Sure.  Jusy say

A;

F;

If you meant to ask something else, it is not clear from your wording.

I did not expect my remark regarding the root x ≈ 1/2015! to generate so many followups.  Had I known that, I would have provided more details to begin with which, in particular,  would have shown that

1/2015! - h*(1/2015!)^2  <  x  <  1/2015!

where

h = 1 + 1/2 + 1/3 + ... + 1/2015 ≈ 8.2,

and would have spared Markiyan some confusion.  The explanation is a bit long.  However, it may be of some interest due to the novel(?) idea of a using a differential equation to solve an algebraic equation.  I have used this method on a few occasions in the past, and I like to think that it's my own invention, although I dare not claim it, since it wouldn't be surprising that others have stumbled upon it as well. I will explain the idea in the context of Markiyan's problem, but the method is general and is not tied to that specific equation.

Let

p(x) = x(x+a1)(x+a2)...(x+an),


where the ai's are distinct positive constants. We want to solve the equation p(x) = t for some x > 0 and  t > 0.  Ultimately we are interested in t = 1, but we let t be an arbitrary parameter for now.

We regard p(x) = t  as an implicit definition of x as a function of t.  We write x(t) for that function.  It follows that p(x(t)) = t, and in particular p(x(1)) = 1.  Therefore, x(1) is the solution of p(x) = 1, and solving Markiyan's problem is equivalent to computing x(1).  For future reference let us note that x(0) = 0.

Upon differentiating p(x(t)) = t with respect to t we obtain p'(x(t)) * x'(t) = 1, therefore x(t) satisfies the differential equation

dx/dt = 1/p'(x),    x(0) = 0.       (*)

We solve the initial value problem (*) by any possible means, and then evaluate x(1) which, as explained above, would be the solution of p(x) = 1.

We choose to solve (*) by the method of power series here.  Experimenting in Maple with a few smallish values of n (recall that p(x) is defined in terms of n) we see that the solution of (*) is an alternating series which begins with

x(t) = (1/A)t - (h/A2) t2 + O(t3)

where

A = a1 a2 ... an,

and

h = 1 + 1/2 + 1/3 + ... + 1/n.

It then follows that x(1), which is the solution of p(x) = 1, is given by the alternating series

x(1) = 1/A - h/A2 + ...


We conclude that

1/A - h/A2  < x(1) < 1/A.


QED

@Carl Love That phrasing struck me as a bit strange too, but I attributed it to a misuse of the English idiom "to think twice" by a non-native speaker.  I assume that it was meant in the spirit of "I will consider the use of piecewise functions more seriously from now on".  I stand to be be corrected, however.

@Markiyan Hirnyk When I copy/paste the lines that you have shown into my Maple 18, things work as intended.  Try a restart.  If that doesn't work, then perhaps the problem is due insufficient memory.  See what happens if you change numpoints=50000 to somethng smaller, perhaps numpoints=5000.  I can't think of another reason.

@Carl Love Thanks again for the explanation.  I seem to have a blind spot somewhere which prevents me from grasping this. I will think it over and ask for clarification if the puzzlement continues.

@Shervin I obtained the ranges by trial-and-error.  Specifically, I plotted the surface in a smallish box, then gradually expanded the box in order to capture the complete surface.  I am sure that this could have been done more systematically but I didn't bother.  Here are the plots corresponding to Mu=0.5 and Mu=0.99 plotted over the x1=0..60, x2=0..10, x3=0..2 range:

As Mu approaches 1, the surface approaches to what looks like the three adjacent faces of a cube.

The following animation shows the surface with Mu = 0 .. 1 in small increments:

Here is the Maple worksheet that produced that: anim.mw

Please note:  I wouldn't use these plots if they are for any serious presentation.  The implicitplot3d() function by its nature produces very rough looking surfaces.  Looking at the results, however, we see that the surface for each Mu value consists of three or four very simple patches.  You can get much better-looking results by figuring out the equations of those patches and using Maple's plot3d() function for plotting perfect representations.

@Carl Love and acer, thanks for your detailed explanations.

What is still bothering me is that I cannot explain the different outcomes of Experiment #1 and Experiment #2 in my post above.  It seems that the two vectors are viewed as identical objects in one experiment and not in the other.

@Markiyan Hirnyk The first condition in the OP's piecewise function is

(`or`(x1 > 55, x2 > 10), x3 >= 2)

In your code this has changed to

(`or`(x1 > 55, x2 > 10), x2 >= 2)

The graph will look different once you fix that.

Then we wil find that the proper plotting range is

x1=0..60, x2=0..10, x3=0..2

Carl, in an attempt to understand the logic of your solution, I did the following two experiments.

Experiment #1:

{ <a,b>, <a,b> };

This works as expected -- a set of two identical items is the same as the set of one of them.

Experiment #2:

<a,b>, <a,b>;

{ % };

I am puzzled by why this set does not collapse to a single item as in Experiment 1.  Do you know the reason why?

Aside: If we replace vectors with lists in these experiments, then the outcomes of the two experiemnts turn out to be the same.  You relied on that in your soluiion.

@rlopez Providing a student with a complete solution to a homework assignment is not unlike "helping" a runner who is training for a race by giving him a ride.

I suspect that during your teaching years you would not have been overjoyed if a student presented you with someone else's work as their own.  There is no good reason to do otherwise now.

@Kitonum How is a graph of (w, p(w)) any different from a graph of (w(z), p(w(z))?  Aren't they the same?  What is wrong with doing it in the simple way that I have shown, especially since w and p are functions of z, and the OP seems to be interested in in the range z=0..26?

@tomleslie I failed to connect this version of the question to the previous one and the corrections that you had suggested, which in retrospect make very good sense.

@reinhardsiegfried The polynomial method you are proposing merely reinvents the series solution.  If the series solution fails for some reason, then your polynomial solution will fail too because it's the same thing.  Linear or nonlinear does not matter.

@reinhardsiegfried OK, if you want a truncated polynomial approxination, then you may want to try this.

mathieulibert Perhaps it can help if you explain how you calculate a, b, c in terms of k1, k2, k3 and the three angles.  I don't know how, and I didn't find an explanation in your worksheet.