^k &\ge \left(1+\frac{1}{2k}\right)^{2k} \end{align*}

Similar as before, we can obtain a contradiction if we keep on deducing.

Therefore, $f(x) < F(m) < e + \epsilon$.

So with all that, for any $\epsilon > 0$, there exist $N\in\N$ and $n,m\in\N$ with $N < n < m$, such that when $n

\[ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x = e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n \] □

Proposition 6.9 $\lim_{x\to0}\frac{a^x - 1}{x} = \log_e a =: lna$ if $a>0$.

Proof Proof is divided into two parts.

(1) Consider the left limit.

Let $t=a^x - 1$, then $x = \log_a(1+x)$. Then

\begin{align*} \lim_{x\to0^+}\frac{a^x - 1}{x} &= \lim{t\to\infty}\frac{t}{\log_a(1+t)}\\ &=\lim{t\to\infty}\frac{1}{\log_a(1+t)^{\frac{1}{t}}}\\ &=\frac{1}{\log_ae} &=\ln a \end{align*}

(2) Consider the right limit.

\[ \lim_{x\to0^-}\frac{a^x - 1}{x} = \lim_{x\to0^-}a^x\frac{a^{-x} - 1}{-x} \]

Let $t = -x$, then the above limit becomes $\lim_{t\to0^+}a^{-t}\frac{a^t - 1}{t}$.

The former term is 0 as $t\to0^+$ since $a^x$ is continuous, and the later term is $\ln e$ by (1).

Therefore, we have the left limit and right limit, with equal quantity $\ln a$, then the limit is $\ln a$. □