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Cell[TextData[{
StyleBox["Learning ",
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StyleBox["Mathematica",
FontSize->16,
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FontSlant->"Italic",
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StyleBox[" Basics\n",
FontSize->16,
FontWeight->"Bold",
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StyleBox["Algebraic Operations II",
FontWeight->"Bold",
FontVariations->{"CompatibilityType"->"Subscript"}]
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"Solve each of the following problems using ",
StyleBox["Mathematica",
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" exclusively."
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"1. ",
StyleBox["Mathematica",
FontSlant->"Italic"],
"'s ",
StyleBox["Solve",
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FontWeight->"Bold"],
" command is where we begin our work today. Ask ",
StyleBox["Mathematica",
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" for its syntax using the ",
StyleBox["?",
FontFamily->"Courier",
FontWeight->"Bold"],
" command."
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"2. As with so many commands in ",
StyleBox["Mathematica",
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", ",
StyleBox["Solve",
FontFamily->"Courier",
FontWeight->"Bold"],
" has a variety of different flavors. The most basic version of the command \
is the first one shown: ",
StyleBox["Solve[",
FontFamily->"Helvetica"],
StyleBox["eqns, vars",
FontSlant->"Italic"],
StyleBox["]",
FontFamily->"Helvetica"],
". Let's use it to solve the equation ",
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FormBox[
RowBox[{
RowBox[{
RowBox[{"5", "x"}], " ", "+", " ", "3"}], " ", "=", " ", "17"}],
TraditionalForm]]],
". Issue the following command and hit ENTER:\n\n",
StyleBox["Solve[5x+3==17,x]",
FontFamily->"Courier",
FontWeight->"Bold"]
}], "Text",
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"3. You should have gotten the answer 14/5. \n\n[It's possible that \
instead of this answer, you got the empty set, which ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" indicates with a pair of braces enclosing nothing, i.e. ",
StyleBox["{}", "Output"],
". If this happened to you, it is almost certainly because your ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" kernel was already running from a previous notebook, and at some earlier \
point during this kernel session the variable ",
StyleBox["x",
FontSlant->"Italic"],
" was assigned a value. If ",
StyleBox["x",
FontSlant->"Italic"],
" has a value (other than 14/5) assigned to it, then the equation that you \
just asked ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" to solve will be a false statement without a solution. To overcome this \
problem, simply remove the value assigned to ",
StyleBox["x",
FontSlant->"Italic"],
" by issuing the command \"",
StyleBox["x=.", "Input"],
"\" as mentioned earlier, then try your ",
StyleBox["Solve", "Input"],
" command again.]\n\nA couple of points about the above exercise are of \
note. First, notice that when declaring an equation in ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" a ",
StyleBox["double equals sign",
FontWeight->"Bold"],
" is used. This is to distinguish ",
StyleBox["equations",
FontWeight->"Bold"],
" from ",
StyleBox["assignment of value",
FontWeight->"Bold"],
" (either the single equals (",
StyleBox["=",
FontFamily->"Courier",
FontWeight->"Bold"],
") or the colon equals (",
StyleBox[":=",
FontFamily->"Courier",
FontWeight->"Bold"],
") depending on whether you want delayed or immediate assignment, remember?)\
\n\n",
StyleBox["==",
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" is a ",
StyleBox["logical",
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" operator. It returns a value of ",
StyleBox["True",
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FontWeight->"Bold"],
" or ",
StyleBox["False",
FontFamily->"Courier",
FontWeight->"Bold"],
". To see this, type in the following equations, hitting ENTER after each \
one individually, so that each resides in a separate cell:\n\n",
StyleBox["5==10\n4==4\n2x==6",
FontFamily->"Courier",
FontWeight->"Bold"]
}], "Text",
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"4. Notice that the first two of these gave predictable results, whereas \
the last one simply returned the equation unevaluated. The reason for this \
is fairly obvious. Since we haven't given ",
StyleBox["x",
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" a value during this session then the third equation remains ",
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FontWeight->"Bold"],
", i.e. it could be ",
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" or ",
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FontFamily->"Courier",
FontWeight->"Bold"],
" depending on the value of ",
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" chosen.\n\nThe second point to be noticed about problem 2 is the form in \
which ",
StyleBox["Mathematica",
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" gives us the solution to the equation. Unlike our \"on paper\" solutions, \
where we would write \"",
StyleBox["x",
FontSlant->"Italic"],
"=14/5\", ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" gives us the result in the form of a ",
StyleBox["replacement rule",
FontWeight->"Bold",
FontSlant->"Italic"],
". (Remember learning about these in the previous notebook. We used \
replacement rules when we wished to do substitutions into algebraic \
expressions without making any \"permanent\" changes to the variables.)\n\nA \
