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942 changes: 942 additions & 0 deletions Calculations-MathML/CMML/Problems/Chapter1/1-3.html
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96 changes: 96 additions & 0 deletions Calculations-MathML/CMML/Problems/Chapter1/13-16.html
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<html>
<table class="pt10_r20">
<tr>
<td class="vtop_zero">13.</td>
<td class="vtop_lten">
The maximum of two numbers $x$ and $y$ is denoted by $\text{max}(x, y)$. Thus $\text{max}(-1,3) = \text{max}(3,3) = 3$, and
$\text{max}(-1,-4) = \text{max}(-4,-1) = -1$. The minimum of $x$ and $y$ is denoted by $\text{min}(x,y)$. Prove that
$$
\begin{align}
\begin{split}
\text{max}(x,y) &= \frac{x+y+|y-x|}{2}, \\
\text{min}(x,y) &= \frac{x+y-|y-x|}{2}.
\end{split}
\end{align}
$$
Derive a formula for $\text{max}(x,y,z)$ and $\text{min}(x,y,z)$, using, for example
$$
\text{max}(x,y,z) = \text{max}(x, \text{max}(y,z)).
$$
</td>
</tr>
</table>
<table class="pr_20">
<tr>
<td class="vtop_zero">14.</td>
<td class="vtop_lten">(a)</td>
<td class="vtop_lten">
Prove that $|a|=|-a|.$ (The trick is not to become confused by too many cases. First prove the statement for $a \geq 0$. Why is it then obvious for $a \leq 0$?)
</td>
</tr>
<tr>
<td class="vtop_zero"></td>
<td class="vtop_lten">(b)</td>
<td class="vtop_lten">
Prove that $-b \leq a \leq b$ if and only if $|a| \leq b.$ In particular, it follows that $-|a| \leq a \leq |a|.$
</td>
</tr>
<tr>
<td class="vtop_zero"></td>
<td class="vtop_lten">(c)</td>
<td class="vtop_lten">Use this fact to give a new proof that $|a+b| \leq |a|+|b|.$</td>
</tr>
</table>
<table class="pt_20">
<tr>
<td class="vtop_zero_cgrey">15.</td>
<td class="vtop_lten">
Prove that if $x$ and $y$ are not both $0$, then $$x^2+xy+y^2 >0, \\ x^4+x^3y+x^2y^2+xy^3+y^4 >0.$$
<span style="padding:0 0 0 3">Hint: Use Problem 1.</span>
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero_cgrey">16.</td>
<td class="vtop_lten">(a)</td>
<td class="vtop_lten">
Show that
<table class="pt20_r20">
<tr>
<td class="vtop_lten">$(x+y)^2 = x^2+y^2$</td>
<td class="vtop_l20">only when $x=0$ or $y=0$,</td>
</tr>
<tr>
<td class="vtop_lten">$(x+y)^3 = x^3+y^3$</td>
<td class="vtop_l20">only when $x=0$ or $y=0$ or $x=-y.$</td>
</tr>
</table>
</td>
</tr>
<tr>
<td class="vtop_zero"></td>
<td class="vtop_lten">(b)</td>
<td class="vtop_lten">
Using the fact that $$x^2+2xy+y^2 = (x+y)^2 \geq 0,$$ show that $4x^2+6xy+4y^2 >0$ unless $x$ and $y$ are both $0$.
</td>
</tr>
<tr>
<td class="vtop_zero"></td>
<td class="vtop_lten">(c)</td>
<td class="vtop_lten">
Use part (b) to find out when $(x+y)^4 = x^4+y^4.$
</td>
</tr>
<tr>
<td class="vtop_zero"></td>
<td class="vtop_lten">(d)</td>
<td class="vtop_lten">
Find out when $(x+y)^5 = x^5+y^5.$ Hint: From the assumption $(x+y)^5$ $=x^5+y^5$ you should be able to derive the equation $x^3+2x^2y+y^3 =$ $0,$ if $x,y \neq 0.$ This implies that $(x+y)^3=x^2y+xy^2=xy(x+y).$
</td>
</tr>
</table>
<p class="r20L44_font16">
You should now be able to make a good guess as to when $(x+y)^n = x^n+y^n;$ the proof is contained in Problem 11&ndash;63.
