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September 14, 1997

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{\bf Example}~~ p. 45, \# 13

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(a) Let $p$ be a positive integer. Prove that:
$$b^{p} - a^{p} = (b - a)(b^{p-1} + b^{p-2}a +
.... + ba^{p-2} + a^{p-1}).$$

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{\it Proof.}~~Expand the RHS by direct
calculation:
\begin{eqnarray*}
RHS = b^{p} &+& b^{p-1}a + b^{p-2}a^{2} + .... +
b^{2}a^{p-2} + ba^{p-1}\\
&-&b^{p-1}a - b^{p-2}a^{2} - ... - b^{2}a^{p-2} -
ba^{p-1} - a^{p} = b^{p} - a^{p}.
\end{eqnarray*}

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(b) Let $p$ and $n$ be positive integers. Use (a)
to show that:
$$n^{p} < {(n+1)^{p+1} - n^{p+1} \over p+1} <
(n+1)^{p}.$$

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{\it Proof.}~~Set $b = n + 1$ and $a = n$ in part
(a). Then
$$(n + 1)^{p} - n^{p} = ((n+1)^{p-1} +
(n+1)^{p-2}n + .... + (n+1)n^{p-2} + n^{p-1}).$$
The RHS above consists of $p+1$ terms and each
term satisfies the condition:
$$n^{p} \leq (n+1)^{i}n^{p-i} \leq (n+1)^{p}.$$
Hence
$$(p+1) n^{p} \leq (n + 1)^{p} - n^{p} \leq (p+1)
(n+1)^{p}$$
and (b) is obtained by dividing through by $p+1$.

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(c) Use induction to prove that:
$$\sum_{k=1}^{n-1} k^{p} < {n^{p+1} \over p+1} <
\sum_{k=1}^{n} k^{p}.$$

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{\it Proof.}~~If $n = 1$ then LHS $= 0$ the
middle term is $1/(p+1)$ and the RHS $= 1$. Thus
the inequality is verified for $n = 1$. Assume
now that the inequality is valid for $n$ we want
to show that it is also valid for $n+1$. For
$n+1$ the LHS is, by assumption
$$LHS = \sum_{k=1}^{n+1-1} k^{p} =
\sum_{k=1}^{n-1} k^{p} + n^{p} < {n^{p+1}
\over p+1} + n^{p}.$$
By (b), we conclude that
$$LHS < {n^{p+1}
\over p+1} + {(n+1)^{p+1} - n^{p+1} \over p+1} =
{(n+1)^{p+1} \over p+1}.$$
On the other hand the RHS (for $n+1$) is by
assumption:
$$RHS = \sum_{k=1}^{n+1} k^{p} =
\sum_{k=1}^{n} k^{p} + (n+1)^{p} > {n^{p+1}
\over p+1} + (n+1)^{p}.$$
Hence, by (b):
$$RHS > {n^{p+1}
\over p+1} + (n+1)^{p} > {n^{p+1}
\over p+1} + {(n+1)^{p+1} - n^{p+1} \over p+1} =
{(n+1)^{p+1} \over p+1}.$$

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We have shown in class (Theorem 1.12, p. 77) that
a monotone function
$f(x)$ defined on a closed interval is integrable.
Thus the function $f(x) = x^{p}, p$ any positive
integer is integrable on the interval $[0, b], b
> 0$. We now show, with the help of the preceding
example that (Theorem 1.23, p. 79-80):
$$\int_{0}^{b} x^{p} dx = {b^{p+1} \over p+1}.$$
{\it Proof.}~~As in the proof of Theorem 1.12, we
consider the partition of $[0, b]$:
$$x_{0} = 0, ~x_{1} = {b \over n},~ ...,~ x_{k} =
{kb
\over n},~ ..., ~x_{n} = b.$$
Thus $f(x) = x^{p}$ is approximated from below by
the step function
$$s_{n} = x_{k-1}^{p} = ({kb \over n})^{p}~{\rm
on}~(x_{k}, x_{k+1})$$
$k = 0, ..., n-1$ and from above by the step
function:
$$t_{n} = x_{k+1}^{p} = ({(k+1)b \over
n})^{p}~{\rm on}~(x_{k}, x_{k+1})$$
$k = 0, ..., n-1$. By the definition of the
integral of a step function we have:
$$\int_{0}^{b} s_{n} dx = {b \over
n}\sum_{k=0}^{n-1} ({kb
\over n})^{p} = {b \over
n}\sum_{k=1}^{n-1} ({kb
\over n})^{p} = {b^{p+1} \over n^{p+1}}
\sum_{k=1}^{n-1} k^{p}$$ (because $kb/n = 0$ if $k
= 0$) and 
$$\int_{0}^{b} t_{n} dx = {b \over
n}\sum_{k=0}^{n-1} ({(k+1)b
\over n})^{p} = {b \over
n}\sum_{k=1}^{n} ({kb
\over n})^{p} = {b^{p+1} \over n^{p+1}}
\sum_{k=1}^{n} k^{p}$$
by the trick of shifting indices (cf. the notes
on September 9). By the preceding example
\begin{eqnarray}
\int_{0}^{b} s_{n} dx < {b^{p+1}
\over n^{p+1}} {n^{p+1} \over p+1} = {b^{p+1} \over
 p+1} < \int_{0}^{b}
t_{n} dx.
\end{eqnarray}
On the other hand,
$$0 \leq \int_{0}^{b} t_{n} dx - \int_{0}^{b}
s_{n} dx = {b^{p+1} \over n^{p+1}}
\sum_{k=1}^{n} k^{p} - {b^{p+1} \over n^{p+1}}
\sum_{k=1}^{n-1} k^{p} = {b^{p+1} \over n^{p+1}}
n^{p} = {b^{p+1} \over n}$$
which tends to $0$ as $n \rightarrow \infty$, i.e.
$$\lim_{n \rightarrow \infty} \int_{0}^{b} t_{n}
dx = \lim_{n
\rightarrow \infty} \int_{0}^{b} s_{n} dx.$$
This together with the estimate (1) imply that
$$\int_{0}^{b} x^{p} dx = \lim_{n
\rightarrow \infty} \int_{0}^{b} t_{n} dx =
\lim_{n
\rightarrow \infty} \int_{0}^{b} s_{n} dx =
{b^{p+1} \over p+1}.$$

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HW p. 70 \# 6, 8, 9; p. 83 \# 21, 22, 23, 24 

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