(* Content-type: application/mathematica *) (*** Wolfram Notebook File ***) (* http://www.wolfram.com/nb *) (* CreatedBy='Mathematica 6.0' *) (*CacheID: 234*) (* Internal cache information: NotebookFileLineBreakTest NotebookFileLineBreakTest NotebookDataPosition[ 145, 7] NotebookDataLength[ 82339, 1791] NotebookOptionsPosition[ 78711, 1670] NotebookOutlinePosition[ 79133, 1688] CellTagsIndexPosition[ 79090, 1685] WindowFrame->Normal ContainsDynamic->False*) (* Beginning of Notebook Content *) Notebook[{ Cell["\<\ I like Mathematica because you don' t need to \"program\" (per se) to use it, \ unless you want to. For instance, you can use it like a graphing calculator. Just type something \ in and press shift-enter. If you want to force a numerical response, \ surround the command with the function N[ ].\ \>", "Text", CellChangeTimes->{{3.415965396328125*^9, 3.415965419546875*^9}, { 3.41596551246875*^9, 3.4159655528125*^9}, {3.41596560334375*^9, 3.41596563425*^9}, {3.415965879578125*^9, 3.415965904359375*^9}, { 3.4159809895757504`*^9, 3.4159809897476254`*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"67", "^", "0.56"}]], "Input", CellChangeTimes->{{3.41596555496875*^9, 3.4159655593125*^9}}], Cell[BoxData["10.534208004287029`"], "Output", CellChangeTimes->{3.4159810316070004`*^9, 3.415983295765625*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{ RowBox[{"Log", "[", "749", "]"}], "-", RowBox[{"Pi", "/", "14"}]}], "\[IndentingNewLine]", RowBox[{"N", "[", RowBox[{ RowBox[{"Log", "[", "749", "]"}], "-", RowBox[{"Pi", "/", "14"}]}], "]"}]}], "Input", CellChangeTimes->{{3.415965572890625*^9, 3.41596560034375*^9}, { 3.415965639875*^9, 3.415965642859375*^9}}], Cell[BoxData[ RowBox[{ RowBox[{"-", FractionBox["\[Pi]", "14"]}], "+", RowBox[{"Log", "[", "749", "]"}]}]], "Output", CellChangeTimes->{3.4159810369507504`*^9, 3.415983296921875*^9}], Cell[BoxData["6.394339508260805`"], "Output", CellChangeTimes->{3.4159810369507504`*^9, 3.415983297*^9}] }, Open ]], Cell["\<\ Variable declaring is simple, just a string of letters and/or numbers \ followed by an equal sign. A semicolon suppresses the output, but the \ command is still performed.\ \>", "Text", CellChangeTimes->{{3.415965594546875*^9, 3.415965596875*^9}, { 3.41596564859375*^9, 3.41596570971875*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{"k1", "=", ".4"}], "\[IndentingNewLine]", RowBox[{ RowBox[{"k2", "=", ".2"}], ";"}]}], "Input", CellChangeTimes->{{3.415965006109375*^9, 3.415965013046875*^9}, 3.41596572040625*^9, 3.415967133203125*^9, {3.4159671655625*^9, 3.415967166203125*^9}}], Cell[BoxData["0.4`"], "Output", CellChangeTimes->{3.4159811015757504`*^9, 3.415983298734375*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData["k2"], "Input", CellChangeTimes->{{3.415968577578125*^9, 3.415968577671875*^9}}], Cell[BoxData["0.2`"], "Output", CellChangeTimes->{3.4159811087007504`*^9, 