7.2 Calculation of Area under Power Functions

7.7   Lemma. Let $r$ be a rational number such that $r\neq -1$. Let $a$ be a real number with $a > 1$. Then

$$A_1^a[t^r]=(a^{r+1}-1)\lim\left\{ {{a^{1\over n}-1}\over {a^{{r+1}\over n}-1}}\right\}.$$

(For the purposes of this lemma, we will assume that the limit exists. In theorem 7.10 we will prove that the limit exists.)

Proof: Let $n$ be a generic element of $\mbox{${\mbox{{\bf Z}}}^{+}$}$. To simplify the notation, I will write

$$p = a^{1\over n}, \mbox{ (so $p > 1)$}.$$

Let

$$P_n=\{1,a^{{1\over n}},a^{{2\over n}},\cdots ,a^{{n\over n}}\} =\{ 1,p,p^2,\cdots,p^n \} = \{x_0,x_1,x_2,\cdots,x_n\}$$

and let

$$S_n=\{1,p,p^2,\cdots,p^{n-1} \} = \{s_1,s_2,s_3\cdots,s_{n} \}.$$

Then for $1\leq i\leq n$

$$x_i-x_{i-1}=p^{i}-p^{ i-1}=p^{ i-1}( p-1),$$

so

$$\mu (P_n)=p^{ n-1}( p - 1)\leq p^n( p-1) =a\left( a^{1\over n} -1 \right).$$

It follows by the $n$th root rule (theorem 6.48) that $\{\mu (P_n)\}\to 0$. Hence it follows from theorem 7.6 that

$$A_1^a[t^r]=\lim\Big(\sum ([t^r],P_n,S_n)\Big).$$ (7.8)

Now

$$\sum ([t^r],P_n,S_n) = \sum_{i=1}^n s_i^r (x_i-x_{i-1})$$

$$= \sum_{i=1}^n (p^{(i-1)})^r p^{ i-1}(p -1)$$

$$= ( p-1) \sum_{i=1}^n\left( p^{r+1} \right)^{(i-1)}$$ (7.9)

$$= (p-1)\left( {p^{ (r+1)n}-1\over p^{r+1}-1}\right) = \left( (p^n)^{r+1} - 1 \right) \left( {p - 1 \over p^{r+1} - 1}\right) \mbox{{\nonumber}}$$

$$= \left( a^{r+1}-1\right) {{(a^{{1\over n}}-1)}\over {(a^{ {{r+1}\over n}}-1)}}.$$

Here we have used the formula for a finite geometric series. Thus, from (7.8)

$$\begin{eqnarray*} A_1^a[t^r]&=&\lim \Big\{ (a^{r+1}-1) {{(a^{{1\over n}}-1)}\ove... ...}\over {a^{ {{r+1}\over n} }-1}} \right\}.\mbox{ $\diamondsuit$} \end{eqnarray*}$$

Now we want to calculate the limit appearing in the previous lemma. In order to do this it will be convenient to prove a few general limit theorems.

7.10   Theorem. Let $\{x_n\}$ be a sequence of positive numbers such that $\{x_n\}\to 1$ and $x_n\neq 1$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. Let $\beta$ be any rational number. Then

$$\Big\{ {{x_n^\beta -1}\over {x_n-1}}\Big\}\to\beta .$$

Proof: Suppose $x_n\neq 1$ for all $n$, and $\{x_n\}\to 1$.

Case 1: Suppose $\beta =0$. Then the conclusion clearly follows.

Case 2: Suppose $\beta\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. Then by the formula for a geometric series

$${{x_n^\beta -1}\over {x_n-1}}=1+x_n+\cdots +x_n^{\beta -1}.$$

By the sum theorem and many applications of the product theorem we conclude that

$$\begin{eqnarray*} \lim\Big\{ {{x_n^\beta -1}\over {x_n-1}}\Big\} &=& \lim\{1\}+\... ... +\lim\{x_n^{\beta -1}\} \\ &=& 1+1+1+\cdots +1 \\ &=& \beta . \end{eqnarray*}$$

