(*********************************************************************** 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). 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[ 77971, 2162]*) (*NotebookOutlinePosition[ 78680, 2187]*) (* CellTagsIndexPosition[ 78636, 2183]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Numerical Methods", "Title", CellFrame->True, TextAlignment->Center, Background->RGBColor[0, 1, 0]], Cell["\<\ This notebook is by Steven Amgott. Please send any questions or \ comments to samgott1@swarthmore.edu. Feel free to use and distribute this \ notebook, but keep this author information in any copy you use or distribute.\ \ \>", "SmallText"], Cell[TextData[{ "In any input cell containing ", StyleBox["xxx", FontColor->RGBColor[1, 0, 1]], " , you must replace it with your input before evaluating the cell. In \ general, anything in ", StyleBox["magenta", FontColor->RGBColor[1, 0, 1]], " is something you can, and possibly should, change." }], "Text"], Cell[CellGroupData[{ Cell["\<\ Initialization. (Can be skipped, if you answer \"Yes\" to the \ initialization request.)\ \>", "Section"], Cell["\<\ The cell in this section is an initialization cell, and will be \ automatically evaluated if you answer \"Yes\" to the initialization request. \ If you do not answer \"Yes\" to this request, you must evaluate it before \ creating the graphs in the second section below.\ \>", "Text"], Cell[BoxData[ \(Needs["\"]\)], "Input", InitializationCell->True], Cell[CellGroupData[{ Cell[TextData[{ "Eliminating some unnecessary warning messages. ", StyleBox["(Should be skipped)", FontColor->RGBColor[1, 0, 1]] }], "Subsubsection"], Cell[BoxData[{ \(\(Off[General::"\"];\)\), "\n", \(\(Off[General::"\"];\)\)}], "Input", InitializationCell->True] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["The Initial Value Problem", "Section"], Cell[TextData[{ "The next cell sets up the initial value problem for which you are going to \ use one of the numerical methods. Make sure to put a period after the value \ you put in for t0 if t0 is an integer. This tells ", StyleBox["Mathematica", FontSlant->"Italic"], " that you are going to use approximate arithmetic rather than exact \ arithmetic, which greatly shortens calculation times." }], "Text"], Cell[BoxData[{\(Clear[f, t, y, t0, y0, tf, exact]\), "\n", RowBox[{\(f[t_, y_]\), ":=", StyleBox[\(\(-t\)\ y\), FontColor->RGBColor[1, 0, 1]]}], "\n", RowBox[{ RowBox[{"t0", "=", StyleBox["0.0", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", RowBox[{ RowBox[{"y0", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", RowBox[{ RowBox[{"tf", "=", StyleBox["1", FontColor->RGBColor[1, 0, 1]]}], ";"}]}], "Input", InitializationCell->True], Cell[TextData[{ "In all the methods below, a semicolon (", StyleBox[";", FontColor->RGBColor[1, 0, 1]], ") has been placed after the Table command in order to supress the display \ of all the calculated intermediate values. If you wish to see these values, \ delete this semicolon. The one to possibly delete is in ", StyleBox["magenta", FontColor->RGBColor[1, 0, 1]], ". (This probably should be done only for small values of ", StyleBox["n", FontColor->RGBColor[1, 0, 1]], ".)" }], "Text"], Cell[CellGroupData[{ Cell["Exact solution (if possible)", "Subsection"], Cell[TextData[{ "The cells below solve the above initial value problem using ", StyleBox["Mathematica", FontSlant->"Italic"], "'s DSolve command, if it is possible to do so." }], "Text"], Cell[BoxData[{ \(Clear[y, t]\), "\n", \(DSolve[{\(y'\)[t] == f[t, y[t]], y[0] \[Equal] y0}, y[t], t]\)}], "Input"], Cell["\<\ Only evaluate the next cell if the preceding cell produced a \ solution. Make sure the preceding cell is the one you evaluated last.