(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. 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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Besides the obvious application to \ ballistics, there is a much more noteworthy connection (for our more pacific \ purposes) to developments and colorful personalities in mathematics and \ physics. Many of these ideas are presented in a compelling paper by Groetsch \ ", StyleBox["[4]", FontWeight->"Bold"], ", who traces a rich history from Tartaglia to Galileo and then employs a \ ", StyleBox["tour de force", FontSlant->"Italic"], " of undergraduate analysis to answer and expand some classical questions \ to the case of projectile motion in a resistive medium.\n\nThe purpose of \ this paper is to indicate how some of Groetsch's ingenious analysis can be \ circumvented with the help of computer algebra and a relatively new symbolic \ object, the Lambert W function, that increasingly seems destined for fame and \ immortality (see FOCUS newsletter ", StyleBox["[1]", FontWeight->"Bold"], "; Corless et al ", StyleBox["[3]", FontWeight->"Bold"], "). We will use these modern tools to simplify the derivation of some of \ Groetsch's results while extending some of them. In particular, we find a \ symbolic solution for the range as a function of the elevation angle in the \ presence of a linear resistance and we also give a partial solution to the \ inverse problem of finding the elevation angle(s) that give rise to given \ range values by obtaining a closed form for the angle generating the ", StyleBox["maximum", FontSlant->"Italic"], " range in terms of the initial velocity and the resistance constant.\n", "\nTwo interesting side issues arise in the process of our development. One \ of them evolves from the need to investigate certain limits involving the \ Lambert W function. To do this we develop a general theorem, which may be of \ interest in its own right, about inverse functions arising from real-analytic \ functions. A second issue relates to the newly emerging area of ", StyleBox["Experimental Mathematics", FontSlant->"Italic"], " (Borwein and Corless ", StyleBox["[2]", FontWeight->"Bold"], ") and raises practical and philosophical questions about use of symbolic \ computation to \"discover\" new results and the extent to which such \ computation can be viewed as an accepted form of proof." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Finding the Range With No Resistance; Struggling With \ Resistance\ \>", "Section"], Cell[TextData[{ "Deriving a formula in the absence of resistance for the range ", Cell[BoxData[ \(TraditionalForm\`R\)]], " as a function of the angle of elevation ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " is a simple exercise in calculus. We race through it here to warm up for \ the more resistant case that follows. 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Cell[BoxData[ \(TraditionalForm\`x\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y\)]], " both 0, we get\n\n\t\t", Cell[BoxData[{ \(TraditionalForm\`x(t) = v\ \(cos(\[Theta])\)*t\), "\[IndentingNewLine]", \(TraditionalForm\`y(t) = \(\(-1\)\/2\) g\ t\^2 + v\ \(sin(\[Theta])\)*\(\(t\)\(.\)\)\)}]], "\n\t\t\nThere are no surprises here -- it is easily shown that these \ parametric equations result in a parabolic path for the projectile. \n\nTo \ get an equation for the range, we set the height ", Cell[BoxData[ \(TraditionalForm\`y(t)\)]], " to 0, compute the nonzero solution for ", Cell[BoxData[ \(TraditionalForm\`t\)]], ", and substitute it in the ", Cell[BoxData[ FormBox[Cell["x(t)"], TraditionalForm]]], " equation. This results in\n\n\t\t", Cell[BoxData[ FormBox[ StyleBox[\(R = \(\(\(2 v\^2\)\/g\) \(sin(\[Theta])\)*\(cos(\[Theta])\) = \ \(v\^2\/g\) \(sin(2 \[Theta])\)\)\), FontSize->18], TraditionalForm]]], ". \n\t\t\nAs the plot of ", Cell[BoxData[ \(TraditionalForm\`R\)]], " vs. ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " (we set ", Cell[BoxData[ \(TraditionalForm\`v = 50\)]], "and ", Cell[BoxData[ \(TraditionalForm\`g = 32.2\)]], ") in Figure 2 indicates, the maximal range occurs when ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Pi]\/4\)]], " and submaximal ranges occur for a pair of \[Theta] values equally spaced \ around ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Pi]\/4\)\(.\)\)\)]], " These results are independent of ", Cell[BoxData[ \(TraditionalForm\`v\)]], " and ", Cell[BoxData[ \(TraditionalForm\`g\)]], " and can be verified analytically using the symmetry of ", Cell[BoxData[ \(TraditionalForm\`sin(2 \[Theta])\)]], " about ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = 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(via constant ", Cell[BoxData[ \(TraditionalForm\`k\)]], ") to the velocity has a propitious beginning. The differential equations \ now become:\n\n\t\t", Cell[BoxData[{ \(TraditionalForm\`x'' = \(-k\)\ x'\), "\[IndentingNewLine]", \(TraditionalForm\`y'' = \(-g\) - k\ \(\(y'\)\(.\)\)\)}]], "\n\t\t\nOne integration on each equation (set ", Cell[BoxData[ \(TraditionalForm\`u = x'\)]], "and coast; set ", Cell[BoxData[ \(TraditionalForm\`w = y'\)]], ", separate, and solve) leads (with initial values ", Cell[BoxData[ \(TraditionalForm\`x' \((0)\) = v\ \(cos(\[Theta])\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y' \((0)\) = v\ \(sin(\[Theta])\)\)]], ")\n\n\t\t", Cell[BoxData[{ FormBox[\(x' = v\ \(cos(\[Theta])\) \[ExponentialE]\^\(\(-k\)\(\ \)\(t\)\(\ \)\)\), TraditionalForm], "\[IndentingNewLine]", FormBox[ RowBox[{\(y'\), "=", StyleBox[\(\(\(\(-g\) + \((g + k\ v\ \(sin(\[Theta])\))\)\ \[ExponentialE]\^\(\(-k\ \)\ t\)\)\/k\)\(.\)\), FontSize->18]}], TraditionalForm]}], FontSize->16], "\n\t\t\n\t\t\n\t\t\nIntegrating again (with initial values ", Cell[BoxData[ \(TraditionalForm\`x(0) = 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y(0) = 0\)]], ") leads to\n\n\t\t", Cell[BoxData[{ FormBox[ StyleBox[\(x( t) = \(v\ \(cos(\[Theta])\) \((1 - \[ExponentialE]\^\(\(-k\)\ t\ \))\)\)\/k\), FontSize->24], TraditionalForm], "\[IndentingNewLine]", FormBox[ StyleBox[\(y( t) = \(\(\(\(-k\)\ t\ g + g + k\ v\ \(sin(\[Theta])\) - \[ExponentialE]\^\(\(-k\)\ t\)\ \ \((g + k\ v\ \(sin(\[Theta])\))\)\)\/k\^2\)\(.\)\)\), FontSize->24], TraditionalForm]}]], "\t\t" }], "Text"], Cell[TextData[{ "Recall that in the \"no resistance\" case we set ", Cell[BoxData[ \(TraditionalForm\`y(t) = 0\)]], ", found a nonzero solution for ", Cell[BoxData[ \(TraditionalForm\`t\)]], ", and evaluated ", Cell[BoxData[ \(TraditionalForm\`x(t)\)]], " at this time ", Cell[BoxData[ \(TraditionalForm\`t\)]], " to express the horizontal range as a function of ", Cell[BoxData[ \(TraditionalForm\`\[Theta], \ g, \ and\ \ v\)]], ". \n\nLooking at the form of ", Cell[BoxData[ \(TraditionalForm\`y(t)\)]], " that results when resistance is present, we see that finding a nonzero \ root may be a daunting, if not impossible, task. Indeed, the presence of ", Cell[BoxData[ \(TraditionalForm\`t\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[ExponentialE]\^\(\(-k\)\ t\)\)]], "in an expression does not bode well for isolating ", Cell[BoxData[ \(TraditionalForm\`t\)]], ". The stage is now set for a dramatic rescue, so let us introduce the new \ function which will come to our aid." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["The Lambert W Function and a Symbolic Solution for the Range", "Section"], Cell[TextData[{ " The Lambert W function can be defined as an inverse of the function ", Cell[BoxData[ \(TraditionalForm\`T(w) = w\ \[ExponentialE]\^w\)]], " (see Figure 3). 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434.875}} -> {-9.02112, 10.5686, 0.0250243, 0.0250243}, {{290.938, 322.875}, \ {509.625, 477.688}} -> {-9.59017, 33.9691, 0.0263016, 0.0851138}, {{290.938, \ 322.875}, {472.563, 440.625}} -> {-9.59017, 30.0259, 0.0263016, 0.0851138}}], Cell[TextData[{ "\t\t\t Figure 4.", StyleBox[" Two inverse functions for ", FontWeight->"Plain"], Cell[BoxData[ FormBox[ StyleBox[\(T(w) = w\[ExponentialE]\^w\), FontWeight->"Plain"], TraditionalForm]]] }], "Text", FontWeight->"Bold"], Cell[TextData[{ "The evolution of ", Cell[BoxData[ \(TraditionalForm\`W\)]], " began with ideas proposed by J.H.Lambert in 1758 and refined by Euler \ over the subsequent two decades. An influential paper by Corless et al ", StyleBox["[3]", FontWeight->"Bold"], " presents a variety of applications, pure and applied, in which ", Cell[BoxData[ \(TraditionalForm\`W\)]], " plays a valuable role. As we will see, the range problem for a projectile \ with linear resistance is another such application.