little thought allows us to justify why ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" chooses to give us the answer in this form:\n\na. ",
StyleBox[" x",
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"=14/5 is an ",
StyleBox["assignment of value",
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". We don't ",
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" to ",
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" ",
StyleBox["x",
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" a value, just to find out ",
StyleBox["which",
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" value makes the equation true.\n\nb. If we plan on using the solution we \
find in later calculations then most likely we'll wish to substitute it into \
another expression, in which case we'd need the solution in replacement rule \
form anyway.\n\nWhew! Let's do something a little more interesting. Let's \
see ",
StyleBox["how good",
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" ",
StyleBox["Mathematica",
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" ",
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" is at solving equations. After all, that first one wasn't much of a \
challenge.\n\nAsk ",
StyleBox["Mathematica",
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" to solve the equation: ",
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".\n\n",
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FontWeight->"Bold",
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Cell[TextData[{
"5. Whoah! That answer was huge. It should have consisted of four \
separate solutions, since the polynomial was fourth degree. If you sift \
through the above gibberish carefully you'll find that you did indeed get \
exactly that. Exactly being the operative word. Notice that the above is an \
",
StyleBox["exact",
FontWeight->"Bold",
FontSlant->"Italic"],
" solution, no decimal approximations are being used.\n\nThis brings up an \
interesting point! It would seem that the best kind of answer you could have \
for any problem is an exact one. However, when an exact answer reaches the \
scale of the one above, it might actually be a more practical or useful \
answer if it ",
StyleBox["were",
FontSlant->"Italic"],
" in decimal form. We can force the issue! Apply ",
StyleBox["N",
FontFamily->"Courier",
FontWeight->"Bold"],
" to the previous output! ",
" (You may use the % shortcut.)"
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Cell[TextData[{
"6. Well now the answers look a little more understandable: two distinct \
real roots, and a complex conjugate pair.\n\nLet's push the solver a little \
harder. Ask ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" to solve the equation:\n\n ",
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Cell[TextData[{
"7. At first this appears to be some complex error message, but in reality \
",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is showing you what function calls it must make in an order to solve the \
equation. It's using an obscure notation for the original polynomial to \
express it as what is called a \"pure function,\" and the Root commands \
wrapped around the pure function says that it's asking for the exact roots of \
the pure function. There are five of these, hence the five calls to the Root \
command. The fact that they remain unevaluated tells us that these functions \
failed to solve the equation ",
StyleBox["exactly",
FontSlant->"Italic"],
". This is not surprising! You may have learned that in general, higher \
order polynomials are impossible to solve by algebraic methods. i.e. the \
variables cannot usually be isolated.\n\nRemember, however, that all ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is saying is that it can't find the answer ",
StyleBox["exactly",
FontWeight->"Bold"],
". You're probably aware of the fact that there are quite a few techniques \
available for solving equations ",
StyleBox["approximately",
FontWeight->"Bold"],
", numerical techniques, usually implemented via computers. Of course, ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" has the ability to access many of these algorithms. For this problem we \
can do this in two ways, though the results are the same:\n\na. We can \
simply apply ",
StyleBox["N", "Input"],
" to the previous output. (You may use the % shortcut.)"
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"b. We can start the problem over again using the ",
StyleBox["NSolve",
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FontWeight->"Bold"],
" command. Reissue your the command in problem 6, but replace ",
StyleBox["Solve",
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FontWeight->"Bold"],
" with ",
StyleBox["NSolve",
FontFamily->"Courier",
FontWeight->"Bold"],
"."
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"8. Using either method a or b, we get three real solutions, and a complex \
conjugate pair, (though curiously not in the same order.) Keep the ",
StyleBox["NSolve",
FontFamily->"Courier",
FontWeight->"Bold"],
" command in mind. It it frequently more useful than the ",
StyleBox["Solve",
FontFamily->"Courier",
FontWeight->"Bold"],
" command, and even though the answers it gives are approximations, they are \
",
StyleBox["extremely accurate",
FontWeight->"Bold",
FontSlant->"Italic"],
" approximations.\n\nMoving on, we will now see that there are even limits \
to the abilities of the ",
StyleBox["NSolve",
FontFamily->"Courier",
FontWeight->"Bold"],
" command. For instance, try using ",
StyleBox["NSolve",
FontFamily->"Courier",
FontWeight->"Bold"],
" on this equation:\n\n ",
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"\n\n(Look up the syntax of the functions in this equation if necessary.)"
