Solving huge equations

Question

I have this expression:

$d=-\frac{\sqrt{\frac{780 (f+130) \left(5 d^2 (13 c+f+130)-5 d (c (f+1300)-325 (f+160))+125 \left(2 (2 c+725) f+5 f^2+104000\right)+13 d^3\right) \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right) \left(300 c^2 (13 d-f)+5 c \left(208 d^2+d (121 f-2600)-5 \left(f^2+1280 f+156000\right)\right)+d \left(52 d^2+5 d (17 f+2600)+25 \left(f^2+290 f+20800\right)\right)\right)}{\left(5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2\right)^3}+\left(-\frac{13 \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right)^2}{4 \left(5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2\right)^2}+\frac{(25 (5 f+728)-780 c) \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right)}{4 \left(5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2\right)}+975 c (f+80)\right)^2}+\frac{13 \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right)^2}{4 \left(5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2\right)^2}+\frac{195 c \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right)}{5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2}-\frac{125 f \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right)}{4 \left(5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2\right)}-\frac{4550 \left(5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3\right)}{5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2}-975 c f-78000 c}{104 \left(-\frac{5 d^2 (52 c+17 f+2600)+5 d \left(c (61 f+5200)+5 \left(f^2+290 f+20800\right)\right)-25 c f (f+80)+52 d^3}{4 \left(5 d (13 c+f-130)-5 ((c+100) f+13000)+13 d^2\right)}-15 c\right)}$

The raw mathematica input for this is:

func = -(((13*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80))^2)/
  (4*(13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000))^2) + 
 (195*c*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80)))/
  (13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000)) - 
 (4550*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80)))/
  (13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000)) - 
 (125*f*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80)))/
  (4*(13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000))) - 78000*c - 975*c*f + 
 Sqrt[(-((13*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80))^2)/
       (4*(13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000))^2)) + ((25*(5*f + 728) - 780*c)*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 
        5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80)))/(4*(13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000))) + 
     975*c*(f + 80))^2 + (780*(f + 130)*(13*d^3 + 5*(13*c + f + 130)*d^2 - 5*(c*(f + 1300) - 325*(f + 160))*d + 
      125*(5*f^2 + 2*(2*c + 725)*f + 104000))*(52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 
      25*c*f*(f + 80))*(300*(13*d - f)*c^2 + 5*(208*d^2 + (121*f - 2600)*d - 5*(f^2 + 1280*f + 156000))*c + 
      d*(52*d^2 + 5*(17*f + 2600)*d + 25*(f^2 + 290*f + 20800))))/(13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000))^3])/
(104*(-15*c - (52*d^3 + 5*(52*c + 17*f + 2600)*d^2 + 5*(c*(61*f + 5200) + 5*(f^2 + 290*f + 20800))*d - 25*c*f*(f + 80))/
   (4*(13*d^2 + 5*(13*c + f - 130)*d - 5*((c + 100)*f + 13000))))))

I want to solve for $func=d$ for real solutions. I'm using

Solve[func==d, d, Reals]

It seems to run for ages - will it complete? Is this the way I should be doing it?

Answer 1

You can get an approximate analytical solution within a domain of other parameters using numerical solution. First get an idea of the nature of your function.

Plot[{func /. {c -> 1.5, f -> 1.5}, d}, {d, -1000, 1000}]

I can see there are two solutions - near 0 and near -500.

FindRoot[Block[{c = 1.5, f = 1.5}, func] == d, {d, 0}]
FindRoot[Block[{c = 1.5, f = 1.5}, func] == d, {d, -500}]

{d -> 0.00831943}

{d -> -375.409}

Let's focus on the solutions near 0.

data = Flatten[Table[{c, f, d /. FindRoot[func - d, {d, 0.01}]}
                   , {c, 0.1,  2, .1}, {f, 0.1, 2, .1}], 1];
ListPlot3D[data]
func1[x_, y_] = Interpolation[data][x, y]

and func1[x,y] is your approximate solution within the domain.