\n\nBy its inverse \ function definition, it follows directly that ", Cell[BoxData[ \(TraditionalForm\`w = W(x)\)]], " is a solution to the equation ", Cell[BoxData[ \(TraditionalForm\`x = w\ \[ExponentialE]\^w\)]], ". As a direct consequence, we have the following results:\n\n\t(1) ", Cell[BoxData[ \(TraditionalForm\`x = \(W(x)\) \[ExponentialE]\^\(W(x)\)\)]], " for all ", Cell[BoxData[ \(TraditionalForm\`x \[Element] \([\(-\(1\/\[ExponentialE]\)\), \ \[Infinity])\)\)]], "\n\t\n\t(2) ", Cell[BoxData[ \(TraditionalForm\`W' \((x)\) = \(W(x)\)\/\(x(1 + W(x))\)\)], FontSize->16], " (follows neatly from the formula for the derivative of an inverse \ function and is a nice calculus exercise).\n\t\n\t(3) A solution for ", Cell[BoxData[ \(TraditionalForm\`t\)]], " in the equation ", Cell[BoxData[ \(TraditionalForm\`a\ t + b + c\ \[ExponentialE]\^\(\(\ \ \)\(d\ t\)\) = 0\)]], " (with ", Cell[BoxData[ \(TraditionalForm\`a\ d \[NotEqual] 0\)]], ") is given by ", Cell[BoxData[ FormBox[ RowBox[{"t", "=", RowBox[{ StyleBox[\(-\(b\/a\)\), FontSize->16], StyleBox["-", FontSize->16], RowBox[{\(1\/d\), StyleBox[\(\(W(\(c\ d\ \[ExponentialE]\^\(-\(\(b\ \ d\)\/a\)\)\)\/a)\)\(.\)\), FontSize->16]}]}]}], TraditionalForm]]], "\n\t\nTo derive (3), we massage the first equation as follows:\n\n\t", Cell[BoxData[ \(TraditionalForm\`\((a\ t\ + b)\) \[ExponentialE]\^\(\(-d\)\ t\) = \(-c\)\)]], " \n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\((\(-d\)\ t - \(d\ b\)\/a)\) \ \[ExponentialE]\^\(\(-d\)\ t\) = \(c\ d\)\/a\)]], "\n\t\n\t", Cell[BoxData[ \(TraditionalForm\`\((\(-d\)\ t - \(d\ b\)\/a)\) \ \[ExponentialE]\^\(\(-d\)\ t - \(d\ b\)\/a\) = \(\(c\ d\)\/a\) \ \[ExponentialE]\^\(-\(\(d\ b\)\/a\)\)\)]], ".\n\t\nSetting ", Cell[BoxData[ \(TraditionalForm\`u = \(-d\)\ t - \(d\ b\)\/a\)]], ", we then have ", Cell[BoxData[ \(TraditionalForm\`u\ \[ExponentialE]\^u = \(\(c\ d\)\/a\) \ \[ExponentialE]\^\(-\(\(d\ b\)\/a\)\)\)]], "from which it follows that ", Cell[BoxData[ FormBox[ StyleBox[\(u = W(\(\(c\ d\)\/a\) \[ExponentialE]\^\(-\(\(d\ b\)\/a\)\))\), FontSize->16], TraditionalForm]]], ". Substituting back for ", Cell[BoxData[ \(TraditionalForm\`u\)]], " and solving for ", Cell[BoxData[ \(TraditionalForm\`t\)]], " gives the desired result. \n\nReturning to the height equation ", Cell[BoxData[ \(TraditionalForm\`y(t) = 0\)]], " in the form of the equation ", Cell[BoxData[ \(TraditionalForm\`\(-k\)\ t\ + 1 + \(\(\(k\)\(\ \)\(v\)\(\ \)\)\/g\) \(sin(\[Theta])\) - \ \[ExponentialE]\^\(\(-k\)\ t\)\ \((1 + \(k\ v\)\/g\ \(sin(\[Theta])\))\) = 0\)]], ", we substitute ", Cell[BoxData[ \(TraditionalForm\`u = \(-1\) - \(\(k\ v\)\/g\) \(sin(\[Theta])\)\)]], " to get the simpler equation ", Cell[BoxData[ \(TraditionalForm\`\(-k\)\ t - u + \[ExponentialE]\^\(\(\ \)\(\(-k\)\ t\)\)\ u = 0\)]], ". Applying result (3) now gives the solution\n\n\t", Cell[BoxData[ \(TraditionalForm\`t = \(1\/k\) \(\((\(-u\) + W(u\ \[ExponentialE]\^\(\(\ \)\(u\)\)))\)\(.\)\)\)]], " \t\n\t\nBefore we find the range function, we deal with an interesting \ side issue. Since ", Cell[BoxData[ \(TraditionalForm\`W\)]], " is the inverse to ", Cell[BoxData[ \(TraditionalForm\`T(w) = w\ \[ExponentialE]\^w\)]], ", it would seem that applying ", Cell[BoxData[ \(TraditionalForm\`W\)]], " to both sides of this equation should give the identity ", Cell[BoxData[ \(TraditionalForm\`w = W(w\ \[ExponentialE]\^w)\)]], ". This is indeed a valid identity, but only for ", Cell[BoxData[ \(TraditionalForm\`w \[GreaterEqual] \(-1\)\)]], "since we restricted the domain of ", Cell[BoxData[ \(TraditionalForm\`T\)]], " to obtain its inverse ", Cell[BoxData[ \(TraditionalForm\`W\)]], ". But a look at the form of ", Cell[BoxData[ \(TraditionalForm\`u = \(-1\) - \(\(k\ v\)\/g\) \(sin(\[Theta])\)\)]], " reveals that this quantity will never exceed -1 (constants are positive \ and ", Cell[BoxData[ \(TraditionalForm\`\[Theta] \[Element] \([0, \[Pi]\/2]\)\)]], "). As a result, our expression for ", Cell[BoxData[ \(TraditionalForm\`t\)]], " does not simplify, though if it did it would simply yield the obvious ", Cell[BoxData[ \(TraditionalForm\`t = 0\)]], " solution, which is no help. It should be noted that ", Cell[BoxData[ \(TraditionalForm\`w\ \[ExponentialE]\^w \[GreaterEqual] \(-\(1\/\ \[ExponentialE]\)\)\)]], " for all real ", Cell[BoxData[ \(TraditionalForm\`w\)]], " and is always in the domain of ", Cell[BoxData[ \(TraditionalForm\`W\)]], ".\n\nSubstituting our ", Cell[BoxData[ \(TraditionalForm\`t\)]], " value into ", Cell[BoxData[ \(TraditionalForm\`x(t)\)]], " gives us the desired range function, ", Cell[BoxData[ FormBox[ StyleBox[\(R(\[Theta]) = \(1\/k\) v\ \(cos(\[Theta])\) \((1 - \ \[ExponentialE]\^\(\(\ \)\(u - \ \ W(u\ \[ExponentialE]\^u)\)\))\)\), FontSize->16], TraditionalForm]]], ". From identity (1) above we may replace ", Cell[BoxData[ FormBox[ StyleBox[\(\[ExponentialE]\^\(\(\ \)\(-\(W( u\ \[ExponentialE]\^u)\)\)\)\), FontSize->16], TraditionalForm]]], " by ", Cell[BoxData[ FormBox[ StyleBox[\(\(W( u\ \[ExponentialE]\^\(\(\ \)\(u\)\))\)\/\(u\ \ \[ExponentialE]\^u\)\), FontSize->16], TraditionalForm]]], " since ", Cell[BoxData[ \(TraditionalForm\`u\ \[ExponentialE]\^\(\(\ \)\(u\)\) \[GreaterEqual] \ \(-\(1\/\[ExponentialE]\)\)\)]], " is always true. This gives the somewhat simpler formula \n\n\t", Cell[BoxData[ FormBox[ StyleBox[\(R(\[Theta]) = \(1\/k\) v\ \(cos(\[Theta])\) \((1 - \(W(u\ \[ExponentialE]\^\(\(\ \)\(u\ \)\))\)\/u)\)\), FontSize->16], TraditionalForm]]], ". \n\t\nSubstituting for ", Cell[BoxData[ \(TraditionalForm\`u\)]], ", we get the range formula in all its elemental glory:\n\n\t", Cell[BoxData[ FormBox[ StyleBox[\(R(\[Theta]) = \(1\/k\) v\ \(cos(\[Theta])\) \((1 - \(W(\((\(-1\) - \(\(k\ v\)\/g\) \ \(sin(\[Theta])\))\)\ \[ExponentialE]\^\(\(\ \)\(\(-1\) - \(\(k\ v\)\/g\) \ \(sin(\[Theta])\)\)\))\)\/\(\(-1\) - \(\(k\ v\)\/g\) \(sin(\[Theta])\)\))\)\), FontSize->16], TraditionalForm]]], ". \n\t" }], "Text"], Cell[TextData[{ StyleBox["Computer Algebra Interlude:", FontVariations->{"Underline"->True}], " Honesty compels us to admit at this point that the idea for using Lambert \ W to find a closed form solution was really ", StyleBox["Mathematica", FontSlant->"Italic"], "'s and not ours. Here is a sequence of commands that gave us a first \ closed form for the range. " }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(x[t_] := v*Cos[\[Theta]] \[ExponentialE]\^\(\(-k\)\ t\);\)\), "\ \[IndentingNewLine]", \(\(y[ t_] := \(\(-k\)\ t\ g\ + g + k\ v\ Sin[\[Theta]] - \(\ \[ExponentialE]\^\(\(-k\)\ t\)\) \((g + k\ v\ Sin[\[Theta]])\)\)\/k\^2;\)\), \ "\[IndentingNewLine]", \(tmax = Solve[y[t] \[Equal] 0, t]\)}], "Input"], Cell[BoxData[ \(InverseFunction::"ifun" \(\(:\)\(\ \)\) "Inverse functions are being used. Values may be lost for multivalued \ inverses."\)], "Message"], Cell[BoxData[ \(Solve::"ifun" \(\(:\)\(\ \)\) "Inverse functions are being used by \!\(Solve\), so some solutions may \ not be found."\)], "Message"], Cell[BoxData[ \({{t \[Rule] \(g + g\ ProductLog[\(-\(\(\[ExponentialE]\^\(\(-1\) - \(k\ \ v\ Sin[\[Theta]]\)\/g\)\ \((g + k\ v\ Sin[\[Theta]])\)\)\/g\)\)] + k\ v\ Sin[\ \[Theta]]\)\/\(g\ k\)}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Simplify[x[t] /. tmax[\([1]\)]]\)], "Input"], Cell[BoxData[ \(\[ExponentialE]\^\(-\(\(g + g\ \ ProductLog[\(-\(\(\[ExponentialE]\^\(\(-1\) - \(k\ v\ Sin[\[Theta]]\)\/g\)\ \ \((g + k\ v\ Sin[\[Theta]])\)\)\/g\)\)] + k\ v\ Sin[\[Theta]]\)\/g\)\)\ v\ \ Cos[\[Theta]]\)], "Output"] }, Open ]], Cell[TextData[{ "Note that \"ProductLog\" is the ", StyleBox["Mathematica", FontSlant->"Italic"], " notation for Lambert W and that some algebra transforms the last output \ to the first form of ", Cell[BoxData[ \(TraditionalForm\`R(\[Theta])\)]], " obtained above. 