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"9. So even ",
StyleBox["NSolve",
FontFamily->"Courier",
FontWeight->"Bold"],
" fails on such a nasty equation, but there is another option. Ask ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" for the syntax of the command ",
StyleBox["FindRoot.",
FontFamily->"Courier",
FontWeight->"Bold"]
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"10. So apparently there's a ",
StyleBox["third",
FontSlant->"Italic"],
" method for solving equations, but this one has a couple of twists. First, \
notice that the description talks about \"",
StyleBox["solution",
FontFamily->"Courier"],
"\" or \"",
StyleBox["root",
FontFamily->"Courier"],
"\", i.e. singular not plural. Apparently ",
StyleBox["FindRoot",
FontFamily->"Courier",
FontWeight->"Bold"],
" is only capable of finding one solution at a time, not the whole set of \
solutions like the previous two commands.\n\nAlso, notice the last part of \
the first description: \"",
StyleBox["starting from the point",
FontFamily->"Arial"],
" ",
StyleBox["x = ",
FontSlant->"Italic"],
Cell[BoxData[
FormBox[
SubscriptBox["x", "0"], TraditionalForm]]],
".\" This means that in order to use this particular method it is necessary \
to have an ",
StyleBox["initial guess",
FontWeight->"Bold"],
" for the solution you are seeking to the equation. The closer your initial \
guess is to the actual solution you seek, the better are the chances of the \
numerical method which ",
StyleBox["FindRoot",
FontFamily->"Courier",
FontWeight->"Bold"],
" uses \"hitting\" the solution you are after.\n\nLet's get down to the \
business of solving the equation from problem 8. Since we'll be using ",
StyleBox["FindRoot",
FontFamily->"Courier",
FontWeight->"Bold"],
", we'll need an initial guess. Unfortunately, the equation is so messy \
it's hard to imagine anything even close to a solution. Where should we \
begin? At times like this a graph is worth a thousand guesses. We've never \
actually used ",
StyleBox["Mathematica",
FontSlant->"Italic"],
"'s graphing features yet, so I'll give you the command you'll need:\n\n",
StyleBox["Plot[ArcTan[x]+E^x-Cos[x]+0.3,{x,-11,2}]",
FontFamily->"Courier",
FontWeight->"Bold"],
"\n\nIn a later notebook we'll discuss the details of this, and other \
graphing commands. For now, just enter the command, and evaluate it."
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"11. Well it looks like there are actually ",
StyleBox["three",
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FontSlant->"Italic"],
" solutions. First we'll try to get the one closest to the origin by using \
an initial guess of ",
StyleBox["x",
FontSlant->"Italic"],
"=0. The command would be:\n\n",
StyleBox["FindRoot[ArcTan[x]+E^x-Cos[x]+0.3==0,{x,0}]",
FontFamily->"Courier",
FontWeight->"Bold"],
"\n\nEnter and evaluate it."
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Cell["\<\
12. You should have gotten an answer of {x -> -0.163763}. Now find the \
other two solutions by using initial guesses of -2 and -4. (You'll need two \
separate commands in different cells, of course.)\
\>", "Text",
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"13. Well that about covers ",
StyleBox["Mathematica",
FontSlant->"Italic"],
"'s abilities with solving equations containing a single variable. We \
haven't yet exhausted the abilities of the ",
StyleBox["Solve",
FontFamily->"Courier",
FontWeight->"Bold"],
" command, however.\n\nLet's have ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" solve the following equation:\n\n",
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"\n \nfor ",
StyleBox["G",
FontSlant->"Italic"],
". The command, (looking back at the syntax in problem 1), would be:\n\n",
StyleBox["Solve[f==(g*m1*m2)/r^2,g]",
FontFamily->"Courier",
FontWeight->"Bold"],
"\n\nNotice that I'm obeying the convention of using lower-case letters for \
user-defined variables. Enter the command and evaluate it."
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"14. Now ",
StyleBox["you",
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" solve the same equation, but this time for ",
StyleBox["r",
FontSlant->"Italic"],
"."
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Cell[TextData[{
"15. Let's move up a notch. What about solving ",
StyleBox["systems",
FontWeight->"Bold"],
" of equations? Let's solve the system:\n\n5",
StyleBox["x",
FontSlant->"Italic"],
" + 3",
StyleBox["y",
FontSlant->"Italic"],
" = 7, 9",
StyleBox["x",
FontSlant->"Italic"],
" - 11",
StyleBox["y",
FontSlant->"Italic"],
" = 4.\n \nThe command \
would be:\n\n",
StyleBox["Solve[{5x+3y==7, 9x-11y==4}",
FontFamily->"Courier",
FontWeight->"Bold"],
StyleBox[",", "Input",
FontFamily->"Courier",
FontWeight->"Bold"],
StyleBox["{x, y}", "Input"],
StyleBox["]",
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FontWeight->"Bold"],
"\n\nEnter then evaluate it."
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"16. How about a non-linear system of equations? Solve the system:\n\n ",
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"17. Oooh! Four very ugly ordered pairs. Notice that ",
StyleBox["Solve",
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