(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 42925, 1718]*) (*NotebookOutlinePosition[ 73603, 2903]*) (* CellTagsIndexPosition[ 73559, 2899]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Discovering Number Theory", "Subtitle"], Cell["\<\ Jeffrey J. Holt University of Virginia John W. Jones Arizona State University\ \>", "Text", TextAlignment->Center, TextJustification->0], Cell["Copyright \[Copyright] 2001 by W. H. Freeman and Company", "Text", TextAlignment->Center, TextJustification->0, FontSize->10] }, Closed]], Cell["\<\ The Euclidean Algorithm and Linear Diophantine Equations\ \>", "Title", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Initializations", "Subsection", Evaluatable->False, TextAlignment->Left, AspectRatioFixed->True], Cell[BoxData[ \(Off[General::spell]; \nOff[General::spell1]; \)], "Input", InitializationCell->True, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["The Division Algorithm Revisited", "Section", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ Below is a statement of the division algorithm, as given in the \ Prelab section of this chapter:\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["Theorem (Division Algorithm)", FontWeight->"Bold"], " Let ", StyleBox["a", FontSlant->"Italic"], " be an integer and ", StyleBox["b", FontSlant->"Italic"], " be a positive integer. Then there exist unique integers ", StyleBox["q", FontSlant->"Italic"], " and ", StyleBox["r", FontSlant->"Italic"], " such that" }], "Text", Evaluatable->False, AspectRatioFixed->True, FontColor->GrayLevel[0]], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`a\ = \ b\ q\ + \ r\)]], "\tand\t0 \[LessEqual] ", StyleBox["r", FontSlant->"Italic"], " < ", StyleBox["b", FontSlant->"Italic"], "." }], "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True, FontColor->GrayLevel[0]], Cell[TextData[{ "Recall that at the end of the last chapter, you used a Mathematica \ function called ", StyleBox["divalg", "Input"], " to compute ", StyleBox["q", FontSlant->"Italic"], " and ", StyleBox["r", FontSlant->"Italic"], " for a given ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ". Here is the definition for ", StyleBox["divalg", "Input"], ":" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(\(divalg[a_, b_] := Module[{q, r}, q = Floor[a/b]; r = a - b\ q; Return[{q, r}]]; \)\)], "Input", InitializationCell->True, AspectRatioFixed->True], Cell[TextData[{ "Here's how it works for ", StyleBox["a", FontSlant->"Italic"], " = 123456789 and ", StyleBox["b", FontSlant->"Italic"], " = 369:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(divalg[123456789, 369]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Thus, for this choice of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", we have ", StyleBox["q", FontSlant->"Italic"], " = 334571 and ", StyleBox["r", FontSlant->"Italic"], " = 90. In Prelab exercise 2, you were asked to compute gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") and gcd(", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], ") for different choices of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ". If we do this computation in this case, we get" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(a\ = \ 123456789; \ \nb\ = \ 369; \nr\ = \ 90; \n{GCD[a, b], GCD[b, r]}\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "We can save ourselves a little typing by combining the computation \ required to compute ", StyleBox["r", FontSlant->"Italic"], " with the computation to compute gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") and gcd(", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], "). We do this by using the fact that ", StyleBox["divalg[a,b][[2]]", "Input"], " will return just ", StyleBox["r.", FontSlant->"Italic"] }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ a = 123456789; b = 369; {GCD[a, b], GCD[b, divalg[a, b][[2]]]}\ \>", "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Mathematica Note", "Subsubtitle"], Cell[TextData[{ "In general, if you have a list such as ", StyleBox["mylist = {1, 2, 4, 8, 16}", "Input"], ", you can select a particular element from the list by using double \ brackets: \n\n ", StyleBox[" mylist[[4]] = 8\n \n", "Input"], StyleBox[ "For those with computer programming experience, this is the analogue of \ working with an array, where we would specify the ", "Text"], StyleBox["n", "Text", FontSlant->"Italic"], StyleBox["th entry of the array ", "Text"], StyleBox["mylist", "Input"], StyleBox[" with ", "Text"], StyleBox["mylist[[n]]", "Input"], StyleBox[".", "Text"] }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Research Question 1", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Compute the values of gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") and gcd(", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], ") as above for different pairs ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], " of your choosing until you have enough data to form a conjecture \ concerning the relationship between gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") and gcd(", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], "). As always, once you have formed your conjecture, try to prove it!" }], "SmallText", CellFrame->True, Evaluatable->False, AspectRatioFixed->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["The Euclidean Algorithm 1", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Now that you have completed Research Question 1, you know that for a pair \ of integers ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", we have" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = gcd(", StyleBox["b", FontSlant->"Italic"], ", ", StyleBox["r", FontSlant->"Italic"], ")," }], "Text", Evaluatable->False, TextAlignment->Center, TextJustification->0, AspectRatioFixed->True], Cell[TextData[{ "where ", StyleBox["r", FontSlant->"Italic"], " is the remainder when ", StyleBox["a", FontSlant->"Italic"], " is divided by ", StyleBox["b", FontSlant->"Italic"], ". (Sure, this gives away the conjecture for Research Question 1, but you \ already knew the conjecture anyway, right?) Thus, for example, if ", StyleBox["a", FontSlant->"Italic"], " = 123456789 and ", StyleBox["b", FontSlant->"Italic"], " = 369 (as in the preceding section), then ", StyleBox["r", FontSlant->"Italic"], " = 90, and we see that" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["gcd(123456789, 369) = gcd(369, 90).", "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "It's clear that computing gcd(369, 90) will be easier than computing \ gcd(123456789, 369). (This will be true using any of the methods discussed \ in the previous chapter.) But rather than trying to compute gcd(369, 90), \ why don't we make the problem even easier by repeating what we did above? If \ now we think of ", StyleBox["a", FontSlant->"Italic"], " = 369 and ", StyleBox["b", FontSlant->"Italic"], " = 90, then we have ", StyleBox["r", FontSlant->"Italic"], " equal to" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell["divalg[369, 90][[2]]", "Input", AspectRatioFixed->True], Cell["\<\ Thus it follows that gcd(369, 90) = gcd(90, 9). Repeating the \ above, we compute\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell["divalg[90, 9][[2]]", "Input", AspectRatioFixed->True], Cell[TextData[{ "It shouldn't be surprising that the remainder in this case is equal to 0. \ After all, it's easy to see that ", Cell[BoxData[ \(TraditionalForm\`90/9\)]], " = 10. Thus 9 is a divisor of 90, so that we have gcd(90, 9) = 9. \ Finally, we just string all of this together to arrive at" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[Cell[BoxData[ \(TraditionalForm \`gcd(123456789, \ 369)\ = \ \(gcd(369, \ 90)\ = \ \(gcd(90, \ 9)\ = \ 9. \)\)\)]]], "Text", Evaluatable->False, TextAlignment->Center, TextJustification->0, AspectRatioFixed->True], Cell[TextData[{ "The process illustrated is called the ", StyleBox["Euclidean Algorithm,", FontSlant->"Italic"], " and it is very efficient for computing gcds. In fact, this is probably \ the method that the Mathematica command ", StyleBox["GCD", "Input"], " uses for computing greatest common divisors." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Exercise 1", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell["Repeat the above procedure to compute gcd(7920, 4536). ", "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Linear Diophantine Equations 1", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "As stated in the Prelab section, one goal of this chapter is to develop a \ systematic method for finding all integer solutions ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " (when there are any) to the linear diophantine equation" }], "Text"], Cell[TextData[{ " ", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ c\)]], "," }], "NumberedEquation"], Cell[TextData[{ "where ", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ", and ", StyleBox["c", FontSlant->"Italic"], " are integer constants. As you discovered when working on the Prelab \ exercises, for a given choice of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", equation (1) may have several solutions or possibly no solutions; it \ depends on the value of ", StyleBox["c", FontSlant->"Italic"], ". If we have specific values for ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", how can we figure out the values of ", StyleBox["c", FontSlant->"Italic"], " for which equation (1) will have a solution? As a first step, let's look \ at the situation for a specific choice of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", say ", StyleBox["a", FontSlant->"Italic"], " = 6 and ", StyleBox["b", FontSlant->"Italic"], " = 4. In this case, the above equation becomes" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`6 x\ + \ 4 y\ = \ c\)]], "," }], "NumberedEquation"], Cell[TextData[{ "and our question is: For what values of ", StyleBox["c", FontSlant->"Italic"], " is there a solution to this equation? One way to approach this is to try \ plugging a bunch of different values for ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " into the left\[Hyphen]hand side of (2), and see what we get out. After \ all, any value that comes out must be a suitable value for ", StyleBox["c", FontSlant->"Italic"], ". (Do you see why?) Below is the Mathematica code to compute the values \ of 6", StyleBox["x", FontSlant->"Italic"], " + 4", StyleBox["y ", FontSlant->"Italic"], "for each choice of ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " satisfying -3 \[LessEqual] ", StyleBox["x", FontSlant->"Italic"], " \[LessEqual] 5 and -3 \[LessEqual] ", StyleBox["y", FontSlant->"Italic"], " \[LessEqual] 5:" }], "Text"], Cell[BoxData[ \(Flatten[Table[6\ x\ + \ 4\ y, \ {x, \(-3\), 5}, {y, \(-3\), 5}]]\)], "Input"], Cell[CellGroupData[{ Cell["Mathematica Note", "Subsubtitle"], Cell[TextData[{ "The ", StyleBox["Flatten", "Input"], " is there to remove extra sets of unnecessary braces. If you really want \ to see what it does, remove the ", StyleBox["Flatten", "Input"], " command and re\[Hyphen]execute the cell." }], "Text"] }, Closed]], Cell[TextData[{ "On the basis of this list, it looks like ", StyleBox["c", FontSlant->"Italic"], " must be an even integer in order for equation (2) to have a solution. We \ can also see that some values appear several times; this indicates that for a \ given value of ", StyleBox["c", FontSlant->"Italic"], " there may be lots of solutions." }], "Text"], Cell[TextData[{ "If we think about it, if ", StyleBox["x", FontSlant->"Italic"], " = ", StyleBox["m", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " = ", StyleBox["n", FontSlant->"Italic"], " are solutions to ", Cell[BoxData[ \(TraditionalForm\`6 x\ + \ 4 y\ = \ c\)]], " then ", StyleBox["x", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\(-m\)\)]], " and ", StyleBox["y", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\(-n\)\)]], " are solutions to ", Cell[BoxData[ \(TraditionalForm\`6 x\ + \ 4 y\ = \ \(-c\)\)]], ". Therefore, if we know the positive values of ", StyleBox["c", FontSlant->"Italic"], " for which there are solutions to equation (2), then we also know the \ story for negative values of ", StyleBox["c", FontSlant->"Italic"], ". (Of course, ", StyleBox["c", FontSlant->"Italic"], " = 0 is easy. Quick, name choices for ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " such that ", Cell[BoxData[ \(TraditionalForm\`6 x\ + \ 4 y\ = \ 0\)]], ".) Thus, we can restrict ourselves to c > 0." }], "Text"], Cell[TextData[{ "Below is a Mathematica function that will compute ", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\)]], " for specified values of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], " by plugging in ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " satisfying ", Cell[BoxData[ \(TraditionalForm\`\(-n\)\)]], " \[LessEqual] ", StyleBox["x", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " and ", Cell[BoxData[ \(TraditionalForm\`\(-n\)\)]], " \[LessEqual] ", StyleBox["y", FontSlant->"Italic"], " \[LessEqual] ", StyleBox["n", FontSlant->"Italic"], " and then removing the repeated and nonpositive values:" }], "Text"], Cell[BoxData[ \(\(findc[a_, b_, n_]\ := \ \n\t Module[{clist}, \n\t\t\t\t\t\t clist\ = \ Table[a\ \ x\ + \ b\ \ y, \ {x, \(-n\), n}, {y, \(-n\), n}]; \n \t\t\t\t\t\tclist\ = \ Rest[Union[Abs[Flatten[clist]]]]; \n \t\t\t\t\t\tReturn[clist]]; \)\)], "Input", InitializationCell->True], Cell[CellGroupData[{ Cell["Mathematica Note", "Subsubtitle"], Cell[TextData[{ "The ", StyleBox["Union", "Input"], " command behaves just as you would expect with sets: it removes duplicate \ entries. Along the way, it also sorts the list." }], "Text"], Cell[TextData[{ "Here we also see ", StyleBox["Abs", "Input"], ", which takes the absolute value of a number. In this case, it is applied \ to a list. The end result is to take the absolute value of every entry in \ the list. Mathematica tries to do something sensible whenever an operation \ is applied to a list. The best way to understand what it will do is by \ experimenting. " }], "Text"], Cell[TextData[{ "Finally, we encounter the ", StyleBox["Rest", "Input"], " command. It removes the first entry of a list. Can you guess why we \ wanted to do that? If not, make a copy of ", StyleBox["findc", "Input"], " in a new cell, give it a new name, and remove ", StyleBox["Rest", "Input"], " from the definition. See if you can spot the difference between the \ original function ", StyleBox["findc", "Input"], " and the new one you created." }], "Text"] }, Closed]], Cell[TextData[{ "Here is what we get when we try ", StyleBox["findc", "Input"], " out on our equation, which corresponds to ", StyleBox["a", FontSlant->"Italic"], " = 6 and ", StyleBox["b", FontSlant->"Italic"], " = 4:" }], "Text"], Cell[BoxData[ \(findc[6, 4, 10]\)], "Input"], Cell[TextData[{ "As we observed above, it looks like ", StyleBox["c", FontSlant->"Italic"], " must be a multiple of 2 in order for equation (2) to have a solution. \ Let's try a different equation, say ", Cell[BoxData[ \(TraditionalForm\`9 x\ + \ 12 y\ = \ c\)]], ". What do we get in this case?" }], "Text"], Cell[BoxData[ \(findc[9, 12, 10]\)], "Input"], Cell[TextData[{ "Hmm, this time it looks like we get solutions only if ", StyleBox["c", FontSlant->"Italic"], " is a multiple of 3. How about ", Cell[BoxData[ \(TraditionalForm\`\(5 x\ + \ 8 y\ = \ c\ \)\)]], "?" }], "Text"], Cell[BoxData[ \(findc[5, 8, 10]\)], "Input"], Cell[CellGroupData[{ Cell["Research Question 2", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Evaluate ", StyleBox["findc[a,b]", "Input"], " for values of ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], " of your choosing until you have enough data to fill in the blank at the \ end of the following conjecture:\n\n\"In order for ", Cell[BoxData[ \(TraditionalForm\`a\ x + b\ y\ = \ c\)]], " to have solutions, ", StyleBox["c", FontSlant->"Italic"], " must be of the form ______.