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\begin{document}


\section{\protect\bigskip Appendix - Evaluation of Scientific Notebook as a
Tool in Mathematics Education}

\paragraph{by Mirek Majewski}

Department of Mathematics and Computer Science, PNG University of
Technology, PMB UNITECH, Lae, Papua New Guinea

Current e-mail: majewski@mupad.com

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of \textquotedblleft Evaluation of Scientific Notebook as a Tool in
Mathematics Education\textquotedblright , PNGJMCE Vol. 3, No. 1, 1997

\subsection{\ Examples}

The following examples show how Scientific Notebook deals with mathematical
objects and performs various operations on them. I have used a very simple
convention here. For example, the expression `choose Insert+Fraction' means:
click on Insert in Scientific Notebook menu, then choose option Fraction.
The expression `'Compute+Plot2D+Rectangular' means: click on Compute in
Scientific Notebook menu, then choose option Plot2D and finally choose
sub-option Rectangular. The expression `copy a formula' means select formula
(highlight it as a block), and while pressing the [Ctrl]\ key, with the
mouse pointer move the selected formula into a new place. This operation
copies selected formula into another place. Without the [Ctrl]\ key the
formula will be moved into another place.

Sometimes, I list operations that are performed and then I show the result.
In such case the character `=' with a number on the top, means that the
formula on its right side was obtained using operations mentioned in the
appropriate list item.

Examples 1 and 2 are taken from the high school textbook \textit{Core
Mathematics for A-level} by L. Bostock and S. Chandler. Example 3 was
formulated in the book \textit{Discrete Mathematics for Computing}, by J.E.
Munro.

\subsection{\ Example 1 (limits, derivative)}

In each of the following examples, set up and simplify the difference
quotient: 
\begin{equation*}
\dfrac{f(x+h)-f(x)}{h}
\end{equation*}%
Then find the limit as $h$ approaches $0,$ thus obtaining $f^{\prime }(x).$

1. $f(x)=x^{3}-12x$

2. $f(x)=x^{2}-4$

3. $f(x)=x^{4}-1$

\subsubsection{Solution}

We will show the solution for the function $f(x)=x^{3}-12x.$ The solution of
other examples will be similar. Below are listed all intermediate steps in
Scientific Notebook that should be performed to show the process of
simplification.

\begin{enumerate}
\item Type the function $f(x)=x^{3}-12x$, leave the input cursor to the
right of the formula, and choose Compute+Definitons+New Definition.

\item Type the difference quotient, leave the input cursor to the right of
the formula and use Compute+Evaluate command.

\item Select the symbol `=' and the expression on the right side and with
the Ctrl key down drag the expression to the right to create a copy.

\item In the last expression select $\boldsymbol{(h+x)}^{3}$\textbf{\ }and
with the Ctrl key down choose Compute+Expand.

\item Repeat step 3, select the expression within parentheses in the last
result and with Ctrl key down choose Compute+Simplify

\item Repeat step 3, select the expression within parentheses in the last
result and with Ctrl key down choose Compute+Factor

\item Leave the input cursor in the last expression and apply
Compute+Simplify.

\item Select all your calculations and press Ctrl-D, finally put the input
cursor in the front of the second, third etc. character `=' and press Enter.
You should obtain the following: 
\begin{eqnarray*}
&&\dfrac{\boldsymbol{f}(\boldsymbol{x+h})-\boldsymbol{f}(\boldsymbol{x})}{%
\boldsymbol{h}}\overset{2}{=}\frac{1}{\boldsymbol{h}}\left( \left( 
\boldsymbol{h+x}\right) ^{\boldsymbol{3}}\boldsymbol{-x}^{\boldsymbol{3}}%
\boldsymbol{-12h}\right) \\
&&\overset{3,4}{=}\frac{1}{\boldsymbol{h}}\left( \boldsymbol{h}^{3}%
\boldsymbol{+x}^{\boldsymbol{3}}\boldsymbol{+3hx}^{\boldsymbol{2}}%
\boldsymbol{+3h}^{\boldsymbol{2}}\boldsymbol{x-x}^{\boldsymbol{3}}%
\boldsymbol{-12h}\right) \\
&&\overset{3,5}{=}\frac{1}{\boldsymbol{h}}\left( \boldsymbol{h}^{\boldsymbol{%
3}}\boldsymbol{-12h+3hx}^{\boldsymbol{2}}\boldsymbol{+3h}^{\boldsymbol{2}}%
\boldsymbol{x}\right) \\
&&\overset{3,6}{=}\frac{1}{\boldsymbol{h}}\left( \boldsymbol{h}\left( 
\boldsymbol{3hx+h}^{\boldsymbol{2}}\boldsymbol{+3x}^{\boldsymbol{2}}%
\boldsymbol{-12}\right) \right) \\
&&\overset{7}{=}\boldsymbol{3hx+h}^{\boldsymbol{2}}\boldsymbol{+3x}^{%
\boldsymbol{2}}\boldsymbol{-12}
\end{eqnarray*}
\end{enumerate}

