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\noindent
{\bf More examples on induction}

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\noindent
{\bf Example 2}
$$1 + 3 + 5 + ... + (2n - 1) = n^{2}$$
\noindent
{\it Proof.}~For $n = 1, LHS = 1, RHS =
1^{2} = 1.$ Assume that identity holds for
$n = k$:
$$1 + 3 + 5 + ... + (2k - 1) = k^{2}.$$
Then for $n = k + 1$,
$$LHS ~= ~1 + 3 + 5 + ... + (2k - 1) +
(2(k+1) - 1) = k^{2} + 2k + 1 = (k + 1)^{2}
= RHS.$$

\bigskip
\noindent
{\bf Example 3}

\smallskip
$$1 = 1$$
$$1 - 4 = - (1 + 2)$$
$$1 - 4 + 9 = 1 + 2 + 3$$
$$1 - 4 + 9 - 16 = - (1 + 2 + 3 + 4)$$
Guess the general formula and prove it.
$$Guess:~~1 - 4 + 9 - 16 + ... +
(-1)^{n-1}n^{2} = (-1)^{n-1}{n(n+1) \over
2}$$

\smallskip
\noindent
{\it Proof.}~For $n = 1$, obviously true.
Assume that the identity holds for $n = k$:
$$1 - 4 + 9 - 16 + ... +
(-1)^{k-1}k^{2} = (-1)^{k-1}{k(k+1) \over
2}.$$
Then
\begin{eqnarray*}
LHS &=& 1 - 4 + 9 - 16 + ...
+ (-1)^{k-1}k^{2} + (-1)^{k}(k+1)^{2}\\
&=& (-1)^{k-1}{k(k+1) \over 2} +
(-1)^{k}(k+1)^{2}\\
&=& (-1)^{k}{2(k+1)^{2} - k(k+1) \over 2}\\
&=& (-1)^{k}{(k+1)(k+2) \over
2}.
\end{eqnarray*}

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\noindent
HW: p.35 \# 1c, 3

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\noindent
{\bf Example 4}
General Properties of Sums:
$$\sum_{k=1}^{n} (a_{n} + b_{n}) =
\sum_{k=1}^{n} a_{n} + \sum_{k=1}^{n}
b_{n}~~~~(additive~property)$$
$$\sum_{k=1}^{n} ca_{n} =
c\sum_{k=1}^{n} a_{n}~~~~(homogeneous~
property)$$
$$\sum_{k=1}^{n} ca_{n} =
c\sum_{k=1}^{n} a_{n}~~~~(telescoping~
property)$$ These properties can be used to
prove:

\smallskip
$$\sum_{k=1}^{n} (2k - 1) =
\sum_{k=1}^{n} \{k^{2} - (k - 1)^{2}\} =
n^{2}$$
by the additive and telescoping property.

\bigskip
$$\sum_{k=1}^{n} k^{2} = ?$$
To figure out what the answer is observe
that
$$k^{2} = {k^{3} - (k - 1)^{3} + 3k - 1
\over 3}.$$
Hence
\begin{eqnarray*}
\sum_{k=1}^{n} k^{2} &=&
\sum_{k=1}^{n} {k^{3} - (k - 1)^{3} + 3k - 1
\over 3} \\
&=& {1 \over 3} \sum_{k=1}^{n}
(k^{3} - (k - 1)^{3}) + \sum_{k=1}^{n}
k - \sum_{k=1}^{n} 1 \\
&=& {n^{3} \over 3} + {n(n+1) \over 2} - {n
\over 3} \\
&=& {n^{3} \over 3} + {n^{2} \over 2} + {n
\over 6}.
\end{eqnarray*}
The main point is this. If we already guess
the correct formula then usually the
validity of the formula can be checked by
induction; however the much more difficult
part is to come up with the correct formula.

