(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 4.0, MathReader 4.0, or any compatible application. The data for the notebook starts with the line containing 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). 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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[ 153489, 3156]*) (*NotebookOutlinePosition[ 154211, 3181]*) (* CellTagsIndexPosition[ 154167, 3177]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Functions from Data", "Title"], Cell[CellGroupData[{ Cell["Introduction", "Section"], Cell[TextData[{ "Suppose you know that an independent variable and a dependent variable are \ related by some underlying function and that you have a collection of ordered \ pairs of data representing that relationship. How do you find a formula for \ the underlying function? There are two options. If the dependent variable \ is completely determined by the independent variable and there are no errors \ in the data, then you can construct a function that goes directly through all \ of the data points. This is called interpolation, and ", StyleBox["Mathematica", FontSlant->"Italic"], " implements this with the ", StyleBox["Interpolation", FontFamily->"Courier"], " and ", StyleBox["InterpolatingFunction", FontFamily->"Courier"], " commands. If the dependent variable is not completely determined by the \ independent variable because other effects on the independent variable are \ being ignored or because of error in the data, then you can construct a \ function that goes near the data points. This is called regression, and ", StyleBox["Mathematica", FontSlant->"Italic"], " implements this with the ", StyleBox["Fit", FontFamily->"Courier"], " command." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Interpolation", "Section"], Cell[TextData[{ "When a sphere moves through a fluid (liquid or gas), a drag force opposes \ the sphere's motion, due to friction and a difference in pressure in front of \ and behind the sphere. This force, ", Cell[BoxData[ \(TraditionalForm\`F\_D\)]], ", depends on the velocity of the sphere (", Cell[BoxData[ \(TraditionalForm\`v\)]], "), the diameter of the sphere (", Cell[BoxData[ \(TraditionalForm\`D\)]], "), the density of the fluid (", Cell[BoxData[ \(TraditionalForm\`\[Rho]\)]], "), and the viscosity of the fluid (", Cell[BoxData[ \(TraditionalForm\`\[Mu]\)]], "). Dimensional analysis can be used to show that these five variables can \ be reduced to two variables without losing any ability to describe the drag \ force. These two dimensionless variables are the drag coefficient, ", Cell[BoxData[ \(TraditionalForm\`C\_D = F\_D\/\(\(1\/2\) \(\[Rho]v\^2\) A\)\)]], ", and the Reynolds number, ", Cell[BoxData[ \(TraditionalForm\`Re = \(\[Rho]\ v\ D\)\/\[Mu]\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`A\)]], " is the cross-sectional area of the sphere. The relationship between ", Cell[BoxData[ \(TraditionalForm\`C\_D\)]], " and Re can be determined experimentally. Below are a list of data points \ for (Re, ", Cell[BoxData[ \(TraditionalForm\`C\_D\)]], ") read from a published graph." }], "Text"], Cell[BoxData[ \(\(reCD = {{0.05875, 492. }, {0.1585, 169.8}, {0.4786, 58.88}, {3.02, 10.86}, {7.015, 5.623}, {15.49, 3.388}, {57.54, 1.479}, {144.5, 0.9204}, {264.9, 0.7194}, {512.9, 0.5623}, {1000. , 0.4786}, {1862. , 0.4365}, {3162. , 0.4074}, {4764. , 0.389}, {8375. , 0.3981}, {15560. , 0.4395}, {26480. , 0.4571}, {34670. , 0.4775}, {58880. , 0.4732}, {100000. , 0.4624}, {170200. , 0.4395}, {231700. , 