\"" }], "SmallText", CellFrame->True, Evaluatable->False, AspectRatioFixed->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["The Euclidean Algorithm 2", "Section", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Finding One Solution", "Subsection"], Cell[TextData[{ "Your conjecture in Research Question 2 gives conditions that ", StyleBox["c", FontSlant->"Italic"], " must satisfy in order for " }], "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ c\)]]], "Text", TextAlignment->Center], Cell[TextData[{ "to have solutions. Suppose that ", StyleBox["c", FontSlant->"Italic"], " does have the form given in your conjecture; in this case, are we \ guaranteed that there will be solutions to our equation? This is a pretty \ general question. Let's look at one special case. Suppose that ", StyleBox["a", FontSlant->"Italic"], " = 408, ", StyleBox["b", FontSlant->"Italic"], " = 126, and ", StyleBox["c", FontSlant->"Italic"], " = gcd(408, 126) = 6. Are there any solutions to the equation" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(408 x\ + \ 126 y\ = \ 6\ \)\)]], "?" }], "Text", TextAlignment->Center, TextJustification->0], Cell["\<\ It turns out that the answer to this question is yes. To find a \ solution, we start by going through the steps of the Euclidean Algorithm to \ show that gcd(408, 126) = 6:\ \>", "Text"], Cell[BoxData[ \(divalg[408, 126]\)], "Input"], Cell["Therefore, 408 = 3 \[CenterDot] 126 + 30. Remember that ", "Text"], Cell["gcd(408, 126) = gcd(126, 30).", "Text", TextAlignment->Center, TextJustification->0], Cell["Now we just repeat:", "Text"], Cell[BoxData[ \(divalg[126, 30]\)], "Input"], Cell["Therefore, 126 = 4 \[CenterDot] 30 + 6.", "Text"], Cell[BoxData[ \(divalg[30, 6]\)], "Input"], Cell["\<\ Therefore, 30 = 5 \[CenterDot] 6 + 0. Since clearly 30 is \ divisible by 6, it follows that gcd(30, 6) = 6. Thus it must be that \ gcd(408, 126) = 6. Now, what does this have to do with solutions to \ \>", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`\(408 x\ + \ 126 y\ = \ 6\ \)\)]], "?" }], "Text", TextAlignment->Center], Cell["\<\ Good question. Let's take a look at the steps in the Euclidean \ Algorithm again:\ \>", "Text"], Cell["(a) 408 = 3 \[CenterDot] 126 + 30.", "Listing"], Cell["(b) 126 = 4 \[CenterDot] 30 + 6.", "Listing"], Cell["(c) 30 = 5 \[CenterDot] 6 + 0.", "Listing"], Cell["Reorganizing the equation in step (b), we have", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`6\ = \ 126 - 4\)]], " \[CenterDot] 30." }], "NumberedEquation"], Cell[TextData[{ "From step (a), we see that 30 = ", Cell[BoxData[ \(TraditionalForm\`408\ - \ 3\ \[CenterDot]\ 126\)]], ". Substituting into equation (3), we get" }], "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm \`6\ \ = \ \(126\ - \ 4\ \[CenterDot]\ 30\n\t = \ \(126\ - \ 4\ \[CenterDot]\ \((408\ - \ 3\ \[CenterDot]\ 126)\)\n \t = \ \(\(-4\)\ \[CenterDot]\ 408\ + \ \((1\ + \ 12)\)\ \[CenterDot]\ 126\n\t = \ \(-4\)\ \[CenterDot]\ 408\ + \ 13\ \[CenterDot]\ 126\)\)\)\)]]], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Therefore, we see that ", Cell[BoxData[ \(TraditionalForm\`x\ = \ \(-4\)\)]], ", ", StyleBox["y", FontSlant->"Italic"], " = 13 is a solution to " }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`408 x\ + \ 126 y\ = \ 6\)]], "." }], "Text", TextAlignment->Center], Cell["\<\ Just to be on the safe side, let's check using Mathematica by \ executing the code below. \ \>", "Text"], Cell[BoxData[ \(408*\((\(-4\))\)\ + \ 126*13\ == \ 6\)], "Input"], Cell[CellGroupData[{ Cell["Mathematica Note", "Subsubtitle"], Cell[TextData[{ "The \"", StyleBox["==", "Input"], "\" means that this code is treated as an equation rather than a \ definition. When you execute this code, Mathematica compares the values on \ both sides of the equation, and returns ", StyleBox["True", "Input"], " if both sides are equal, and ", StyleBox["False", "Input"], " otherwise." }], "Text"] }, Closed]], Cell[TextData[{ "The ", StyleBox["True", "Input"], " means that the left and right sides of the equation are equal. Let's \ look at another example. Suppose that ", StyleBox["a", FontSlant->"Italic"], " = 1232 and ", StyleBox["b", FontSlant->"Italic"], " = 573, and we want to find a solution to" }], "Text"], Cell[TextData[{ "1232 ", StyleBox["x", FontSlant->"Italic"], " + 573 ", StyleBox["y", FontSlant->"Italic"], " = ", StyleBox["d", FontSlant->"Italic"], "," }], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "where ", StyleBox["d", FontSlant->"Italic"], " = gcd(1232, 573). First we compute ", StyleBox["d", FontSlant->"Italic"], " using the Euclidean Algorithm:" }], "Text"], Cell[BoxData[ \(divalg[1232, 573]\)], "Input"], Cell["1. 1232 = 2 \[CenterDot] 573 + 86.", "Listing"], Cell[BoxData[ \(divalg[573, 86]\)], "Input"], Cell["2. 573 = 6 \[CenterDot] 86 + 57.", "Listing"], Cell[BoxData[ \(divalg[86, 57]\)], "Input"], Cell["3. 86 = 1 \[CenterDot] 57 + 29.", "Listing"], Cell[BoxData[ \(divalg[57, 29]\)], "Input"], Cell["4. 57 = 1 \[CenterDot] 29 + 28.", "Listing"], Cell[BoxData[ \(divalg[29, 28]\)], "Input"], Cell["5. 29 = 1 \[CenterDot] 28 + 1.", "Listing"], Cell[BoxData[ \(divalg[28, 1]\)], "Input"], Cell["6. 28 = 28 \[CenterDot] 1 + 0.", "Listing"], Cell[TextData[{ "We see that ", StyleBox["d", FontSlant->"Italic"], " = gcd(1232, 573) = 1, and so we are looking for a solution to " }], "Text"], Cell[TextData[{ "1232 ", StyleBox["x", FontSlant->"Italic"], " + 573 ", StyleBox["y", FontSlant->"Italic"], " = 1." }], "Text", TextAlignment->Center, TextJustification->0], Cell["Here's a summary of the above steps:", "Text"], Cell["1. 1232 = 2 \[CenterDot] 573 + 86.", "Listing"], Cell["2. 573 = 6 \[CenterDot] 86 + 57.", "Listing"], Cell["3. 86 = 1 \[CenterDot] 57 + 29.", "Listing"], Cell["4. 57 = 1 \[CenterDot] 29 + 28.", "Listing"], Cell["5. 29 = 1 \[CenterDot] 28 + 1.", "Listing"], Cell["6. 28 = 28 \[CenterDot] 1 + 0.", "Listing"], Cell["Now we work our way back up the chain:", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm \`1\ \ = \ \ \(29\ - \ 1\ \[CenterDot]\ 28\n\t = \ \ \(29\ - \ 1\ \[CenterDot]\ \((57\ - \ 1\ \[CenterDot]\ 29)\)\ \ \n \t = \ \ \(\(-\ 1\)\ \[CenterDot]\ 57\ + \ 2\ \[CenterDot]\ 29\n \t = \ \ \(\(-\ 1\)\ \[CenterDot]\ 57\ + \ 2\ \[CenterDot]\ \((86\ - \ 1\ \[CenterDot]\ 57)\)\n\t = \ \ \(2\ \[CenterDot]\ 86\ - \ 3\ \[CenterDot]\ 57\n\t = \ \ \(2\ \[CenterDot]\ 86\ - \ 3\ \[CenterDot]\ \((573\ - \ 6\ \[CenterDot]\ 86)\)\n \t = \ \ \(\(-\ 3\)\ \[CenterDot]\ 573\ + \ 20\ \[CenterDot]\ 86\n \t = \ \ \(\(-\ 3\)\ \[CenterDot]\ 573\ + \ 20\ \[CenterDot]\ \((1232\ - \ 2\ \[CenterDot]\ 573)\)\n\t = \ \ 20\ \[CenterDot]\ 1232\ - \ 43\ \[CenterDot]\ 573. \)\)\)\)\)\)\)\)\)]]], "Text", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "So, ", StyleBox["x", FontSlant->"Italic"], " = 20 and ", StyleBox["y", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\(-43\)\)]], " should be a solution to ", Cell[BoxData[ \(TraditionalForm\`1232 x\ + \ 573 y\ = \ 1\)]], ". As before, let's check:" }], "Text"], Cell[BoxData[ \(20*1232\ - \ 43*573\ \ == \ 1\)], "Input"], Cell["Solving the linear equation", "Text"], Cell[TextData[Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ gcd(a, \ b)\)]]], "Text", TextAlignment->Center], Cell[TextData[{ "is useful in a variety of places in number theory. Indeed, the fact that \ this equation always has a solution is so handy that we give it a special \ name: ", StyleBox["The", FontSlant->"Italic"], " ", StyleBox["GCD Trick", FontSlant->"Italic"], ". We'll have several opportunities to exploit the GCD Trick as the course \ progresses. Mathematica has a built\[Hyphen]in command that implements the \ procedure illustrated above. Given ", StyleBox["a", FontSlant->"Italic"], " and ", StyleBox["b", FontSlant->"Italic"], ", the command ", StyleBox["ExtendedGCD[a, b]", "Input"], " computes both gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") and a pair (", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], ") which satisfies the above equation. For example, if we start with ", StyleBox["a", FontSlant->"Italic"], " = 1232 and ", StyleBox["b", FontSlant->"Italic"], " = 573 as above, we get:" }], "Text"], Cell[BoxData[ \(ExtendedGCD[1232, \ 573]\)], "Input"], Cell[TextData[{ "The output comes in the form {gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], "), {", StyleBox["x", FontSlant->"Italic"], ", ", StyleBox["y", FontSlant->"Italic"], "}}. Note that it gives the same answer as was found above." }], "Text"] }, Closed]], Cell[CellGroupData[{ Cell["Finding All Solutions", "Subsection"], Cell["\<\ So far, so good. The method that we used in the two previous \ examples is quite general, and will work to find a solution to any equation \ of the form\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ d\)]], "," }], "NumberedEquation", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "where ", StyleBox["d", FontSlant->"Italic"], " = gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], "). Although we have made some progress towards our original goal of \ finding all solutions to ", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ c\)]], ", we still have a long way to go. The next question that we shall \ consider is the following: