I've just posted a comment to this ancient Post of yours

But unfortunately just yet another proof that ChatGPT often talks nonsense (I wonder about what it says in Politics, History and so on).

Anyway  I forwarded your post to one of my daughters who teaches math in French high-schools (equivalent to US 11th and 12th grades).
She sent me a mail containing what ChatGPT answered to the question "I don't understand the calculus 5.9-5.11"... and this answer definitely worths a read.

Here it is (in French) but I provide you a litteral translation


ChatGPT acknowledges that is indeed a good question (Bonne question!).
Next line: "Let's see why this seems confusing"  (ChapGPT is about to prove us it was right, I am delighted in advance.)

Step 1 (Etape 1): Write the numbers with the seme number digits right to the comma (hum...)
Next line: To ease the comparison or to substract, one can write 5,90 instead of 5,9 (Yep...)

Step 2 (Etape 2) realize the substraction
                          5,90 - 5,11 = ...  = 0.79 (Good, ChatGPT really is a champion)
Next line: But be careful (Mais attention)
Next line: It's done the wrong way around! (C'est fait dans le mauvais sens)  (what the Hell ChatGPT has in its artificial mind ???)

Let's ChatGPT explains itself
Next line: What you wabt do do is: (Tu veux faire:)
                          5,9 - 5,11 = 5,90 - 5,11  = -0,21 (WT...!!!)
Last line: ChaptGPT ends its explanation saying "Because you substract a larger number (5,11) to a smaller (5, 9)"
(Car tu soustrais...).

I am confused, does ChatGPt thinks 5,11 is larger than 5,9 because the former has decimal digits than the latter?

BTW: I adore this pure mathematical claim "It's done the wrong way around!"

@Alfred_F 

You will find in the attached file several plots which should help you understand why your latler command is a dead end.

To illustrate my purpose I defined the function

f := a -> x -> cos(a*x)+cos(x)

ad looked to what happens when `a` is equal to Pi or to approximations of Pi obtained by continued fractions.

Let C_n the nth approximation of Pi got this way.
The period of f(C_n) increases with `n` and plotting f(C_n) over a perior will show absolutly nothing.

So I have thought plotting f(C_n) over limited ranges R(m) of length 4*Pi defined this way:

R := m -> 4*Pi*(m-1) .. 4*Pi*m

Plots compare f(C_n) to f(Pi) over different ranges corresponding to increasing values of m.

The two last plots show that up to m = 10⁹  (that is x ~ 1.2x10¹⁰) there is practically no difference between f(C₁₀) and f(Pi) but that differences appear for m ~ 3x10⁹ .  (For the record |C₁₀ - Pi| is about 10^-12.)

What does that mean? That if you want to see differences (I'm not even talking here of periodicity) between f(a) where `a` is an irrational number and some accute rational repentations of `a` you will have to look very far... in fact at the infinity when the rational representations become more and more accute.

So, to your question
At what "closeness" of a(n) to e does it become apparent that for irrational a, i.e., a = e, the function y(x) is not periodic.
I will answer: it never becomes apparent that ...

Indeed, given some irrational number A, we know that f(R(A, e)) is periodic for any rational approximation R(A, e) of A with accuracy `e`.
The plots in the attached file show that you fave to look beyond some threshold T(e) to see differences between f(A) and f(R(A, e)). Can you conclude from this that f(A) is non perioric?
Of course not because f(R(A, e' > e)) is almost indistinguishable from f(A) up to T(e') > T(e): the farther you look your can always use a more accute rational representation of A such that no difference between the plots may be visible.

Empirical observations suggest that f(Pi) and f(R(Pi, e)) are almost indistinguishable up to x ~ 1/e

periodicity.mw

@nm 

Let's do this

# With u(x) = v(x)^2

usys := {(1/2)*diff(u(x), x)=g, u(x__0) = v__0^2}:
usol := dsolve(usys);
vsol := solve(v(x)^2 = rhs(usol), v(x))

vsol makes you think there are two solutions, which is wrong, the first one corresponds to v__0 > 0 and the second one to v__0 < 0.

In the excerpt you give problem 30 has some untold implicit hypotheses (for instance x__0 > 0, v__0 > 0, g > 0, V > V__0, x > x__0).
Let's do this:

sys := {v(x)*diff(v(x),x) = g, v(x__0) = v__0};
SOL := (x0, v0) -> dsolve(eval(sys, {x__0=x0, v__0=v0})):
print():

`<|>`(
  `<,>`([` `, x__0=-1, x__0=+1])
  ,
  Matrix(3, 2, (i, j) -> `if`(i=1, v__0=(-1)^j, SOL((-1)^(i-1), (-1)^j)))
)

The ANSwer of problem 30 is quite restrictive as it does not enable finding those four solutions above (in fact ANSwer provides spurious solutions)

ANSwer := (x__0, v__0) -> v^2-v__0^2 = 2*g*(x-x__0);
`<|>`(
  `<,>`([` `, x__0=-1, x__0=+1])
  ,
  Matrix(3, 2, (i, j) -> `if`(i=1, v__0=(-1)^j, allvalues(isolate(ANSwer((-1)^(i-1), (-1)^j), v))))
)

@acer 

Not yet, but it's a good idea.
The main barrier is that it takes quite a bit of time if you want to do it properly: explanations, references, documentation. Probably the worksheet has to be written differently and 2D mode used to make the stuff attractive.