3.415983300484375*^9}] }, Open ]], Cell[TextData[{ "Simple plotting is also easy. The command is Plot[ f(x) , {x,xmin,xmax} \ ], where f(x) is the function you want to plot, x is the independant variable \ you're plotting against, and xmin - xmax is the range you want to plot. ", StyleBox["Mathematica", FontSlant->"Italic"], " is pretty good at picking appropriate ranges and zooms to display, but you \ can define them explicitly if you'd like (see the help files)." }], "Text", CellChangeTimes->{{3.415965726171875*^9, 3.41596586834375*^9}, { 3.41596861390625*^9, 3.415968655328125*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Plot", "[", RowBox[{ RowBox[{ RowBox[{"x", "^", "2"}], "+", "x", "-", "1"}], ",", RowBox[{"{", RowBox[{"x", ",", "0", ",", "10"}], "}"}]}], "]"}]], "Input", CellChangeTimes->{{3.415965736671875*^9, 3.41596574846875*^9}}], Cell[BoxData[ GraphicsBox[{{}, {}, {Hue[0.67, 0.6, 0.6], LineBox[CompressedData[" 1:eJwVzns01HkYBnCXaHeoY8p2Qcxg3VW0nSJ831LUkORaZicZ16VIG12kLHYt NTXJYSIrpaRSK5QZ5UdKM2ZclhOaGSumwY7JmEpFrP3uH895zue85z3noTKT AqK1NDQ0onH+79IzTF2nqHMeSa5J+QsLKuJ5L8U3lPITquWTJoWjKmKhrSmL QjmJ4rzuvTrSoSJKm1lmOpRcZG2guWt7sYo45J5y+LPZFfRVqibIziqi92tc wZRZFZqhT016hU0SiyUj1u/MuMhXdXCvEUtJDDQ/TVGOc5Eggef4JVFJ3LnF aZ2o4aHH/asL7/srCb+jfuEKzydoPF1BurlcSRR+yysciyVQuXZvdkPRBLHs Kl9L3f8MxX1s5p7kKIjy9rLkCTMB8kswsIk+PU585NTbDZ0XIJEbrzI8ZJzw ihXKemYEaEHcz7dfP04otT8H83rbUVZ6omqzbIzY7O7nmpsjQtr0bzutvMeI 7gdzWtaqLiQ/Pt3trztKnG13oDmF9yInl8p3gxEy4v6DXMmhtWLUVju8iGss IRTLFOdpEWJk3LJj1pAQE9apNA+bAjFi3BgypDDFxDU3vXLZjBg9HK3jela+ JvIF52PCXkhQfpy8pdlxgEiVXVR7/ziInPcfEF2yf0WU6QmzyDveoFFJ+GWz Vx1ELPrrivKoDFFCTRmPwosIjeufeJJzMsROmfwtM4VNcBaZSNsr8P3M9+dG 9bMIviDGtKpPhjLqr81Oc5KRTdDc9WjXt4iiVaj70e0aGo+zvve3phxlKIr9 OcxHKD4/vambPYqaw7i/zP0jQIlyW1nd/X+QOC/jYU+hFJGo2dZDm6eQgiG4 Gnt3Cl0wENqVH/yA9tDasmMHZhFTEkRoiz6hUp2cRyU3tCH5RTYzyXoWib8y aPM/k6DmiH7gql3zKPI7Vkl/gwGUpyUo1odqwMilosRXEkPILJvb+1GpCUG/ urEe6a0Gjd438OSUNkgbSLy8Y2vA6ljMiSA7HfAxaZyl06jwdrooRO+pLizt r3K92G8BNHl8en7yN8BOfHAkkG8Ffm/Z9qamJLjTwVAUDNiCopWZwa/Wg4DO y+5RUgdYS+5b4xm+BFoe+2gJrdYDCU1HSD8sBUbg5X2NfU7g/S7ss3mwAThY Mf4MbNkAA8+iNMMcyPDHzr113dyNUP/78wOUaTKIsu/4D1RsghidlDednctA 6nyXltrlApFH0+r1i5fDyL3llxwUW2BdrH2nb4QhaEnuTows8YCuZQ56KS7f QevW2vgFV4DG+5klDtor4AfGkEtr1FZI07TzFw2ugN0v4oW1D7cBSXpqZ2Ld SlAeSbgr19kO8lvUgrLTq6DpzIXbcb47oMBm9SJu8GrIVs+/ZnG9gF0c8zKC YgTx1Oph4aqdEFbA6NQfM4IKmv4d8ald8GX7zZmyJmMgDezJUuv4gEy+7cSz HBM49FldKMrwhevuTsYdEWvg8KjX7ttWflD9awo51tEU/E1XXVjauAdygiN9 tKdN4UTN1Z3Vt/1ho7S8KqfDDGKcvlfn5O0Fjm1JdYcPBZagkxtOMwPgbDOL 48OlgNqZl2seEAi7jYNvtltSgUeeZJl4BsHWxJd0ci4V4rb3e8/bBEOZCdNS NE0F+1JZqNA6BCxbKwv96Oaw3mpbvcwoFNiOz+x6msxBFNChWrF0H8R8WPSX wM4Cnu9b/CXeeD8wa95Tci9awMr35qf+Ng4Dfvs3WU/mLCC2hJmZtIkOhzyH 2QLsQsejLZewDZ5w/+jDlugXadRh76uO501hVz+knP2CPXZRpLac//9/Q/ov