5.4 Logarithms.

5.57   Notation ($A_a^bf$, $A_a^b[f(t){]}$.) Let $f$ be a bounded function from the interval $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$. We will denote the area of $S_a^bf$ by $A_a^bf$. Thus

$$A_a^bf=\alpha\Big(\{(x,y)\in\mbox{{\bf R}}^2\colon a\leq x\leq b \mbox{ and } 0\leq y\leq f(x)\}\Big)$$

We will sometimes write $A_a^b[f(t)]$ instead of $A_a^bf$. Thus, for example

$$A_a^b[t^2] =\alpha\Big(\{(x,y)\in\mbox{{\bf R}}^2\colon a\leq x\leq b \mbox{ and } 0\leq y\leq x^2\}\Big)$$

We will also write $I_a^b([f(t)],P)$ and $O_a^b([f(t)],P)$ for $I_a^b(f,P)$ and $O_a^b(f,P)$ respectively.

5.58   Lemma. ^5.2 Let $a,b,$ and $c$ be real numbers such that $0<a<b$ and $c>0$. Then

$$A_{ac}^{bc}\Big[{1\over t}\Big] = A_a^b\Big[{1\over t}\Big].$$

Proof: Let $P=\{x_0,x_1,\cdots ,x_n\}$ be a partition of $[a,b]$, and let

$$cP=\{cx_0,cx_1,\cdots ,cx_n\}$$

be the partition of $[ca,cb]$ obtained by multiplying the points of $P$ by $c$.

Then

$$\alpha(I_{ac}^{bc}(\Big[{1\over t}\Big],cP)) = \sum_{i=1}^n{1\over cx_i}(cx_i - cx_{i-1}) = \sum_{i=1}^n{1\over cx_i}\cdot c(x_i - x_{i-1}) \mbox{{}}$$

$$= \sum_{i=1}^n {1\over x_i}(x_i - x_{i-1}) = \alpha(I_a^b(\Big[{1\over t}\Big],P))$$ (5.59)

and

$$\alpha(O_{ac}^{bc}(\Big[{1\over t}\Big],cP)) = \sum_{i=1}^n{1\over cx_{i-1}} (cx_i - cx_{i-1}) = \sum_{i=1}^n{1\over cx_{i-1}}\cdot c(x_i - x_{i-1}) \mbox{{}}$$

$$= \sum_{i=1}^n {1\over x_{i-1}}(x_i - x_{i-1}) = \alpha(O_a^b(\Big[{1\over t}\Big],P))$$ (5.60)

We know that

$$\alpha(I_{ac}^{bc}(\Big[{1\over t}\Big],cP)) \leq A_{ac}^{bc}... ...ver t}\Big] \leq \alpha(O_{ac}^{bc}(\Big[{1\over t}\Big],cP)).$$

Hence by (5.59) and (5.60) we have

$$\alpha(I_a^b(\Big[{1\over t}\Big],P)) \leq A_{ac}^{bc}\Big[{1\over t}\Big] \leq \alpha(O_{a}^{b}(\Big[{1\over t}\Big],P))$$

for every partition $P$ of $[a,b]$. It follows from this and the last statement of theorem 5.40 that

$$A_{ac}^{bc}\Big[{1\over t}\Big] = A_a^b\Big[{1\over t}\Big].\mbox{ $\diamondsuit$}$$

5.61   Exercise. A From lemma 5.58 we see that

$$A_a^b\Big[{1\over t}\Big]=A_{ac}^{bc}\Big[{1\over t}\Big]$$

whenever $0<a<b$, and $c>0$. Use this result to show that for $a\geq 1$ and $b\geq 1$

$$A_1^{ab}\Big[{1\over t}\Big]=A_1^a\Big[{1\over t}\Big]+A_1^b\Big[{1\over t}\Big].$$ (5.62)

5.63   Definition ($L(x)$.) We will define a function $L\colon[1,\infty )\to\mbox{{\bf R}}$ by

$$L(a)=A_1^a\displaystyle { \Big[ {1\over t}\Big]} \mbox{ for all }a\in [1,\infty ).$$

By exercise 5.61A we have

$$L(ab)=L(a)+L(b) \mbox{ for all }a\geq 1,\; b\geq 1.$$ (5.64)

In this section we will extend the domain of $L$ to all of $\mbox{${\mbox{{\bf R}}}^{+}$}$ in such a way that (5.64) holds for all $a,b\in\mbox{${\mbox{{\bf R}}}^{+}$}$.