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\(TraditionalForm\`\[Pi]\/4\)]], " as resistance increases. Also, the symmetry about the maximal range angle \ for smaller ranges that was present in the no-resistance case disappears \ dramatically." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ A Limit Theorem for a Class of Inverse Functions and Its \ Application\ \>", "Section"], Cell[TextData[{ "The theorem that follows was initially motivated by efforts to show that \ the range formula for ", Cell[BoxData[ \(TraditionalForm\`R(\[Theta])\)]], " given above reduces to the classical \"no resistance\" result as the \ resistance ", Cell[BoxData[ \(TraditionalForm\`k\)]], " goes to zero. We will apply it to the Lambert W function for this \ purpose. Since the theorem holds for a much wider class of functions, we \ state and prove it in a more general context.\n\n", StyleBox["Theorem 1.", FontVariations->{"Underline"->True}], " Let ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " be a nonconstant real-analytic function in a neighborhood of ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], ". Suppose ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " has a local extremum at ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], ". Then there is a neighborhood of ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " such that ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is strictly monotonic on ", Cell[BoxData[ \(TraditionalForm\`\([x\_0, x\_0 + h]\)\)]], " and on ", Cell[BoxData[ \(TraditionalForm\`\([x\_0 - h, x\_0]\)\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`g(x)\)]], " be the inverse function of ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " on ", Cell[BoxData[ \(TraditionalForm\`\([x\_0, x\_0 + h]\)\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`y(x) = g(f(x))\)]], " is defined on ", Cell[BoxData[ \(TraditionalForm\`\([x\_0 - k, x\_0]\)\)]], " for some ", Cell[BoxData[ \(TraditionalForm\`k > 0\)]], "and ", Cell[BoxData[ FormBox[ StyleBox[\(lim\+\(x \[Rule] \(x\_0\^-\)\)\(y(x) - x\_0\)\/\(x - \ x\_0\)\ = \ \(-1. \)\), FontSize->16], TraditionalForm]]], " Note that this is the left-side derivative of ", Cell[BoxData[ \(TraditionalForm\`y(x)\)]], " because ", Cell[BoxData[ \(TraditionalForm\`y(x\_0) = \(\(x\_0\)\(.\)\)\)]], "\n\n", StyleBox["Proof.", FontVariations->{"Underline"->True}], " We have ", Cell[BoxData[ \(TraditionalForm\`f( x) = \[Sum]\+\(i = 0\)\%\[Infinity]\(\( a\_i\)(x - x\_0)\)\^i\)]], " in a neighborhood of ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], ". Let ", Cell[BoxData[ \(TraditionalForm\`N\)]], " be the least index ", Cell[BoxData[ \(TraditionalForm\`N > 0\)]], " such that ", Cell[BoxData[ \(TraditionalForm\`a\_N \[NotEqual] 0\)]], " (such indices exist because ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is nonconstant). So ", Cell[BoxData[ \(TraditionalForm\`f(x) = a\_0 + \[Sum]\+\(i = N\)\%\[Infinity]\(\( a\_i\)(x - x\_0)\)\^i\)]], ". By noting that ", Cell[BoxData[ \(TraditionalForm\`f(x) = a\_0 + \(\((x - x\_0)\)\^N\) \(F(x)\)\)]], ", where ", Cell[BoxData[ \(TraditionalForm\`F( x) = \[Sum]\+\(i = N\)\%\[Infinity]\(\( a\_i\)(x - x\_0)\)\^\(i - N\ \)\)]], ", we see that ", Cell[BoxData[ \(TraditionalForm\`F(x)\)]], " has the same radius of convergence as ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " and as a consequence must be continuous at ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`F(x\_0) = a\_N \[NotEqual] 0\)]], ", ", Cell[BoxData[ \(TraditionalForm\`F(x)\)]], " has constant sign in a neighborhood of ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], " and it is straightforward to see that ", Cell[BoxData[ \(TraditionalForm\`N\)]], " must be even in order for ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " to have a local extremum at ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], ". Since the derivative ", Cell[BoxData[ \(TraditionalForm\`f' \((x)\)\)]], " is real-analytic and not identically zero, it is nonzero in some deleted \ neighborhood of ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], ". This implies that ", Cell[BoxData[ \(TraditionalForm\`f(x)\)]], " is strictly monotonic on the left and right parts of this deleted \ neighborhood separately. Using the notation ", Cell[BoxData[ \(TraditionalForm\`y = y(x)\)]], ", we have, for any ", Cell[BoxData[ \(TraditionalForm\`x \[Element] \([x\_0 - k, x\_0]\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`f(y) = \(\(f(g(f(x)))\)\(=\)\(f(x)\)\(\ \)\)\)]], " since ", Cell[BoxData[ \(TraditionalForm\`f(x) \[Element] \([x\_0, x\_0 + h]\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`f\[SmallCircle]g\)]], "is the identity function on this domain. Hence ", Cell[BoxData[ \(TraditionalForm\`a\_0 + \(\((y - x\_0)\)\^N\) \(F(y)\) = a\_0 + \(\((x - x\_0)\)\^N\) \(F(x)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\((y - x\_0)\)\^N\/\((x - x\_0)\)\^N = \ \(F(x)\)\/\(F(y)\)\)]], ", Taking ", Cell[BoxData[ \(TraditionalForm\`N\)]], "th roots of both sides and noting that ", Cell[BoxData[ \(TraditionalForm\`\(y - x\_0\)\/\(x - x\_0\)\)]], " is negative, we have ", Cell[BoxData[ \(TraditionalForm\`\(y - x\_0\)\/\(x - x\_0\) = \(-\@\(\(F(x)\)\/\(F(y)\ \)\)\%N\)\)]], ". The desired result, ", Cell[BoxData[ FormBox[ StyleBox[\(lim\+\(x \[Rule] \(x\_0\^-\)\)\(y(x) - x\_0\)\/\(x - \ x\_0\)\ = \ \(-1\)\), FontSize->16], TraditionalForm]]], ", follows directly by bringing the limit inside the radical on the right \ hand side.\n" }], "Text"], Cell[TextData[{ "By applying this theorem to the function ", Cell[BoxData[ \(TraditionalForm\`f(u) = u\ *\[ExponentialE]\^\(\(\ \)\(u\)\)\)]], " at ", Cell[BoxData[ \(TraditionalForm\`\(\(u\_0 = \(-1\)\)\(,\)\)\)]], " the inverse function ", Cell[BoxData[ \(TraditionalForm\`g\)]], " is Lambert W and ", Cell[BoxData[ \(TraditionalForm\`y(u) = W(u\ *\[ExponentialE]\^\(\(\ \)\(u\)\))\)]], " on intervals with right hand enpoint -1. This results in the following \ limit, which we will use twice in what follows.\n\n", StyleBox["Corollary.", FontVariations->{"Underline"->True}], " ", Cell[BoxData[ \(TraditionalForm\`lim\_\(u \[Rule] \ \(-\(1\^-\)\)\)\(W(u*\[ExponentialE]\^u) + 1\)\/\(u + 1\) = \(-1\)\)]], ".", StyleBox["\n", FontFamily->"Helvetica"], " " }], "Text"], Cell[TextData[{ "Returning to our projectile with resistance, we now seek a more analytical \ check of our work by investigating the limiting value for our range formula \ as the resist ance ", Cell[BoxData[ \(TraditionalForm\`k\)]], " goes to 0. Again setting ", Cell[BoxData[ \(TraditionalForm\`u = \(-1\) - \(\(k\ v\)\/g\) \(sin(\[Theta])\)\)]], " and noting that ", Cell[BoxData[ FormBox[ StyleBox[\(1\/k = \(-\ \(\(v\ \(sin(\[Theta])\)\)\/\(g(u + 1)\)\)\)\), FontSize->14], TraditionalForm]]], ", we have from our earlier range formula \n\n\t", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{\(R(\[Theta])\), "=", RowBox[{\(\(1\/k\) v\ \(cos(\[Theta])\) \((1 - \(W(u*\[ExponentialE]\^\(\(\ \ \)\(u\)\))\)\/u)\)\), "=", RowBox[{\(\(-\(\(\(v\^2\) \(sin(\[Theta])\) \(cos(\[Theta])\)\)\ \/\(g(1 + u)\)\)\) \((1 - \(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\)\/u)\)\), "=", RowBox[{\(v\^2\/g\), \(sin(\[Theta])\), \(cos(\[Theta])\), RowBox[{ FractionBox[ RowBox[{\(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\), "-", RowBox[{"u", Cell[""]}]}], \(u*\((u + 1)\)\)], "."}]}]}]}]}], FontSize->16], TraditionalForm]]], "\n\n\nSince ", Cell[BoxData[ \(TraditionalForm\`u\)]], " approaches -1 from below as ", Cell[BoxData[ \(TraditionalForm\`k\)]], " approaches 0 from above, the limit we seek to evaluate becomes\n\n\t", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(lim\_\(u \[Rule] \(-\(1\^-\)\)\)R(\[Theta])\), FontSize->16], StyleBox["=", FontSize->16], RowBox[{ StyleBox[ FractionBox[ StyleBox[\(v\^2\), FontSize->16], "g"], FontSize->16], StyleBox[\(sin(\[Theta])\), FontSize->16], StyleBox[\(cos(\[Theta])\), FontSize->16], StyleBox["*", FontSize->16], StyleBox[" ", FontSize->16], RowBox[{ StyleBox[\(lim\_\(u \[Rule] \(-\(1\^-\)\)\)\), FontSize->16], RowBox[{ StyleBox[ FractionBox[ RowBox[{\(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\), "-", RowBox[{"u", Cell[""]}]}], \(u(u + 1)\)], FontSize->16], " ", "."