Are there any solutions to equation (4) other than \ the one guaranteed by the GCD Trick and found using the ", StyleBox["reverse Euclidean Algorithm", FontSlant->"Italic"], " method? Let's go with a simple example, " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`7 x\ + \ 2 y\ = \ 1\)]], "." }], "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "It's clear that gcd(7, 2) = 1, and we get the solution ", Cell[BoxData[ \(TraditionalForm\`x\_0\ = \ 1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`y\_0\ = \ \(-3\)\)]], " by reversing the steps of the Euclidean Algorithm. One way to look for \ other solutions is to use Mathematica to do a simple search. Solving for ", StyleBox["y", FontSlant->"Italic"], " in terms of ", StyleBox["x", FontSlant->"Italic"], ", we get the equation " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ StyleBox["y", FontSlant->"Italic"], " = ", Cell[BoxData[ \(TraditionalForm\`\((1\ - \ 7 x)\)/2\)]], "." }], "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "If we plug in an integer value of ", StyleBox["x", FontSlant->"Italic"], ", and the corresponding value of ", StyleBox["y", FontSlant->"Italic"], " is also an integer, then the pair ", Cell[BoxData[ \(TraditionalForm\`\((x, \ y)\)\)]], " is a solution to our equation. (Be sure that you understand why before \ moving forward.) Here's the Mathematica code to do the work. In this case, \ we are just using values of ", StyleBox["x", FontSlant->"Italic"], " satisfying ", Cell[BoxData[ \(TraditionalForm\`\(-10\) \[LessEqual] x \[LessEqual] 10\)]], ". " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Do[If[IntegerQ[\((1 - 7\ x)\)/2], Print[{x, \((1 - 7\ x)\)/2}]], {x, \(-10\), 10}]\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Mathematica Note", "Subsubtitle"], Cell[TextData[{ StyleBox["IntegerQ[t]", "Input"], " returns ", StyleBox["True", "Input"], " if ", StyleBox["t", "Input"], " is an integer, and ", StyleBox["False", "Input"], " otherwise." }], "Text"] }, Closed]], Cell[TextData[{ "As you can see, our solution ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " = 1, ", Cell[BoxData[ \(TraditionalForm\`y\_0\ = \ \(-3\)\)]], " is among those found. Let's try a different equation, say " }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`8 x\ + \ 3 y\ = \ 1\)]], "." }], "Text", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell[TextData[{ "In this case, it's easy to see that gcd(8, 3) = 1. The Euclidean \ Algorithm yields the solution ", Cell[BoxData[ \(TraditionalForm\`x\_0\ = \ \(-1\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`y\_0\ = \ 3\)]], ". Solving for ", StyleBox["y", FontSlant->"Italic"], " in terms of ", StyleBox["x", FontSlant->"Italic"], ", we have ", Cell[BoxData[ \(TraditionalForm\`y\ = \ \((1\ - \ 8 x)\)/3\)]], ". Here's the test:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(Do[If[IntegerQ[\ \((1 - 8\ x)\)/3], Print[{x, \((1 - 8\ x)\)/3}]], {x, \(-10\), 10}]\)], "Input", AspectRatioFixed->True], Cell["\<\ As you might guess, we are working towards forming a conjecture, \ and it will be handy to have a function that will automatically look for \ solutions. Here's one provided for your convenience:\ \>", "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(sols[a_, b_, d_, n_] := Do[If[IntegerQ[\((d - a\ x)\)/b], Print[{x, \((d - a\ x)\)/b}]], {x, \(-n\), n}]\)], "Input", InitializationCell->True, AspectRatioFixed->True], Cell[TextData[{ "The function ", StyleBox["sols", "Input"], " will search for solutions to ", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ d\)]], ", trying values of ", StyleBox["x", FontSlant->"Italic"], " satisfying ", Cell[BoxData[ \(TraditionalForm\`\(-n\) \[LessEqual] x \[LessEqual] n\)]], ". Let's check this function against the preceding output:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(sols[8, 3, 1, 10]\)], "Input", AspectRatioFixed->True], Cell[TextData[{ "Looks fine. Here's one more equation: ", Cell[BoxData[ \(TraditionalForm\`11 x\ + \ 4 y\ = \ 1\)]], ". As we can see, gcd(11, 4) = 1, and the Euclidean Algorithm produces the \ solution ", Cell[BoxData[ \(TraditionalForm\`x\_0\ = \ \(-1\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`y\_0\ = \ 3\)]], ". Here's the result of a search:" }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[BoxData[ \(sols[11, 4, 1, 20]\)], "Input", AspectRatioFixed->True], Cell[CellGroupData[{ Cell["Research Question 3", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Suppose that gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = 1, and that ", StyleBox["(", "Text"], Cell[BoxData[ \(TraditionalForm\`x\_0\)], "Text"], StyleBox[",", "Text"], Cell[BoxData[ FormBox[ StyleBox[\(\ y\_0\), "Text"], TraditionalForm]], "Text"], StyleBox[")", "Text"], " is the solution to \n\n\t\t\t\t", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ 1\)]], "\n\nproduced by the Euclidean Algorithm. Find a general formula for all \ solutions ", Cell[BoxData[ \(TraditionalForm\`\((x, \ y)\)\)]], " to the above equation, giving ", StyleBox["x", FontSlant->"Italic"], " in terms of ", Cell[BoxData[ \(TraditionalForm\`x\_0\)], "Text"], " and ", StyleBox["y", FontSlant->"Italic"], " in terms of ", Cell[BoxData[ \(TraditionalForm\`y\_0\)], "Text"], ". Be sure to prove that your formula works and that it accounts for ", StyleBox["all", FontSlant->"Italic"], " solutions." }], "SmallText", CellFrame->True, Evaluatable->False, TextAlignment->Left, TextJustification->1, AspectRatioFixed->True] }, Closed]], Cell[CellGroupData[{ Cell["Research Question 4", "Subsubsection", Evaluatable->False, AspectRatioFixed->True], Cell[TextData[{ "Now suppose that gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = ", StyleBox["d", FontSlant->"Italic"], " > 1, and that ", StyleBox["(", "Text"], Cell[BoxData[ \(TraditionalForm\`x\_0\)], "Text"], StyleBox[",", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\ y\_0\)\)], "Text"], StyleBox[")", "Text"], " is the solution to \n\n\t\t\t\t", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ d\)]], "\t\t\t\n\nproduced by the Euclidean Algorithm. Find a general formula for \ all solutions ", Cell[BoxData[ \(TraditionalForm\`\((x, \ y)\)\)]], " to the above equation, giving ", StyleBox["x", FontSlant->"Italic"], " in terms of", StyleBox[" ", "Text"], Cell[BoxData[ \(TraditionalForm\`x\_0\)], "Text"], StyleBox[" ", "Text"], "and ", StyleBox["y", FontSlant->"Italic"], " in terms of", StyleBox[" ", "Text"], Cell[BoxData[ \(TraditionalForm\`y\_0\)], "Text"], ". \n\n", StyleBox["Hint:", FontWeight->"Bold"], " The equations ", Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = d\)]], " and ", StyleBox["(", "Text"], Cell[BoxData[ \(TraditionalForm\`a/d\)], "Text"], StyleBox[") ", "Text"], StyleBox["x", "Text", FontSlant->"Italic"], StyleBox[" + (", "Text"], Cell[BoxData[ \(TraditionalForm\`b/d\)], "Text"], StyleBox[") ", "Text"], StyleBox["y", "Text", FontSlant->"Italic"], StyleBox[" = 1", "Text"], " have the same set of solutions." }], "SmallText", CellFrame->True, Evaluatable->False, TextAlignment->Left, AspectRatioFixed->True] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Linear Diophantine Equations 2", "Section", Evaluatable->False, AspectRatioFixed->True], Cell["\<\ Now we're getting somewhere. With the research questions in the \ preceding section complete, we now know all of the solutions to the equation \ \ \>", "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ d\)]], "," }], "Text", TextAlignment->Center], Cell[TextData[{ "where ", StyleBox["d", FontSlant->"Italic"], " = gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], "). Therefore, it remains to determine all of the solutions to the \ equation" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`a\ x\ + \ b\ y\ = \ k\ d\)]], "," }], "Text", TextAlignment->Center], Cell[TextData[{ "where ", StyleBox["k", FontSlant->"Italic"], " is an integer. To get a feel for what is going on, let's look at an \ example. In the preceding section, we looked at the equation ", Cell[BoxData[ \(TraditionalForm\`7 x\ + \ 2 y\ = \ 1\)]], ". Let's change this a bit, say to" }], "Text"], Cell[TextData[{ Cell[BoxData[ \(TraditionalForm\`7 x\ + \ 2 y\ = \ 5\)]], "." }], "Text", TextAlignment->Center], Cell[TextData[{ "Using the Euclidean Algorithm, we found that ", Cell[BoxData[ \(TraditionalForm\`x\ = \ 1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`y\ = \ \(-3\)\)]], " is a solution to ", Cell[BoxData[ \(TraditionalForm\`7 x\ + \ 2 y\ = \ 1\)]], ". Thus, it is easy to see that ", Cell[BoxData[ \(TraditionalForm\`x\)]], " = 1\[CenterDot]5 = 5 and ", Cell[BoxData[ \(TraditionalForm\`\(\ y\)\)]], " = ", Cell[BoxData[ \(TraditionalForm\`\(-3\)\)]], "\[CenterDot]5 = ", Cell[BoxData[ \(TraditionalForm\`\(-15\)\)]], " is a solution to the equation ", Cell[BoxData[ \(TraditionalForm\`7 x\ + \ 2 y\ = \ 5\)]], ". What about the other solutions? Let's put ", StyleBox["sols", "Input"], " to work to find some others:" }], "Text"], Cell[BoxData[ \(sols[7, 2, 5, 10]\)], "Input"], Cell[CellGroupData[{ Cell["Research Question 5", "Subsubsection"], Cell[TextData[{ "Suppose that gcd(", StyleBox["a", FontSlant->"Italic"], ", ", StyleBox["b", FontSlant->"Italic"], ") = ", StyleBox["d", FontSlant->"Italic"], " and that ", StyleBox["(", "Text"], Cell[BoxData[ \(TraditionalForm\`x\_0\)], "Text"], StyleBox[",", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\ y\_0\)\)], "Text"], StyleBox[")", "Text"], " is a solution to ", Cell[BoxData[ \(TraditionalForm\`a\ x + b\ y\ = \ \(d . \)\)]], " Find the general form of all solutions ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], " to\n\n\t\t\t\t", StyleBox["a", FontSlant->"Italic"], " ", StyleBox["x", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " ", StyleBox["y", FontSlant->"Italic"], " = ", StyleBox["k", FontSlant->"Italic"], " ", StyleBox["d", FontSlant->"Italic"], ",\t\t\n\ngiving ", StyleBox["x", FontSlant->"Italic"], " in terms of", StyleBox[" ", "Text"], Cell[BoxData[ \(TraditionalForm\`x\_0\)], "Text"], " and ", StyleBox["y", FontSlant->"Italic"], " in terms of", StyleBox[" ", "Text"], Cell[BoxData[ \(TraditionalForm\`y\_0\)], "Text"], "." }], "SmallText", CellFrame->True, Evaluatable->False, AspectRatioFixed->True] }, Closed]] }, Closed]] }, FrontEndVersion->"Microsoft Windows 3.0", ScreenRectangle->{{0, 800}, {0, 544}}, AutoGeneratedPackage->None, WindowToolbars->"EditBar", CellGrouping->Manual, WindowSize->{530, 383}, WindowMargins->{{1, Automatic}, {Automatic, 5}}, PrintingCopies->1, PrintingPageRange->{Automatic, Automatic}, PageHeaders->{{Cell[ TextData[ { CounterBox[ "Page"]}], "PageNumber"], Inherited, Cell[ TextData[ { ValueBox[ "FileName"]}], "Header"]}, {Cell[ TextData[ { ValueBox[ "FileName"]}], "Header"], Inherited, Cell[ TextData[ { CounterBox[ "Page"]}], "PageNumber"]}}, PrintingOptions->{"PrintingMargins"->{{0, 79.1875}, {0, 0}}, "PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "PrintCellBrackets"->True, "PrintRegistrationMarks"->False, "PrintMultipleHorizontalPages"->False, "FacingPages"->True, "Magnification"->1}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, 128}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, CharacterEncoding->"XAutomaticEncoding", Magnification->1, StyleDefinitions -> Notebook[{ Cell[CellGroupData[{ Cell["Style Definitions", "Subtitle"], Cell["\<\ Modify the definitions below to change the default appearance of all cells in \ a given style. Make modifications to any definition using commands in the Format menu. \ \>", "Text"], Cell[CellGroupData[{ Cell["Style Environment Names", "Section"], Cell[StyleData[All, "Working"], PageWidth->WindowWidth, ScriptMinSize->9], Cell[StyleData[All, "Presentation"], PageWidth->WindowWidth, ScriptMinSize->12, FontSize->16], Cell[StyleData[All, "Condensed"], PageWidth->WindowWidth, CellBracketOptions->{"Margins"->{1, 1}, "Widths"->{0, 5}}, ScriptMinSize->8, FontSize->11], Cell[StyleData[All, "Printout"], PageWidth->PaperWidth, ScriptMinSize->5, FontSize->10, PrivateFontOptions->{"FontType"->"Outline"}] }, Closed]], Cell[CellGroupData[{ Cell["Notebook Options", "Section"], Cell["\<\ The options defined for the style below will be used at the Notebook level. \ \>", "Text"], Cell[StyleData["Notebook"], PageHeaders->{{Cell[ TextData[ { CounterBox[ "Page"]}], "PageNumber"], None, Cell[ TextData[ { ValueBox[ "FileName"]}], "Header"]}, {Cell[ TextData[ { ValueBox[ "FileName"]}], "Header"], None, Cell[ TextData[ { CounterBox[ "Page"]}], "PageNumber"]}}, StyleMenuListing->None] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Headings", "Section"], Cell[CellGroupData[{ Cell[StyleData["Title"], CellMargins->{{12, Inherited}, {20, 40}}, CellGroupingRules->{"TitleGrouping", 0}, PageBreakBelow->False, TextAlignment->Center, CounterIncrements->"Title", CounterAssignments->{{"Section", 0}, {"Equation", 0}, {"Figure", 0}, { "Subtitle", 0}, {"Subsubtitle", 0}}, FontFamily->"Helvetica", FontSize->24, FontWeight->"Bold"], Cell[StyleData["Title", "Presentation"], CellMargins->{{24, 10}, {20, 40}}, TextAlignment->Center, LineSpacing->{1, 0}, FontSize->24], Cell[StyleData["Title", "Condensed"], CellMargins->{{8, 10}, {4, 8}}, TextAlignment->Center, FontSize->24], Cell[StyleData["Title", "Printout"], CellMargins->{{2, 10}, {15, 30}}, TextAlignment->Center, FontSize->24] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subtitle"], CellDingbat->None, CellMargins->{{12, Inherited}, {10, 15}}, CellGroupingRules->"InputGrouping", PageBreakBelow->False, TextAlignment->Center, CounterIncrements->"Subtitle", FontFamily->"Helvetica", FontSize->16, FontWeight->"Plain", FontSlant->"Italic", FontTracking->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], Cell[StyleData["Subtitle", "Presentation"], CellDingbat->None, CellMargins->{{24, 10}, {15, 20}}, TextAlignment->Center, LineSpacing->{1, 0}, FontSize->16, FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subtitle", "Condensed"], CellDingbat->None, CellMargins->{{8, 10}, {4, 4}}, TextAlignment->Center, FontSize->16, FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subtitle", "Printout"], CellDingbat->None, CellMargins->{{2, 10}, {10, 15}}, TextAlignment->Center, FontSize->16, FontSlant->"Italic", FontColor->RGBColor[0, 0, 1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubtitle"], CellDingbat->"\[LightBulb]", CellMargins->{{12, Inherited}, {10, 20}}, CellGroupingRules->"OutputGrouping", PageBreakBelow->False, TextJustification->1, FontFamily->"Times", FontSize->14, FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subsubtitle", "Presentation"], CellDingbat->"\[LightBulb]", CellMargins->{{24, 10}, {10, 20}}, LineSpacing->{1, 0}, FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subsubtitle", "Condensed"], CellDingbat->"\[LightBulb]", CellMargins->{{8, 10}, {8, 12}}, FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Subsubtitle", "Printout"], CellDingbat->"\[LightBulb]", CellMargins->{{2, 10}, {8, 10}}, FontSlant->"Plain", FontColor->RGBColor[0, 0, 1]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Section"], CellDingbat->"\[FilledSquare]", CellMargins->{{25, Inherited}, {8, 24}}, CellGroupingRules->{"SectionGrouping", 30}, PageBreakBelow->False, CellFrameLabelMargins->6, CounterIncrements->"Section", CounterAssignments->{{"Subsection", 0}, {"Subsubsection", 0}}, FontFamily->"Helvetica", FontSize->18, FontWeight->"Bold"], Cell[StyleData["Section", "Presentation"], CellMargins->{{40, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->18], Cell[StyleData["Section", "Condensed"], CellMargins->{{18, Inherited}, {6, 12}}, FontSize->18], Cell[StyleData["Section", "Printout"], CellMargins->{{13, 0}, {7, 22}}, FontSize->18] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsection"], CellDingbat->"\[FilledSmallSquare]", CellMargins->{{22, Inherited}, {8, 20}}, CellGroupingRules->{"SectionGrouping", 40}, PageBreakBelow->False, CellFrameLabelMargins->6, CounterIncrements->"Subsection", CounterAssignments->{{"Subsubsection", 0}}, FontFamily->"Times", FontSize->16, FontWeight->"Bold"], Cell[StyleData["Subsection", "Presentation"], CellMargins->{{36, 10}, {11, 32}}, LineSpacing->{1, 0}, FontSize->16], Cell[StyleData["Subsection", "Condensed"], CellMargins->{{16, Inherited}, {6, 12}}, FontSize->16], Cell[StyleData["Subsection", "Printout"], CellMargins->{{9, 0}, {7, 22}}, FontSize->16] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Subsubsection"], CellDingbat->"\[GraySquare]", CellMargins->{{22, Inherited}, {8, 18}}, CellGroupingRules->{"SectionGrouping", 50}, PageBreakBelow->False, CellFrameLabelMargins->6, CounterIncrements->"Subsubsection", FontFamily->"Times", FontSize->14, FontWeight->"Bold", FontColor->RGBColor[1, 0, 0]], Cell[StyleData["Subsubsection", "Presentation"], CellDingbat->"\[GraySquare]", CellMargins->{{34, 