To find the limit:

\begin{enumerate}
\item Type $G(x,h)=3x^{2}+3hx+h^{2}-12$ and apply Compute+Definitons+New
Definition. Since this moment $h$ will be recognized as a variable in the
formula $G.$

\item Type in maths mode the expression $\lim\limits_{h\rightarrow 0}G(x,h).$
Type an `=' and copy this expression to the right, select $G(x,h)$ in the
right side and apply [Ctrl]+Compute+Evaluate.

\item Put the round bracket around the obtained polynomial $\boldsymbol{3hx+h%
}^{\boldsymbol{2}}+\boldsymbol{3x}^{\boldsymbol{2}}-\boldsymbol{12}$, leave
the input cursor to the right of the formula and apply Compute+Evaluate.
Your final result should be as follows:
\end{enumerate}

\QTP{Body Math}
\begin{equation*}
\lim\limits_{h\rightarrow 0}G(x,h)\overset{2}{=}\lim\limits_{h\rightarrow 0}(%
\boldsymbol{3hx+h}^{\boldsymbol{2}}+\boldsymbol{3x}^{\boldsymbol{2}}-%
\boldsymbol{12})\overset{3}{=}3x^{2}-12
\end{equation*}

This completes our solution.$\blacksquare $

\subsection{\ Example 2 (partial fractions, graphs)}

\begin{itemize}
\item 
\begin{enumerate}
\item Given that: 
\begin{equation*}
\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}=\dfrac{A}{2-x}+\dfrac{Bx+C}{(x^{2}+1)}
\end{equation*}
determine the values of A,B and C.

\item A curve has the equation: 
\begin{equation*}
y=\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}
\end{equation*}
Determine the equation of tangent line to the curve at the point $(1,3)$.

\item Prove that the area of the region bounded by the curve, the X-axis and
the line $x=1$ is $\dfrac{7}{2}\ln 2-\dfrac{\pi }{4}.$
\end{enumerate}
\end{itemize}

\subsubsection{Solution}

\paragraph{Part 1}

Let: 
\begin{equation*}
\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}=\dfrac{A}{2-x}+\dfrac{Bx+C}{(x^{2}+1)}.
\end{equation*}

Multiply both sides of the equation by $(2-x)(x^{2}+1),$ to get:

\begin{equation*}
\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}(2-x)(x^{2}+1)=\left( \dfrac{A}{2-x}+\dfrac{%
Bx+C}{(x^{2}+1)}\right) (2-x)(x^{2}+1)
\end{equation*}

Select the left side of the equation, with [Ctrl]\ key down and use
Compute+Evaluate. Select the right side of the equation, with [Ctrl]\ down
use Compute+Simplify. You should obtain the following result:

\begin{equation*}
\boldsymbol{7x-x}^{2}\boldsymbol{=A+2C+2Bx-Cx+Ax}^{2}\boldsymbol{-Bx}^{2}
\end{equation*}

Having subtracted $7x-x^{2}$ from both sides of equation, you should get:

\begin{equation*}
(7x-x^{2})-(7x-x^{2})=(Ax^{2}+A+2Bx-Bx^{2}+2C-Cx)-(7x-x^{2}).
\end{equation*}

Select the left side of the equation, with [Ctrl]\ down choose
Compute+Evaluate. Repeat the same with the right side of the equation.
Result should be following:

\begin{equation*}
\boldsymbol{0=A+2C-7x+2Bx-Cx+x}^{2}\boldsymbol{+Ax}^{2}\boldsymbol{-Bx}^{2}
\end{equation*}