Case 3: Suppose $\beta\in\mbox{{\bf Z}}^-$. Let $\gamma =-\beta$. Then $\gamma\in\mbox{${\mbox{{\bf Z}}}^{+}$}$, so by Case 2 we get

$$\begin{eqnarray*} \lim\Big\{ {{x_n^\beta-1}\over {x_n-1}}\Big\} &=& \lim\Big\{ {... ...r {x_n-1}}\Big\}\\ &=& {1\over {-1}}\cdot\gamma =-\gamma=\beta. \end{eqnarray*}$$

Case 4: Suppose $\beta \displaystyle { ={p\over q} }$ where $q\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ and $p\in\mbox{{\bf Z}}$. Let $y_n=x_n^{ {1\over q} }$. Then

$${{x_n^\beta -1}\over {x_n-1}} = {{x_n^{ {p\over q}}-1}\over ... ...g)}\over \displaystyle {\Big({{y_n^q-1}\over {y_n-1}}\Big)} }.$$

Now if we could show that $\{y_n\}\to 1$, it would follow from this formula that

$$\lim\Big\{ {{x_n^\beta -1}\over {x_n-1}}\Big\}= {\lim {\left\... ...\left\{ {{y_n^q-1}\over {y_n-1}} \right\}}} ={p\over q}=\beta .$$

The next lemma shows that $\{y_n\}\to 1$ and completes the proof of theorem 7.10.

7.11   Lemma. Let $\{x_n\}$ be a sequence of positive numbers such that $\{x_n\}\to 1$, and $\{x_n\} \not = 1$ for all $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$. Then for each $q$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$, $${ \{x_n^{{1\over q}}\} \to 1}$$.

Proof: Let $\{x_n\}$ be a sequence of positive numbers such that $\{x_n\}\to 1$. Let $${ y_n=x_n^{{1\over q}}}$$ for each $n$ in $\mbox{${\mbox{{\bf Z}}}^{+}$}$. We want to show that $\{y_n\}\to 1$. By the formula for a finite geometric series

$$1+y_n+\cdots +y_n^{q-1}={{(1-y_n^q)}\over {1-y_n}}={{(1-x_n)}\over {1-y_n}}$$

so

$$(1-y_n)={{(1-x_n)}\over {1+y_n+\cdots +y_n^{q-1}}}.$$

Now

$$0 \leq \vert 1 - y_n\vert ={\vert 1-x_n\vert \over \vert 1+y_... ...n\vert \over 1+y_n+\cdots +y_n^{q-1}} \leq \vert 1 - x_n\vert.$$

Since $\{x_n\}\to 1$, we have $\lim\{\vert 1 - x_n\vert\} = 0$, so by the squeezing rule $\lim \{\vert 1 - y_n\vert\} = 0$, and hence

$$\lim \{ y_n\} = 1.\mbox{ $\diamondsuit$}$$

7.12   Lemma (Calculation of $A_1^b{[}t^r{]}$.) Let $b$ be a real number with $b > 1$, and let $r\in\mbox{{\bf Q}}\setminus \{-1\}$. Then

$$A_1^b[t^r] = {b^{r+1}-1\over r+1}.$$

Proof: By lemma 7.7,

$$A_1^b[t^r]=(b^{r+1}-1)\lim\Big\{ {{b^{{1\over n}}-1}\over {b^{ {{r+1}\over n}}-1}}\Big\}.$$

By theorem 7.10,

$$\lim\left\{ {{b^{{1\over n}}-1}\over {b^{ {{r+1}\over n} }-1}... ...r n}}-1} \over {b^{ {1\over n} }-1}}\right\}} ={1\over {r+1}},$$

and putting these results together, we get

$$A_1^b[t^r]={{b^{r+1}-1}\over {r+1}}.\mbox{ $\diamondsuit$}$$

7.13   Lemma. Let $r\in\mbox{{\bf Q}}$, and let $a,c \in \mbox{${\mbox{{\bf R}}}^{+}$}$, with $1 < c$. Then

$$A_a^{ca}[t^r] = a^{r+1} A_1^{c}[t^r].$$

Proof: If

$$P = \{ x_0,x_1,\ldots,x_n\}$$

is a partition of $[1,c]$, let

$$aP = \{ ax_0,ax_1,\ldots,ax_n\}$$

be the partition of $[a,ca]$ obtained by multiplying the points of $P$ by $a$. Then

$$\mu(aP) = a\mu(P).$$ (7.14)