10}, {11, 26}}, LineSpacing->{1, 0}, FontSize->14, FontColor->RGBColor[1, 0, 0]], Cell[StyleData["Subsubsection", "Condensed"], CellDingbat->"\[GraySquare]", CellMargins->{{17, Inherited}, {6, 12}}, FontSize->14, FontColor->RGBColor[1, 0, 0]], Cell[StyleData["Subsubsection", "Printout"], CellDingbat->"\[GraySquare]", CellMargins->{{9, 0}, {7, 14}}, FontSize->14, FontColor->RGBColor[1, 0, 0]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Body Text", "Section"], Cell[CellGroupData[{ Cell[StyleData["Text"], CellMargins->{{12, 10}, {7, 7}}, TextJustification->1, LineSpacing->{1, 3}, CounterIncrements->"Text", FontFamily->"Times", FontSize->12], Cell[StyleData["Text", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, TextJustification->1, LineSpacing->{1, 5}], Cell[StyleData["Text", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, TextJustification->1, LineSpacing->{1, 1}], Cell[StyleData["Text", "Printout"], CellMargins->{{2, 2}, {6, 6}}, TextJustification->1] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SmallText"], CellFrame->True, CellMargins->{{12, 10}, {7, 7}}, CellGroupingRules->"OutputGrouping", TextJustification->1, LineSpacing->{1, 3}, CounterIncrements->"SmallText", FontFamily->"Times", FontSize->12, FontSlant->"Plain"], Cell[StyleData["SmallText", "Presentation"], CellFrame->True, CellMargins->{{14, 24}, {8, 8}}, TextJustification->1, LineSpacing->{1, 5}, FontSize->12, FontSlant->"Plain"], Cell[StyleData["SmallText", "Condensed"], CellFrame->True, CellMargins->{{14, 24}, {5, 5}}, TextJustification->1, LineSpacing->{1, 2}, FontSize->12, Background->GrayLevel[1]], Cell[StyleData["SmallText", "Printout"], CellFrame->True, CellMargins->{{14, 24}, {5, 5}}, TextJustification->1, FontSize->12, Background->GrayLevel[1]] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Input/Output", "Section"], Cell["\<\ The cells in this section define styles used for input and output to the kernel. Be careful when modifying, renaming, or removing these styles, because the front end associates special meanings with these style names. \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Input"], CellMargins->{{45, 10}, {5, 7}}, Evaluatable->True, CellGroupingRules->"InputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultInputFormatType, AutoItalicWords->{}, FormatType->InputForm, ShowStringCharacters->True, NumberMarks->True, CounterIncrements->"Input", FontSize->12, FontWeight->"Bold"], Cell[StyleData["Input", "Presentation"], CellMargins->{{72, Inherited}, {8, 10}}, LineSpacing->{1, 0}], Cell[StyleData["Input", "Condensed"], CellMargins->{{40, 10}, {2, 3}}], Cell[StyleData["Input", "Printout"], CellMargins->{{39, 0}, {4, 6}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Output"], CellMargins->{{47, 10}, {7, 5}}, CellEditDuplicate->True, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Output", FontSize->12], Cell[StyleData["Output", "Presentation"], CellMargins->{{72, Inherited}, {10, 8}}, LineSpacing->{1, 0}], Cell[StyleData["Output", "Condensed"], CellMargins->{{41, Inherited}, {3, 2}}], Cell[StyleData["Output", "Printout"], CellMargins->{{39, 0}, {6, 4}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Message"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Message", StyleMenuListing->None, FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["Message", "Presentation"], CellMargins->{{72, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Message", "Condensed"], CellMargins->{{41, Inherited}, {Inherited, Inherited}}], Cell[StyleData["Message", "Printout"], CellMargins->{{39, Inherited}, {Inherited, Inherited}}, FontColor->GrayLevel[0]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Print"], CellMargins->{{45, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"OutputGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GroupPageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, CellLabelMargins->{{23, Inherited}, {Inherited, Inherited}}, DefaultFormatType->DefaultOutputFormatType, AutoItalicWords->{}, FormatType->InputForm, CounterIncrements->"Print", StyleMenuListing->None, FontSize->12], Cell[StyleData["Print", "Presentation"], CellMargins->{{72, Inherited}, {Inherited, Inherited}}, LineSpacing->{1, 0}], Cell[StyleData["Print", "Condensed"], CellMargins->{{41, Inherited}, {Inherited, Inherited}}], Cell[StyleData["Print", "Printout"], CellMargins->{{39, Inherited}, {Inherited, Inherited}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Graphics"], CellMargins->{{4, Inherited}, {Inherited, Inherited}}, CellGroupingRules->"GraphicsGrouping", CellHorizontalScrolling->True, PageBreakWithin->False, GeneratedCell->True, CellAutoOverwrite->True, ShowCellLabel->False, DefaultFormatType->DefaultOutputFormatType, FormatType->InputForm, CounterIncrements->"Graphics", ImageMargins->{{43, Inherited}, {Inherited, 0}}, StyleMenuListing->None, FontSize->12], Cell[StyleData["Graphics", "Presentation"], ImageMargins->{{62, Inherited}, {Inherited, 0}}], Cell[StyleData["Graphics", "Condensed"], ImageSize->{175, 175}, ImageMargins->{{38, Inherited}, {Inherited, 0}}], Cell[StyleData["Graphics", "Printout"], ImageSize->{250, 250}, ImageMargins->{{30, Inherited}, {Inherited, 0}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["CellLabel"], StyleMenuListing->None, FontFamily->"Helvetica", FontSize->12, FontColor->RGBColor[0, 0, 1]], Cell[StyleData["CellLabel", "Presentation"], FontSize->12], Cell[StyleData["CellLabel", "Condensed"], FontSize->12], Cell[StyleData["CellLabel", "Printout"], FontFamily->"Courier", FontSize->12, FontSlant->"Italic", FontColor->GrayLevel[0]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Automatic Numbering", "Section"], Cell["\<\ The following styles are useful for numbered equations, figures, etc. They automatically give the cell a FrameLabel containing a reference to a particular counter, and also increment that counter. \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["NumberedEquation"], CellFrameLabels->{{None, Cell[ TextData[ {"(", CounterBox[ "NumberedEquation"], ")"}]]}, {None, None}}, DefaultFormatType->DefaultInputFormatType, TextAlignment->Center, CounterIncrements->"NumberedEquation", FormatTypeAutoConvert->False, FontSize->12], Cell[StyleData["NumberedEquation", "Presentation"]], Cell[StyleData["NumberedEquation", "Condensed"]], Cell[StyleData["NumberedEquation", "Printout"]] }, Open ]], Cell[CellGroupData[{ Cell[StyleData["NumberedFigure"], CellMargins->{{4, Inherited}, {Inherited, Inherited}}, CellFrameLabels->{{None, None}, {Cell[ TextData[ {"Figure ", CounterBox[ "NumberedFigure"]}]], None}}, CounterIncrements->"NumberedFigure", ImageMargins->{{43, Inherited}, {Inherited, 0}}, FormatTypeAutoConvert->False], Cell[StyleData["NumberedFigure", "Presentation"]], Cell[StyleData["NumberedFigure", "Condensed"]], Cell[StyleData["NumberedFigure", "Printout"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["NumberedTable"], CellMargins->{{44, Inherited}, {Inherited, Inherited}}, CellFrameLabels->{{None, None}, {Cell[ TextData[ {"Table ", CounterBox[ "NumberedTable"]}]], None}}, TextAlignment->Center, CounterIncrements->"NumberedTable", FormatTypeAutoConvert->False], Cell[StyleData["NumberedTable", "Presentation"]], Cell[StyleData["NumberedTable", "Condensed"]], Cell[StyleData["NumberedTable", "Printout"]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Formulas and Programming", "Section"], Cell[CellGroupData[{ Cell[StyleData["InlineFormula"], ShowCellBracket->False, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, ScriptLevel->1, SingleLetterItalics->True, StyleMenuListing->None, FontSize->12, GridBoxOptions->{ColumnWidths->Automatic}], Cell[StyleData["InlineFormula", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}], Cell[StyleData["InlineFormula", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}], Cell[StyleData["InlineFormula", "Printout"], CellMargins->{{0, 0}, {6, 6}}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Research Question"], CellDingbat->"\[GraySquare]", CellMargins->{{22, Inherited}, {8, 20}}, CellGroupingRules->"InputGrouping", PageBreakBelow->False, CellFrameLabelMargins->6, LineSpacing->{1, 3}, CounterIncrements->"Text", FontFamily->"Times", FontSize->14, FontWeight->"Bold"], Cell[StyleData["Research Question", "Presentation"], CellDingbat->"\[GraySquare]", CellMargins->{{36, 10}, {11, 32}}, CellGroupingRules->"InputGrouping", LineSpacing->{1, 