# Having trouble finding roots with FindRoot

I am trying to numerically solve a system of three nonlinear equations with FindRoot.

But FindRoot seems to be having trouble with one the variables and it keeps insisting that the variable should be a list. Why is this happening? Is there someway I can tell the function how to understand that the variable is just another numerical unknown?

The code I am running is as shown in the image.

### Edit

When I provide a starting point for all variables, Findroot retunrns the following error:

FindRoot::cvmit: Failed to converge to the requested accuracy or precision within 100 iterations."

and the problem once again seems to be the isolated variable.

\$Assumptions =
b ∈ Reals && L ∈ Reals && c + L > 0 && c - L < 0 &&
n ∈ Integers && L > 0;
FindRoot[
{(7.2 *A *b (E^(-b *(7.2 + c)) π + π Cos[0.4363323129985824*c] -
7.2*b*Sin[0.4363323129985824*c]))/(162.8601631620949* b^2 + π^3) == -1.19489,
(7.2*A*b (-2 E^(-b *(7.2 + c)) π + 2 π Cos[0.8726646259971648*c] -
7.2*b*Sin[0.8726646259971648*c]))/
(325.7203263241898* b^2 + 8 π^3) == -0.779047,
(7.2*A*b (3 E^(-b*(7.2 + c)) π + 3 π Cos[1.3089969389957472*c] -
7.2*b*Sin[1.3089969389957472* c]))/
(488.58048948628465*b^2 + 27 π^3) == -0.16887},
{{A, 5}, {b, 3}, {c, 2.38333333333333333333}}]


• Mathematica just wants you to specify an initial guess for the variable b. Mar 17, 2018 at 19:16
• Btw.: Welcome on Mathematica.StackExchange! It is a good habit to post the actual code as copyable text (copy the code in InputForm and paste it here) so that other users can experiment with it. Mar 17, 2018 at 19:22
• See MaxIterations. But more importantly you should find a good initial guess. Mar 17, 2018 at 19:38
• Example solutions: {A -> 10.6972, b -> 0.445259, c -> 2.97381} or {A -> 11.505, b -> 0.411923, c -> 17.4012} or {A -> -666.251, b -> 3.47452, c -> 0.273296}. Mar 17, 2018 at 19:41
• For some initial values for your problem, FindRoot follows a path for which the function gets closer to zero but never reaches it. For simpler examples, consider FindRoot[(x - 2)/(x^2 + 4), {x, 0}] and FindRoot[(x - 2)/(x^2 + 4), {x, 6}] or perhaps Needs@"OptimizationUnconstrainedProblems"; FindRootPlot[(x - 2)/(x^2 + 4), {x, 6}]. See also this tutorial. For multivariate problems, the picture can be more complicated. Mar 18, 2018 at 13:15

You can first solve the first equation for A, insert it into the two other equations and than raster the area of the remanining b and c with FindRoot in small steps.

sol1 = First@Solve[equations[[1]], A] // Simplify

eq2 = equations[[2 ;; 3]] /. sol1 // Simplify


With small raster steps of r and s you have a good change to get most roots.

(tab = DeleteCases[
Flatten[Table[
Check[FindRoot[
eq2, {{b, .05 + r, r, .1 + r}, {c, .05 + s,
s, .1 + s}}], {}], {r, 0, 50, .1}, {s, 0, 50, .1}] //
Quiet, 1], {}]) // Timing

(*   {236.907, {{b -> 0.445259, c -> 2.97381}, {b -> 0.411923,
c -> 17.4012}, {b -> 0.411808, c -> 31.8013}, {b -> 0.411808,
c -> 46.2013}, {b -> 3.47452, c -> 0.273296}, {b -> 3.47452,
c -> 14.6733}, {b -> 3.47452, c -> 29.0733}, {b -> 3.47452,
c -> 43.4733}}}   *)


Find the corresponding A with A /. sol1 and check, how good left side minus right side of equations is near zero for this solutions.

solutions = {A /. sol1, b, c} /. tab

(*   {{10.6972, 0.445259, 2.97381},
{11.505, 0.411923, 17.4012},
{11.508, 0.411808, 31.8013},
{11.508, 0.411808, 46.2013},
{-666.251, 3.47452,0.273296},
{-666.251, 3.47452, 14.6733},
{-666.251, 3.47452, 29.0733},
{-666.251, 3.47452, 43.4733}}   *)

(((equations[[All, 1]] - equations[[All, 2]]) /. sol1) /. # &) /@ tab

(*   {{-2.22045*10^-16, -8.88178*10^-16, -1.66533*10^-16},
{0.,-4.44089*10^-16, -8.88178*10^-16},
{0., 1.11022*10^-16, 3.88578*10^-16},
{0., 2.22045*10^-15, 2.16493*10^-15},
{0., 1.88738*10^-15, 1.16573*10^-15},
{0., -1.45439*10^-14, -5.27633*10^-14},
{2.22045*10^-16, -4.60743*10^-14, 1.18044*10^-13},
{0., 6.00631*10^-14, -7.31637*10^-14}}   *)


First some commments.