</p>
</html>
122 changes: 122 additions & 0 deletions Calculations-MathML/CMML/Problems/Chapter1/17-19.html
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<html>
<table class="pt10_r20">
<tr>
<td class="vtop_zero">17.</td>
<td class="vtop_lten">
<table class="pzero">
<tr>
<td class="vtop_zero">(a)</td>
<td class="vtop_lten">
Find the smallest possible value of $2x^2-3x+4$. Hint: "Complete the square," i.e., write $2x^2-3x+4 = 2(x-3/4)^2+$ ?
</td>
</tr>
<tr>
<td class="vtop_zero">(b)</td>
<td class="vtop_lten">
Find the smallest possible value of $x^2-3x+2y^2+4y+2$.
</td>
</tr>
<tr>
<td class="vtop_zero">(c)</td>
<td class="vtop_lten">
Find the smallest possible value of $x^2-4xy+5y^2-4x-6y+7$.
</td>
</tr>
</table>
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero">18.</td>
<td class="vtop_zero">
<table class="pzero">
<tr>
<td class="vtop_lten">(a)</td>
<td class="vtop_lten">
Suppose that $b^2-4c \geq 0.$ Show that the numbers
$$
\frac{-b+\displaystyle\sqrt{{b}^2-4c}}{2} \ , \ \ \ \ \ \ \ \frac{-b-\displaystyle\sqrt{{b}^2-4c}}{2}
$$
both satisfy the equation $x^2+bx+c=0$.
</td>
</tr>
<tr>
<td class="vtop_lten">(b)</td>
<td class="vtop_lten">
Suppose that $b^2-4c < 0.$ Show that there are no numbers $x$ satisfying $x^2+bx+c=0;$ in fact $x^2+bx+c > 0$ for all $x.$ Hint: Complete the square.
</td>
</tr>
<tr>
<td class="vtop_lten">(c)</td>
<td class="vtop_lten">
Use this fact to give another proof that if $x$ and $y$ are not both $0$, then $x^2+xy+y^2 > 0.$
</td>
</tr>
<tr>
<td class="vtop_lten">(d)</td>
<td class="vtop_lten">
For which numbers $\alpha$ is it true that $x^2 + \alpha xy + y^2 > 0$ whenever $x$ and $y$ are not both $0$?
</td>
</tr>
<tr>
<td class="vtop_lten">(e)</td>
<td class="vtop_lten">
Find the smallest possible value of $x^2+bx+c$ and of $ax^2+bx+c$ for $a > 0.$
</td>
</tr>
</table>
</td>
</tr>
</table>
<table class="pt20_r20_b10">
<tr>
<td class="vtop_zero">19.</td>
<td class="vtop_lten">
The fact that $a^2 \geq 0$ for all numbers $a$, elementary as it may seem, is the fundamental idea upon which most important inequalities are ultimately based. The great&ndash;granddaddy of all inequalities is the <i>Schwartz inequality</i>.

$$x_1y_1+x_2y_2 \leq \displaystyle\sqrt{{x_1}^2 + {x_2}^2}\sqrt{{y_1}^2 + {y_2}^2}.$$

(A more general form occurs in Problem 2&ndash;21.) The three proofs of the Schwarz inequality outlined below have only one thing in common&mdash;their reliance on the fact that $a^2 \geq 0$ for all $a$.
<table class="pt10">
<tr>
<td class="vtop_zero">(a)</td>
<td class="vtop_lten">
Prove that if $x_1=\lambda y_1$ and $x_2=\lambda y_2$ for some number $\lambda \geq 0,$ then equality holds in the Schwarz inequality. Prove the same thing if $y_1=y_2=0.$ Now suppose that $y_1$ and $y_2$ are not both $0,$ and that there is no number $\lambda$ such that $x_1=\lambda y_1$ and $x_2= \lambda y_2.$ Then
$$
\begin{eqnarray}
0 &<& (\lambda y_1 - x_1)^2 + (\lambda y_2 - x_2)^2 \\
&=& \lambda^2({y_1}^2 + {y_2}^2) - 2\lambda (x_1y_1 + x_2y_2)+({x_1}^2+{x_2}^2).
\end{eqnarray}
$$
Using Problem 18, complete the proof of the Schwarz inequality.