5.65   Theorem. Let $a,b,c$ be real numbers such that $a\leq b\leq c$, and let $f$ be a bounded function from $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$. Then

$$A_a^cf=A_a^bf+A_b^cf.$$ (5.66)

Proof: We want to show

$$\alpha (S_a^cf)=\alpha (S_a^bf)+ \alpha (S_b^cf).$$

Since $S_a^cf=S_a^bf\cup S_b^cf$ and the sets $S_a^bf$ and $S_b^cf$ are almost disjoint, this conclusion follows from our assumption about additivity of area for almost disjoint sets.

I now want to extend the definition of $A_a^bf$ to cases where $b$ may be less than $a$. I want equation (5.66) to continue to hold in all cases. If $c=a$ in (5.66), we get

$$0=A_a^af=A_a^bf+A_b^af$$

i.e.,

$$A_b^af=-A_a^bf.$$

Thus we make the following definition:

5.67   Definition. Let $a,b$ be real numbers with $a\leq b$ and let $f$ be a bounded function from $[a,b]$ to $\mbox{{\bf R}}_{\geq 0}$. Then we define

$$A_b^af=-A_a^bf \mbox{ or } A_b^a[f(t)]=-A_a^b[f(t)].$$

5.68   Theorem. Let $a,b,c$ be real numbers and let $f$ be a bounded non-negative real valued function whose domain contains an interval containing $a,\;b$, and $c$. Then

$$A_a^cf=A_a^bf+A_b^cf.$$

Proof: We need to consider the six possible orderings for $a,\;b$ and $c$. If $a\leq b\leq c$ we already know the result. Suppose $b\leq c\leq a$. Then $A_b^af=A_b^cf+A_c^af$ and hence $-A_a^bf=A_b^cf-A_a^cf$, i.e., $A_a^cf=A_a^bf+A_b^cf$. The remaining four cases are left as an exercise.

5.69   Exercise. Prove the remaining four cases of theorem 5.68.

5.70   Definition (Logarithm.) If $a$ is any positive number, we define the logarithm of $a$ by

$$\ln (a)=L(a)=A_1^a\Big[ {1\over t}\Big].$$

5.71   Theorem (Properties of Logarithms.) For all $a,b\in\mbox{${\mbox{{\bf R}}}^{+}$}$ and all $r\in\mbox{{\bf Q}}$ we have

$$L(ab) = L(a)+L(b)$$

$$L\Big({a\over b}\Big) = L(a)-L(b)$$

$$L(a^{-1}) = -L(a)$$

$$L(a^r) = rL(a)$$ (5.72)

$$L(1) = 0.$$ (5.73)

Proof: Let $a,b,c\in\mbox{${\mbox{{\bf R}}}^{+}$}$. From lemma 5.58 we know that if $a\leq c$ then

$$A_a^c\Big[ {1\over t}\Big] =A_{ba}^{bc}\Big[{1\over t}\Big]$$ (5.74)

If $c<a$ we get

$$A_a^c\Big[{1\over t}\Big]=-A_c^a\Big[{1\over t}\Big]=-A_{bc}^{ba}\Big[{1\over t}\Big]=A_{ba}^{bc}\Big[{1\over t}\Big]$$

so equation (5.74) holds in all cases. Let $a,b$ be arbitrary elements in $\mbox{${\mbox{{\bf R}}}^{+}$}$. Then

$$\begin{eqnarray*} L(ab)&=&A_1^{ab}\Big[{1\over t}\Big]=A_1^a\Big[{1\over t}\Big]... ...=&A_1^a\Big[{1\over t}\Big]+A_1^b\Big[{1\over t}\Big]=L(a)+L(b). \end{eqnarray*}$$

Also

$$L(1)=A_1^1\Big[{1\over t}\Big]=0,$$

so

$$0=L(1)=L(a\cdot a^{-1})=L(a)+L(a^{-1})$$

and it follows from this that

$$L(a^{-1})=-L(a).$$

Hence

$$L\Big({a\over b}\Big)=L(a\cdot b^{-1})=L(a)+L(b^{-1})=L(a)-L(b).$$

5.75   Lemma. For all $n\in\mbox{{\bf Z}}_{\geq 0}$, $L(a^n)=nL(a)$.

Proof: The proof is by induction on $n$. For $n=0$ the lemma is clear. Suppose now that the lemma holds for some $n\in\mbox{{\bf Z}}_{\geq 0}$, i.e., suppose that $L(a^n)=nL(a)$. Then

$$L(a^{n+1})=L(a^n\cdot a)=L(a^n)+L(a)=nL(a)+L(a)=(n+1)L(a).$$

The lemma now follows by induction.