If

$$S = \{s_1,s_2,\ldots,s_n\}$$

is a sample for $P$, let

$$aS = \{as_1,as_2,\ldots,as_n\}$$

be the corresponding sample for $aP$. Then

$$\begin{eqnarray*} \sum([t^r],aP,aS) &=& \sum_{i=1}^n (as_i)^r(ax_i - ax_{i-1})\\... ...um_{i=1}^n s_i^r (x_i - x_{i-1})\\ &=& a^{r+1} \sum([t^r],P,S). \end{eqnarray*}$$

Let $\{P_n\}$ be a sequence of partitions of $[1,c]$ such that $\{\mu (P_n)\}\to 0$, and for each $n\in\mbox{${\mbox{{\bf Z}}}^{+}$}$ let $S_n$ be a sample for $P_n$. It follows from (7.14) that $\{ \mu( aP_n)\} \to 0$. By the area theorem for monotonic functions (theorem 7.6), we have

$$\left\{ \sum([t^r],P_n,S_n) \right\} \to A_1^c[t^r] \mbox{ and } \left\{ \sum([t^r],aP_n,aS_n) \right\} \to A_a^{ca}[t^r].$$

Thus

$$\begin{eqnarray*} A_a^{ca}[t^r] &=& \lim \{ \sum([t^r],aP_n,aS_n) \} \\ &=& \li... ...r],P_n,S_n) \} \\ &=& a^{r+1}A_1^c[t^r]. \mbox{ $\diamondsuit$} \end{eqnarray*}$$

7.15   Theorem (Calculation of $A_a^b{[}t^r{]}$.) Let $a,b\in\mbox{${\mbox{{\bf R}}}^{+}$}$ with $a<b$, and let $r\in\mbox{{\bf Q}}$. Then

$$A_a^b[t^r]=\cases{ {\displaystyle {b^{r+1}-a^{r+1}}\over {r+1}} &if $r\neq -1$\vspace{1ex}\cr \ln (b)-\ln (a) &if $r=-1$.\cr}$$

Proof: The result for the case $r=-1$ was proved in theorem 5.76. The case $r \not = -1$ is done in the following exercise.

7.16   Exercise. A Use the two previous lemmas to prove theorem 7.15 for the case $r \not = -1$.

Remark: In the proof of lemma 7.7, we did not use the assumption $r\neq -1$ until line (7.9). For $r=-1$ equation (7.9) becomes

$$\sum([t^{-1}],P_n,S_n)=n(a^{{1\over n}}-1).$$

Since in this case $${\{\sum ([t^{-1}],P_n,S_n)\}\to A_1^a[{1\over t}]=\ln (a)}$$,we conclude that

$$\lim\{n(a^{{1\over n}}-1)\}=\ln (a) \mbox{ for all } a>1.$$ (7.17)

This formula give us method of calculating logarithms by taking square roots. We know $${2^n(a^{ {1\over {2^n}}}-1)}$$ will be near to $\ln(a)$ when $n$ is large, and $${ a^{ {1\over {2^n}}}}$$ can be calculated by taking $n$ successive square roots. On my calculator, I pressed the following sequence of keys

$$2\underbrace{\sqrt{} \sqrt{} \cdots \sqrt{} }_{\mbox{15 times... ...\;\; \times 2\;\; \cdots \;\; \times 2} _{\mbox{15 times}}\;=\;$$

and got the result $0.693154611$. My calculator also says that
$\mbox{ln} (2)=0.69314718$. It appears that if I know how to calculate square roots, then I can calculate logarithms fairly easily.

7.18   Exercise. A Let $r$ be a non-negative rational number, and let $b\in\mbox{{\bf R}}^+$. Show that

$${ A_0^b[t^r]={{b^{r+1}}\over {r+1}}}.$$

Where in your proof do you use the fact that $r\geq 0$?