\bigskip
\noindent
HW: p.40 \#7, 8

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\noindent
{\bf Solutions}
\noindent
(\# 7)~~Find the sum: $\sum_{i=1}^{n}
i^{3}$.
\medskip
\noindent
Since
$$(i+1)^{4} - i^{4} = 4i^{3} + 6i^{2} + 4i +
1$$
hence
$$\sum_{i=1}^{n}
i^{3} = {1 \over 4}
\{\sum_{i=1}^{n} ((i+1)^{4} - i^{4}) - 6
\sum_{i=1}^{n} i^{2} - 4 \sum_{i=1}^{n} i -
n.$$
By the telescoping property:
$${1 \over 4}
\sum_{i=1}^{n} ((i+1)^{4} - i^{4}) =
{1 \over 4} \{(n+1)^{4} - 1\},$$
and by the previous examples:
$${1 \over 4}
\{6 \sum_{i=1}^{n} i^{2}\} = {3 \over 2}
\{{n^{3} \over 3} + {n^{2} \over 2} + {n
\over 6}\}$$
and
$${1 \over 4}
\{4 \sum_{i=1}^{n} i\} =
\{{n^{2} \over 2} + {n \over 2}\}.$$
Thus
$$\sum_{i=1}^{n}
i^{3} = {1 \over 4} \{(n+1)^{4} - 1\} - {3 \over 2}
\{{n^{3} \over 3} + {n^{2} \over 2} + {n
\over 6}\} - \{{n^{2} \over 2} + {n \over
2}\} - {n \over 4} = {n^{4} \over 4} +
{n^{3} \over 2} + {n \over 6}.$$

\bigskip
\noindent
(\# 8)~~(a)~~$\sum_{k=0}^{n} = (1 -
x^{n+1})/(1 - x), ~x \neq 1.$

\medskip
\noindent
Observe that, by the telescoping property:
$$(1 - x) \sum_{k=0}^{n} x^{k} =
\sum_{k=0}^{n} (x^{k} - x^{k+1}) = 1 -
x^{n+1}.$$
The formula now follows by dividing both
sides by $1 - x$.

\medskip
\noindent
(b)~~If $x= 1$ then
$$\sum_{k=0}^{n} x^{k} = 1^{0} + 1^{1} +
... + 1^{n} = n + 1.$$

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\noindent
{\bf Lemma}~If $a \ge 0$ then $|x| \leq
a$ if and only if $- a \leq x \leq a$.

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\noindent
{\bf Theorem}(Triangle
Inequality)~~For any real numbers $a_{1},
..., a_{n}$ we have:
$$|\sum_{i=1}^{n} \leq a_{i}| \leq
\sum_{i=1}^{n} |a_{i}|^{2}.$$
\noindent
{\it Proof.}~Theorem is clearly true if $n
= 1$. If $n = 2$, we have, by Lemma:
$$- |a_{1}| \leq a_{1} \leq |a_{1}|,~-
|a_{2}| \leq a_{2} \leq |a_{2}|.$$ 
Adding the two inequalities we obtain:
$$- (|a_{1}| + |a_{2}|) \leq a_{1} +
a_{2}\leq |a_{1}| + |a_{2}|.$$
By Lemma this last inequality is equivalent
to
$$|a_{1} + a_{2}| \leq |a_{1}| + |a_{2}|.$$
The case of general $n$ follows by
induction.~~QED

\bigskip
\noindent
{\bf Theorem}~(Cauchy-Schwarz
Inequality)~~If
$a_{1}, ..., a_{n}$ and $b_{1}, ..., b_{n}$
are arbitrary real numbers. Then
$$(\sum_{i=1}^{n} a_{i}b_{i})^{2} \leq
(\sum_{i=1}^{n} a_{i}^{2})(\sum_{i=1}^{n}
b_{i}^{2}).$$
In fact, we have:
$$(\sum_{i=1}^{n} a_{i}^{2})(\sum_{i=1}^{n}
b_{i}^{2}) - (\sum_{i=1}^{n}
a_{i}b_{i})^{2} = \sum_{1 \leq i < j \leq
n} (a_{i}b_{j} - a_{j}b_{i})^{2}.$$
\noindent
{\it Proof.}~Theorem is clearly true if $n
= 1$. For $n = 2$,
$$(a_{1}b_{1} + a_{2}b_{2})^{2} =
(a_{1}b_{1})^{2} + (a_{2}b_{2})^{2} +
2a_{1}b_{1}a_{2}b_{2}$$
while
$$
(a_{1}^{2} + a_{2}^{2})(b_{1}^{2} +
b_{2}^{2}) = (a_{1}b_{1})^{2} +
(a_{2}b_{2})^{2} + a_{1}^{2}b_{2}^{2} +
a_{2}^{2}b_{1}^{2}.
$$
Thus
$$(a_{1}^{2} + a_{2}^{2})(b_{1}^{2} +
b_{2}^{2}) -  (a_{1}b_{1} +
a_{2}b_{2})^{2} = a_{1}^{2}b_{2}^{2} +
a_{2}^{2}b_{1}^{2} - 2a_{1}b_{1}a_{2}b_{2}
= (a_{1}b_{2} - a_{2}b_{1})^{2}$$
is non-negative and is zero if and only if
$a_{1}b_{2} = a_{2}b_{1}$.