\ \>", \ "Text"], Cell[BoxData[ \(N[exact = \((y[t] /. %)\) /. t \[Rule] tf]\)], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Euler method", "Section"], Cell[TextData[{ "This cell implements the Euler method for the differential equation y' = \ f(t,y) and initial conditions defined in ", StyleBox["The Initial Value Problem", FontColor->RGBColor[0, 0, 1]], " above. The line with \"exact\" only makes sense if an exact solution was \ found in the ", StyleBox["Exact Solution (if possible)", FontColor->RGBColor[0, 0, 1]], " subsection of that section." }], "Text"], Cell[BoxData[{\(Clear[n, h, j, t, y]\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", RowBox[{\(Table[y[j + 1] = y[j] + h\ f[t[j], y[j]], {j, 0, n - 1}]\), StyleBox[";", FontColor->RGBColor[1, 0, 1]]}], "\n", \(y[n]\), "\n", \(y[n] - exact\)}], "Input"], Cell[CellGroupData[{ Cell["Animation of increasing n in the Euler Method", "Subsection"], Cell[BoxData[{ \(Clear[m, t, y]\), "\n", \(\(t[0] = t0;\)\), "\n", \(\(y[0] = y0;\)\), "\n", \(\(solution\ = \ NDSolve[{\(y'\)[t] \[Equal] f[t, y[t]], y[t0] \[Equal] y0}, y, {t, t0, tf}];\)\), "\n", \(Animate[ Show[ListPlot[ Prepend[Table[{t[j + 1] = N[t0 + \(\(tf - t0\)\/m\) \((j + 1)\)], y[j + 1] = y[j] + \(tf - t0\)\/m\ f[t[j], y[j]]}, {j, 0, m - 1}], {t0, y0}], PlotStyle \[Rule] {PointSize[ .015], RGBColor[1, 0, 0]}, DisplayFunction \[Rule] Identity], Plot[Evaluate[y[t]\ /. \ solution], {t, t0, tf}, DisplayFunction \[Rule] Identity], DisplayFunction \[Rule] $DisplayFunction], {m, 5, 100, 5}]\)}], "Input"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Improved Euler method", "Section"], Cell[TextData[{ "This cell implements the improved Euler (Heun) method for the differential \ equation y' = f(t,y) and initial conditions defined in ", StyleBox["The Initial Value Problem", FontColor->RGBColor[0, 0, 1]], " above. The line with \"exact\" only makes sense if an exact solution was \ found in the ", StyleBox["Exact Solution (if possible)", FontColor->RGBColor[0, 0, 1]], " subsection of that section." }], "Text"], Cell[BoxData[{\(Clear[n, h, j, t, y]\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", RowBox[{\(Table[ y[j + 1] = y[j] + \(h\/2\) \((f[t[j], y[j]] + f[t[j + 1], y[j] + h\ f[t[j], y[j]]])\), {j, 0, n - 1}]\), StyleBox[";", FontColor->RGBColor[1, 0, 1]]}], "\n", \(y[n]\), "\n", \(y[n] - exact\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Three term Taylor method", "Section"], Cell[TextData[{ "This cell implements the three term Taylor method method for the \ differential equation y' = f(t,y) and initial conditions defined in ", StyleBox["The Initial Value Problem", FontColor->RGBColor[0, 0, 1]], " above. The line with \"exact\" only makes sense if an exact solution was \ found in the ", StyleBox["Exact Solution (if possible)", FontColor->RGBColor[0, 0, 1]], " subsection of that section." }], "Text"], Cell[BoxData[{\(Clear[ft, fy, n, h, j, t, y]\), "\n", \(ft[t_, y_] = \[PartialD]\_t\ f[t, y];\), "\n", \(fy[t_, y_] = \[PartialD]\_y\ f[t, y];\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", RowBox[{\(Table[ y[j + 1] = y[j] + h\ f[t[j], y[j]] + \(h\^2\/2\) \((ft[t[j], y[j]] + fy[t[j], y[j]]\ f[t[j], y[j]])\), {j, 0, n - 1}]\), StyleBox[";", FontColor->RGBColor[1, 0, 1]]}], "\n", \(y[n]\), "\n", \(y[n] - exact\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Runge-Kutta method", "Section"], Cell[TextData[{ "This cell implements the 4th order