If $n\in\mbox{{\bf Z}}^-$ then $-n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ and

$$L(a^n)=L\Big( (a^{-n})^{-1}\Big)=-L(a^{-n})=-(-n)L(a)=nL(a).$$

Thus equation (5.72) holds whenever $r\in\mbox{{\bf Z}}$. If $p\in\mbox{{\bf Z}}$ and $n\in\mbox{{\bf Z}}\setminus\{0\}$, then

$$pL(a)=L(a^p)=L\Big( (a^{{p\over n}})^n\Big)=nL\Big(a^{{p\over n}}\Big)$$

so

$$L\Big( a^{{p\over n}}\Big)={p\over n}L(a).$$

Thus (5.72) holds for all $r\in Q$. $\diamondsuit$

5.76   Theorem. Let $a$ and $b$ be numbers such that $0<a<b$. Then

$$A_a^b \Big[ {1\over t}\Big] = \ln({b\over a}) = \ln(b) - \ln(a).$$

Proof: By lemma 5.58

$$A_a^b \Big[ {1\over t}\Big] = A_{aa^{-1}}^{ba^{-1}} \Big[ {1\... ...ig] = \ln({b\over a}) = \ln(b) - \ln(a).\mbox{ $\diamondsuit$}$$

Logarithms were first introduced by John Napier (1550-1632) in 1614. Napier made up the word logarithm from Greek roots meaning ratio number, and he spent about twenty years making tables of them. As far as I have been able to find out, the earliest use of $\ln$ for logarithms was by Irving Stringham in 1893[15, vol 2, page 107]. The notation $\log(x)$ is probably more common among mathematicians than $\ln(x)$, but since calculators almost always calculate our function with a key called ``ln'', and calculate something else with a key called ``log'', I have adopted the ``ln'' notation. (Napier did not use any abbreviation for logarithm.) Logarithms were seen as an important computational device for reducing multiplications to additions. The first explicit notice of the fact that logarithms are the same as areas of hyperbolic segments was made in 1649 by Alfons Anton de Sarasa (1618-1667), and this observation increased interest in the problem of calculating areas of hyperbolic segments.

5.77   Entertainment (Calculate $\ln (2)$.) Using any computer or calculator, compute $\ln (2)$ accurate to 10 decimal places. You should not make use of any special functions, e.g., it is not fair to use the ``ln" key on your calculator. There are better polygonal approximations to $A_1^2\displaystyle { \Big[{1\over t}\Big]}$ than the ones we have discussed.

The graph of the logarithm function is shown below.

We know that $\ln (1)=0$ and it is clear that $\ln$ is strictly increasing.

If $0<r<s$, then

$$\ln (s)-\ln (r)=A_r^s\Big[{1\over t}\Big]>(s-r){1\over s}>0.$$

From the fact that $\ln (a^n)=n\ln (a)$ for all $n\in\mbox{{\bf Z}}$, it is clear that $\ln$ takes on arbitrarily large positive and negative values, but the function increases very slowly. Let

$$P = \{1,{4\over 3},{5\over 3},{6\over 3} \}$$

be the regular partition of $[1,2]$ into three subintervals.

Then

$$\begin{eqnarray*} \ln(2) &=& A_1^2 \Big[{1\over t}\Big] \geq \alpha(I_1^2(\Big[... ...\over 6} = {1\over 4} + {1\over 5} + {1\over 6} = {37 \over 60}. \end{eqnarray*}$$

Now

$$\ln (2)=A_1^2\Big[{1\over t}\Big] \leq \alpha \Big(B(1,2\colon 0,1)\Big)=1,$$

and

$$\ln (4)=\ln(2^2)=2\ln (2) \geq 2\cdot{37\over {60}}>1,$$

i.e.,

$$\ln (2) \leq 1 \leq \ln (4).$$ (5.78)

There is a unique number $e\in [2,4]$ such that $\ln (e)=1$. The uniqueness is clear because $\ln$ is strictly increasing.

The existence of such a number was taken as obvious before the nineteenth century. Later we will introduce the intermediate value property which will allow us to prove that such a number $e$ exists. For the time being, we will behave like eighteenth century mathematicians, and just assert that such a number $e$ exists.

5.79   Definition ($e$.) We denote the unique number in $\mbox{${\mbox{{\bf R}}}^{+}$}$ whose logarithm is $1$ by $e$.

5.80   Exercise. A Prove that $2\leq e\leq 3$. (We already know $2\leq e$.)

5.81   Entertainment (Calculate $e$.) Using any computing power you have, calculate $e$ as accurately as you can,e.g., as a start, find the first digit after the decimal point.