}]}]}]}], TraditionalForm]]], "\n\t\nWe now show that the limit on the right hand side is 2, so that our \ results are fully consistent with the no resistance case discussed earlier. \ Indeed,\n\t\n", Cell[BoxData[ FormBox[ StyleBox[ RowBox[{ RowBox[{\(lim\_\(u \[Rule] \(-\(1\^-\)\)\)\), FractionBox[ RowBox[{\(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\), "-", RowBox[{"u", Cell[""]}]}], \(u(u + 1)\)]}], "=", RowBox[{ RowBox[{\(lim\_\(u \[Rule] \(-\(1\^-\)\)\)\), FractionBox[ RowBox[{\(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\), "-", \((\(-1\))\), "-", RowBox[{\((u + 1)\), Cell[""]}]}], \(u(u + 1)\)]}], "=", RowBox[{ RowBox[{ RowBox[{\(lim\_\(u \[Rule] \(-\(1\^-\)\)\)\), FractionBox[ RowBox[{\(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\), "-", RowBox[{\((\(-1\))\), Cell[""]}]}], \(u(u + 1)\)]}], "+", "1"}], "=", RowBox[{ RowBox[{"-", RowBox[{\(lim\_\(u \[Rule] \(-\(1\^-\)\)\)\), FractionBox[ RowBox[{\(W(u*\[ExponentialE]\^\(\(\ \)\(u\)\))\), "-", RowBox[{\((\(-1\))\), Cell[ ""]}]}], \(u - \((\(-1\))\)\)]}]}], "+", "1."}]}]}]}], FontSize->16], TraditionalForm]]], "\n\nApplying the Corollary to Theorem 1, the final limit has a value of ", Cell[BoxData[ \(TraditionalForm\`\(-1\)\)]], "and an overall limit of 2 as claimed. The result is the no resistance \ range formula ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(R(\[Theta])\), FontSize->16], StyleBox["=", FontSize->16], RowBox[{ StyleBox[ RowBox[{ FractionBox[ StyleBox[\(v\^2\), FontSize->16], "g"], "*", "2", \(sin(\[Theta])\), \(cos(\[Theta])\)}], FontSize->16], StyleBox["=", FontSize->16], RowBox[{ StyleBox[\(v\^2\/g\), FontSize->18], StyleBox[\(sin(2 \[Theta])\), FontSize->14]}]}]}], TraditionalForm]]], ".\t\n\t\nIt is worth noting that ", StyleBox["Mathematica", FontSlant->"Italic"], " seems to offer a much more direct proof as it symbolically evaluates the \ limit we started with. The first command that follows defines Lambert's W \ function, built in to ", StyleBox["Mathematica", FontSlant->"Italic"], " as the function ProductLog. The second command then evaluates, \ symbollically, the desired one-sided limit." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[{ \(\(W = ProductLog;\)\), "\n", \(Limit[\(W[u*Exp[u]] - u\)\/\(u \((u + 1)\)\), u \[Rule] \(-1\), Direction \[Rule] 1]\)}], "Input", CellLabel->"In[8]:=", CellLabelAutoDelete->False, ShowCellTags->True], Cell[BoxData[ \(2\)], "Output", CellLabel->"Out[9]=", CellLabelAutoDelete->False] }, Open ]], Cell[TextData[{ "The authors are not aware of precisely what method ", StyleBox["Mathematica", FontSlant->"Italic"], " uses to evaluate this limit and we have not found an \"easy\" proof (this \ was our motivaton for Theorem 1). Despite our general enthusiasm for machine \ computation, we would not be comfortable with publication use of a \"", StyleBox["Mathematica", FontSlant->"Italic"], " proof\" at this time. It is interesting to speculate, however, on how \ mathematical standards for proof might change if and when \"experimental \ mathematics\" and symbolic computation become more widely appreciated and \ accepted. " }], "Text", PageBreakBelow->True] }, Open ]], Cell[CellGroupData[{ Cell["The Inverse Range Problem", "Section"], Cell[TextData[{ "In the spirit of Groetsch ", StyleBox["[4]", FontWeight->"Bold"], ", we now consider the inverse problem of what can be proved about angle(s) \ ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " that satisfy ", Cell[BoxData[ \(TraditionalForm\`R(\[Theta]) = r\)]], " for a given range value ", Cell[BoxData[ \(TraditionalForm\`r\)]], ". While finding a closed form inverse relationship does not seem feasible \ at this point (perhaps it awaits the invention of a few more functions), we \ can use the form of ", Cell[BoxData[ \(TraditionalForm\`R(\[Theta])\)]], " to prove much of what is apparent from our graphs and our physical \ intuition. The notation and the lemma that follow will streamline our efforts \ in this regard, which involve extensive use of the chain rule and some \ surprisingly compact and helpful identities. \n\n", StyleBox["Notation.", FontVariations->{"Underline"->True}], "\n", StyleBox[" ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`u\ \ = \ \(-1\) - \(v\/g\) k\ \ sin\ \((\[Theta])\)\)]], " \n ", Cell[BoxData[ \(TraditionalForm\`y\ \ = \ W(u\ e\^u)\)]], " \n ", Cell[BoxData[ \(TraditionalForm\`R\ \ = \ v\ \(cos(\ \[Theta]\ )\) \(1\/k\) \((1 - y\/u)\)\)]], ", where we treat ", Cell[BoxData[ \(TraditionalForm\`k\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " as the independent variables.\n \n", StyleBox["Lemma.", FontVariations->{"Underline"->True}], StyleBox[" \n \n", FontVariations->{"CompatibilityType"->0}], "(a) ", Cell[BoxData[ \(TraditionalForm\`\(-1\) < y < 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`0 < y\/u < 1\)]], " for all ", Cell[BoxData[ \(TraditionalForm\`0 < \[Theta] < \[Pi]\/2\)]], " and hence for all ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ".", StyleBox["\n\n(b) ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ FormBox[ RowBox[{\(\(d\ y\)\/\(d\ u\)\), "=", RowBox[{\(\(\((u + 1)\) y\)\/\(\((y + 1)\) u\)\), StyleBox["=", FontSize->16], \(\((1 + 1\/u)\)\/\((1 + 1\/y)\) < 0\)}]}], TraditionalForm]]], " for all ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ".\n\n(c) For ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(d\)\(\ \)\)\^2\) y\)\/\(d\ u\^\(\(\ \ \)\(2\)\)\) < 0\)]], " and ", Cell[BoxData[ FormBox[Cell[TextData[Cell[BoxData[ FormBox[ RowBox[{\(-1\), "<", FormBox[\(\(d\ y\)\/\(d\ u\) < 0\), "TraditionalForm"]}], TraditionalForm]]]]], TraditionalForm]]], ".", StyleBox["\n\n(d) ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`d\/\(d\ u\)\ \((1 - y\/u)\)\ = \(-\ \(1\/\(u + 1\)\)\)\ \(d\ y\)\/\(d\ u\)\ \((1 - y\/u)\)\)]], StyleBox["\n\n(e) ", FontVariations->{"CompatibilityType"->0}], Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_k R\ \ = \(-\(1\/k\^2\)\)\ v\ \(cos(\ \[Theta])\)\ \((1 - y\/u)\) \((1 + \(d\ y\)\/\(d\ u\))\)\)]], StyleBox["\n\n(f) ", FontVariations->{"CompatibilityType"->0}], " ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta] R\ \ = \(\(-\(v\/k\)\) sin\ \((\ \[Theta])\)\ \((1 - y\/u)\)\ \((1 + \ \(d\ y\)\/\(d\ u\)\ \(\(cos\^2\)(\ \ \[Theta])\)\/\(sin\^2\ \((\ \[Theta])\)\))\) = \(-\(v\/k\)\) sin\ \((\ \[Theta])\)\ \(\(cot\^2\)(\ \[Theta])\)\ \((1 - y\/u)\)\ \(\((\(tan\^2\)(\ \[Theta]) + \ \(d\ y\)\/\(d\ \ u\))\)\(.\)\)\)\)]], "\n\n\n", StyleBox["Proofs.", FontVariations->{"Underline"->True}], "\n\n(a) Since ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"-", FormBox[\(1\/\[ExponentialE] < u\ \[ExponentialE]\^u\), "TraditionalForm"]}], "<", "0"}], TraditionalForm]]], " for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ", we get ", Cell[BoxData[ \(TraditionalForm\`\(-1\) < W(u\ \[ExponentialE]\^u) = y < 0\)]], " and consequently ", Cell[BoxData[ \(TraditionalForm\`0 < y\/u < 1\)]], ".\n\n(b) Since ", Cell[BoxData[ \(TraditionalForm\`W\)]], " is the inverse to ", Cell[BoxData[ \(TraditionalForm\`T(u) = u\ \[ExponentialE]\^u\)]], " on ", Cell[BoxData[ \(TraditionalForm\`\([\(-1\), \ \[Infinity])\)\)]], ", for any ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], "we get ", Cell[BoxData[ \(TraditionalForm\`T(y) = T(W(u\ \[ExponentialE]\^u))\)]], ", which says ", Cell[BoxData[ \(TraditionalForm\`y\ \[ExponentialE]\^y = u\ \[ExponentialE]\^u\)]], ". Implicit differentiation then gives ", Cell[BoxData[ FormBox[ RowBox[{ StyleBox[\(\(d\ y\)\/\(d\ u\)\), FontSize->16], StyleBox["=", FontSize->16], RowBox[{ StyleBox[\(\(\((u + 1)\) e\^u\)\/\(\((y + 1)\) e\^y\)\), FontSize->16], StyleBox["=", FontSize->16], RowBox[{ StyleBox[\(\(\((u + 1)\) y\)\/\(\((y + 1)\) u\)\), FontSize->16], "."}]}]}], TraditionalForm]]], " Using ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], " and (a), we see that ", Cell[BoxData[ FormBox[ StyleBox[\(\(d\ y\)\/\(d\ u\)\), FontSize->16], TraditionalForm]]], " has 3 negative factors and a positive factor and must be negative.