m+mgG8w+yML2Hg5Nv+xCB8+hlR4H/rWAco9i1uMtdBgNnfJJwH76c2DXa+y8 bv7+E9jvp6nkOeyelpMp+dg3hv8tADc6MCsk99qwKSENpQLsjJ/KjNctWEBa rW2txJ0OliPHbd2wDXibPs1j88P8N+3CblpCd6F44P2+moGR2OFXh4ko7LK1 kXlF2NyQyY532Nsqt3AqsBe33SIbILzPzPBWDXZy5rFgZ+y15BctQuy8kT2D qdg9uaVdA9guvzGoV7BTtVIH5diaxzOiG7GN0vwm3mMP1TVUDWI3fbCaWcDu Q7qTC9j/Aa4pDos= "]]}}, AspectRatio->NCache[GoldenRatio^(-1), 0.6180339887498948], Axes->True, AxesOrigin->{0, 0}, PlotRange->{{0, 10}, {-0.9999997959183257, 108.99999571428575`}}, PlotRangeClipping->True, PlotRangePadding->{ Scaled[0.02], Scaled[0.02]}]], "Output", CellChangeTimes->{3.4159811450913754`*^9, 3.415983304875*^9}] }, Open ]], Cell[TextData[{ "Now, the more complex functions have some tricky syntax (as with any \ program). The function NDSolve[ ] solves differential equations numerically. \ The format you need to use is as follows, if you want to solve dy/dx = x, \ y(0) = 2:\n\nNDSolve[{ y'[x] == x , y[0] == 2 }, {y[x]} , {x,xmin,xmax}]\n\n\ So, the first item is a list of your equations (including boundary/initial \ conditions) surrounded by curly brackets. Be sure to call your dependent \ variable y[x] to tell ", StyleBox["Mathematica", FontSlant->"Italic"], " it's a function of the independent variable. The second item is a list of \ your dependent variables. Finally, the third item is your independent \ variable and the range of numerical solutions you want." }], "Text", CellChangeTimes->{{3.4159659179375*^9, 3.41596593740625*^9}, { 3.415965967671875*^9, 3.41596621503125*^9}, {3.415968776859375*^9, 3.41596878590625*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"numSoln", "=", RowBox[{"NDSolve", "[", "\[IndentingNewLine]", RowBox[{ RowBox[{"{", RowBox[{ RowBox[{ RowBox[{ RowBox[{"\[Xi]1", "'"}], "[", "t", "]"}], "\[Equal]", RowBox[{"k1", "*", RowBox[{"(", RowBox[{"1", "-", RowBox[{"\[Xi]1", "[", "t", "]"}]}], ")"}], "*", RowBox[{ RowBox[{"(", RowBox[{"2", "-", RowBox[{"\[Xi]1", "[", "t", "]"}], "-", RowBox[{"\[Xi]2", "[", "t", "]"}]}], ")"}], "^", "0.5"}]}]}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"\[Xi]2", "'"}], "[", "t", "]"}], "\[Equal]", RowBox[{"k2", "*", RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"\[Xi]1", "[", "t", "]"}], "-", RowBox[{"\[Xi]2", "[", "t", "]"}]}], ")"}], "^", "0.5"}], "*", RowBox[{ RowBox[{"(", RowBox[{"2", "-", RowBox[{"\[Xi]1", "[", "t", "]"}], "-", RowBox[{"\[Xi]2", "[", "t", "]"}]}], ")"}], "^", "1.5"}]}]}], ",", "\[IndentingNewLine]", RowBox[{ RowBox[{"\[Xi]1", "[", "0", "]"}], "\[Equal]", "0"}], ",", RowBox[{ RowBox[{"\[Xi]2", "[", "0", "]"}], "\[Equal]", "0"}]}], "}"}], ",", "\[IndentingNewLine]", RowBox[{"{", RowBox[{ RowBox[{"\[Xi]1", "[", "t", "]"}], ",", RowBox[{"\[Xi]2", "[", "t", "]"}]}], "}"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "100"}], "}"}]}], "]"}]}]], "Input", CellChangeTimes->{{3.4159650245625*^9, 3.4159652065*^9}, 3.415965323671875*^9}], Cell[BoxData[ RowBox[{"{", RowBox[{"{", RowBox[{ RowBox[{ RowBox[{"\[Xi]1", "[", "t", "]"}], "\[Rule]", RowBox[{ TagBox[ RowBox[{"InterpolatingFunction", "[", RowBox[{ RowBox[{"{", RowBox[{"{", RowBox[{"0.