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\begin{flushleft} 
Recitation 7\hfill \course\\
17 September 1998 \hfill \semester \\
\medskip

\bf Reminders: \rm  

1.  The project on hypocycloids is due on Friday, 9/18/98.
2. The first exam will be in CB114 from 7:30--9:30pm, Tuesday 22
September. 3. There will be a quiz on Friday over section 1.6.
4. The third graded homework assignment will be due on Tuesday 22
September in recitation. The assignment is \S1.6, \#58, p. 93 \#16 and
Appendix C, \# 24.
 
Below is a selection of problems related to mathematical induction and
the
trigonometric functions.
These problems will not be collected or graded. However, you should
understand how to work each of these problems.  You should begin
working on these problems in groups in recitation. You will probably
want to finish these problems outside of class. 
 If you have questions,
please ask your TA or instructor. If you find a problem difficult,
consider working similar problems from the text for additional practice.

\begin{enumerate}
\item  Use mathematical induction to prove that for each $n =1,2, \dots$,
$$
\sum _{k=1 }^n k = n(n+1) /2.
$$


\item   \begin{enumerate}  \item Find a simple formula for
$$
\sum_{k=1 }^n (k+1)^2 - k^2  = 2^2- 1 + (3^2 - 2^2) + \dots + n^2 -
(n-1)^2 + (n+1)^2 - n^2 .
$$
\item Using your answer to part a), find another proof of the formula
in the previous problem.  To do this you should simplify each summand
on the left.
	\end{enumerate}
\item Use mathematical induction to prove that
$$
\sum _{k=1 } ^n k^2 = n(n+1)  ( 2n+1)/6.
$$

\item Use mathematical induction to prove the following simple form of
the binomial theorem: For $n=1,2, \dots,$ we have
$$
(a+h)^n = a^n + n a^{n-1 }h + h^2 P_n(h)
$$
where $P_n(h)$ is a polynomial in $h$. The coefficients of this
polynomial depend on $a$.

\item Let $f(x) = x^2$  and compose $f$ with itself $n$ times to
form  $g_n(x) = f\circ f\circ \dots
\circ f$.  Use mathematical induction to prove that
$$
g_n(x) = x^{(2^n)}. 
$$

\item Work problems 15, 16 and 18 on page 93 of Stewart.  


\item Appendix C, \# 21, 23, 25, 27, 29, 30, 33, 37, 46, 48. I
apologize that we have not directly discussed the material from
appendix C,  yet. We will do this on Friday.



\end{enumerate}

Since the text only briefly mentions mathematical induction, the main
ideas of mathematical induction from Wednesday's lecture appear below.

\em Principle of mathematical induction \rm Suppose that $P_n$ is a
sequence of statements depending on a natural number $n =1,2, \dots$. If we can
show that:
\begin{itemize}
\item $P_1$ is true 
\item If $P_n $ is true, then $P_{n+1}$ is true.
\end{itemize}
Then, we can conclude that all the statements $P_n$ are true.

The reason this  works is that if we know $P_1$ is true, then the
second step allows us to conclude $P_2$ is true. Now that we know
$P_2$ is true, then the second step allows us to 
conclud $P_3$ is true.  If we repeat this $n-1$ times, we know that
$P_{n}$ is true.

This principle is useful  because it allows us to prove an  infinite
number  of statements are  true  in just two easy steps!

Below are several examples to illustrate how to use this
principle. Also, refer to pages 88 and 91 of the text.

\em Example \rm Show that for $n= 1, 2, 3, \dots$, the number $4^n-1$
is a multiple of $3$.

\em Solution \rm Step 1. We need to show this is true when $n =1$.  This is
easy since $4^1 -1  =4-1 =3$ and 3 is divisible by 3.

Step 2.  We suppose that $4^n-1$ is a multiple of $3$. This means that
for some whole number $N$, $ 4^n -1 = 3N$. Now  
$4^{n+1}-1$. If we add and subtract $4$, we have
$$
4^{n+1} -1 = 4^{n+1}-4 + 4 -1 = 4 ( 4^n -1 ) + 3.
$$
Now we use our assumption that $4^n-1$ is a multiple of $3$ to replace
$4^n -1$ by $3N$ and obtain that
$$
4^{n+1} -1 = 4 \cdot 3 N + 3 = 3 (4N+1).
$$
Thus we have shown that $4^{n+1} -1 $ is a multiple of $3$. 


\medskip

\em Example \rm Show that for $n =1,2, \dots$, we have
$$
\sum_{k=1 } ^ n 2k = n (n+1).
$$

\em Solution \rm  Step 1.  If $n=1$, then $n (n+1) = 1 \cdot
2 = 2$. Also, 
$$
\sum _{k=1} ^ 1 2 k = 2.
$$
Thus both sides are equal if $n =1$. 

Step 2. Now suppose that the formula is true for $n$ and consider the sum
$$
\sum_{k=1} ^{n+1} 2k = \sum_{k=1 } ^ n 2k + 2(n+1).
$$
We use our assumption that $ \sum_{k=1} ^n 2k = n(n+1)$ to conclude 
that
$$
\sum_{k=1 }^{n+1} 2k = n(n+1) + 2(n+2).$$
Simplifying this last expression gives
$$
n(n+1) + 2 (n+1) = n^2 + n + 2n + 2= n^2 + 3n + 2 = (n+2 ) (n+1).
$$
Since $(n+ 2) (n+1) = ( n+1+ 1) (n+1)$, we have shown
that  the formula
$$
\sum _{k=1} ^ { n+1} 2k = (n+1+1) ( n+1)
$$
is true for $n+1$. This completes the proof by induction.

\medskip 


Finally, let me explain the use of the $ \sum$ for summation.
The notation 
$$
\sum_{k = 1} ^ n f(k)$$
means to evaluate the function $f(k)$ at $k =1, 2, \dots ,n$ and add
up the results.
In other words:
$$
\sum_{k=1} ^n f(k) = f(1) + f(2) + \dots + f(n).
$$
For example:
$$
\sum _{k=1} ^ 4 k^ 2 = 1 + 4 + 9 + 16,
$$
$$
\sum _{k=1} ^n 2k-1 = 1 + 3 + 5+ \dots + 2n -1 ,
$$
and
$$
 \sum_{k=1} ^n 1 = n.
$$

\end{flushleft}
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