\n\n(c) \ Note that by the Corollary to Theorem 1, ", Cell[BoxData[ \(TraditionalForm\`lim\+\(x \[Rule] \(-\(1\^-\)\)\)\(y + 1\)\/\(u + 1\)\ \ = \ \(-1\)\)]], ". Using (b), ", Cell[BoxData[ \(TraditionalForm\`lim\+\(u \[Rule] \(-\(1\^-\)\)\)\(d\ y\)\/\(d\ u\) = \ \(lim\+\(u \[Rule] \(-\(1\^-\)\)\)\((\(u + 1\)\/\(y + 1\))\) \(lim\+\(x \ \[Rule] \(-\(1\^-\)\)\)\((y\/u)\)\)\ = \ \(-1\)\)\)]], ". Now consider the second derivative, which we also obtain by implicit \ differentiation ( of ", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ u\) = \((1 + 1\/x)\)\/\((1 + \ 1\/y)\)\)]], ") and substitution back in for ", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ u\)\)]], ", ", Cell[BoxData[ FormBox[ StyleBox[\(\(\(\(\(d\)\(\ \)\)\^2\) y\)\/\(d\ u\^\(\(\ \)\(2\)\)\) = \ \(\(\(-\(1\/u\^2\)\) \((1 + 1\/y)\) + \(1\/y\^2\) \((1 + 1\/u)\) y'\)\/\((1 + \ 1\/y)\)\^2 = \(\(\(-\(1\/u\^2\)\) \((1 + 1\/y)\)\^2 + \(1\/y\^2\) \((1 + \ 1\/u)\)\^2\)\/\((1 + 1\/y)\)\^3 = \(\((u + 1)\)\^2 - \((y + \ 1)\)\^2\)\/\(\(u\^2\) \(\(y\^2\)(1 + 1\/y)\)\^3\)\)\)\), FontSize->14], TraditionalForm]]], ". Note that the denominator is negative for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ". To prove that the numerator is always positive, let ", Cell[BoxData[ \(TraditionalForm\`G(u) = \((u + 1)\)\^2 - \((y + 1)\)\^2\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`G' \((u)\) = \(2 \((u + 1)\) - 2 \((y + 1)\) y' = \(2 \((u + 1)\) - 2 \((y + 1)\) \((u + 1)\) y\/\(\((y + 1)\) u\) = 2 \((u + 1)\) \((1 - y\/u)\)\)\)\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`G' \((u)\)\)]], " is easily seen to be negative for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`G(\(-1\)) = 0\)]], ", this implies ", Cell[BoxData[ \(TraditionalForm\`G(u) > 0\)]], " for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ". Thus the numerator of ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(d\)\(\ \)\)\^2\) y\)\/\(d\ u\^\(\(\ \ \)\(2\)\)\)\)]], " is always positive. Thus ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(d\)\(\ \)\)\^2\) y\)\/\(d\ u\^\(\(\ \ \)\(2\)\)\)\)]], " is always negative, for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ". Finally, since ", Cell[BoxData[ \(TraditionalForm\`lim\+\(u \[Rule] \(-\(1\^-\)\)\)\(d\ y\)\/\(d\ u\)\ \ = \ \(-1\)\)]], "and ", Cell[BoxData[ \(TraditionalForm\`\(\(\(\(d\)\(\ \)\)\^2\) y\)\/\(d\ u\^\(\(\ \ \)\(2\)\)\)\)]], "is negative for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ", we have ", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ u\) > \(-1\)\)]], " for ", Cell[BoxData[ \(TraditionalForm\`u < \(-1\)\)]], ".\n\n(d) By product rule, we have ", Cell[BoxData[ \(TraditionalForm\`d\/\(d\ u\)\ \((1 - y\/u)\)\ = \ \(-\ \(1\/u\)\)\ \(d\ y\)\/\(d\ u\) + \(\(y\/u\^2\ \)\(.\)\)\)]], " From (b), ", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ u\) = \(\((u + 1)\) y\)\/\(\((y + \ 1)\) u\)\)]], ", so ", Cell[BoxData[ \(TraditionalForm\`y\/u\ = \((y + 1)\)\/\((u + 1)\)\ \(d\ y\)\/\(d\ \ u\)\)]], ". Thus we have ", Cell[BoxData[ \(TraditionalForm\`d\/\(d\ u\)\ \((1 - y\/u)\)\ = \(\(-\ \(1\/u\)\)\ \(d\ y\)\/\(d\ u\) + \(1\/u\) \ \((y + 1)\)\/\((u + 1)\)\ \(d\ y\)\/\(d\ u\) = \(\(-\ \(1\/u\)\) \(\(d\ y\)\/\ \(d\ u\)\) \((1 - \(y + 1\)\/\(u + 1\))\) = \(\(-\ \(1\/u\)\) \(\(d\ y\)\/\(d\ \ u\)\) \((\(u - y\)\/\(u + 1\))\) = \(-\ \(1\/\(u + 1\)\)\) \(\(d\ y\)\/\(d\ u\)\) \(\((1 - y\/u)\)\(.\)\)\)\)\)\)]], " \n\n(e) By product rule and chain rule, we have ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_k\ \((\ v\ cos\ \[Theta]\ \(1\/k\) \((1 - y\/u)\))\)\ = \ v\ cos\ \(\(\[Theta]\ [\(-\ \(1\/k\^2\)\) \((1 - y\/u)\)\ + \ 1\/k\ \(\(\(d\)\(\ \)\)\/\(d\ u\)\) \((1 - y\/u)\)\ \[PartialD]\_k\ u]\)\(.\)\)\)]], " Using (d) and the fact that ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_k\ u\ = \(\(\(-\(v\/g\)\)\ sin\ \[Theta]\)\(\ \)\(=\)\(\ \)\(\(u + 1\)\/k\)\(\ \)\)\)]], ", we have ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_k\ \((\ v\ cos\ \[Theta]\ \(1\/k\) \((1 - y\/u)\))\)\ = \ \(v\ cos\ \[Theta]\ [\(-\ \(1\/k\^2\)\) \ \((1 - y\/u)\)\ - \ \(1\/k\) 1\/\(u + 1\)\ \(d\ y\)\/\(d\ u\)\ \((1 - y\/u)\)\ \(u + 1\)\/k] = v\ cos\ \(\[Theta]\ [\(-\ \(1\/k\^2\)\) \((1 - y\/u)\)\ ]\)[ 1 + \ \(d\ y\)\/\(d\ u\)]\)\)]], " as claimed.\n\n(f) By product rule and chain rule, we have ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta]\ \((\ v\ cos\ \[Theta]\ \(1\/k\) \((1 - y\/u)\))\)\ = \ \(\(\(v\/k\)\ [\(-\ sin\)\ \[Theta]\ \((1 - y\/u)\) + \ cos\ \[Theta]\ \(\(\(d\)\(\ \)\)\/\(d\ u\)\) \((1 - y\/u)\)\ \[PartialD]\_\[Theta]\ u]\)\(.\)\)\)]], " Using (d) and the fact that ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta]\ u\ = \(\(-\(v\/g\)\)\ k\ cos\ \[Theta]\ = \ \(\((u + 1)\) \(\(cos\ \[Theta]\)\/\(sin\ \[Theta]\)\)\(\ \)\)\)\)]], ", we have ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta]\ \((\ v\ cos\ \[Theta]\ \(1\/k\) \((1 - y\/u)\))\)\ = \ \(\(v\/k\)\ [\(-\ sin\)\ \[Theta]\ \((1 - y\/u)\) + \ cos\ \[Theta]\ \(-1\)\/\(u + 1\)\ \(d\ y\)\/\(d\ u\)\ \((1 - y\/u)\)\ \((u + 1)\)\ \(cos\ \[Theta]\)\/\(sin\ \[Theta]\)] = \(-\(v\/k\)\ \) sin\ \[Theta]\ \((1 - y\/u)\)\ [ 1 + \ \(d\ y\)\/\(d\ u\)\ \(\(cos\^2\) \[Theta]\)\/\(sin\^2\ \ \[Theta]\)]\)\)]], " and the second form follows directly.\n\n", StyleBox["Theorem 2.", FontVariations->{"Underline"->True}], " For fixed angle ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " the range values decrease as the resistance ", Cell[BoxData[ \(TraditionalForm\`k\)]], " increases. Also, for any resistance ", Cell[BoxData[ \(TraditionalForm\`k > 0\)]], " there is a maximum range value, M, that occurs for a unique angle ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max < \[Pi]\/4\)]], ". Finally, for all nonnegative ranges ", Cell[BoxData[ \(TraditionalForm\`r < M\)]], ", there are exactly two angles ", Cell[BoxData[ \(TraditionalForm\`\[Theta] \[Element] \([0, \[Pi]\/2]\)\)]], " that satisfy ", Cell[BoxData[ \(TraditionalForm\`R(\[Theta]) = r\)]], ", one on either side of ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], ". \n\n", StyleBox["Proof.", FontVariations->{"Underline"->True}], " By parts (e) and (a) of the lemma, the partial of ", Cell[BoxData[ \(TraditionalForm\`R\)]], " with respect to ", Cell[BoxData[ \(TraditionalForm\`k\)]], " is negative, so the range decreases with increasing ", Cell[BoxData[ \(TraditionalForm\`k\)]], ". Looking at the (", Cell[BoxData[ \(TraditionalForm\`\(tan\^2\)(\ \[Theta]) + \ \(d\ y\)\/\(d\ u\)\)]], ") factor of ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta] R\)]], " given in part (f) of the lemma, we note that ", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ u\)\)]], " is an increasing function of ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " by (b) of the lemma (", Cell[BoxData[ \(TraditionalForm\`\(d\^2\ y\)\/\(d\ u\^2\) < 0\)]], ") and the fact that ", Cell[BoxData[ FormBox[Cell["u"], TraditionalForm]]], " is a decreasing function of ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], ". Since ", Cell[BoxData[ \(TraditionalForm\`tan\^2\)]], "is also increasing in ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], ", ", Cell[BoxData[ \(TraditionalForm\`\((\(tan\^2\) \[Theta] + \ \(d\ y\)\/\(d\ \ u\))\)\)]], " is an increasing function of \[Theta]. Since ", Cell[BoxData[ \(TraditionalForm\`\((\(tan\^2\) \[Theta] + \ \(d\ y\)\/\(d\ \ u\))\)\)]], " goes to ", Cell[BoxData[ \(TraditionalForm\`\(-1\)\)]], " as ", Cell[BoxData[ \(TraditionalForm\`\[Theta] \[Rule] \(0\^+\)\)]], "and goes to \[Infinity] as ", Cell[BoxData[ \(TraditionalForm\`\[Theta] \[Rule] \(\((\[Pi]\/2)\)\^-\)\)]], ", we conclude that ", Cell[BoxData[ \(TraditionalForm\`\((\(tan\^2\) \[Theta] + \ \(d\ y\)\/\(d\ u\))\) = 0\)]], " at a unique ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_0\)]], " strictly between 0 and ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ". Noting that for ", Cell[BoxData[ \(TraditionalForm\`0 < \[Theta] < \[Pi]\/2\)]], ", we have ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta]\ R = 0\)]], " if and only if ", Cell[BoxData[ \(TraditionalForm\`\((\(tan\^2\) \[Theta] + \ \(d\ y\)\/\(d\ u\))\) = 0\)]], " and we use ", Cell[BoxData[ \(TraditionalForm\`\(-1\) < \(d\ y\)\/\(d\ u\) < 0\)]], " (part (c) of the lemma) to conclude that the unique ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " where ", Cell[BoxData[ \(TraditionalForm\`\((\(tan\^2\) \[Theta]\_max + \ \(d\ y\)\/\(d\ \ u\))\) = 0\)]], " must satisfy ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max < \(\(\[Pi]\/4\)\(.