Runge-Kutta method for the differential \ equation y' = f(t,y) and initial conditions defined in ", StyleBox["The Initial Value Problem", FontColor->RGBColor[0, 0, 1]], " above. The line with \"exact\" only makes sense if an exact solution was \ found in the ", StyleBox["Exact Solution (if possible)", FontColor->RGBColor[0, 0, 1]], " subsection of that section." }], "Text"], Cell[BoxData[{\(Clear[n, h, j, k1, k2, k3, k4, t, y]\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", \(k1[t_, y_] := f[t, y]\), "\n", \(k2[t_, y_] := f[t + .5\ h, y + .5\ h\ k1[t, y]]\), "\n", \(k3[t_, y_] := f[t + .5\ h, y + .5\ h\ k2[t, y]]\), "\n", RowBox[{\(k4[t_, y_] := f[t + h, y + h\ k3[t, y]]\), "\t\t\t\t"}], "\n", RowBox[{\(Table[ y[j + 1] = y[j] + \(h\/6\) \((\ k1[t[j], y[j]] + 2\ k2[t[j], y[j]] + 2 k3[t[j], y[j]] + k4[t[j], y[j]])\), {j, 0, n - 1}]\), StyleBox[";", FontColor->RGBColor[1, 0, 1]]}], "\n", \(y[n]\), "\n", \(y[n] - exact\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Sometimes things can go wrong...", "Section"], Cell["\<\ An interesting example to use with the Runge-Kutta method is the \ following:\ \>", "Text"], Cell[BoxData[{\(Clear[g, t, x, y, t0, tf, exactg]\), "\n", RowBox[{ StyleBox[\(g[t_, y_]\), FontColor->GrayLevel[0]], StyleBox[":=", FontColor->GrayLevel[0]], " ", \(y - 1\/\((1 + t)\)\^2 - 1\/\(1 + t\)\)}], "\n", RowBox[{ RowBox[{ StyleBox["t0", FontColor->GrayLevel[0]], StyleBox["=", FontColor->GrayLevel[0]], "0.0"}], ";"}], "\n", RowBox[{ RowBox[{ StyleBox["y0", FontColor->GrayLevel[0]], StyleBox["=", FontColor->GrayLevel[0]], "1"}], StyleBox[";", FontColor->GrayLevel[0]]}], "\n", RowBox[{ RowBox[{"tf", "=", StyleBox["20", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(DSolve[{\(x'\)[t] \[Equal] g[t, x[t]], x[0] \[Equal] y0}, x[t], t]\), "\n", \(N[ exactg = \((x[t] /. %)\) /. t \[Rule] tf]\)}], "Input"], Cell[TextData[{ "Let's try using the Runge-Kutta method in the next cell with n = ", StyleBox["10 ", FontColor->RGBColor[1, 0, 1]], "and n = ", StyleBox["1,000", FontColor->RGBColor[1, 0, 1]], " and look at the error." }], "Text"], Cell[BoxData[{\(Clear[n, h, j, k1, k2, k3, k4, t, y]\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", \(k1[t_, y_] := g[t, y]\), "\n", \(k2[t_, y_] := g[t + .5\ h, y + .5\ h\ k1[t, y]]\), "\n", \(k3[t_, y_] := g[t + .5\ h, y + .5\ h\ k2[t, y]]\), "\n", RowBox[{\(k4[t_, y_] := g[t + h, y + h\ k3[t, y]]\), "\t\t\t\t"}], "\n", \(Table[ y[j + 1] = y[j] + \(h\/6\) \((\ k1[t[j], y[j]] + 2\ k2[t[j], y[j]] + 2 k3[t[j], y[j]] + k4[t[j], y[j]])\), {j, 0, n - 1}];\), "\n", \(y[ n]\), "\n", \(y[n] - exactg\)}], "Input"], Cell["\<\ With 10 steps the error is horrendous. Even with 1000 steps we \ would not expect anywhere near this large an error. (Compare with our \ original example.)\ \>", "Text"], Cell[CellGroupData[{ Cell["Why this happens.", "Subsection"], Cell["\<\ We can see why we get such errors by looking at the general \ solution to the differential equation and a plot of the direction \ field.\ \>", "Text"], Cell[BoxData[ \(DSolve[\(x'\)[t] \[Equal] g[t, x[t]], x[t], t]\)], "Input"], Cell["\<\ We had chosen initial conditions which made the exponential term \ vanish. However when we numerically generate solutions, the errors that \ occur move us from our actual solution curve to nearby solution curves. We \ move further away each step when there are a small number of steps, and stay \ closer when there are a large number of steps. 