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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

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

(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

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


\begin{displaymath}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\}\end{displaymath}

and let

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

Then for $1\leq i\leq n$

\begin{displaymath}x_i-x_{i-1}=p^{i}-p^{ i-1}=p^{ i-1}( p-1),\end{displaymath}


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

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
\end{displaymath} (7.8)

$\displaystyle \sum ([t^r],P_n,S_n)$ $\textstyle =$ $\displaystyle \sum_{i=1}^n s_i^r (x_i-x_{i-1})$  
  $\textstyle =$ $\displaystyle \sum_{i=1}^n (p^{(i-1)})^r p^{ i-1}(p -1)$  
  $\textstyle =$ $\displaystyle ( p-1) \sum_{i=1}^n\left( p^{r+1} \right)^{(i-1)}$ (7.9)
  $\textstyle =$ $\displaystyle (p-1)\left( {p^{ (r+1)n}-1\over p^{r+1}-1}\right)
= \left( (p^n)^{r+1} - 1 \right) \left( {p - 1 \over p^{r+1} - 1}\right) \mbox{{\nonumber}}$  
  $\textstyle =$ $\displaystyle \left( a^{r+1}-1\right) {{(a^{{1\over n}}-1)}\over {(a^{ {{r+1}\over

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

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{ $\diamondsuit$}

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

\begin{displaymath}\Big\{ {{x_n^\beta -1}\over {x_n-1}}\Big\}\to\beta .\end{displaymath}

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

\begin{displaymath}{{x_n^\beta -1}\over {x_n-1}}=1+x_n+\cdots +x_n^{\beta -1}.\end{displaymath}

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

\lim\Big\{ {{x_n^\beta -1}\over {x_n-1}}\Big\} &=& \lim\{1\}+\...
+\lim\{x_n^{\beta -1}\} \\
&=& 1+1+1+\cdots +1 \\
&=& \beta .

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

\lim\Big\{ {{x_n^\beta-1}\over {x_n-1}}\Big\} &=& \lim\Big\{ {...
&=& {1\over {-1}}\cdot\gamma =-\gamma=\beta.

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 ...
\displaystyle {\Big({{y_n^q-1}\over {y_n-1}}\Big)} }.\end{displaymath}

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

\begin{displaymath}\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 .\end{displaymath}

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}}}^{+}$}$, $\displaystyle { \{x_n^{{1\over q}}\} \to 1}$.

Proof: Let $\{x_n\}$ be a sequence of positive numbers such that $\{x_n\}\to 1$. Let $\displaystyle { y_n=x_n^{{1\over q}}}$ 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

\begin{displaymath}1+y_n+\cdots +y_n^{q-1}={{(1-y_n^q)}\over {1-y_n}}={{(1-x_n)}\over {1-y_n}}\end{displaymath}


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


\begin{displaymath}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

\begin{displaymath}\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

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

Proof: By lemma 7.7,

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

By theorem 7.10,

\begin{displaymath}\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}},\end{displaymath}

and putting these results together, we get

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

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

\begin{displaymath}A_a^{ca}[t^r] = a^{r+1} A_1^{c}[t^r]. \end{displaymath}

Proof: If

\begin{displaymath}P = \{ x_0,x_1,\ldots,x_n\} \end{displaymath}

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

\begin{displaymath}aP = \{ ax_0,ax_1,\ldots,ax_n\} \end{displaymath}

be the partition of $[a,ca]$ obtained by multiplying the points of $P$ by $a$. Then
\mu(aP) = a\mu(P).
\end{displaymath} (7.14)


\begin{displaymath}S = \{s_1,s_2,\ldots,s_n\}\end{displaymath}

is a sample for $P$, let

\begin{displaymath}aS = \{as_1,as_2,\ldots,as_n\} \end{displaymath}

be the corresponding sample for $aP$. Then

\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).

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

\begin{displaymath}\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].


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{ $\diamondsuit$}

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

{\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}\end{displaymath}

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

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

Since in this case $\displaystyle {\{\sum ([t^{-1}],P_n,S_n)\}\to A_1^a[{1\over t}]=\ln
(a)}$,we conclude that
\lim\{n(a^{{1\over n}}-1)\}=\ln (a) \mbox{ for all } a>1.
\end{displaymath} (7.17)

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

\begin{displaymath}2\underbrace{\sqrt{} \sqrt{} \cdots \sqrt{} }_{\mbox{15 times...
...\;\; \times 2\;\; \cdots \;\; \times 2} _{\mbox{15 times}}\;=\;\end{displaymath}

and got the result $0.693154611$. My calculator also says that
(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}}}.\end{displaymath}

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

next up previous index
Next: 8. Integrable Functions Up: 7. Still More Area Previous: 7.1 Area Under a   Index
Ray Mayer 2007-09-07