\)\)\)]], " The fact that this unique ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " maximizes ", Cell[BoxData[ \(TraditionalForm\`R\)]], " follows since ", Cell[BoxData[ FormBox[ FormBox[\(\(\[PartialD]\_\[Theta] R\)\(\ \)\), "TraditionalForm"], TraditionalForm]]], "is positive to the left of ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " and negative to its right. That there are exactly two elevation angles \ that achieve any submaximal range also follows from what was just said about \ the signs of ", Cell[BoxData[ \(TraditionalForm\`\[PartialD]\_\[Theta] R\)]], " and the fact that ", Cell[BoxData[ \(TraditionalForm\`R\)]], " is zero when ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = 0\)]], " and when ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Pi]\/2\)]], "." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["A Closed Form for the Maximum Range Angle", "Section"], Cell[TextData[{ "The efforts of the previous section provide a surprising additional \ dividend. Using some interesting algebra and our old friend Lambert W, we are \ able to find a formula for the angle that maximizes the range for given \ values of the resistance ", Cell[BoxData[ \(TraditionalForm\`k\)]], ", the initial velocity ", Cell[BoxData[ \(TraditionalForm\`v\)]], ", and the acceleration of gravity ", Cell[BoxData[ \(TraditionalForm\`g\)]], ". In fact, the key parameter turns out to be ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = \(k\ v\)\/g\)]], ", a notation we will use for the remainder of this section. A similar \ result using a \"log-like\" function has recently been offered by Groetsch \ [", StyleBox["6", FontWeight->"Bold"], "]. With s", "ome further analysis are able to show that the maximum range function is \ continuous and decreasing in ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Alpha]\)\(.\)\)\)]], " \n\n", StyleBox["Theorem 3.", FontVariations->{"Underline"->True}], " The angle ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " that gives the maximum range for a given ", Cell[BoxData[ \(TraditionalForm\`v, g, k\)]], " depends only on ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = \(k\ v\)\/g\)]], " and is given by the formula ", Cell[BoxData[ FormBox[ RowBox[{\(\[Theta]\_max\), "=", RowBox[{"ArcSin", "[", FractionBox[ StyleBox[\(\[Alpha]W(\(\[Alpha]\^2 - 1\)\/\[ExponentialE])\), FontSize->16], StyleBox[\(\[Alpha]\^2 - 1 - W(\(\[Alpha]\^2 - 1\)\/\[ExponentialE])\), FontSize->16]], "]"}]}], TraditionalForm]]], " for ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[NotEqual] 1\)]], ", and ", Cell[BoxData[ FormBox[ RowBox[{\(\[Theta]\_max\), "=", RowBox[{"ArcSin", "[", StyleBox[\(1\/\(\[ExponentialE] - 1\)\), FontSize->16], "]"}]}], TraditionalForm]]], " for ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 1\)]], " . \n\n", StyleBox["Proof.", FontVariations->{"Underline"->True}], " ", StyleBox[" ", FontWeight->"Bold"], "For simplicity, just write ", Cell[BoxData[ \(TraditionalForm\`\[Theta] = \[Theta]\_max\)]], ". From Theorem 2, we have ", Cell[BoxData[ \(TraditionalForm\`\(\(sin\^2\) \[Theta]\)\/\(\(cos\^2\) \[Theta]\) + \ \ \(d\ y\)\/\(d\ u\) = 0\)]], ". Using ", Cell[BoxData[ \(TraditionalForm\`\(d\ y\)\/\(d\ u\) = \(\((u + 1)\) y\)\/\(u(y + 1)\)\ \)]], " and ", Cell[BoxData[ \(TraditionalForm\`sin\ \[Theta] = \(-\(\(u + 1\)\/\[Alpha]\)\)\)]], ", this can be rewritten as ", Cell[BoxData[ \(TraditionalForm\`\(\(1\/\[Alpha]\^2\) \((u + 1)\)\^2\)\/\(1 - \(1\/\ \[Alpha]\^2\) \((u + 1)\)\^2\) = \(-\(\(\((u + 1)\) y\)\/\(u( y + 1)\)\)\)\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(u(u + 1)\)\/\(\[Alpha]\^2 - \((u + \ 1)\)\^2\) = \(-\(y\/\((y + 1)\)\)\)\)\)\)]], ". Taking reciprocals, ", Cell[BoxData[ \(TraditionalForm\`\(\(\ \)\(\(\[Alpha]\^2 - \((u + 1)\)\^2\)\/\(u(u + \ 1)\) = \(-\(\(y + 1\)\/y\)\)\)\)\)]], ". Adding 1 to both sides, ", Cell[BoxData[ \(TraditionalForm\`\(\[Alpha]\^2 - \((u + 1)\)\^2 + u(u + 1)\)\/\(u(u + \ 1)\) = \(-\(1\/y\)\)\)]], ". So ", Cell[BoxData[ \(TraditionalForm\`\(\[Alpha]\^2 - \((u + 1)\)\)\/\(u(u + 1)\) = \(-\(1\ \/y\)\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`y = \(-\(\(u(u + 1)\)\/\(\[Alpha]\^2 - 1 - u\)\)\)\)]], ". Now, using ", Cell[BoxData[ \(TraditionalForm\`y\ e\^y = u\ e\^u\)]], ", we get ", Cell[BoxData[ \(TraditionalForm\`\(-\(\(u(u + 1)\)\/\(\[Alpha]\^2 - 1 - u\)\)\) e\^\(-\(\(u(u + 1)\)\/\(\[Alpha]\^2 - 1 - u\)\)\) = u\ e\^u\)]], ". Then ", Cell[BoxData[ \(TraditionalForm\`\(-\(\((u + 1)\)\/\(\[Alpha]\^2 - 1 - u\)\)\) = e\^\(u + \(u(u + 1)\)\/\(\[Alpha]\^2 - 1 - u\)\)\)]], ", and so ", Cell[BoxData[ \(TraditionalForm\`\(-\(\((u + 1)\)\/\(\[Alpha]\^2 - 1 - u\)\)\) = e\^\(\(\(\[Alpha]\^2\) u\)\/\(\[Alpha]\^2 - 1 - u\)\)\)]], ". \n\nCase ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 1\)]], ": We have ", Cell[BoxData[ \(TraditionalForm\`\(u + 1\)\/u = \[ExponentialE]\^\(-1\)\)]], ". Solving for ", Cell[BoxData[ \(TraditionalForm\`u\)]], " leads to ", Cell[BoxData[ FormBox[ RowBox[{"u", "=", StyleBox[\(\(-\[ExponentialE]\)\/\(\[ExponentialE] - 1\)\), FontSize->16]}], TraditionalForm]]], ". Then ", Cell[BoxData[ FormBox[ RowBox[{\(sin\ \[Theta]\_max\), "=", RowBox[{\(-\(\(u + 1\)\/\[Alpha]\)\), "=", StyleBox[\(1\/\(\[ExponentialE] - 1\)\), FontSize->16]}]}], TraditionalForm]]], ", as claimed.\n\nCase ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[NotEqual] 1\)]], ": To solve this for ", Cell[BoxData[ \(TraditionalForm\`u\)]], ", make the substitution ", Cell[BoxData[ \(TraditionalForm\`t = 1\/\(\[Alpha]\^2 - 1 - u\)\)]], " (and so ", Cell[BoxData[ \(TraditionalForm\`u = \[Alpha]\^2 - 1 - 1\/t\)]], ") to get ", Cell[BoxData[ FormBox[ RowBox[{\(\(-\((\[Alpha]\^2 - 1\/t)\)\) t\), "=", SuperscriptBox[ StyleBox["\[ExponentialE]", FontSize-> 14], \(\(\ \)\(\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1 - 1\/t)\) t\)\)]}], TraditionalForm]]], ". So ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\(\[Alpha]\^2\) t\), "-", "1", "+", SuperscriptBox[ StyleBox["\[ExponentialE]", FontSize->14], \(\(\ \)\(\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\) t - \[Alpha]\^2\)\)]}], "=", "0"}], TraditionalForm]]], ". This equation is in the form given in result (3) of the section where \ the Lambert W function was introduced with ", Cell[BoxData[ FormBox[ RowBox[{\(a = \[Alpha]\^2\), ",", " ", \(b = \(-1\)\), ",", " ", RowBox[{"c", "=", SuperscriptBox[ StyleBox["\[ExponentialE]", FontSize->14], \(\(\ \)\(-\[Alpha]\^2\)\)]}], ",", " ", \(d = \(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\)}], TraditionalForm]]], ". Result (3) therefore gives ", Cell[BoxData[ FormBox[ RowBox[{"t", "=", RowBox[{\(1\/\[Alpha]\^2 - \(1\/\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\)\) \(W(\(\(e\^\(-\[Alpha]\^2\)\) \ \(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\) e\^\(\(\(\[Alpha]\^2\)(\[Alpha]\^2 - \ 1)\)\/\[Alpha]\^2\)\)\/\[Alpha]\^2)\)\), "=", RowBox[{\(1\/\[Alpha]\^2\), "-", RowBox[{\(1\/\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\)\), RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}]}]}]}], TraditionalForm]]], ". Using ", Cell[BoxData[ \(TraditionalForm\`u = \[Alpha]\^2 - 1 - 1\/t\)]], " again, we get ", Cell[BoxData[ FormBox[ RowBox[{"u", "=", RowBox[{ RowBox[{\(\[Alpha]\^2\), "-", "1", "-", FractionBox["1", RowBox[{\(1\/\[Alpha]\^2\), "-", RowBox[{\(1\/\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\)\), RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}]}]]}], "=", RowBox[{\(\[Alpha]\^2\), "-", "1", "-", FractionBox[\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\), RowBox[{\(\[Alpha]\^2\), "-", "1", "-", RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}]]}]}]}], TraditionalForm]]], ". Then ", Cell[BoxData[ FormBox[ RowBox[{\(sin\ \[Theta]\), "=", RowBox[{\(-\(\(u + 1\)\/\[Alpha]\)\), "=", RowBox[{\(-\(1\/\[Alpha]\)\), RowBox[{"(", RowBox[{\(\[Alpha]\^2\), "-", FractionBox[\(\(\[Alpha]\^2\)(\[Alpha]\^2 - 1)\), RowBox[{\(\[Alpha]\^2\), "-", "1", "-", RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}]]}], ")"}]}]}]}], TraditionalForm]]], ", which yields the formula ", Cell[BoxData[ FormBox[ RowBox[{\(sin(\ \[Theta])\), "=", FractionBox[ RowBox[{"\[Alpha]", " ", RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}], RowBox[{\(\[Alpha]\^2\), "-", "1", "-", RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}]]}], TraditionalForm]]], ", as desired.