0.4046}, {264800. , 0.3733}, {271000. , 0.3467}, {285100. , 0.2472}, {302000. , 0.1778}, {338800. , 0.1047}, {398100. , 0.09772}, {512900. , 0.1}, {1.778\ *10^6, 0.1778}, {2.291*10^6, 0.1862}, {5.012*10^6, 0.1862}};\)\)], "Input"], Cell[TextData[{ "Since these numbers vary greatly in magnitude, we will take the logarithm, \ base 10, of each of the values of Re and ", Cell[BoxData[ \(TraditionalForm\`C\_D\)]], "." }], "Text"], Cell[BoxData[ \(logReCDData = Map[Log[10, #] &, reCD, {2}]\)], "Input"], Cell["Here is a log-log plot of the data points.", "Text"], Cell[BoxData[ \(\(loglogPlot = ListPlot[logReCDData, AxesLabel \[Rule] {"\", \*"\"\\""}, AxesOrigin \[Rule] {\(-2\), \(-1\)}, AspectRatio \[Rule] Automatic, GridLines \[Rule] Automatic];\)\)], "Input"], Cell[TextData[{ "The abrupt changes at ", Cell[BoxData[ \(TraditionalForm\`log(Re) \[TildeTilde] 3\)]], " and ", Cell[BoxData[ \(TraditionalForm\`log(Re) \[TildeTilde] 5.5\)]], " are due to changes in the nature of the fluid flow, as shown in the \ diagram below. At low Reynolds numbers, less than about ", Cell[BoxData[ \(TraditionalForm\`10\^3\)]], ", the fluid wraps around the sphere. At intermediate Reynolds numbers, \ between about ", Cell[BoxData[ \(TraditionalForm\`10\^3\)]], " and ", Cell[BoxData[ \(TraditionalForm\`10\^5.5\)]], ", the fluid has a broad wake behind the sphere. At high Reynolds numbers, \ greater than about ", Cell[BoxData[ \(TraditionalForm\`10\^5.5\)]], ", a turbulent boundary layer forms around the sphere and the wake \ diminishes. The dimples on a golf ball and the seams on a spinning baseball \ are designed to induce a turbulent boundary layer, and thus to reduce the \ drag coefficient." }], "Text"], Cell[GraphicsData["Bitmap", "\<\ CF5dJ6E]HGAYHf4PAg9QL6QYHg<0oooo1@00003C 0?ooo`00gP3oool500000=P0oooo003K0?ooo`<00000g@3oool00=P0oooo 0`00003P0?ooo`003`3oool2000000050?ooo`00000000000000003oool0 0P0000040?ooo`030000003oool0000000`0oooo00<000000?ooo`3oool0 Z03oool200000><0oooo000?0?ooo`070000003oool0oooo0?ooo`000000 oooo000000020?ooo`070000003oool0oooo0?ooo`000000oooo0000000= 0?ooo`030000003oool0oooo0:H0oooo00<000000?ooo`3oool06P3oool@ 00000;T0oooo000?0?ooo`070000003oool0oooo0?ooo`000000oooo0000 00020?ooo`080000003oool0oooo0000003oool000000?ooo`0000040?oo o`X00000Y@3oool2000001<0oooo2P0000390?ooo`003`3oool01`000000 oooo0?ooo`3oool000000?ooo`0000000P3oool020000000oooo0?ooo`00 0000oooo0000003oool00000303oool00`000000oooo0?ooo`2Q0?ooo`<0 00003P3oool700000=<0oooo000?0?ooo`8000000P3oool00`000000oooo 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Edward Donley, UMAP Module 712, The Drag Force on a Sphere \ (UMAP Journal, Volume 12, no. 1, Spring 1991, pp. 47-80.", StyleBox[", ", FontFamily->"Times New Roman"], ButtonBox["http://www.math.iup.edu/projects/CalcDEMma/drag/drag.html", ButtonData:>{ URL[ "http://www.math.iup.edu/projects/CalcDEMma/drag/drag.html"], None}, ButtonStyle->"Hyperlink"], StyleBox[".", FontFamily->"Times New Roman"] }], "Text"], Cell[TextData[{ "Should we use interpolation or regression to construct a function that \ follows the trend of this data? We need to examine two issues to answer this \ question. ", Cell[BoxData[ \(TraditionalForm\`C\_D\)]], " is completely determined from Re. We have not ignored any other \ variables that might effect the value of ", Cell[BoxData[ \(TraditionalForm\`C\_D\)]], ". Secondly, the measurement error in the data points is small, so these \ data points are truly on the graph of the underlying function. Our data and \ model satisfy the