Select the right side of the last equation and choose
Compute+Polynomials+Collect. Select $x$\ as the variable.\textbf{\ }Here is
the result:

\begin{eqnarray*}
0 &=&Ax^{2}+A+2Bx-Bx^{2}+2C-Cx-7x+x^{2} \\
&=&\boldsymbol{\allowbreak A+2C+x}(\boldsymbol{2B-C-7})+\boldsymbol{x}^{%
\boldsymbol{2}}(\boldsymbol{A-B+1})
\end{eqnarray*}

Finally define a matrix with 3 rows and one column and copy to its cells
coefficients of the above polynomial. You should obtain a system of
equations:

\begin{equation*}
\begin{array}{l}
\boldsymbol{A+2C}=0 \\ 
2B-C-7=0 \\ 
\boldsymbol{A-B+1}=0%
\end{array}%
.
\end{equation*}

Leave the input cursor in the matrix and choose Compute+Solve+Exact. You
should obtain:

\begin{equation*}
\begin{array}{l}
\boldsymbol{A+2C}=0 \\ 
2B-C-7=0 \\ 
\boldsymbol{A-B+1}=0%
\end{array}%
,\text{Solution is}:[\boldsymbol{A=2,B=3,C=-1}]
\end{equation*}

This completes the first part of the example:

\begin{equation*}
\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}=\dfrac{2}{2-x}+\dfrac{3x-1}{(x^{2}+1)}
\end{equation*}

\paragraph{Part 2}

Copy the equation 2 to the line below:

\begin{equation*}
y=\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}
\end{equation*}

Leave the input cursor to the right of the formula and choose
Compute+Definitions+New Definition. This way you defined a new function $y$
for further calculations. In maths mode type the fraction $\dfrac{dy}{dx},$
leave the input cursor at the end of the fraction and choose
Compute+Evaluate. Here is the result you should get:

\begin{equation*}
\frac{dy}{dx}=\allowbreak \frac{\boldsymbol{2x-7}}{\boldsymbol{x-2x}^{%
\boldsymbol{2}}\boldsymbol{+x}^{\boldsymbol{3}}-\boldsymbol{2}}\boldsymbol{+}%
\frac{\boldsymbol{7x-x}^{\boldsymbol{2}}}{\boldsymbol{5x}^{\boldsymbol{2}}-%
\boldsymbol{4x-4x}^{\boldsymbol{3}}\boldsymbol{+x}^{\boldsymbol{4}}%
\boldsymbol{+4}}\boldsymbol{+}\frac{\boldsymbol{14x}^{\boldsymbol{2}}-%
\boldsymbol{2x}^{\boldsymbol{3}}}{\boldsymbol{x-4x}^{\boldsymbol{2}}%
\boldsymbol{+2x}^{\boldsymbol{3}}\boldsymbol{-2x}^{\boldsymbol{4}}%
\boldsymbol{+x}^{5}\boldsymbol{-2}}
\end{equation*}

Finally type: $m=\left[ \dfrac{dy}{dx}\right] _{x=1},$ leave the input
cursor to the right of the formula and choose Compute+Evaluate. This will
calculate the value of the derivative for $x=1:$

\begin{equation*}
m=\left[ \dfrac{dy}{dx}\right] _{x=1}=\allowbreak \frac{5}{2}
\end{equation*}

As we know the equation of a tangent line to the curve is given by the
equation: $y=m(x-x_{0})+y_{0}.$ Thus you obtain the equation of the normal
line: 
\begin{equation*}
y=\dfrac{5}{2}(x-1)+3
\end{equation*}

\paragraph{Part 3}

First let's plot the graph of the curve: 
\begin{equation*}
y=\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}
\end{equation*}

To obtain the graph leave the input cursor to the right of the equation, and
choose Compute+Plot2D+Rectangular.