`", ",", "100.`"}], "}"}], "}"}], ",", "\<\"<>\"\>"}], "]"}], False, Editable->False], "[", "t", "]"}]}], ",", RowBox[{ RowBox[{"\[Xi]2", "[", "t", "]"}], "\[Rule]", RowBox[{ TagBox[ RowBox[{"InterpolatingFunction", "[", RowBox[{ RowBox[{"{", RowBox[{"{", RowBox[{"0.`", ",", "100.`"}], "}"}], "}"}], ",", "\<\"<>\"\>"}], "]"}], False, Editable->False], "[", "t", "]"}]}]}], "}"}], "}"}]], "Output", CellChangeTimes->{3.4159813592320004`*^9, 3.4159833225*^9}] }, Open ]], Cell[TextData[{ "Note that ", StyleBox["Mathematica", FontSlant->"Italic"], " returned InterpolatingFunctions for the two extents vs time. Note also \ that I stored these in the variable numSoln. This is ", StyleBox["Mathematica", FontSlant->"Italic"], "'s way of holding onto these numerical solutions until you want them. Look \ at the function \[Xi]1[t] without calling these solutions:" }], "Text", CellChangeTimes->{{3.415966238140625*^9, 3.415966352*^9}, 3.415966388765625*^9, {3.415983332109375*^9, 3.415983333734375*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"\[Xi]1", "[", "t", "]"}]], "Input", CellChangeTimes->{{3.4159663544375*^9, 3.415966354765625*^9}, 3.41596638646875*^9}], Cell[BoxData[ RowBox[{"\[Xi]1", "[", "t", "]"}]], "Output", CellChangeTimes->{3.4159813907945004`*^9, 3.41598333703125*^9}] }, Open ]], Cell["Nothing. Now, let's use slash-dot to call the solutions.", "Text", CellChangeTimes->{{3.41596635634375*^9, 3.415966380734375*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"\[Xi]1", "[", "t", "]"}], "/.", "numSoln"}]], "Input", CellChangeTimes->{{3.41596639334375*^9, 3.415966415109375*^9}}], Cell[BoxData[ RowBox[{"{", RowBox[{ TagBox[ RowBox[{"InterpolatingFunction", "[", RowBox[{ RowBox[{"{", RowBox[{"{", RowBox[{"0.`", ",", "100.`"}], "}"}], "}"}], ",", "\<\"<>\"\>"}], "]"}], False, Editable->False], "[", "t", "]"}], "}"}]], "Output", CellChangeTimes->{3.4159814037632504`*^9, 3.41598333840625*^9}] }, Open ]], Cell["\<\ Now, here' s where the syntax gets strange. If we want a numerical value for \ \[Xi]1[10], we need to use ANOTHER slash-dot to replace t with 10.\ \>", "Text", CellChangeTimes->{{3.41596642234375*^9, 3.41596646803125*^9}, { 3.415968869421875*^9, 3.4159688704375*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{ RowBox[{"\[Xi]1", "[", "t", "]"}], "/.", "numSoln"}], "/.", RowBox[{"{", RowBox[{"t", "\[Rule]", "10"}], "}"}]}]], "Input", CellChangeTimes->{{3.4159664713125*^9, 3.415966476171875*^9}, 3.415967178140625*^9, {3.415978820870944*^9, 3.41597882149586*^9}}], Cell[BoxData[ RowBox[{"{", "0.9644155225013635`", "}"}]], "Output", CellChangeTimes->{3.4159814646382504`*^9, 