Hello guys, can you help me with my problem in this thread?

http://community.wolfram.com/groups/-/m/t/455321

in that discussion, the formula is just an example of my work using ParametricNDSolveValue, the problem surfaced when FindRoot have an answer of complex value. is there anyway to force FindRoot to present only Real numbers?

However, here you can download the obsolete add-on package which did the job I actually need. Works for version 6

http://library.wolfram.com/infocenter/MathSource/6771/

At a certain point it says

"The RealOnly package is obsolete. Use instead functionality provided by \
Reduce[equations, variables, Reals] and by Assuming, Refine, and similar \
functions."

I tried several combinations of Reduce or Assuming without success :-(

Attachments:

Luther, no, probably not an error on your part. Some integrals do not converge in the limit x -> infinity. For instance, solutions for an oscillating system will cycle around a stable state (and have "imaginary" parts). Solutions for an oscillating system, driven at resonant frequency (without proper damping) will diverge! In engineering, this can have catastrophic results.

Have an example, or a description of the system you are trying to model?

Interesting discussion. My experience in the complex domain was limited to various engineering discussions and formal methods such as Laplace transforms. Imaginary components were a phase shift. Again in the distant past my Fortran integrals using the simple numerical models and were all in the real domain. Once I went into management my brain atrophied. I relied on the smart PhDs working for me to get it right!!! LOL.

My take away is that I CAN ignore the imaginary component, even if it is very large. On the other hand, I remember some old physics scattering problems in which the imaginary parts represented phase shifts that caused the output wave functions to oscillate with interference patterns...and these were measured and seemed to be real effects.

My integrals are similar to Elliptical integrals, but there are additional terms in the denominator of the integrand function that make an analytical solution impossible. But the numerical integration works very nicely, except that over a certain limited range and size of variable these pesky imaginary components spring up. They may mean something that I simply cannot throw away or at least I do so at some risk to the validity of my results.

Take a charged ring...so much charge per unit area of the ring. If I put a test charge in the vicinity of the first ring, I find a mutual force between the ring and the test charge. If the test charge is on the ring axis, the solution is trivial. If the test charge is off-axis, the mutual force requires the solution to an elliptical integral. If I replace the test charge with another coaxial ring, my result is an elliptical integral. If I extend the ring into a disk, the resulting integral is similar to an elliptical integral but cannot be solved analytically because now the denominator of the integrand contains additional spatial variables, at least if am correctly recalling my efforts at having Mathematica solve that integral. I switched to numerical integration so long ago I forget how I got there!!!

If I further modify the geometry from disks to spheres using cylindrical symmetry, the integrand is only a little more complex and the same limits on finding an analytical solution exist. Using spheres I also do not have additional relative orientation geometries to make thing even more intractable. Now, as I fool around with the numerical integration making the sizes and distances of the spheres different, these imaginary components to the solutions begin to show up...not always but under certain circumstances. Even though the problem is a static problem, I am loath to simply throw the imaginary components away because it is possible that there are some phase effects that the integrals are showing me. It could be a scattering problem if I were looking at the actual dynamics as I change the separation distances.

That is my dilemma. What is real and what is not. If I use Plot and use the separation distance as the variable, I get some strange discontinuities at certain distances and sizes. I am trying to track these down, also. At first I thought it was noise, but now I am not so sure. On the surface, it a straight forward physics problem and I was only looking at it to learn how to use Mathematica. Only if the results are not artifacts, that simple problem may not be so simple and straight forward as I had expected. Capiche?

My take away is that I CAN ignore the imaginary component, even if it is very large.

Not always, especially if it is large. You can safely ignore imaginary components whose magnitude is small in comparison to your expected/desired numerical precision. On the other hand, a large imaginary component in a result you'd expect to be real is likely a symptom that you've wandered off onto the wrong branch of a multivalued function. The result may be mathematically correct, but in a sense that doesn't apply to your physical problem.

I'm a physicist. I'm used to mathematics needing guidance to properly represent physical reality. It starts with very simple high school problems. Consider dropping a ball: how long will it take to hit the floor. Well, it's the solution to an algebra problem:

In[24]:= Solve[ 1/2 a t^2 == h, t]

Out[24]= {{t -> -((Sqrt[2] Sqrt[h])/Sqrt[a])}, {t -> (
   Sqrt[2] Sqrt[h])/Sqrt[a]}}

Oops, there are two solutions! However, the ball can't hit the floor before you drop it, so we throw out the negative one. This is completely normal.