Nevertheless, I will think about it.

@acer 

That's much better, thanks a lot.

@acer 

indeed works weel with Maple 2015 indeed works well with Maple 2015.
The real definition of Dist was something like

Dist := proc(N::{posint, name}, alpha::{Vector, list})

and thus aimed to work with different alpha types.
So I just have to remove the Vector type and your code will be perfect.
Nevertheless the need to put L between quotes in X2 := RandomVariable(Dist(N, 'L')) is quite

For information: your codes 1 and 2 generate this error in Maple 2015

Error, (in Statistics:-PDF) bad index into Vector

I have 3 questions concerning your last code

1. The need to put L between quotes in X2 := RandomVariable(Dist(N, 'L')) doesn't feel quite natural or comfortable.
   I assume there is no other way to proceed with Maple 2015, is there?
    
2. You write value(eval(res2,1)) whereas value(eval(res2)) does the same thing: what is the need to specify the evaluation level here?
    
3. As this question is related to a post I'm about to deliver, would indeed works weel with Maple 2015 work with, let us say, Maple 2025?

In any case, thank you very much for your contribution.

By the way: in your Maple 2025 codes you write alpha::evaln in the parameter sequence. I had always thought in this sequence N::T refered to to expected type T of name N... but evaln is not a type, so "How can  alpha::evaln be authorized?"

I assume there are points on the border that are placed in the same way as the red dots, aren't there?

If it is so and you replace the origin by a quarter of circle with a small radius,  then there exists an homeomorphism from this truncated domain to a square (more simply you can unfold and rescale the truncated domain to get a square).
So you find yourself in the situation you addressed in the example contained in your worksheet.

In fact not really the same because the origin (more precisely circle C) has been replaced by a side of the square which thus contains N boundary points.
So, after having decomposed the matrix the way you did, simpliy remove on each matrices N-1 rows and columns corresponding to N-1 of these boundary points .

It seems to me that as long as:

1. the points are vertices of a structured mesh made of squares (possibly curved squares),
2. and there exists an homeoporphism from the domain you consider into a square (possibly after having removed some singularities),

then you can apply your decomposition strategy.

Note that if your domain was the whole disk D, replacing the origin by a small radius circle would create a fictitious boundary containing N points (there still exists an homeomorphism from D\{0} into the unit square), then you should remove N rows and columns in the matrices the decomposition gives.

@Ahmed111 

You're wrong, add these commands at the end of the file I sent you to see that "Maple DOESN'T treat both I and J as the imaginary unit"  (at least with my Maple 2015 version [see top this same file])

is(I=J)
                          false

sqrt(J);
sqrt(I);
                     1  (1/2)   1    (1/2)
                     - 2      + - I 2     
                     2          2         
                              (1/2)
                             I     
J*conjugate(J);
I*conjugate(I)
                               1
                                _
                              I I

Because I use Maple 2015 and the simplification rules will not do the same thing that those of your Maple version, I did not go further 
the expression of f2.
Nevertheless you will found how to "factor out I = sqrt(-1) from square roots".
up_to_f2.mw

Use this in your own code and Maple version to see if this helps.

@Alfred_F 

You write "the three circles aren't enough to define exactly one sphere. There are obviously a whole host/set of spheres that meet the requirements": so why did you ask the question?


"This leads to the question: What is the geometric locus of all sphere centers?"
Ill posed problem: radii of the intersection circles doesn't enable finding this locus.

To understand why ;simply take the anologuous 2D situation of a disk D (it will be simpler to consider a disk than a circle) intersecting axes Ox and Oy.
Its imprints are two segments Sx and Sy whose mid points have coordinates (a, 0) and (b, 0). Thus the center of D is (a, b).
Let Lx and Ly the lentghts of 2_*p and 2_*q respectively. Then the square radius of D is R² = b² + p² = a² + q² [eq1]
Can we move Sx along Ox and Sy along Sy to other locations, let's say (a_*m, 0) and (b_*n, 0) in order that they still are imprints of a disk D' of a possibly different radius R'?
Because of the relation b²n² + p² = a²m² + q² [eq2] not all couple (m, n) is allowable.
Using [eq1], [eq2] rewritties n² = 1+ (m² -1)_*(p/q)²  [eq3].

One special case if p = q which leads to m = n or  m = -n.
Then the locus of the disk centers are the two main diagonals y = x and y = -x.