</td>
</tr>
<tr>
<td class="vtop_zero">(b)</td>
<td class="vtop_lten">
Prove the Schwarz inequality by using $2xy \leq x^2+y^2$ (how is this derived?) with
$$
x = \frac{x_i}{\displaystyle\sqrt{{x_1}^2 + {x_2}^2}} \ , \ \ \ \ \ \ \ \
y = \frac{y_i}{\displaystyle\sqrt{{y_1}^2 + {y_2}^2}} \ ,
$$
first for $i=1$, and then for $i=2.$
</td>
</tr>
<tr>
<td class="vtop_zero">(c)</td>
<td class="vtop_lten">
Prove the Schwarz inequality by first proving that $$({x_1}^2+{x_2}^2)({y_1}^2+{y_2}^2) = (x_1y_1 + x_2y_2)^2+(x_1y_2 - x_2y_1)^2.$$
</td>
</tr>
<tr>
<td class="vtop_zero">(d)</td>
<td class="vtop_lten">
Deduce, from each of these proofs, that equality holds only when $y_1=y_2=0$ or when there is a number $\lambda \geq 0$ such that $x_1=\lambda y_1$ and $x_2=\lambda y_2.$
</td>
</tr>
</table>
</td>
</tr>
</table>
</html>
72 changes: 72 additions & 0 deletions Calculations-MathML/CMML/Problems/Chapter1/20-23.html
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<html>
<table class="pt10_r20">
<tr>
<td class="vtop_zero">
In our later work, three facts about inequalities will be crucial. Although proofs will be supplied at the appropriate point in the text, a personal assault on these problems is infinitely more enlightening than a perusal of a completely worked-out proof. The statements of these propositions involve some weird numbers, but their basic message is simple: if $x$ is close enough to $x_0,$ and $y$ is close enough to $y_0,$ then $x+y$ will be close to $x_0+y_0,$ and $xy$ will be close to $x_0y_0,$ and $1/y$ will be close to $1/y_0.$ The symbol "$\varepsilon$" which appears in these propositions is the fifth letter of the Greek alphabet ("epsilon"), and could just as well be replaced by a less intimidating Roman letter; however, tradition has made the use of $\varepsilon$ sacrosanct in the contexts to which these theorems apply.
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero">20.</td>
<td class="vtop_lten">
Prove that if $\vert x-x_0 \rvert < \dfrac{\varepsilon}{2}$ &nbsp;&nbsp; and &nbsp;&nbsp; $\lvert y-y_0 \rvert < \dfrac{\varepsilon}{2},$ then
<p style="padding-left:20%">
$
\begin{align}
\lvert (x+y) - (x_0+y_0) \rvert &< \varepsilon, \\
\lvert (x-y) - (x_0-y_0) \rvert &< \varepsilon.
\end{align}
$
</p>
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero_cgrey">21.</td>
<td class="vtop_lten">
Prove that if
$$
\lvert x-x_0 \rvert < \text{min} \bigg(\frac{\varepsilon}{2(\lvert y_0 \rvert +1)} \ , 1 \bigg) \ \ \ \ \text{and} \ \ \ \ \lvert y-y_0 \rvert < \frac{\varepsilon}{2(\lvert x_0 \rvert +1)} \ ,
$$
then $\lvert xy-x_0y_0 \rvert < \varepsilon.$
<br>
(The notation "min" was defined in Problem 13, but the formula provided by that problem is irrelevent at the moment; the first inequality in the hypothesis just means that
$$
\vert x-x_0 \rvert < \frac{\varepsilon}{2(\lvert y_0 \rvert+1)} \ \ \ \ \text{and} \ \ \ \ \lvert x-x_0 \rvert < 1;
$$
at one point in the argument you will need the first inequality, and at another point you will need the second. One more word of advice: since the hypotheses only provide information about $x-x_0$ and $y-y_0,$ it is almost a foregone conclusion that the proof will depend upon writing $xy-x_0y_0$ in a way that involves $x-x_0$ and $y-y_0.$)
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero_cgrey">22.</td>
<td class="vtop_lten">
Prove that if $y_0 \neq 0$ and
<div style="margin:10 0 0 70">
$
\displaystyle
\lvert y-y_0 \rvert < \text{min} \bigg( \displaystyle\frac{\lvert y_0 \rvert}{2} \ , \frac{\varepsilon \lvert y_0 \rvert ^2}{2} \ \bigg), \ \text{then} \ y \neq 0, \ \text{and} \ \Bigg\lvert \frac{1}{y} - \frac{1}{y_0} \Bigg\rvert< \varepsilon.