3.415983340546875*^9}] }, Open ]], Cell["\<\ This would be very inconvenient to do constantly, so I'm going to store the \ numerical solutions in other functions. Again, the syntax is bizarre and may \ seem needlessly complicated, but hey every program has its nuances. The most important thing is to NOT double-define your variables and create a \ recursion loop. Especially if you're in a computer lab, because a loud \ \"DING\" accompanies each of the infinite number of errors you'll generate.\ \>", "Text", CellChangeTimes->{{3.415966485546875*^9, 3.41596661940625*^9}, { 3.415968885703125*^9, 3.415968910296875*^9}}, Background->RGBColor[1, 1, 0]], Cell[BoxData[{ RowBox[{ RowBox[{"\[Xi]1a", "[", "ta_", "]"}], ":=", RowBox[{ RowBox[{ RowBox[{"\[Xi]1", "[", "t", "]"}], "/.", RowBox[{"numSoln", "[", RowBox[{"[", "1", "]"}], "]"}]}], "/.", RowBox[{"{", RowBox[{"t", "\[Rule]", "ta"}], "}"}]}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{"\[Xi]2a", "[", "ta_", "]"}], ":=", RowBox[{ RowBox[{ RowBox[{"\[Xi]2", "[", "t", "]"}], "/.", RowBox[{"numSoln", "[", RowBox[{"[", "1", "]"}], "]"}]}], "/.", RowBox[{"{", RowBox[{"t", "\[Rule]", "ta"}], "}"}]}]}]}], "Input", CellChangeTimes->{{3.41596662578125*^9, 3.415966658625*^9}, { 3.415966694671875*^9, 3.415966695359375*^9}}], Cell["\<\ Going through those lines, y[ x_ ] : = defines a new usermade function. I \ called \[Xi]1[t] with the first A/.B, but added [[1]] to extract the item \ from the list (i.e. to get rid of the curly brackets). The second B/.C \ replaces t with ta to avoid those infinite variable recursions. At the end of the day, we can directly get numbers for \[Xi]1[t] and \ \[Xi]2[t].\ \>", "Text", CellChangeTimes->{{3.4159667086875*^9, 3.415966866109375*^9}, { 3.415968946453125*^9, 3.415968956859375*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{"\[Xi]1a", "[", "2", "]"}], "\[IndentingNewLine]", RowBox[{"\[Xi]2a", "[", "2", "]"}]}], "Input", CellChangeTimes->{{3.415966892546875*^9, 3.41596690290625*^9}}], Cell[BoxData["0.619295223532827`"], "Output", CellChangeTimes->{3.4159816152476254`*^9, 3.415983352125*^9}], Cell[BoxData["0.30756677285171014`"], "Output", CellChangeTimes->{3.4159816152476254`*^9, 3.41598335215625*^9}] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Plot", "[", RowBox[{ RowBox[{"\[Xi]1a", "[", "t", "]"}], ",", RowBox[{"{", RowBox[{"t", ",", "0", ",", "5"}], "}"}]}], "]"}]], "Input", CellChangeTimes->{{3.415965331421875*^9, 3.4159653848125*^9}, 3.415966906328125*^9, 3.4159816283882504`*^9}], Cell[BoxData[ GraphicsBox[{{}, 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Often you can use \ functions like Maximize[ ] or NMaximize[ ] or Solve[ ] to accomplish this, \ but we're in a tricky spot because our function is ultimately a complex \ numerical interpolation. FindRoot is the most robust means I know of to find \ the maximum value of X[ ] in this sort of case, so let's go with that.