0}, FontSize->14], Cell[StyleData["Research Question", "Condensed"], CellDingbat->"\[GraySquare]", CellMargins->{{16, Inherited}, {6, 12}}, CellGroupingRules->"InputGrouping", FontSize->14], Cell[StyleData["Research Question", "Printout"], CellDingbat->"\[GraySquare]", CellMargins->{{9, 0}, {7, 22}}, CellGroupingRules->"InputGrouping", FontSize->14] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Listing"], CellMargins->{{63, 10}, {7, 7}}, TextJustification->1, LineSpacing->{1, 3}, CounterIncrements->"Text", FontFamily->"Times", FontSize->12], Cell[StyleData["Listing", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, TextJustification->1, LineSpacing->{1, 5}], Cell[StyleData["Listing", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, TextJustification->1, LineSpacing->{1, 1}], Cell[StyleData["Listing", "Printout"], CellMargins->{{2, 2}, {6, 6}}, TextJustification->1] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["DisplayFormula"], ShowCellBracket->False, CellMargins->{{10, Inherited}, {Inherited, Inherited}}, CellHorizontalScrolling->True, TextAlignment->Center, ScriptLevel->0, SingleLetterItalics->True, StyleMenuListing->None, FontSize->18, UnderoverscriptBoxOptions->{LimitsPositioning->True}, GridBoxOptions->{ColumnWidths->Automatic}], Cell[StyleData["DisplayFormula", "Presentation"], TextAlignment->Center, LineSpacing->{1, 5}, FontSize->18], Cell[StyleData["DisplayFormula", "Condensed"], TextAlignment->Center, LineSpacing->{1, 1}, FontSize->18], Cell[StyleData["DisplayFormula", "Printout"], TextAlignment->Center, FontSize->18] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["Program"], ShowCellBracket->False, CellMargins->{{10, 4}, {0, 8}}, CellHorizontalScrolling->True, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["Program", "Presentation"], CellMargins->{{24, 10}, {10, 10}}, LineSpacing->{1, 5}, FontSize->15], Cell[StyleData["Program", "Condensed"], CellMargins->{{8, 10}, {6, 6}}, LineSpacing->{1, 1}, FontSize->10.5], Cell[StyleData["Program", "Printout"], CellMargins->{{0, 0}, {6, 6}}, FontSize->9.5] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Styles for Headers and Footers", "Section"], Cell[StyleData["Header"], CellMargins->{{0, 0}, {4, 1}}, StyleMenuListing->None, FontSize->10, FontSlant->"Italic"], Cell[StyleData["Footer"], CellMargins->{{0, 0}, {0, 4}}, StyleMenuListing->None, FontSize->9, FontSlant->"Italic"], Cell[StyleData["PageNumber"], CellMargins->{{0, 0}, {4, 1}}, StyleMenuListing->None, FontFamily->"Times", FontSize->10, FontWeight->"Bold"] }, Closed]], Cell[CellGroupData[{ Cell["Palette Styles", "Section"], Cell["\<\ The cells below define styles that define standard ButtonFunctions, for use in palette buttons. \ \>", "Text"], Cell[StyleData["Paste"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, After]}]&)}], Cell[StyleData["Evaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["EvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionMove[ FrontEnd`InputNotebook[ ], All, Cell, 1], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluate"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluate[ FrontEnd`InputNotebook[ ], All]}]&)}], Cell[StyleData["CopyEvaluateCell"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionCreateCell[ FrontEnd`InputNotebook[ ], All], FrontEnd`NotebookApply[ FrontEnd`InputNotebook[ ], #, All], FrontEnd`SelectionEvaluateCreateCell[ FrontEnd`InputNotebook[ ], All]}]&)}] }, Closed]], Cell[CellGroupData[{ Cell["Hyperlink Styles", "Section"], Cell["\<\ The cells below define styles useful for making hypertext ButtonBoxes. The \"Hyperlink\" style is for links within the same Notebook, or between Notebooks. \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Hyperlink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`NotebookLocate[ #2]}]&), Active->True, ButtonNote->ButtonData}], Cell[StyleData["Hyperlink", "Presentation"]], Cell[StyleData["Hyperlink", "Condensed"]], Cell[StyleData["Hyperlink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell["\<\ The following styles are for linking automatically to the on-line help system. \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["MainBookLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontWeight->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "MainBook", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["MainBookLink", "Presentation"]], Cell[StyleData["MainBookLink", "Condensed"]], Cell[StyleData["MainBookLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["AddOnsLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Times", FontWeight->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "AddOns", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["AddOnsLink", "Presentation"]], Cell[StyleData["AddOnsLink", "Condensed"]], Cell[StyleData["AddOnLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["RefGuideLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Courier", FontWeight->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "RefGuideLink", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["RefGuideLink", "Presentation"]], Cell[StyleData["RefGuideLink", "Condensed"]], Cell[StyleData["RefGuideLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["GettingStartedLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Times", FontWeight->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "GettingStarted", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["GettingStartedLink", "Presentation"]], Cell[StyleData["GettingStartedLink", "Condensed"]], Cell[StyleData["GettingStartedLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["OtherInformationLink"], StyleMenuListing->None, ButtonStyleMenuListing->Automatic, FontFamily->"Times", FontWeight->"Plain", FontColor->RGBColor[0, 0, 1], FontVariations->{"Underline"->True}, ButtonBoxOptions->{ButtonFunction:>(FrontEndExecute[ { FrontEnd`HelpBrowserLookup[ "OtherInformation", #]}]&), Active->True, ButtonFrame->"None"}], Cell[StyleData["OtherInformationLink", "Presentation"]], Cell[StyleData["OtherInformationLink", "Condensed"]], Cell[StyleData["OtherInformationLink", "Printout"], FontColor->GrayLevel[0], FontVariations->{"Underline"->False}] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Placeholder Styles", "Section"], Cell["\<\ The cells below define styles useful for making placeholder objects in palette templates. \ \>", "Text"], Cell[CellGroupData[{ Cell[StyleData["Placeholder"], Editable->False, Selectable->False, StyleBoxAutoDelete->True, Placeholder->True, StyleMenuListing->None], Cell[StyleData["Placeholder", "Presentation"]], Cell[StyleData["Placeholder", "Condensed"]], Cell[StyleData["Placeholder", "Printout"]] }, Closed]], Cell[CellGroupData[{ Cell[StyleData["SelectionPlaceholder"], Editable->False, Selectable->False, StyleBoxAutoDelete->True, StyleMenuListing->None, DrawHighlighted->True], Cell[StyleData["SelectionPlaceholder", "Presentation"]], Cell[StyleData["SelectionPlaceholder", "Condensed"]], Cell[StyleData["SelectionPlaceholder", "Printout"]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["FormatType Styles", "Section"], Cell["\<\ The cells below define styles that are mixed in with the styles of most cells. If a cell's FormatType matches the name of one of the styles defined below, then that style is applied between the cell's style and its own options. \ \>", "Text"], Cell[StyleData["CellExpression"], PageWidth->Infinity, CellMargins->{{6, Inherited}, {Inherited, Inherited}}, ShowCellLabel->False, ShowSpecialCharacters->False, AllowInlineCells->False, StyleMenuListing->None, FontFamily->"Courier", FontSize->12, FontColor->GrayLevel[0]], Cell[StyleData["InputForm"], AllowInlineCells->False, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["OutputForm"], PageWidth->Infinity, TextAlignment->Left, LineSpacing->{1, -5}, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["StandardForm"], LineSpacing->{1.25, 0}, StyleMenuListing->None, FontFamily->"Courier"], Cell[StyleData["TraditionalForm"], LineSpacing->{1.25, 0}, SingleLetterItalics->True, TraditionalFunctionNotation->True, DelimiterMatching->None, StyleMenuListing->None], Cell["\<\ The style defined below is mixed in to any cell that is in an inline cell within another. \ \>", "Text"], Cell[StyleData["InlineCell"], TextAlignment->Left, ScriptLevel->1, StyleMenuListing->None], Cell[StyleData["InlineCellEditing"], StyleMenuListing->None, Background->RGBColor[1, 