FindRoot
documentation reports that if the equation and the initial point are reals, the solutions are searched in the real domain. However, in the following case I get a complex solution
FindRoot[eq, {h, 1.7}]
{h -> -0.990042 - 0.689686 I}
Same for
Assuming[Reals, FindRoot[eq, {h, 1.7}]]
Where eq
is the following. What can I do to force Mathematica to return only real solutions?
eq = 0.2888960456513873` \[Sqrt](0.4149486782073932` + (0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 2.2076945564517114` h \[Sqrt](0.4149486782073932` +
(0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 6.735521430523215` h^2 \[Sqrt](0.4149486782073932` +
(0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 10.54153050379155` h^3 \[Sqrt](0.4149486782073932` +
(0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 8.93280406600778` h^4 \[Sqrt](0.4149486782073932` +
(0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 3.8752891330788843` h^5 \[Sqrt](0.4149486782073932` +
(0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 0.666148416209407` h^6 \[Sqrt](0.4149486782073932` +
(0.5637605604986459` +
h)^2) \[Sqrt](0.4756674436221575` + (0.9900423417575865` +
h)^2) + 0.4167441738314279` \[Sqrt](0.4756674436221575` +
(0.9900423417575865` +
h)^2) \[Sqrt](1 - (0.4444444444444444`
(-0.022551321792606827` + (0.5637605604986459` +
h)^2)^2)/(0.4149486782073932` + (0.5637605604986459` +
h)^2))
Answer
It will be more convenient to define the function eq
this way :
eq[h_] := the formula
Real solutions
For a general technique of using FindRoot
the way you would like I recommend to read this post : First positive root. However in order to demonstrate how it works for real numbers we should have a different equation since this one has no real solutions :
NSolve[ eq[h] == 0, h, Reals]
{}
Sometimes it is reasonable to restrict the search for roots to a special region, e.g. if we add an inequality 0 < h < 3
then we don't have to specify the domain Reals
because the system understands that h
must be real :
NSolve[ eq[h] == 0 && 0 < h < 3, h ]
General remarks on using NSolve
you could find here : Solve an equation in R+ since they are valid for Solve
as well as for NSolve
.
For the function f
we can find a global minimum of the real part :
nmin = { h /. #[[2]], #[[1]]}& @ NMinimize[ Re @ eq[h], h]
{-1.83596, 0.100607}
For an idea what we can expect let's plot the real and imaginary parts of f
:
Plot[{Re @ eq[h], Im @ eq[h]}, {h, -3, 2}, PlotStyle -> Thick, PlotRange -> {-1, 4},
Epilog -> {Red, PointSize[0.015], Point[nmin]},
PlotLegends -> {Placed["Expressions", {Left, Center}]}]
Even though eq
has no real roots we can see that one can expect real solutions e.g. for eq[h] - 3/2
near h == -2
and h == 0
, using FindRoot
with appropriate starting points we find :
FindRoot[ eq[h] - 3/2, #]& /@ {{h, -2}, {h, 0}}
{{h -> -2.15793}, {h -> 0.14209}}
For deeper understandig of the behavior of the eq
function we should take a look at the complex plane.
Complex solutions
The complex roots can be rewritten as pairs of real numbers :
pts = {Re @ #, Im @ #}& /@ (h /. NSolve[ eq[h] == 0, h])
{{-1.86925, 0.135055}, {-1.86925, -0.135055}, {-1.08654, 0.84989},
{-1.08654, -0.84989}, {-0.990042, 0.689686}, {-0.990042, -0.689686},
{-0.0268207, 0.391104}, {-0.0268207, -0.391104}}
Let's visualize the function eq
in the complex plane :
GraphicsRow[
Table[ Show[ ContourPlot @@@ {
{ g[eq[x + I y]], {x, -2.5, 0.5}, {y, -1.5, 1.5},
PlotLabel -> Style[ g[HoldForm @ eq[x + I y]], Blue, 25],
ColorFunction -> "DeepSeaColors", Epilog -> { Darker @ Green, PointSize[0.03],
Point[pts] } },
{ Re @ eq[x + I y] == 0, {x, -2.5, 0.5}, {y, -1.5, 1.5},
ContourStyle -> {Red, Thick}},
{ Im @ eq[x + I y] == 0, {x, -2.5, 0.5}, {y, -1.5, 1.5}, PlotPoints -> 50,
MaxRecursion -> 3, ContourStyle -> {Cyan, Thick}}}],
{g, {Re, Im}} ] ]
The green points denote the complex roots, you can see that there are no real roots. The red lines are solutions to Re[ eq[h] ] == 0
while the cyan lines denote solutions to Im[ eq[h] ] == 0
. One can see a that there are some branch cuts on the plots, i.e. the definition of the function eq
is reliable only if we use it in appropriate regions in the complex plane.
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