$
</div>
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero_cgrey">23.</td>
<td class="vtop_lten">
Replace the question marks in the following statement by expressions involving $\varepsilon, \ x_0$ and $y_0$ so that the conclusion will be true:
<p>
If $y_0 \neq 0$ and
$$\lvert y-y_0 \rvert < \ ? \ \ \ \ \text{and} \ \ \ \ \lvert x-x_0 \rvert < \ ?$$
then $y \neq 0$ and
$$\Bigg\lvert \frac{x}{y} \ - \frac{x_0}{y_0} \ \Bigg\rvert < \varepsilon.$$
This problem is trivial in the sense that its solution follows from Problem 21 and 22 with almost no work at all (notice that $x/y=x \cdot 1/y$). The crucial point is not to become confused; decide which of the two problems should be used first, and don't panic if your answer looks unlikely.
</p>
</td>
</tr>
</table>
</html>
90 changes: 90 additions & 0 deletions Calculations-MathML/CMML/Problems/Chapter1/24-25.html
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<html>
<table class="pt10_r20">
<tr>
<td class="vtop_zero_cgrey">24.</td>
<td class="pl_10">
This problem shows that the actual placement of parentheses in a sum is irrelevant. The proofs involve "mathematical induction"; if you are not familiar with such proofs, but still want to tackle this problem, it can be saved until after Chapter 2, where proofs by induction are explained.
<br>
Let us agree, for definiteness, that $a_1+ \cdots + a_k$ will denote
$$a_1 + (a_2 + (a_3 + \cdots + (a_{n-2} + (a_{n-1}+a_n))) \cdots).$$
Thus $a_1+a_2+a_3$ denotes $a_1+(a_2+a_3),$ and $a_1+a_2+a_3+a_4$ denotes $a_1+(a_2+(a_3+a_4)),$ etc.
<table class="pt_15">
<tr>
<td class="vtop_zero">(a)</td>
<td class="vtop_lten">
Prove that
$$(a_1+ \cdots + a_k)+a_{k+1} = a_1+ \cdots + a_{k+1}.$$
Hint: Use induction on $k$.
</td>
</tr>
<tr>
<td class="vtop_zero">(b)</td>
<td class="vtop_lten">
Prove that if $k \leq n,$ then
$$(a_1+ \cdots + a_k)+(a_{k+1}+ \cdots + a_n) = a_1+ \cdots + a_n.$$
Hint: Use part (a) to give a proof by induction on $k$.
</td>
</tr>
<tr>
<td class="vtop_zero">(c)</td>
<td class="vtop_lten">
Let $s(a_1, \dots, a_k)$ be some sum formed from $a_1, \dots, a_k.$ Show that
$$s(a_1, \dots, a_k) = a_1 + \cdots + a_k.$$
Hint: There must be two sums $s'(a_1, \dots, a_l)$ and $s''(a_{l+1}, \dots, a_k)$ such that
$$s(a_1, \dots, a_k) = s'(a_1, \dots, a_l) + s''(a_{l+1}, \dots, a_k).$$
</td>
</tr>
</table>
</td>
</tr>
</table>
<table class="pt20_r20">
<tr>
<td class="vtop_zero_cgrey">25.</td>
<td class="pl_10">
Suppose that we interpret "number" to mean either $0$ or $1$, and + and $\cdot$ to be the operations defined by the following two tables.
<div style="padding:0 0 0 50">
<div style="float:left">
<table style="border-collapse:collapse">
<tr>
<td style="padding:10 19 10 19">$+$</td>
<td style="padding:10 19 10 19">$0$</td>
<td style="padding:10 19 10 19">$1$</td>
</tr>
<tr>
<td style="padding:10 19 0 19">$0$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$0$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$0$</td>
</tr>
<tr>
<td style="padding:10 19 0 19">$1$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$1$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$0$</td>
</tr>
</table>
</div>
<div style="float:left; padding:0 0 0 50">
<table style="border-collapse:collapse">
<tr>
<td style="padding:10 19 10 19">$\cdot$</td>
<td style="padding:10 19 10 19">$0$</td>
<td style="padding:10 19 10 19">$1$</td>
</tr>
<tr>
<td style="padding:10 19 0 19">$0$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$0$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$0$</td>
</tr>
<tr>
<td style="padding:10 19 0 19">$1$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$1$</td>
<td style="border:1 solid grey; padding:10 19 10 19">$0$</td>
</tr>
</table>
</div>
</div>
</td>
</tr>
</table>
<p style="padding-left:92; font-size:16">Check that properties P1&ndash;P9 all hold, even though $1+1=0.$</p>
</html>
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