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 Discussion on answer by bbgodfrey: NM…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Dec 27, 2024 07:03
@bbgodfrey: Sorry, you still don't answer my comment " got the range {0, 6 Pi/5}, but then u[0] == 3*Pi/5 should be. This produces ArcLength equal to 12.8527".
user64494
Dec 27, 2024 01:01
@bbgodfrey: Sorry, somewhat more accurate execution gives ArcLength is equal to 9.94234. Details on demand.
user64494
Dec 27, 2024 01:01
I got the range {0, 6 Pi/5}, but then u[0] == 3*Pi/5 should be. This produces ArcLength equal to 12.8527.
user64494
Dec 27, 2024 01:01
Sorry for the typos. It should be "The question arises; what is found by you?" in my first comment to your answer.
user64494
Dec 27, 2024 01:01
I repeat, you don't define h[u, v], Is it h[u_, v_] = {(R + r Cos[u]) Cos[v], (R + r Cos[u]) Sin[v], r Sin[u]}; from the question?
user64494
Dec 27, 2024 01:01
1. Sorry, "Your last comment is, I believe` incorrect.' are ungrounded words. 2. Integrate[eq2*2*2 u'[v], v] results in 4*Integrate[(Derivative[2][u][v] == ((4 + 2*Cos[u[v]])*(-32 + g^2 - 32*Cos[u[v]] - 8*Cos[u[v]]^2)*Sin[u[v]])/(2*g^2))*Derivative[1][u][v], v], not eq'.
user64494
Dec 27, 2024 01:01
Next, should not be ArcLength[%%, {v, 3*Pi/4, 6 Pi/5}] instead of ArcLength[%%, {v, 0, 6 Pi/5}] (see the question under consideration)? The former produces 4.97607.
user64494
Dec 27, 2024 01:01
Thank you. 1. Unfortunately, the result of Plot[f[g], {g, -6, 6}, AxesLabel -> {g, "f"}, LabelStyle -> {10, Bold, Black}, PlotRange -> All] looks fantastically (compare with the plots in thethred linked by me. ). 2. I am not a specialist in ODEs. Can you kindly elaborate your "eq1 is equivalent to eq2 in the sense that it is a first integral of eq2", giving us details? 3, h[u, v] is not specified by tou so I see empty plots.
user64494
Dec 27, 2024 01:01
Executing you code, after f = ParametricNDSolveValue[{eq2, u[0] == 0, eq1 /. v -> 0}, u[6 Pi/5] - 3 Pi/5, {v, 0, 6 Pi/5}, g]; Plot[f[g], {g, -6, 6}, AxesLabel -> {g, "f"}, LabelStyle -> {10, Bold, Black}] I obtain "ParametricNDSolveValue::ndinnt: Initial condition -((0.166673 (r+R) Sqrt[-35.9971+(r+R)^2])/r) is not a number or a rectangular array of numbers." I think you omitted the specification of R and r.
user64494
Dec 27, 2024 01:01
BTW, I don't see your eq2 in the document linked by you.
user64494
Dec 27, 2024 01:01
One more question. D[Derivative[1][u][ v] == ((R + r Cos[u[v]]) Sqrt[-g^2 + (R + r Cos[u[v]])^2])/(g r), v] // FullSimplify produces Derivative[2][u][v] == ((g^2 - 2*R^2 - 2*r*Cos[u[v]]*(2*R + r*Cos[u[v]]))*Sin[u[v]]* Derivative[1][u][v])/(g*Sqrt[-g^2 + (R + r*Cos[u[v]])^2]), not u''[v] == ((2 + Cos[u[v]]) (-32 + g^2 - 32 Cos[u[v]] - 8 Cos[u[v]]^2) Sin[u[v]])/g^2. Are eq2 and eq1 equivalent?
user64494
Dec 27, 2024 01:01
Can you kindly elaborate your claim "The crossing at g = 0 is spurious", giving us details?
user64494
Dec 27, 2024 01:01
Also see that thread. There are other references (not only linked by you) there.
user64494
Dec 27, 2024 01:01
I'd like tp quote the linked document "The majority of geodesics on the torus are not æsthetically pleasing. They are aperiodic and cover either the entire surface (if the geodesic is unbounded) or the outer region of the surface bounded by the barrier curves (if bounded). The rare exceptions are the geodesics which return to their starting point after just a few circuits around the z axis". The question arises; whay is found be you?
 