\vspace{1pt}$y=\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}$\FRAME{dtbpFX}{2.9222in}{%
1.9476in}{0pt}{}{}{Plot}{\special{language "Scientific Word";type
"MAPLEPLOT";width 2.9222in;height 1.9476in;depth 0pt;display
"USEDEF";plot_snapshots FALSE;mustRecompute FALSE;lastEngine "MuPAD";xmin
"-5";xmax "5";xviewmin "-5.01";xviewmax "5.01";yviewmin
"-50.3289377545843";yviewmax "49.8710621662674";plottype 4;numpoints
100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis
\TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function
\TEXUX{$\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}$};linecolor "black";linestyle
1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range
"-5,5";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle
"Line";}}

Click on the obtained plot to activate it and then click on the blue
rectangle on the bottom right side of the plot. Scientific Notebook\
displays a dialog box where in Items Plotted in option [Variables and
Intervals] you can declare plot options. Choose the domain interval [0,1].
You should obtain a picture like this below:

$y=\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}$\FRAME{dtbpFX}{3.0727in}{2.0487in}{0pt}{%
}{}{Plot}{\special{language "Scientific Word";type "MAPLEPLOT";width
3.0727in;height 2.0487in;depth 0pt;display "USEDEF";plot_snapshots
FALSE;mustRecompute FALSE;lastEngine "MuPAD";xmin "0";xmax "1";xviewmin
"-0.001";xviewmax "1.001";yviewmin "-0.003";yviewmax "3.003";plottype
4;numpoints 100;plotstyle "patch";axesstyle "normal";xis \TEXUX{x};yis
\TEXUX{y};var1name \TEXUX{$x$};var2name \TEXUX{$y$};function
\TEXUX{$\dfrac{7x-x^{2}}{(2-x)(x^{2}+1)}$};linecolor "black";linestyle
1;pointstyle "point";linethickness 1;lineAttributes "Solid";var1range
"0,1";num-x-gridlines 100;curveColor "[flat::RGB:0000000000]";curveStyle
"Line";rangeset"X";}}

Now you can calculate the area between the curve, X-axis, and $x=1.$ This
can be done as follows.

Type the integral$\int_{0}^{1}ydx$, leave the insertion beam at the end of
integral and choose Compute+Evaluate and then again Compute+Evaluate
Numerically

\begin{eqnarray*}
\int_{0}^{1}ydx &=&\allowbreak \int_{\boldsymbol{0}}^{\boldsymbol{1}}\frac{%
\boldsymbol{x}^{\boldsymbol{2}}\boldsymbol{-7x}}{\boldsymbol{x-2x}^{%
\boldsymbol{2}}\boldsymbol{+x}^{3}\boldsymbol{-2}}\,\boldsymbol{dx} \\
&=&1.\,\allowbreak 640\,6\allowbreak
\end{eqnarray*}

This completes our solution. You can also calculate the above integral by
integrating the partial fractions obtained in the first part of the solution
and collecting obtained integrals of partial fractions.$\blacksquare $

\paragraph{NOTE:}

Most of the above Compute commands can be obtained by using the appropriate
icons on the toolbox:

\begin{itemize}
\item Compute+Evaluate =\FRAME{itbpF}{0.3001in}{0.2793in}{0in}{}{}{eval.wmf}{%
\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
TRUE;display "PICT";valid_file "F";width 0.3001in;height 0.2793in;depth
0in;original-width 0.2707in;original-height 0.2499in;cropleft "0";croptop
"1";cropright "1";cropbottom "0";filename
'C:/Susan/NewWeb/images/eval.wmf';file-properties "XNPEU";}}

\item Compute+Definitons+New Definition =\FRAME{itbpF}{0.3001in}{0.2793in}{%
0in}{}{}{newdef.wmf}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width
0.3001in;height 0.2793in;depth 0in;original-width 0.2707in;original-height
0.2499in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'C:/Susan/NewWeb/images/newdef.wmf';file-properties "XNPEU";}}

\item Compute+Plot2D+Rectangular =\FRAME{itbpF}{0.3001in}{0.2793in}{0in}{}{}{%
2dplot.wmf}{\special{language "Scientific Word";type
"GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width
0.3001in;height 0.2793in;depth 0in;original-width 0.2707in;original-height
0.2499in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename
'C:/Susan/NewWeb/images/2dplot.wmf';file-properties "XNPEU";}}

\item To copy a formula into another location you can use the Edit, Copy -
Paste options of menu or simply by selecting a formula and dragging it with
right mouse button depressed.\newpage
\end{itemize}

\subsection{\ Example 3 (Discrete Mathematics)}

Use the binary search to determine if 2.773 is in the sorted sequence: $\ln
10,$ $\ln 11,$ $\ln 12,...,\ln 29,$ where each member of the sequence is
given correct to three decimal places.