\n\n\nIn Figure 6 we show a plot of the maximum angle \ function as a function of \[Alpha]. 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0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW 0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW 0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW 0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW 0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00o`3ooooW0?ooo`00\ \>"], ImageRangeCache->{{{179.438, 438.125}, {615.125, 455.625}} -> {-74.3567, \ 3.49944, 0.256581, 0.0065212}}], Cell[TextData[{ "\t\t\t Figure 6.", StyleBox[" Maximum elevation angle as a function of ", FontWeight->"Plain"], Cell[BoxData[ FormBox[ StyleBox[\(\[Alpha] = \(k\ v\)\/g\), FontWeight->"Plain"], TraditionalForm]]], "\n" }], "Text", FontWeight->"Bold"], Cell[TextData[{ "We close this section by establishing two properties of the ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " function. The first is the physically nonsurprising fact that this \ function is continuous for all ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[GreaterEqual] 0\)]], ". Inspecting the formula for", StyleBox[" ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], StyleBox[" ", FontWeight->"Bold"], "from Theorem 3, it suffices to show that ", StyleBox[" ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], StyleBox[" ", FontWeight->"Bold"], "is continuous at ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 1\)]], ". First we need to show that ", Cell[BoxData[ \(TraditionalForm\`lim\+\(\[Alpha] \[Rule] \(0\^+\)\)\[Theta]\_max = \ \[Pi]\/4\)]], ". The key is that if we write ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{"W", "=", RowBox[{"W", "(", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12], ")"}]}], ","}], TraditionalForm]]], " then ", Cell[BoxData[ FormBox[ RowBox[{\(W\ e\^W\), "=", StyleBox[ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], FontSize->12]}], TraditionalForm]]], " (using result (1) for Lambert W). Now expand the left side as a power \ series about ", Cell[BoxData[ \(TraditionalForm\`W = \(-1\)\)]], ". (This power series converges for all real numbers ", Cell[BoxData[ \(TraditionalForm\`W\)]], ".) So ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ RowBox[{ RowBox[{"-", FractionBox["1", StyleBox["\[ExponentialE]", FontSize->14]]}], "+", RowBox[{ FractionBox["1", RowBox[{"2", StyleBox["\[ExponentialE]", FontSize->14]}]], \(\((W + 1)\)\^2\)}], "+", RowBox[{ FractionBox["2", RowBox[{"6", StyleBox["\[ExponentialE]", FontSize->14]}]], \(\((W + 1)\)\^3\)}], "+"}], "..."}], "=", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]]}], TraditionalForm]]], ". Thus ", Cell[BoxData[ \(TraditionalForm\`\(1\/2\) \((W + 1)\)\^2 + \(2\/6\) \((W + 1)\)\^3 + \ .. = \[Alpha]\^2\)]], ". If we denote ", Cell[BoxData[ \(TraditionalForm\`H(W) = 1\/2 + \(2\/6\) \((W + 1)\) + ... \)]], ", then ", Cell[BoxData[ \(TraditionalForm\`H(W)\)]], " also converges for all real numbers, and in particular is continuous at \ ", Cell[BoxData[ \(TraditionalForm\`W = \(-1\)\)]], ". Note ", Cell[BoxData[ \(TraditionalForm\`\[Alpha]\^2 = \(\((W + 1)\)\^2\) \(H(W)\)\)]], ". From this, we get ", Cell[BoxData[ \(TraditionalForm\`lim\+\(\[Alpha] \[Rule] \(0\^+\)\)\(W + \ 1\)\/\[Alpha] = \(\@\(lim\+\(\[Alpha] \[Rule] \(0\^+\)\)\((W + 1)\)\^2\/\ \[Alpha]\^2\) = \(\@\(lim\+\(W \[Rule] \(-\(1\^+\)\)\)\((W + \ 1)\)\^2\/\(\(\((W + 1)\)\^2\) \(\(H(W)\)\(.\)\)\)\) = \(\@\(lim\+\(W \[Rule] \ \(-\(1\^+\)\)\)1\/\(\(H(W)\)\(.\)\)\) = \@2\)\)\)\)]], ". Then we have ", Cell[BoxData[ FormBox[ RowBox[{\(lim\+\(\[Alpha] \[Rule] \(0\^+\)\)sin(\ \[Theta]\_max)\), "=", RowBox[{ RowBox[{\(lim\+\(\[Alpha] \[Rule] \(0\^+\)\)\), FractionBox[\(\[Alpha]W(\(\[Alpha]\^2 - 1\)\/e)\), RowBox[{\(\[Alpha]\^2\), "-", "1", "-", RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]}]]}], "=", RowBox[{ RowBox[{\(lim\+\(\[Alpha] \[Rule] \(0\^+\)\)\), FractionBox[ RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}], RowBox[{"\[Alpha]", "-", FractionBox[ RowBox[{"1", "+", RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]}], "\[Alpha]"]}]]}], "=", \(\(-1\)\/\(0 - \@2\) = 1\/\@2\)}]}]}], TraditionalForm]]], ". This implies ", Cell[BoxData[ \(TraditionalForm\`lim\+\(\[Alpha] \[Rule] \(0\^+\)\)\[Theta]\_max = \ \[Pi]\/4\)]], ", which is what we expect as the resistance ", Cell[BoxData[ \(TraditionalForm\`k \[Rule] 0. \)]], " We must also show that ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(lim\+\(\[Alpha] \[Rule] 1\)\), " ", FractionBox[ RowBox[{"\[Alpha]", " ", RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]}], RowBox[{\(\[Alpha]\^2\), "-", "1", "-", RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]}]]}], "=", FractionBox["1", RowBox[{ StyleBox["\[ExponentialE]", FontSize->14], "-", "1"}]]}], TraditionalForm]]], " . This can be done by rewriting the limit as ", Cell[BoxData[ FormBox[ RowBox[{\(lim\+\(\[Alpha] \[Rule] 1\)\), " ", FractionBox["\[Alpha]", RowBox[{ FractionBox[\(\[Alpha]\^2 - 1\), RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]], "-", "1"}]]}], TraditionalForm]]], " and using L'Hopitals Rule on the part ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(lim\+\(\[Alpha] \[Rule] 1\)\), " ", FractionBox[\(\[Alpha]\^2 - 1\), RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]]}], "=", RowBox[{ RowBox[{\(lim\+\(\[Alpha] \[Rule] 1\)\), " ", FractionBox[\(2 \[Alpha]\), RowBox[{\(W'\), RowBox[{"(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}], "2", FractionBox["\[Alpha]", StyleBox["\[ExponentialE]", FontSize->14]]}]]}], "=", RowBox[{ FractionBox[ StyleBox["\[ExponentialE]", FontSize->14], \(W' \((0)\)\)], "=", RowBox[{ FractionBox[ StyleBox["\[ExponentialE]", FontSize->14], "1"], "=", StyleBox["\[ExponentialE]", FontSize->14]}]}]}]}], TraditionalForm]]], ". Thus ", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(lim\+\(\[Alpha] \[Rule] 1\)\), " ", FractionBox["\[Alpha]", RowBox[{ FractionBox[\(\[Alpha]\^2 - 1\), RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]], "-", "1"}]]}], "=", FractionBox["1", RowBox[{ StyleBox["\[ExponentialE]", FontSize->14], "-", "1"}]]}], TraditionalForm]]], ", as desired. We note in passing that ", StyleBox["Mathematica", FontSlant->"Italic"], " had no trouble (other than being \"slow\") in computing ", Cell[BoxData[ \(TraditionalForm\`lim\+\(\[Alpha] \[Rule] \(0\^+\)\)sin(\ \ \[Theta]\_max) = 1\/\@2\)]], ", but was unable to evaluate ", Cell[BoxData[ \(TraditionalForm\`lim\+\(\[Alpha] \[Rule] 1\)\(\(sin(\ \ \[Theta]\_max)\)\(.\)\)\)]], "\n\n\nThe second property of ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " confirms our physical intuition that increasing either the resistance or \ the initial velocity will decrease the maximum range angle. More precisely \ (and pleasingly), ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], " is a decreasing function of ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Alpha]\)\(.