\ \>", "Text", CellChangeTimes->{{3.415967273609375*^9, 3.41596737828125*^9}, { 3.41596901603125*^9, 3.41596902778125*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{"Xmax", "=", RowBox[{"FindRoot", "[", "\[IndentingNewLine]", RowBox[{ RowBox[{ RowBox[{"D", "[", RowBox[{ RowBox[{"X", "[", "tb", "]"}], ",", "tb"}], "]"}], "\[Equal]", "0"}], ",", RowBox[{"{", RowBox[{"tb", ",", "3"}], "}"}]}], "]"}]}]], "Input", CellChangeTimes->{{3.415967380765625*^9, 3.41596741275*^9}, { 3.415967499640625*^9, 3.4159675206875*^9}, 3.4159817591695004`*^9}], Cell[BoxData[ RowBox[{"{", RowBox[{"tb", "\[Rule]", "3.4016541664573254`"}], "}"}]], "Output", CellChangeTimes->{3.4159818119507504`*^9, 3.415983371359375*^9}] }, Open ]], Cell[TextData[{ "What I did was ask for the derivative of X[tb] with respect to tb (D[X[tb], \ tb), set it equal to zero, and asked ", StyleBox["Mathematica", FontSlant->"Italic"], " to find the root (solution) to that equation near t=3 (eyeballed from the \ graph)." }], "Text", CellChangeTimes->{{3.415967416921875*^9, 3.41596748265625*^9}, { 3.4159675266875*^9, 3.415967530859375*^9}}, Background->RGBColor[1, 1, 0]], Cell["\<\ Finally, let' s find the compositions of all species at that maximum.\ \>", "Text", CellChangeTimes->{{3.4159675325625*^9, 3.415967543265625*^9}}, Background->RGBColor[1, 1, 0]], Cell[CellGroupData[{ Cell[BoxData[{ RowBox[{ RowBox[{"A", "[", "tb", "]"}], "/.", RowBox[{"Xmax", "[", RowBox[{"[", "1", "]"}], "]"}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{"B", "[", "tb", "]"}], "/.", RowBox[{"Xmax", "[", RowBox[{"[", "1", "]"}], "]"}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{"X", "[", "tb", "]"}], "/.", RowBox[{"Xmax", "[", RowBox[{"[", "1", "]"}], "]"}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{"Y", "[", "tb", "]"}], "/.", RowBox[{"Xmax", "[", RowBox[{"[", "1", "]"}], "]"}]}], "\[IndentingNewLine]", RowBox[{ RowBox[{"Z", "[", "tb", "]"}], "/.", RowBox[{"Xmax", "[", RowBox[{"[", "1", "]"}], "]"}]}]}], "Input", CellChangeTimes->{{3.415967558921875*^9, 3.41596757396875*^9}}], Cell[BoxData["0.22310666180047045`"], "Output", CellChangeTimes->{3.4159818286226254`*^9, 3.415983373609375*^9}], Cell[BoxData["0.7764652330815043`"], "Output", CellChangeTimes->{3.4159818286226254`*^9, 3.415983373625*^9}], Cell[BoxData["0.3302519094805634`"], "Output", CellChangeTimes->{3.4159818286226254`*^9, 3.415983373671875*^9}], Cell[BoxData["0.44664142871896617`"], "Output", CellChangeTimes->{3.4159818286226254`*^9, 3.415983373703125*^9}], Cell[BoxData["1.2235347669184957`"], "Output", CellChangeTimes->{3.4159818286226254`*^9, 3.41598337371875*^9}] }, Open ]], Cell["\<\ We could also be asked for the effluent at a given conversion. 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