Let us assume that p < q, a > p and b > q (the reasoning which comes can easily be extend to the case p > q by interverting p and q), just to locate the center of disk D in the first quadrant.
When m = 0 the segment Sx corresponds to the interval x = [-p, p] and by relation [eq3] the vertical imprint Sy of the disk is the seqment [Z-q, Z+q] where Z is the positive square root of 1- (p/q)² (which is strictly larger than 0). In this case the radius of the disk is 2_*q (Z-q < 0 and Z+q > 0).
So the point [0, Z > 0] is on the locus if the disks centers.
The simple change of coordinates x --> x, y --> (y-Z)_*(p/q) shows that the locus of the disk centers is made of two two main diagonals (a point established before).
So, in the original referential the locus is made of two straight lines passing through the point (0, Z) and of slopes arctan(b/a) and -arctan(b/a).

Note that the sole values of the imprint radii p and q are not enough to determine this locus.

Extending this reasonning to the the sphere case  shows the locii are straight lines (the 4 main diagonals of the 8 octants of the ad hoc transformed referential).
This transformation is of the form x --> (x-a_x)/p_x, y --> (y-a_y)/p_y, z --> (z-a_z)/p_z, where the a_i denote the coordinates of the imprints  mid point and the p_i the imprint sizes.
By the inverse transformation one gets the equations of the locci.

Lets take a sphere S of radius r whose center C has coordinates (a, a, a).
So the intersections of S with each of the three canonic planes are circles of identical  radii (provided of course that the intersections do exist, that is that r > a). Let u this common radius
Now move the center at  another place (A, A, A) (A > a), rgere is always a value or R > r such that R²-A²=u².
So the radius and the coordinates of the sphere cannot be uniqueky determined by the radii of the intersection circles.

"the sphere's radius r and the coordinates of its center can be determined?": NO

One has exactly the same result in 2D for a circle intersecting two orthogonal lines.

@KIRAN SAJJAN 

saying "When I change the value of j, only one line is plotted".
If you make a zoom you will see there are 4 curves. Change for instance Grt from 0.1 to 2 to convince yourself the code gives different resullts.
For instance

By the way I wonder by what mistery you can have past 12 hours working on the code I posted only 8 hours ago?

@KIRAN SAJJAN 

I only took Br=0.1 has an illustration of what you can do for the other values in BrVals.

Feel free to go further when and proceed as described in the second part of the worksheet for those other Br values.
I don't think it is an impossible task for you, you have just to change the value of j here
As I told you I don't have more time to spend on this problem and your initial worksheet symetry_paper_work.mw proves you know perfectly how to browse all Br values in BrVals, build a table of results and display them...so I don't feel you need me any longer.

@KIRAN SAJJAN 

I guess this is closer to what you have in mind:

• The first part of the worksheet 2_domains_sand15 explains the strategy I use.
  The whole solution (the "two-domains" solution) depends on two parameters A and B where A is the common unknown value of u₁(0) and u₂(0) and B is the common unknown value of Theta₁(0) and Theta₂(0).
  In domain D₁ = {y, y in [-1, 0]} the differential system is well posed and can be solved numerically a soon as numerical values a and b have been assigned to A and B. Let S₁(y ; a, b) the solution of the problem 1 on domain D₁.
  In domain D₂ = {y, y in [0, +1]} the differential system can also be solved numerically and let S₂(y ; a, b) its solution.
  Let Jump the vector whose components are

  Jump1 := (D(u1))(0)     - Z1*(D(u2))(0),
  Jump2 := (D(Theta1))(0) - Z4*(D(Theta2))(0)
  

  For an arbitrary choice of a and b there is no reason for Jump₁ = Jump₂ = 0  when S₁(0 ; a, b) and S₂(0 ; a, b) are used to assess D(u₁))(0), D(u₂))(0), D(Theta₁)(0) and D(Theta₂)(0).
   
• Second part of the worksheet 2_domains_sand15.
  I wrote a procedure named TwoDomainsSolver(a, b) wich gathers all the steps presented in the first part and returns the L₂ norm of the Jump vector.
  Then all you have to do is to use an optimization function to findthe couple (a*, b*) wich minimizes the output of TwoDomainsSolver(a, b) over some (a, b) domain.
  Once (a*, b*) is found you can draw the "left" and "right" solutions.

I provide you two worksheets corresponding to two different data
2_domains_sand15.mw
2_domains_sand15_other_data.mw

For other data, for instance higher values of the Grashof number, you may get this error

Error, (in dsolve/numeric/bvp) initial Newton iteration is not converging

A quite common one but an error which is always borring and may be relatively difficult to fix, even if some recipes are given here Numerical Solution of Difficult ODE Boundary Value Problems

One example is given in 2_domains_sand15_Gr=15.mw where this error appears for Gr=15.
You will find in this worksheet a way to fix it (read carefully the last comment lines).

I'm sory to tell you that I'm done with this thread because I am a lot of other things to do.
I hope this response has been helpful.