I was able to get a solution copying your code and running FindRoot without using any assumptions.

FindRoot[{(7.2*A*
b (E^(-b*(7.2 + c)) π + π Cos[0.4363323129985824*c] -
7.2*b*Sin[0.4363323129985824*c]))/(162.8601631620949*
b^2 + π^3) == -1.19489, (7.2*A*
b (-2 E^(-b*(7.2 + c)) π +
2 π Cos[0.8726646259971648*c] -
7.2*b*Sin[0.8726646259971648*c]))/(325.7203263241898*b^2 +
8 π^3) == -0.779047, (7.2*A*
b (3 E^(-b*(7.2 + c)) π +
3 π Cos[1.3089969389957472*c] -
7.2*b*Sin[1.3089969389957472*c]))/(488.58048948628465*b^2 +
27 π^3) == -0.16887}, {{A, 5}, {b, 3}, {c, 2.3}}]

During evaluation of FindRoot::cvmit: Failed to converge
to the requested accuracy or precision within 100 iterations.
(* {A -> 4.83787, b -> 32.8529, c -> 2.04422} *)


I rather like using FindMinimum when the root is not exact. I moved the right hand side of each equation to the left hand side so that at a solution it would be minimum.

I squared and summed each term. If the solution is perfect the sum would be zero. The solutions are similar except b is much larger.

Module[
{
eq1 = (7.2*A*
b (E^(-b*(7.2 + c)) π + π Cos[0.4363323129985824*c] -
7.2*b*Sin[0.4363323129985824*c]))/(162.8601631620949*
b^2 + π^3) + 1.19489,
eq2 = (7.2*A*
b (-2 E^(-b*(7.2 + c)) π +
2 π Cos[0.8726646259971648*c] -
7.2*b*Sin[0.8726646259971648*c]))/(325.7203263241898*b^2 +
8 π^3) + 0.779047,
eq3 = (7.2*A*
b (3 E^(-b*(7.2 + c)) π +
3 π Cos[1.3089969389957472*c] -
7.2*b*Sin[1.3089969389957472*c]))/(488.58048948628465*b^2 +
27 π^3) + 0.16887
},
FindMinimum[eq1^2 + eq2^2 + eq3^2,
{{A, 4.8}, {b, 1000}, {c, 2.09}}
]
]
(* {0.00265477, {A -> 4.80874, b -> 3.89195*10^7, c -> 2.10254}} *)


Below is the equation (with a little rounding).

If you study these equations you will see that when b gets large (say above 1000 or so) the equations simplify (define eq1 through eq3 as in the above Module).

Limit[eq1, b -> Infinity]
(* ConditionalExpression[1.19489 - 0.31831 A Sin[0.436332 c],
A ∈ Reals && 5. c > -36.] *)

Limit[eq2, b -> Infinity]
(* ConditionalExpression[0.779047 - 0.159155 A Sin[0.872665 c],
A ∈ Reals && 5. c > -36.] *)

Limit[eq3, b -> Infinity]
(* ConditionalExpression[0.16887 - 0.106103 A Sin[1.309 c],
A ∈ Reals && 5. c > -36.] *)


Now we have three equations and are overdetermined so we need to use an optimization.

Module[
{
eq1l = 1.19489 - 0.31830988618379064 A Sin[0.4363323129985824 c],
eq2l = 0.779047 - 0.15915494309189532 A Sin[0.8726646259971647 c],
eq3l = 0.16887 - 0.10610329539459688 A Sin[1.3089969389957472 c]
},
FindMinimum[eq1l^2 + eq2l^2 + eq3l^2,
{{A, 4.8}, {c, 2.09}}
]
]
(* {0.00265477, {A -> 4.80874, c -> 2.10254}} *)


Note that the A and c values are the identical to the first optimization.

Bottom line, I think we know A and c but we only know that b` is a large number.