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0.0908578}}], Cell["\<\ The (almost) actual solution curve to our initial value problem is \ in magenta, and approaches a horizontal asymptote. The other curves result \ when we deviate slightly at some point from our solution curve, and the \ exponential eventually takes over. The direction fields show how any \ deviation from the original curve will eventually pull us away from it.\ \>", \ "Text"], Cell["\<\ The moral of the story is that we cannot always blindly accept the \ results of numerical solvers. Some understanding of the expected general \ form of the solution (possibly from a direction field) and how we get errors \ due to the software and hardware is important in deciding how much to trust \ the output of the solver.\ \>", "Text"] }, Closed]] }, Closed]], Cell[CellGroupData[{ Cell["Example of use with systems of equations", "Section"], Cell[TextData[{ "The only thing we need to do to use the methods with a system of \ differential equations is to properly define the initial value problem. We \ view y as a vector, and give f as a vector valued function in terms of y. If \ the system is linear, we can use matrices to define f(t,y). The example \ below is from a compartment model for lead in the body. Again, if t0 is an \ integer, it is best to follow the value you put in by a period to switch ", StyleBox["Mathematica", FontSlant->"Italic"], " into numerical evaluation mode. Also, you do not need to define the \ values of constants (such as a01) separately if you enter them directly into \ the definition of f, in which case you can delete the items in blue." }], "Text"], Cell[BoxData[{ RowBox[{"Clear", "[", RowBox[{"f", ",", "t", ",", "y", ",", "t0", ",", "tf", ",", "y0", StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a01", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a02", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a12", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a13", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a21", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a31", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["iL", FontColor->RGBColor[0, 0, 1]]}], "]"}], "\n", StyleBox[\(a01 = .0211;\), FontColor->RGBColor[0, 0, 1]], "\n", StyleBox[\(a02 = .0162;\), FontColor->RGBColor[0, 0, 1]], "\n", StyleBox[\(a12 = .0124;\), FontColor->RGBColor[0, 0, 1]], "\n", StyleBox[\(a13 = .000035;\), FontColor->RGBColor[0, 0, 1]], "\n", StyleBox[\(a21 = .0111;\), FontColor->RGBColor[0, 0, 1]], "\n", StyleBox[\(a31 = .0039;\), FontColor->RGBColor[0, 0, 1]], "\n", StyleBox[\(iL = 49.3;\), FontColor->RGBColor[0, 0, 1]], "\n", RowBox[{\(f[t_, y_]\), ":=", RowBox[{ RowBox[{ StyleBox[ RowBox[{"(", GridBox[{ {\(-\((a01 + a21 + a31)\)\), "a12", "a13"}, {"a21", \(-\((a02 + a12)\)\), "0"}, {"a31", "0", "a13"} }], ")"}], FontColor->RGBColor[1, 0, 1]], ".", "y"}], "+", StyleBox[\({iL, 0, 0}\), FontColor->RGBColor[1, 0, 1]]}]}], "\n", RowBox[{ RowBox[{"t0", "=", StyleBox["0.0", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", RowBox[{ RowBox[{"y0", "=", StyleBox[\({0, 0, 0}\), FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", RowBox[{ RowBox[{"tf", "=", StyleBox["1", FontColor->RGBColor[1, 0, 1]]}], ";"}]}], "Input"], Cell[TextData[{ "If the system is not linear, we can no longer use matrices to define \ f(t,y). Whether or not the system is linear, we can use another approach. \ ", StyleBox["Mathematica", FontSlant->"Italic"], " refers to the jth coordinate of the vector y by the notation y[[j]]. \ Below is the syntax for our compartment model for lead in the body, but this \ time without using matrices to define f(t,y). You can evaluate either the \ previous cell or