two conditions needed to warrent using interpolation. If \ either of these conditions were not met, then we should have used regression. \ Interpolation constructs a function that passes through all of the data \ points. However, the data in this graph do not follow the pattern of any \ standard elementary function. So, we must use a piecewise defined function. \ One of the simplest choices would be to use a piecewise linear function; that \ is, we can just connect the dots on the graph. ", StyleBox["Mathematica", FontSlant->"Italic"], " can use higher degree piecewise polynomials, though, that captures some \ of the curvature in the data. This algorithm is built into the ", StyleBox["Interpolation", FontFamily->"Courier"], " command and the polynomial degree can be set using the ", StyleBox["InterpolationOrder", FontFamily->"Courier"], " option." }], "Text"], Cell[BoxData[ \(logCD = Interpolation[logReCDData, InterpolationOrder \[Rule] 2]\)], "Input"], Cell["Here is the graph of the interpolating function.", "Text"], Cell[BoxData[ \(\(functionPlot = Plot[logCD[x], {x, Min[\(Transpose[logReCDData]\)[\([1]\)]], Max[\(Transpose[logReCDData]\)[\([1]\)]]}, AxesLabel \[Rule] {"\", \*"\"\\""}, AxesOrigin \[Rule] {\(-2\), \(-1\)}, AspectRatio \[Rule] Automatic, GridLines \[Rule] Automatic, PlotRange \[Rule] All];\)\)], "Input"], Cell["\<\ As you can see in the following graph, the interpolating function \ goes through all of the data points.\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(Show[loglogPlot, functionPlot, PlotRange \[Rule] All];\)\)], "Input"], Cell[GraphicsData["PostScript", "\<\ %! %%Creator: Mathematica 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Water, at 20\[Degree]C, has density 0.998 \ gm/", Cell[BoxData[ \(TraditionalForm\`cm\^3\)]], " and viscosity ", Cell[BoxData[ \(TraditionalForm\`1.00\[Cross]10\^\(-2\)\)]], "gm/cm sec.", " Describe the nature of this fluid flow, in relation to the three types \ of flows described above." }], "Text", Background->GrayLevel[0.750011]] }, Closed]], Cell[CellGroupData[{ Cell["Linear Regression", "Section"], Cell["\<\ Best Buy has a chain of retail stores supported by a central \ computing center. The following data shows the number of stores in the chain \ each month and the central computing power used during that month, measured \ in millions of instructions per second (MIPS).\ \>", "Text"], Cell[BoxData[{ \(\(bestBuy = {{"\<8/1/1996\>", 268, 145.2}, {"\<9/1/1996\>", 268, 150.5}, {"\<10/1/1996\>", 269, 153.1}, {"\<11/1/1996\>", 272, 136.1}, {"\<12/1/1996\>", 272, 152.6}, {"\<1/1/1997\>", 272, 143.5}, {"\<2/1/1997\>", 272, 145.9}, {"\<3/1/1997\>", 272, 148.3}, {"\<4/1/1997\>", 273, 159.1}, {"\<5/1/1997\>", 274, 165.1}, {"\<6/1/1997\>", 275, 167.5}, {"\<7/1/1997\>", 277, 179.1}, {"\<8/1/1997\>", 279, 192.9}, {"\<9/1/1997\>", 280, 197.9}, {"\<10/1/1997\>", 282, 212}, {"\<11/1/1997\>", 284, 191.1}, {"\<12/1/1997\>", 284, 219.5}, {"\<1/1/1998\>", 284, 196.1}, {"\<2/1/1998\>", 284, 212}, {"\<3/1/1998\>", 284, 226.6}, {"\<4/1/1998\>", 286, 214.1}, {"\<5/1/1998\>", 288, 206.6}, {"\<6/1/1998\>", 291, 218.3}, {"\<7/1/1998\>", 294, 235}, {"\<8/1/1998\>", 297, 266.4}, {"\<9/1/1998\>", 301, 218.9}, {"\<10/1/1998\>", 305, 254.8}, {"\<11/1/1998\>", 311, 269.1}, {"\<12/1/1998\>", 311, 297.3}, {"\<1/1/1999\>", 311, 274.5}, {"\<2/1/1999\>", 311, 279.4}, {"\<3/1/1999\>", 311, 292.3}, {"\<4/1/1999\>", 312, 305.7}, {"\<5/1/1999\>", 313, 304.9}, {"\<6/1/1999\>", 314, 347.8}, {"\<7/1/1999\>", 317, 358}, {"\<8/1/1999\>", 332, 