\medskip
Assuming that the Theorem holds for all $n
\leq k$ and consider the case $n = k + 1$:
$$(\sum_{i=1}^{k+1} a_{i}b_{i})^{2} \leq
(\sum_{i=1}^{k+1}
a_{i}^{2})(\sum_{i=1}^{k+1} b_{i}^{2}) ?$$
Write the LHS as
$$((\sum_{i=1}^{k} a_{i}b_{i}) +
a_{k+1}b_{k+1})^{2} = (\sum_{i=1}^{k}
a_{i}b_{i})^{2} + (a_{k+1}b_{k+1})^{2} +
2a_{k+1}b_{k+1}\sum_{i=1}^{k} a_{i}b_{i}$$
and write the RHS as
\begin{eqnarray*}
&&((\sum_{i=1}^{k}
a_{i}^{2}) + a_{k+1}^{2})((\sum_{i=1}^{k}
b_{i}^{2}) + b_{k+1}^{2})\\
&&= (\sum_{i=1}^{k}
a_{i}^{2})(\sum_{i=1}^{k} b_{i}^{2})
+ a_{k+1}^{2}(\sum_{i=1}^{k}
b_{i}^{2}) + b_{k+1}^{2}(\sum_{i=1}^{k}
a_{i}^{2}) + a_{k+1}^{2} b_{k+1}^{2}.
\end{eqnarray*}
Then the difference
\begin{eqnarray*}
&&RHS - LHS \\
&&= (\sum_{i=1}^{k}
a_{i}^{2})(\sum_{i=1}^{k} b_{i}^{2}) -
(\sum_{i=1}^{k} a_{i}b_{i})^{2} + a_{k+1}^{2}(\sum_{i=1}^{k}
b_{i}^{2}) + b_{k+1}^{2}(\sum_{i=1}^{k}
a_{i}^{2}) - 2a_{k+1}b_{k+1}\sum_{i=1}^{k}
a_{i}b_{i}\\
&&= \sum_{1 \leq i < j \leq k} (a_{i}b_{j}
- a_{j}b_{i})^{2} + 
a_{k+1}^{2}(\sum_{i=1}^{k} b_{i}^{2}) +
b_{k+1}^{2}(\sum_{i=1}^{k} a_{i}^{2}) -
2a_{k+1}b_{k+1}\sum_{i=1}^{k} a_{i}b_{i}\\
&&=  \sum_{1 \leq i < j \leq k} (a_{i}b_{j}
- a_{j}b_{i})^{2} + \sum_{i=1}^{k}
(a_{i}b_{k+1} - a_{k+1}b_{i})^{2}\\
&&= \sum_{1 \leq i < j \leq k+1} (a_{i}b_{j}
- a_{j}b_{i})^{2}.
\end{eqnarray*}
We have used the induction hypothesis (i.e.
the case $n = k$ is valid) and the
elementary identity:

\begin{eqnarray*}
\sum_{i=1}^{k}
(a_{i}b_{k+1} - a_{k+1}b_{i})^{2} =
b_{k+1}^{2} \sum_{i=1}^{k} a_{i}^{2} -
2a_{k+1}b_{k+1} \sum_{i=1}^{k} a_{i}b_{i} +
a_{k+1}^{2} \sum_{i=1}^{k} b_{i}^{2}
\end{eqnarray*}
QED

\bigskip
\noindent
HW p. 43 \# 1 (h), 1 (i), 1 (j); p.44 \# 2,
3, 4

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