the next cell, but only the one you evaluate last will be \ used for the sections below." }], "Text"], Cell[BoxData[{ RowBox[{"Clear", "[", RowBox[{"f", ",", "t", ",", "y", ",", "t0", ",", "tf", StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a01", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a02", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a12", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a13", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a21", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["a31", FontColor->RGBColor[0, 0, 1]], StyleBox[",", FontColor->RGBColor[0, 0, 1]], StyleBox["iL", FontColor->RGBColor[0, 0, 1]]}], "]"}], "\n", RowBox[{ StyleBox[\(a01 = .0211\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontColor->RGBColor[0, 0, 1]], StyleBox[\(a02 = .0162\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontColor->RGBColor[0, 0, 1]], StyleBox[\(a12 = .0124\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontColor->RGBColor[0, 0, 1]], StyleBox[\(a13 = .000035\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontColor->RGBColor[0, 0, 1]], StyleBox[\(a21 = .0111\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontColor->RGBColor[0, 0, 1]], StyleBox[\(a31 = .0039\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], StyleBox["\n", FontColor->RGBColor[0, 0, 1]], StyleBox[\(iL = 49.3\), FontColor->RGBColor[0, 0, 1]], StyleBox[";", FontColor->RGBColor[0, 0, 1]], "\n", RowBox[{\(f[t_, y_]\), ":=", StyleBox[\({\(-\((a01 + a21 + a31)\)\) y[\([1]\)] + a12\ y[\([2]\)] + a13\ y[\([3]\)] + iL, a21\ y[\([1]\)] - \((a02 + a12)\)\ y[\([2]\)], a31\ y[\([1]\)] - a13\ y[\([3]\)]}\), FontColor->RGBColor[1, 0, 1]]}]}], "\n", RowBox[{ RowBox[{"t0", "=", StyleBox["0.", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", RowBox[{ RowBox[{"y0", "=", StyleBox[\({0, 0, 0}\), FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", RowBox[{ RowBox[{"tf", "=", StyleBox["1", FontColor->RGBColor[1, 0, 1]]}], ";"}]}], "Input"], Cell["\<\ The methods below have been copied without change from above, \ except the difference from the exact answer has been left out.\ \>", "Text"], Cell[CellGroupData[{ Cell["Euler", "Subsection"], Cell[BoxData[{\(Clear[n, h, j, t, y]\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", RowBox[{\(Table[y[j + 1] = y[j] + h\ f[t[j], y[j]], {j, 0, n - 1}]\), StyleBox[";", FontColor->RGBColor[1, 0, 1]]}], "\n", \(y[n]\)}], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Runge-Kutta", "Subsection"], Cell[BoxData[{\(Clear[n, h, j, k1, k2, k3, k4, t, y]\), "\n", RowBox[{ RowBox[{"n", "=", StyleBox["10", FontColor->RGBColor[1, 0, 1]]}], ";"}], "\n", \(h = \(tf - t0\)\/n;\), "\n", \(t[j_] := N[t0 + h\ j]\), "\n", \(y[0] = y0;\), "\n", \(k1[t_, y_] := f[t, y]\), "\n", \(k2[t_, y_] := f[t + .5\ h, y + .5\ h\ k1[t, y]]\), "\n", \(k3[t_, y_] := f[t + .5\ h, y + .5\ h\ k2[t, y]]\), "\n", RowBox[{\(k4[t_, y_] := f[t + h, y + h\ k3[t, y]]\), "\t\t\t\t"}], "\n", RowBox[{\(Table[ y[j + 1] = y[j] + \(h\/6\) \((\ k1[t[j], y[j]] + 2\ k2[t[j], y[j]] + 2 k3[t[j], y[j]] + k4[t[j], y[j]])\), {j, 0, n - 1}]\), StyleBox[";", FontColor->RGBColor[1, 0, 1]]}], "\n", \(y[n]\)}], "Input"] }, Closed]] }, Closed]] }, Open ]] }, FrontEndVersion->"4.0 for Microsoft Windows", ScreenRectangle->{{0, 1024}, {0, 720}}, AutoGeneratedPackage->None, WindowToolbars->"EditBar", WindowSize->{592, 671}, WindowMargins->{{10, Automatic}, {Automatic, 1}} ] (*********************************************************************** Cached data follows. 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