393.5}, {"\<9/1/1999\>", 337, 428.2}, {"\<10/1/1999\>", 344, 432.6}, {"\<11/1/1999\>", 344, 453.8}, {"\<12/1/1999\>", 354, 531.8}, {"\<1/1/2000\>", 354, 472.3}, {"\<2/1/2000\>", 357, 436.5}, {"\<3/1/2000\>", 357, 470.4}, {"\<4/1/2000\>", 358, 425}, {"\<5/1/2000\>", 361, 416.6}, {"\<6/1/2000\>", 362, 485.1}, {"\<7/1/2000\>", 365, 488.5}};\)\), "\[IndentingNewLine]", \(TableForm[bestBuy, TableHeadings \[Rule] {{}, {"\", "\", \ "\"}}]\)}], "Input"], Cell["\<\ Let's look for a relationship between the number of stores and the \ computer usage. First, we will extract the second and third columns from the \ data set.\ \>", "Text"], Cell[BoxData[ \(bestBuyData = Transpose[{\(Transpose[bestBuy]\)[\([2]\)], \(Transpose[ bestBuy]\)[\([3]\)]}]\)], "Input"], Cell[BoxData[ \(\(dataPlot = ListPlot[bestBuyData, PlotStyle \[Rule] {PointSize[0.02]}, AxesLabel \[Rule] {"\", "\"}];\)\)], "Input"], Cell[TextData[{ "Notice that an interpolating function would be an awful choice for this \ data. There are other factors effecting computer usage besides the number of \ stores. Can you think of any of these other factors? These factors cause \ the data in the plot to deviate from a nice smooth trend. If you ignore these \ deviations, though, it looks like the relationship between the number of \ stores and the computer usage is linear. ", StyleBox["Mathematica'", FontSlant->"Italic"], "'s ", StyleBox["Fit", FontFamily->"Courier"], " command allows you to find a curve that best fits a set of data. But you \ have to decide in advance the terms from which the function will be \ constructed. ", StyleBox["Fit", FontFamily->"Courier"], " will pick the coefficients for those terms to best approximate the data. \ If we let ", Cell[BoxData[ \(TraditionalForm\`x\)]], " represent the number of stores, then our linear function will be of the \ form ", Cell[BoxData[ \(TraditionalForm\`a\_1 + \(a\_2\) x\)]], ". So our terms will be 1 and ", Cell[BoxData[ \(TraditionalForm\`x\)]], ". ", StyleBox["Fit", FontFamily->"Courier"], " will find the values for ", Cell[BoxData[ \(TraditionalForm\`a\_1\)]], " and ", Cell[BoxData[ \(TraditionalForm\`a\_2\)]], "." }], "Text"], Cell[BoxData[ \(computerUsage[x_] = Fit[bestBuyData, {1, x}, x]\)], "Input"], Cell[TextData[{ "If you want additional statistical information about the regression line, \ use the ", StyleBox["Regress", FontFamily->"Courier"], " command in the", StyleBox[" Statistics`LinearRegression`", FontFamily->"Courier"], " package that comes with ", StyleBox["Mathematica", FontSlant->"Italic"], "." }], "Text"], Cell["Here is a graph of the data with the regression line.", "Text"], Cell[BoxData[{ \(\(linePlot = Plot[computerUsage[x], {x, Min[\(Transpose[bestBuyData]\)[\([1]\)]], Max[\(Transpose[bestBuyData]\)[\([1]\)]]}, PlotStyle \[Rule] {RGBColor[0, 0, 1]}, DisplayFunction \[Rule] Identity];\)\), "\[IndentingNewLine]", \(\(Show[dataPlot, linePlot, DisplayFunction \[Rule] $DisplayFunction];\)\)}], "Input"], Cell["\<\ In order to determine if the line really captures the trend of the \ data, we need to look at the differences between the actual data values and \ the predicted linear values to be sure. These differences are called \ residuals, represented by the vertical red lines in the graph below.\ \>", \ "Text"], Cell[BoxData[{ \(\(residualPlot = Graphics[ Map[{RGBColor[1, 0, 0], Line[{{#[\([\)\(1\)\(]\)], #[\([\)\(2\)\(]\)]}, {#[\([\)\(1\)\ \(]\)], computerUsage[#[\([\)\(1\)\(]\)]]}}]} &, bestBuyData]];\)\), "\n", \(\(Show[linePlot, residualPlot, dataPlot, AxesLabel \[Rule]