Now, for algebra problems there are "root isolation" methods that find all solutions rigorously, and Mathematica uses them. So, you throw out the solutions you don't want and you have the solutions for your problem. However, once you go beyond algebra, you generally can't have confidence that your method, even if it has found a solution, has found every solution to your mathematical problem, nor can you be certain that the solution it has found is the solution that you want for your physical problem. You're in the realm of "experimental verification".

Comparing results from Integrate and NIntegrate is a good sort of mathematical experiment, as they have different sorts of problems. You've already encountered Integrate's branch cut trouble. NIntegrate has the problem that if its sampling of points in the function in question accidentally misses the exciting parts, it may yield an entirely bogus result.

Mathematica will not do your thinking for you.

The basic mathematical issue is that the grand principles of algebra and analysis operate in the complex domain. Often, you have to reason through the complex domain to get a real answer. Any attempt to restrict Mathematica to operate in the real domain will generally impact its ability to break a problem down into subproblems. It is, however, often possible to ask for a real result:

In[20]:= Solve[x^3 == -10, Element[x, Reals]]

Out[20]= {{x -> Root[10 + #1^3 &, 1]}}

In[21]:= % // N

Out[21]= {{x -> -2.15443}}

The Root object is yet another way to express "the real cube root of -10".

Yes, that is the answer. Thanks!

However,

psurd2 = Plot[y = Surd[x, 3]; y, {x, -10, 10}]

Thank you, John! That solves a mystery.

Another way to look at it: for negative real numbers, there are several solutions coming from the complex roots. For instance, in the example above...

in:= solutions = Solve[x^3 == -10, x] // N
out:= {{x -> 1.07722 + 1.8658 I}, {x -> -2.15443}, {x -> 
   1.07722 - 1.8658 I}}

These solutions all have the same magnitude r[x]:

in:= Clear[r]
r[x_] := Sqrt[Re[x]^2 + Im[x]^2]
r@Values@solutions
out:= {{2.15443}, {2.15443}, {2.15443}}

If you don't care to differentiate from among these solutions (though in my opinion, you'd be missing out on the best part!!) you could keep a single, real-valued magnitude:

in:= -1*Flatten@Union@%
out:= {-2.15443}

This yields a similar result as as plotting psurd[] above...

Plot[Piecewise[{{-1*Sqrt[Re[x]^2 + Im[x]^2]^(1/3), x < 0}, {x^(1/3), 
    x > 0}}], {x, -10, 10}]

And thinking some more about Caitlin's observation:

Consider that -1 = Exp[Pi I], then an nth root of -1 is Exp[Pi I / n].

The n nth roots of unity are given by Exp[2 Pi I k / n] with k from 1 to n. (de Moivres numbers)

Multiplying the nth root of -1 by these gives the n nth roots of -1 as Exp[(2k+1) Pi I / n], again with k from 1 to n.

Since any negative number -N (N positive) can be written as -1*N, we have the nth root as the principal root of N multiplied by these n roots of -1.

Here, for example, are nine 9th roots of -1:

In[1]:= rootOfMinusOne[k_, n_] := Exp[(2 k + 1) Pi I/n]

In[2]:= roots9 = Table[rootOfMinusOne[k, 9], {k, 9}]

Out[2]= {E^((I \[Pi])/3), E^((5 I \[Pi])/9), E^((
 7 I \[Pi])/9), -1, E^(-((7 I \[Pi])/9)), E^(-((5 I \[Pi])/9)), E^(-((
  I \[Pi])/3)), E^(-((I \[Pi])/9)), E^((I \[Pi])/9)}

In[3]:= #^9 & /@ roots9

Out[3]= {-1, -1, -1, -1, -1, -1, -1, -1, -1}

In[4]:= cPlot[lst_, opts___] := 
 ListPlot[{Re[#], Im[#]} & /@ lst, AspectRatio -> 1, 
  AxesLabel -> {"Re", "Im"}, opts]

In[5]:= cplt = cPlot[roots9, PlotLabel -> "The nine 9th roots of -1"]

If we have to cite Euler every time we use his discoveries we will all spend half our time writing citations. ;-)

Happy Pi Day!

Thanks.

What I would like to obtain, if it exists, is a general command that sets Mathematica to work only on the real branches of solutions. In the old Derive program there is such a command : "Branch ? Real"

I know that I could arrange every plot or calculation for every single case, for example in the plot I mentioned I could do

Plot[Sign[x]*Abs[x]^(1/3), {x, -10, 10}]

and I would obtain my result.

But I'd like something general, so to set it up and go ahead without particular arrangements of the functions involved.

Yes, that was my first try to solve the problem......

Ah . . . It chooses one of the complex roots:

In[16]:= x /. Solve[x^3 == -10, x] // N

Out[16]= {1.07722 + 1.8658 I, -2.15443, 1.07722 - 1.8658 I}