\)\)\)]], " To show this, it is convenient to look at the reciprocal ", Cell[BoxData[ \(TraditionalForm\`1\/\(sin(\ \[Theta]\_max)\) = \(\[Alpha]\^2 - \ 1\)\/\(\[Alpha]\ W\) - 1\/\[Alpha]\)]], ". The derivative of this (again using ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " for ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], ") is \n", Cell[BoxData[ FormBox[ RowBox[{\(\(d\/\(d\ \[Alpha]\)\) \((1\/\(sin(\ \[Theta])\))\)\), "=", RowBox[{ RowBox[{ RowBox[{\(1\/\(\(\[Alpha]\^2\) W\^2\)\), "[", RowBox[{\(2 \[Alpha]\ \[Alpha]\ W\), "-", RowBox[{\((\[Alpha]\^2 - 1)\), RowBox[{"(", RowBox[{"W", "+", RowBox[{"\[Alpha]", FractionBox["W", RowBox[{ FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], \((1 + W)\)}]], "2", FractionBox["\[Alpha]", StyleBox["\[ExponentialE]", FontSize->14]]}]}], ")"}]}]}], "]"}], "+", \(1\/\[Alpha]\^2\)}], "=", \(\(\(1\/\(\(\[Alpha]\^2\) \(\(W\^2\)(1 + W)\)\)\)[ 2 \( \[Alpha]\^2\) W\^2 + W + 1]\)\(.\)\)}]}], TraditionalForm]]], " Note that we used the fact ", Cell[BoxData[ FormBox[ RowBox[{\(\(d\ W\)\/\(d\ \[Alpha]\)\), "=", RowBox[{ RowBox[{\(\(\(d\)\(\ \)\)\/\(d\ \[Alpha]\)\), RowBox[{"W", "(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}]}], "=", RowBox[{ FractionBox["W", RowBox[{ RowBox[{"(", FractionBox[\(\[Alpha]\^2 - 1\), StyleBox["\[ExponentialE]", FontSize->14]], ")"}], \((W + 1)\)}]], "2", FractionBox["\[Alpha]", StyleBox["\[ExponentialE]", FontSize->14]]}]}]}], TraditionalForm]]], " by the Chain Rule. The above expression for ", Cell[BoxData[ \(TraditionalForm\`\(d\/\(d\ \[Alpha]\)\) \((1\/\(sin(\ \ \[Theta])\))\)\)]], " is clearly positive for ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] > 0\)]], " and ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[NotEqual] 1\)]], " (because ", Cell[BoxData[ \(TraditionalForm\`W + 1 > 0\)]], ") . Since we already know ", Cell[BoxData[ \(TraditionalForm\`1\/\(sin(\ \[Theta])\)\)]], " is a continuous function on ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[GreaterEqual] 0\)]], ", then ", Cell[BoxData[ \(TraditionalForm\`1\/\(sin(\ \[Theta])\)\)]], " is an increasing function of \[Alpha]. Since ", Cell[BoxData[ \(TraditionalForm\`1\/\(sin(\ \[Theta])\)\)]], " is also positive function, then ", Cell[BoxData[ \(TraditionalForm\`sin(\[Theta])\)]], " is a decreasing function of ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Alpha]\)\(.\)\)\)]], " Since ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " is between 0 and ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/2\)]], ", ", StyleBox[" ", FontWeight->"Bold"], Cell[BoxData[ \(TraditionalForm\`\[Theta]\_max\)]], StyleBox[" ", FontWeight->"Bold"], "is therefore a decreasing function of ", Cell[BoxData[ \(TraditionalForm\`\(\(\[Alpha]\)\(.\)\)\)]] }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Concluding Comments", "Section"], Cell[TextData[{ "Earlier we mentioned the phrase ", StyleBox["Experimental Mathematics", FontSlant->"Italic"], ", which is relevant to this paper in several ways. It was only in working \ with ", StyleBox["Mathematica", FontSlant->"Italic"], " that we realized that the Lambert W function offered a way of getting a \ closed form solution for the range function in the presence of linear \ resistance. Also, symbolic and graphical outputs suggested some of our formal \ mathematical efforts and swiftly confirmed others. In our belief that this \ may become an increasingly useful approach for doing mathematical research, \ we quote part of the statement of philosophy of the ", StyleBox["Journal of Experimental Mathematics", FontSlant->"Italic"], ", a journal started in 1992." }], "Text", CellMargins->{{2, Inherited}, {Inherited, Inherited}}], Cell[TextData[{ StyleBox["Experiment has always been, and increasingly is, an important \ method of mathematical discovery. .... Yet this tends to be concealed by the \ tradition of presenting only elegant, well-rounded and rigorous results. \n\n\ While we value the theorem-proof method of exposition, and while we do not \ depart from the established view that a result can only become part of \ mathematical knowledge once it is supported by a logical proof, we consider \ it anomalous that an important component of the process of mathematical \ creation is hidden from public discussion. It is to our loss that most of us \ in the mathematical community are almost always unaware of how new results \ have been discovered. ....", FontFamily->"Times New Roman", FontSize->11], StyleBox["\n\nExperimental Mathematics", FontFamily->"Times New Roman", FontSize->11, FontSlant->"Italic"], StyleBox[" was founded in the belief that theory and experiment feed on \ each other, and that the mathematical community stands to benefit from a more \ complete exposure to the experimental process. The early sharing of insights \ increases the possibility that they will lead to theorems: an interesting \ conjecture is often formulated by a researcher who lacks the techniques to \ formalize a proof, while those who have the techniques at their fingertips \ have been looking elsewhere. Even when the person who had the initial insight \ goes on to find a proof, a discussion of the heuristic process can be of \ help, or at least of interest, to other researchers. There is value not only \ in the discovery itself, but also in the road that leads to it. ", FontFamily->"Times New Roman", FontSize->11], "\n" }], "Text", CellMargins->{{47, 91}, {Inherited, Inherited}}, FontSize->10], Cell[TextData[{ "One might question the need for detailed and sometimes complex proofs of \ results which which are either physically \"obvious\" or obtainable by a \ simple sequence of commands in a computer algebra system. We would respond by \ not only reaffirming the importance of careful proof, but also by emphasizing \ the thrill of the chase and the satisfaction when results work out so cleanly \ and lead to more general results. In particular, Theorem 1 evolved from the \ specific need to check results of our \"experimental\" computer \ investigation. And we were both surprised and pleased at the concise final \ form of Theorem 3 and the crisp identities that came up in the process of its \ proof.\n\nThe success of the Lambert W function in obtaining closed form \ solutions suggests several directions for further projectile research. We \ have already mentioned, without much optimism, the inverse problem of finding \ what ", Cell[BoxData[ \(TraditionalForm\`\[Theta]\)]], " value(s) give rise to a given range ", Cell[BoxData[ \(TraditionalForm\`\(\(R\)\(.\)\)\)]], " One might also consider the more general overall problem of firing a \ projectile at a target located on an inclined plane (Groetsch, ", StyleBox["[5]", FontWeight->"Bold"], ") or the sad but likely fact that resistance contributes a term that is \ probably not ", StyleBox["linear", FontSlant->"Italic"], " in terms of the velocity. ", "Regardless of its possible help in attacking these and other projectile \ problems, we have become believers in Lambert W and are confident that its \ role in mathematics will continue to expand." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["References", "Section"], Cell[TextData[{ "1. ---, Time for a New Elementary Function?, ", StyleBox["FOCUS; Newsletter Math. Assoc. Amer.", FontSlant->"Italic"], " 20 (2000) 2. \n2. Jonathan M. Borwein and Robert M. Corless, Emerging \ Tools for Experimental Mathematics, ", StyleBox["Amer. Math Monthly", FontSlant->"Italic"], " 106 (1999) 889-909.\n3. R. M. Corless, G. H. Gonnet, D. E. G. Hare, D. \ J. Jeffrey, and D. E. Knuth, On the Lambert W Function, ", StyleBox["Adv. Comput. Math.", FontSlant->"Italic"], " 5 (1996) 329-359.\n4. C. W. Groetsch, Tartaglia's Inverse Problem in a \ Resistive Medium, ", StyleBox["Amer. Math. Monthly", FontSlant->"Italic"], " 103 (1996) 546-551.\n5. C. W. Groetsch, Halley's Gunnery Rule, ", StyleBox["The College Math. Journal", FontSlant->"Italic"], " 28 (1997) 49-50.\n6. C. W. Groetsch, Tartaglia's Bet, forthcoming in ", StyleBox["Cubo Matematica Educacional", FontSlant->"Italic"], ".\n " }], "Text"] }, Open ]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1280}, {0, 941}}, WindowToolbars->{}, WindowSize->{1272, 886}, WindowMargins->{{0, Automatic}, {Automatic, 0}}, PrintingCopies->1, PrintingPageRange->{Automatic, Automatic}, Magnification->1 ] (******************************************************************* 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. 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