0.749996, 0.8]] }, Closed]] }, Open ]] }] ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. ***********************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1731, 51, 45, 0, 46, "Subtitle"], Cell[1779, 53, 151, 8, 109, "Text"], Cell[1933, 63, 137, 3, 30, "Text"] }, Closed]], Cell[2085, 69, 129, 5, 86, "Title", Evaluatable->False], Cell[CellGroupData[{ Cell[2239, 78, 108, 3, 49, "Subsection", Evaluatable->False], Cell[2350, 83, 132, 3, 50, "Input", InitializationCell->True] }, Closed]], Cell[CellGroupData[{ Cell[2519, 91, 99, 2, 35, "Section", Evaluatable->False], Cell[2621, 95, 169, 5, 70, "Text", Evaluatable->False], Cell[2793, 102, 465, 19, 70, "Text", Evaluatable->False], Cell[3261, 123, 321, 14, 70, "Text", Evaluatable->False], Cell[3585, 139, 497, 21, 70, "Text", Evaluatable->False], Cell[4085, 162, 181, 5, 70, "Input", InitializationCell->True], Cell[4269, 169, 224, 10, 70, "Text", Evaluatable->False], Cell[4496, 181, 81, 2, 70, "Input"], Cell[4580, 185, 769, 34, 70, "Text", Evaluatable->False], Cell[5352, 221, 138, 3, 70, "Input"], Cell[5493, 226, 606, 24, 70, "Text", Evaluatable->False], Cell[6102, 252, 113, 5, 70, "Input"], Cell[CellGroupData[{ Cell[6240, 261, 39, 0, 70, "Subsubtitle"], Cell[6282, 263, 630, 16, 70, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[6949, 284, 92, 2, 70, "Subsubsection", Evaluatable->False], Cell[7044, 288, 851, 36, 70, "SmallText", Evaluatable->False] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[7944, 330, 92, 2, 35, "Section", Evaluatable->False], Cell[8039, 334, 278, 11, 70, "Text", Evaluatable->False], Cell[8320, 347, 339, 18, 70, "Text", Evaluatable->False], Cell[8662, 367, 638, 23, 70, "Text", Evaluatable->False], Cell[9303, 392, 124, 3, 70, "Text", Evaluatable->False], Cell[9430, 397, 587, 17, 70, "Text", Evaluatable->False], Cell[10020, 416, 63, 1, 70, "Input"], Cell[10086, 419, 155, 5, 70, "Text", Evaluatable->False], Cell[10244, 426, 61, 1, 70, "Input"], Cell[10308, 429, 376, 9, 70, "Text", Evaluatable->False], Cell[10687, 440, 253, 7, 70, "Text", Evaluatable->False], Cell[10943, 449, 379, 10, 70, "Text", Evaluatable->False], Cell[CellGroupData[{ Cell[11347, 463, 83, 2, 70, "Subsubsection", Evaluatable->False], Cell[11433, 467, 72, 0, 70, "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[11554, 473, 97, 2, 35, "Section", Evaluatable->False], Cell[11654, 477, 317, 9, 70, "Text"], Cell[11974, 488, 122, 5, 70, "NumberedEquation"], Cell[12099, 495, 1140, 44, 70, "Text"], Cell[13242, 541, 115, 4, 70, "NumberedEquation"], Cell[13360, 547, 998, 35, 70, "Text"], Cell[14361, 584, 101, 2, 70, "Input"], Cell[CellGroupData[{ Cell[14487, 590, 39, 0, 70, "Subsubtitle"], Cell[14529, 592, 261, 7, 70, "Text"] }, Closed]], Cell[14805, 602, 373, 10, 70, "Text"], Cell[15181, 614, 1264, 51, 70, "Text"], Cell[16448, 667, 831, 35, 70, "Text"], Cell[17282, 704, 333, 7, 70, "Input", InitializationCell->True], Cell[CellGroupData[{ Cell[17640, 715, 39, 0, 70, "Subsubtitle"], Cell[17682, 717, 196, 5, 70, "Text"], Cell[17881, 724, 406, 8, 70, "Text"], Cell[18290, 734, 481, 12, 70, "Text"] }, Closed]], Cell[18786, 749, 253, 10, 70, "Text"], Cell[19042, 761, 48, 1, 70, "Input"], Cell[19093, 764, 335, 9, 70, "Text"], Cell[19431, 775, 49, 1, 70, "Input"], Cell[19483, 778, 249, 8, 70, "Text"], Cell[19735, 788, 48, 1, 70, "Input"], Cell[CellGroupData[{ Cell[19808, 793, 92, 2, 70, "Subsubsection", Evaluatable->False], Cell[19903, 797, 566, 20, 70, "SmallText", Evaluatable->False] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[20518, 823, 92, 2, 35, "Section", Evaluatable->False], Cell[CellGroupData[{ Cell[20635, 829, 42, 0, 49, "Subsection"], Cell[20680, 831, 168, 5, 33, "Text"], Cell[20851, 838, 113, 2, 29, "Text"], Cell[20967, 842, 550, 16, 90, "Text"], Cell[21520, 860, 162, 6, 33, "Text"], Cell[21685, 868, 197, 4, 52, "Text"], Cell[21885, 874, 49, 1, 30, "Input"], Cell[21937, 877, 73, 0, 33, "Text"], Cell[22013, 879, 94, 2, 33, "Text"], Cell[22110, 883, 35, 0, 33, "Text"], Cell[22148, 885, 48, 1, 30, "Input"], Cell[22199, 888, 55, 0, 33, "Text"], Cell[22257, 890, 46, 1, 30, "Input"], Cell[22306, 893, 230, 5, 52, "Text"], Cell[22539, 900, 138, 5, 33, "Text"], Cell[22680, 907, 106, 3, 33, "Text"], Cell[22789, 912, 54, 0, 33, "Listing"], Cell[22846, 914, 52, 0, 33, "Listing"], Cell[22901, 916, 52, 0, 33, "Listing"], Cell[22956, 918, 62, 0, 33, "Text"], Cell[23021, 920, 123, 4, 25, "NumberedEquation"], Cell[23147, 926, 188, 5, 33, "Text"], Cell[23338, 933, 438, 10, 86, "Text"], Cell[23779, 945, 195, 8, 33, "Text"], Cell[23977, 955, 132, 5, 33, "Text"], Cell[24112, 962, 115, 3, 33, "Text"], Cell[24230, 967, 71, 1, 30, "Input"], Cell[CellGroupData[{ Cell[24326, 972, 39, 0, 48, "Subsubtitle"], Cell[24368, 974, 367, 10, 70, "Text"] }, Closed]], Cell[24750, 987, 332, 11, 50, "Text"], Cell[25085, 1000, 241, 13, 33, "Text"], Cell[25329, 1015, 202, 8, 33, "Text"], Cell[25534, 1025, 50, 1, 30, "Input"], Cell[25587, 1028, 54, 0, 33, "Listing"], Cell[25644, 1030, 48, 1, 30, "Input"], Cell[25695, 1033, 52, 0, 33, "Listing"], Cell[25750, 1035, 47, 1, 30, "Input"], Cell[25800, 1038, 51, 0, 33, "Listing"], Cell[25854, 1040, 47, 1, 30, "Input"], Cell[25904, 1043, 51, 0, 33, "Listing"], Cell[25958, 1045, 47, 1, 30, "Input"], Cell[26008, 1048, 50, 0, 33, "Listing"], Cell[26061, 1050, 46, 1, 30, "Input"], Cell[26110, 1053, 50, 0, 33, "Listing"], Cell[26163, 1055, 155, 5, 33, "Text"], Cell[26321, 1062, 194, 10, 33, "Text"], Cell[26518, 1074, 52, 0, 33, "Text"], Cell[26573, 1076, 54, 0, 33, "Listing"], Cell[26630, 1078, 52, 0, 33, "Listing"], Cell[26685, 1080, 51, 0, 33, "Listing"], Cell[26739, 1082, 51, 0, 33, "Listing"], Cell[26793, 1084, 50, 0, 33, "Listing"], Cell[26846, 1086, 50, 0, 33, "Listing"], Cell[26899, 1088, 54, 0, 33, "Text"], Cell[26956, 1090, 1087, 20, 181, "Text"], Cell[28046, 1112, 337, 14, 33, "Text"], Cell[28386, 1128, 64, 1, 30, "Input"], Cell[28453, 1131, 43, 0, 33, "Text"], Cell[28499, 1133, 123, 2, 29, "Text"], Cell[28625, 1137, 1082, 38, 129, "Text"], Cell[29710, 1177, 57, 1, 30, "Input"], Cell[29770, 1180, 325, 14, 52, "Text"] }, Closed]], Cell[CellGroupData[{ Cell[30132, 1199, 43, 0, 49, "Subsection"], Cell[30178, 1201, 225, 6, 52, "Text", Evaluatable->False], Cell[30406, 1209, 163, 6, 25, "NumberedEquation", Evaluatable->False], Cell[30572, 1217, 734, 22, 90, "Text", Evaluatable->False], Cell[31309, 1241, 176, 7, 33, "Text", Evaluatable->False], Cell[31488, 1250, 552, 17, 71, "Text", Evaluatable->False], Cell[32043, 1269, 224, 10, 33, "Text", Evaluatable->False], Cell[32270, 1281, 679, 21, 90, "Text", Evaluatable->False], Cell[32952, 1304, 150, 3, 70, "Input"], Cell[CellGroupData[{ Cell[33127, 1311, 39, 0, 70, "Subsubtitle"], Cell[33169, 1313, 215, 9, 70, "Text"] }, Closed]], Cell[33399, 1325, 302, 10, 70, "Text", Evaluatable->False], Cell[33704, 1337, 176, 7, 33, "Text", Evaluatable->False], Cell[33883, 1346, 549, 20, 52, "Text", Evaluatable->False], Cell[34435, 1368, 152, 3, 30, "Input"], Cell[34590, 1373, 267, 6, 52, "Text", Evaluatable->False], Cell[34860, 1381, 209, 5, 50, "Input", InitializationCell->True], Cell[35072, 1388, 460, 15, 53, "Text", Evaluatable->False], Cell[35535, 1405, 76, 2, 30, "Input"], Cell[35614, 1409, 452, 14, 52, "Text", Evaluatable->False], Cell[36069, 1425, 77, 2, 30, "Input"], Cell[CellGroupData[{ Cell[36171, 1431, 92, 2, 45, "Subsubsection", Evaluatable->False], Cell[36266, 1435, 1205, 45, 70, "SmallText", Evaluatable->False] }, Closed]], Cell[CellGroupData[{ Cell[37508, 1485, 92, 2, 31, "Subsubsection", Evaluatable->False], Cell[37603, 1489, 1682, 65, 70, "SmallText", Evaluatable->False] }, Closed]] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell[39346, 1561, 97, 2, 35, "Section", Evaluatable->False], Cell[39446, 1565, 171, 4, 52, "Text"], Cell[39620, 1571, 128, 5, 33, "Text"], Cell[39751, 1578, 269, 12, 33, "Text"], Cell[40023, 1592, 131, 5, 33, "Text"], Cell[40157, 1599, 331, 9, 52, "Text"], Cell[40491, 1610, 128, 5, 33, "Text"], Cell[40622, 1617, 829, 28, 72, "Text"], Cell[41454, 1647, 50, 1, 30, "Input"], Cell[CellGroupData[{ Cell[41529, 1652, 44, 0, 45, "Subsubsection"], Cell[41576, 1654, 1321, 60, 144, "SmallText", Evaluatable->False] }, Closed]] }, Closed]] } ] *) (*********************************************************************** End of Mathematica Notebook file. ***********************************************************************)