\subsubsection{Solution}

\paragraph{Step 1.}

To model the above sequence we use a function $s(i)=\ln (9+i),$ where $%
i=1,2,...,20.$ Type the formula for the sequence and use Compute+Define+New
Definition.

\paragraph{Step 2. (The binary search algorithm)}

Let's remind a definition for the Binary Search algorithm. It can be like
this:

\QTP{Body Math}
INPUT: $a,$ $s=<s_{1},s_{2},...,s_{n}>,$ here $n=20$

\QTP{Body Math}
OUTPUT: report

\QTP{Body Math}
METHOD:

$p\leftarrow 1,$

$q\leftarrow n$

\QTP{Body Math}
REPEAT

$m\leftarrow \left\lfloor \dfrac{p+q}{2}\right\rfloor $

if $a<s_{m}$ then $q\leftarrow (m-1)$ else $p\leftarrow (m+1)$

\QTP{Body Math}
UNTIL $s_{m}=a$ or $p>q$

If $s_{m}=a$ then report: $x_{m}=a$ else report 'nor found'

Let us step through the example to observe the working of the algorithm.
First let's identify and define (\textbf{Compute+}Define) the basic
variables and functions:

\QTP{Body Math}
$a=2.773$

$s(i)=\ln (9+i),$ defined above,

\QTP{Body Math}
$m(p,q)=\left\lfloor \dfrac{p+q}{2}\right\rfloor $

Finally we will define a function $L(x,y)$ where $L$ is a function that uses
MuPAD to evaluate if $x<y$. In order to do this type in the Windows Notepad
the following MuPAD code

\textbf{less:=proc(x,y)}

\textbf{begin}

\textbf{\ \ \ bool(x \TEXTsymbol{<} y)}

\textbf{end:}

and save this code as \textsc{less.mu} somewhere on the hard disk of your
computer. Now, in SNB in the Compute menu find option Definitions and then
Define MuPAD Name. Here in the dialog box write less(x,y) as a MuPAD name,
L(x,y) as Scientific Notebook name and in the bottom part of the dialog
browse to the place where you saved the less.mu file. In my computer it
looks like in the enclosed picture.

\FRAME{dtbpF}{9.1489cm}{7.82cm}{0pt}{}{}{Figure}{\special{language
"Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display
"PICT";valid_file "T";width 9.1489cm;height 7.82cm;depth 0pt;original-width
3.5743in;original-height 3.0511in;cropleft "0";croptop "1";cropright
"1";cropbottom "0";tempfilename 'HUSKLD00.wmf';tempfile-properties "XPR";}}

Now, you can try if your new function works, for example after typing $%
L(1,2) $ and pressing Ctrl+Enter you should get $L(1,2)=\allowbreak true$
and $L(3.67,1.34)=\allowbreak false$.

Now let's put the appropriate elements of the algorithm into a table and
evaluate each of them. In order to evaluate values in sixth column you we
use Compute+Check Equality.

\QTP{Body Math}
$%
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
$m$ & $s(m)$ & $a<s(m)$ & $p$ & $q$ & $s(m)=a$ & $p>q$ \\ \hline
&  &  & $1$ & $20$ &  &  \\ \hline
$m(1,20)=\allowbreak 10$ & $s(10)=2.\,\allowbreak 944$ & $%
L(a,2.944)=\allowbreak true$ & $1$ & $9$ & $(a=2.944)$ is false & $false$ \\ 
\hline
$m(1,9)=\allowbreak 5$ & $s(5)=\allowbreak 2.\,\allowbreak 639$ & $%
L(a,2.639)=\allowbreak false$ & $\allowbreak 6$ & $9$ & $(a=2.639)$ is false
& $false$ \\ \hline
$m(6,9)=\allowbreak 7$ & $s(7)=\allowbreak 2.\,\allowbreak 773$ & $%
L(a,2.773)=\allowbreak false$ & $8$ & $9$ & $(a=2.773)$ is true & $false$ \\ 
\hline
&  &  &  &  &  &  \\ \hline
\end{tabular}%
\ $\newline
\linebreak OUTPUT: $s(7)=a.$

This completes the example.$\blacksquare $

\end{document}