2
$\begingroup$

I have 6 equations

0.344027==0.5 (a+b)-0.5 (a-b) Cos[2 (-1.3439-0.0174533 c)]
0.679511==0.5 (a+b)-0.5 (a-b) Cos[2 (0.20944 -0.0174533 c)]
0.436543==0.5 (a+b)-0.5 (a-b) Cos[2 (-0.733038-0.0174533 c)]
0.324024==0.5 (a+b)-0.5 (a-b) Cos[2 (1.18682 -0.0174533 c)]
0.304968==0.5 (a+b)-0.5 (a-b) Cos[2 (-1.5708-0.0174533 c)]
0.676049==0.5 (a+b)-0.5 (a-b) Cos[2 (-0.174533-0.0174533 c)]

I could pick 3 equations from the list and got exact answers. However, I want my variables (a,b,c) to have values such that the right hand sides of all 6 equations approach the left hand side as much as possible. I'm not really sure what I can use in this case to resolve this.

Edit: I used Kuba's method to solve my problem. However, using FindFit (per belisarius' helpful suggestion) also gave me the same result. This works because all 6 equations have a pair of x and y that could be fitted through multivariable FindFit-- though if I were to be given random equations with a,b,c, NMinimize might be the only method to find them.

FindFit example in MMA's documentation Mathematica FindFit multivariable

Using FindFit to my example

Original equations:

0.344027==0.5 (a+b)-0.5 (a-b) Cos[2 (-1.3439-0.0174533 c)]
0.679511==0.5 (a+b)-0.5 (a-b) Cos[2 (0.20944 -0.0174533 c)]
0.436543==0.5 (a+b)-0.5 (a-b) Cos[2 (-0.733038-0.0174533 c)]
0.324024==0.5 (a+b)-0.5 (a-b) Cos[2 (1.18682 -0.0174533 c)]
0.304968==0.5 (a+b)-0.5 (a-b) Cos[2 (-1.5708-0.0174533 c)]
0.676049==0.5 (a+b)-0.5 (a-b) Cos[2 (-0.174533-0.0174533 c)]

FindFit

yData = {0.344027, 0.679511, 0.436543, 0.324024, 0.304968, 0.676049};
xData = {-1.3439, 0.20944, -0.733038, 1.18682, -1.5708, -0.174533};
data = {xData, yData, Table[0, {6}]} // Transpose;
model = y - (0.5 (a + b) - 0.5 (a - b) Cos[2 (x - 0.0174533 c)]);
fit = FindFit[data, model, {a, b, c}, {x, y}]
model /. fit /. {y -> yData, x -> xData}
Show[Plot3D[model /. fit, {x, -2, 2}, {y, 0, 1},
  AxesLabel -> Automatic,
  MeshShading -> {{None, None}, {None, None}}], 
 ListPointPlot3D[data, PlotStyle -> Directive[PointSize[Medium], Red]],
 ListPlot3D[data, VertexColors -> Hue]]
$\endgroup$
4
$\begingroup$
ClearAll[a, b, c]
NMinimize[
 #.# &[Subtract @@@ {
    0.344027 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-1.3439 - 0.0174533 c)],
    0.679511 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (0.20944 - 0.0174533 c)],
    0.436543 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-0.733038 - 0.0174533 c)],
    0.324024 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (1.18682 - 0.0174533 c)],
    0.304968 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-1.5708 - 0.0174533 c)],
    0.676049 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-0.174533 - 0.0174533 c)]
    }], {a, b, c}]
{0.0059057, {a -> 0.291325, b -> 0.677473, c -> 3.2186}}
$\endgroup$
  • $\begingroup$ Thank you so much for your help Kuba. I have a question: it seems like you were trying to find the dot product of the list of the (subtraction) equations and minimize it to find the best roots. Is this a standard math practice, and if so does it have a name or is there anywhere I could learn more about it? I'm just a college student so my math skills are not very strong. $\endgroup$ – seismatica Jun 21 '14 at 8:06
  • 1
    $\begingroup$ @seismatica Hi, I'm glad I could help. Before dot product there is subtraction so at the end we are minimizing sum of squared differences, like in standard lsq method but via black box NMinimize :) $\endgroup$ – Kuba Jun 21 '14 at 14:23
3
$\begingroup$

You may use FindFit

l = {0.344027 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-1.3439 - 0.0174533 c)],
     0.679511 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (0.20944 - 0.0174533 c)],
     0.436543 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-0.733038 - 0.0174533 c)],
     0.324024 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (1.18682 - 0.0174533 c)],
     0.304968 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-1.5708 - 0.0174533 c)],
     0.676049 == 0.5 (a + b) - 0.5 (a - b) Cos[2 (-0.174533 - 0.0174533 c)]};
l1 = (List@@@l) /. {x1_,x2_} :>Flatten@{Cases[x2,Cos[2 x__] :> (List @@ x /.c :> 1), 2], x1};
ff = FindFit[l1, .5 u - .5 v Cos[2 (x + y c)], {u, v, c}, {x, y}]
Solve[{a + b == u, a - b == v} /. ff, {a, b}]

(*
  {u -> 0.968798, v -> -0.386148, c -> 3.2186}
  {{a -> 0.291325, b -> 0.677473}}
*)
$\endgroup$
  • $\begingroup$ Thank you so much for your help belisarius. I've edited my question with the FindFit method you suggested in a way that feels more intuitive to me. $\endgroup$ – seismatica Jun 21 '14 at 12:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.