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4

Actually, they are equal. Redefine s = First@Solve[pt[t] == qt[t], {a, b}] (* {a -> 1/2 (-3 Csc[t] + 3 Sec[t] - 2 Tan[t]), b -> 1/2 (3 Csc[t] + 3 Sec[t] - 2 Tan[t])} *) to eliminate the extra set of brackets. Then, for any value of t FullSimplify[(pt[t] /. s) - (qt[t] /. s)] (* {0, 0} *)


3

New way The OP mentioned ContourPlot but its behavior is V10 makes my original solution practically unusable except for a very rough plot. Another approach is to solve the equation for all the roots in a given region. From the ContourPlot, one can see there are two types, ones that cross y == -5 and ones that cross y == 5. We can use NDSolve to solve the ...


2

This function prints out the solutions to all polynomials of degree d with coefficients of unit magnitude. printAllSolutions[d_] := Do[ With[{p = FromDigits[c~Prepend~1, x]}, Print[Expand[p] -> (x /. NSolve[p == 0, x])] ], {c, Tuples[{-1, +1}, d]} ] I made one optimization, which is that the leading coefficient is always 1. If it is ...


2

Append has no side effect You may try AppendTo a = {x^2 + y^2 == 1}; AppendTo[a, x + y == 0] Solve[a, {x, y}] resulting in (* {{x -> -(1/Sqrt[2]), y -> 1/Sqrt[2]}, {x -> 1/Sqrt[2], y -> -(1/Sqrt[2])}} *) or consider simply Append[a, x + y == 0] // Solve


2

I may have misunderstood the intent but post this in case it is helpful. Note: I have tried to mimic the spiral, clockwise rotation with displaced centre of spiral I have displayed tangent to curve, the slope and angle. Parametrization seems the most useful approach: sp[t_, m_, a_, v_] := v + a {Exp[Tan[m] t] Cos[t], -Exp[Tan[m] t] Sin[t]} der[t_, ...


2

Just use NMaximize yRx1[a_, b_] := y /. #2 & @@ NMaximize[{Rx1[y, a, b], y > 0}, y] Example: yRx1[1, 1.5] (*0.903322*)


1

Try this exponential derivative operator: expD[f_, x_] := Module[{x0}, Sum[SeriesCoefficient[f, {x, x0, i}], {i, 0, \[Infinity]}] /. {x0 -> x} ] Examples: expD[x^2, x] (* (1 + x)^2 *) expD[Sin[x], x] (* Sin[1 + x] *) expD[Exp[x], x] (* Exp[1 + x] *)


1

For a graphical study I would solve the two equations independently. First one gives you one up to three x-solutions with y as additional parameter. Second one give one up to three y-solutions with x as additional parameter. Next I would plot the solutions as ParametricPlot[] with different colors on top of each other. So something like solx=Solve[(rA-rB) ...


1

I'm guessing you are looking for the list of replacements that need to be made going down to level 0 instead of just showing the level 0 results. Are you looking for something like this? assignments[d] = a + b; assignments[e] = b c; assignments[f] = d e; assignments[x_] := x; FixedPointList[Map[assignments, #, {-1}] &, {f}][[;; -2, 1]] {f, d e, b ...


1

Maybe I missed the point, but if you just copy and paste your expressions (and insert spaces to give multiplication) you get: d = a + b; e = b c; f = d e which gives you b (a + b) c, which is equivalent to your given f. Is this all you were asking?


1

One approach is to Solve using cond1 and cond2 only, then keep only those answers that satisfy cond3 and cond4. For instance, DiracPoint[{theta_, phi_}] := Module[{c1, c2, c3, x, y, cond1, cond2}, {c1, c2, c3} = 1 - 3 Sin[theta]^2 Cos[phi - 2 Pi (# - 1)/3]^2 & /@ {1, 2, 3}; Cases[If[0 <= ((c2 + c3)^2 - c1^2)/(4 c2 c3) <= 1, cond1 = ...



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