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Linear Equation matches polynom

Posted 11 years ago
 Hi, i have a Problem finding a nice solution with mathematica. The Problem: I want to find all coordinates where a linear equation matches a polynom. I tried it with findroot, but that didnt bring the result i need. My Solution so far is shown below. The problem with findroot is that i have to give a starting point and that i only get one point close to the starting point. Maybe somebody can give me a hint what to do. Thank You!           Remove["Global`*"]Points = {{0,     0}, {1, -1}, {2, -2}, {3, -4}, {4, -3}, {5, -4}, {6, -1}, {7, \-4}, {8, 2}, {9, 3}, {10, -1}, {11, -4}};f = Interpolation[Points,    InterpolationOrder -> 1](*Polynom from Points*);\ = 45 °; (*Angle of Linear*)fg[x_, Pos_] := -x*Tan[\] + Points[][[2]] +   Points[][[1]]*   Tan[\] (*Linear equation throug point (Points), gradient \from \  *)Plot[{fg[x, 9], f}, {x, 0, 11}, Epilog -> Map[Point, Points], ImageSize -> Large, AspectRatio -> 1] (*Plot*)sol = FindRoot[fg[x, 9] == f, {x, 8}, AccuracyGoal -> 8,    PrecisionGoal -> 8];solution = x /. sol(*Check*)f == fg[solution, 9]f
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Posted 11 years ago
 I hope this helps, I am rewriting the problem a little bit.First the list of points:points = {{0, 0}, {1, -1}, {2, -2}, {3, -4}, {4, -3}, {5, -4}, {6, -1}, {7, -4}, {8, 2}, {9, 3}, {10, -1}, {11, -4}};Next the piecewise linear interpolation function:f = Interpolation[points, InterpolationOrder -> 1]Visualize them together in a plot:Plot[f[x], {x, 0, 11}, Epilog -> {AbsolutePointSize[6], Red, Point[points]}]Next the 'fg' function:fg[x_, pos_] := -x*Tan[Pi/4] + points[[pos, 2]] + points[[pos, 1]]*Tan[Pi/4]And visualize again, everything together:Plot[Join[{f[x]}, Table[fg[x, i], {i, 1, 12}]], {x, 0, 11}, Epilog -> {AbsolutePointSize[6], Red, Point[points]}]Now solve for one of the intersections:FindRoot[fg[x, 9] == f[x], {x, 8.5}]  (* returns {x->8.} *)
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