 Discussion on answer by azerbajdzan: …

Imported from a comment discussion on mathematica.stackexchang...
user64494
Nov 21, 2024 16:10
@xzczd: In 14.1 `Integrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]` performs `1/12 (Sqrt[Sqrt[x] + x] (-3 + 2 Sqrt[x] + 8 x) +
3 ArcCoth[Sqrt[Sqrt[x] + x]/Sqrt[x]])` without any condition. The command of Maple 2024 `FunctionAdvisor(branch_cuts, sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, plot = 2.)` produces the following branch cuts `[sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, x < 0, And(x = alpha + 1/2 - sqrt(4*alpha + 1)/2, alpha in RealRange(-infinity, Open(0))), And(x = alpha + 1/2 + sqrt(4*alpha + 1)/2, alpha in …
user64494
Nov 21, 2024 16:08
@xczd: In 14.1 `Integrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]` performs `1/12 (Sqrt[Sqrt[x] + x] (-3 + 2 Sqrt[x] + 8 x) +
3 ArcCoth[Sqrt[Sqrt[x] + x]/Sqrt[x]])` without any condition. The command of Maple 2024 `FunctionAdvisor(branch_cuts, sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, plot = 2.)` produces the following branch cuts `[sqrt(sqrt(x) + x)*(-3 + 2*sqrt(x) + 8*x)/12 + arccoth(sqrt(sqrt(x) + x)/sqrt(x))/4, x < 0, And(x = alpha + 1/2 - sqrt(4*alpha + 1)/2, alpha in RealRange(-infinity, Open(0))), And(x = alpha + 1/2 + sqrt(4*alpha + 1)/2, alpha in …
user64494
Nov 21, 2024 00:38
Sorry, you wrote "If you provide a more complicated example I may try to do the same with it".
user64494
Nov 21, 2024 00:38
Can you kidly support your claim " Exactly the same RegionFunction works also for your code" by a code and a plot? And the same with Num[x_?NumericQ] := NIntegrate[Sqrt[y + (y + I)^(1/5)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I}].
user64494
Nov 21, 2024 00:38
This is a fake, not a solution How about Num[x_?NumericQ] := NIntegrate[Sqrt[y + (y)^(1/2)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I}] and Num[x_?NumericQ] := NIntegrate[Sqrt[y + (y + I)^(1/2)], {y, 0, x}]; ComplexContourPlot[Im[sNum[x]] == 0, {x, -2 - I, 2 + I}]?
 

 Discussion on question by babyK: Samp…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Nov 5, 2024 00:40
@DavidlG.Stork: "You said "No computer system can operate with these [irrational numbers]." That is provably false.". The exact quote is "Any real irrational number is an infinite non-periodic decimal. No computer system can operate with these". Sorry, I prefer arguments over ungrounded claims. That's all.
user64494
Nov 5, 2024 00:40
In fact, RandomReal[{0,1},WorkingPrecision->10] produces a random rational from the set Table[j/10^10,{j,0,10^10}]. See the documentation for info.
user64494
Nov 5, 2024 00:40
@DavidlG.Stork: Yes, Mathematica can operate with some irrational numbers from a countable set. However, all the real irrational numbers form an uncountable set. BTW, I am PhD in math for ages and I am too old for your remarks. Do you understand me?
user64494
Nov 5, 2024 00:40
Any real irrational number is an infinite non-periodic decimal. No computer system can operate with these.
 

 Discussion on question by yode: How t…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Jan 31, 2024 12:41
@yode: I repeat my question "What did you try on your own?". Waiting for a serious reply of you.
user64494
Jan 31, 2024 12:41
@yode: What did you try on your own? Did you at least look in the link from the above comment of me?
user64494
Jan 31, 2024 12:41
Look in mathoverflow.net/questions/314189/… for info.
user64494
Jan 31, 2024 12:41
This is rather math, not Mathematica.
 

 Discussion between user64494 and Alex…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Jan 30, 2024 06:14
@AlexTrounev: I prefer arguments and formulas over ungrounded words "call standard utility InitializePDECoefficients with parameters "DiffusionCoefficients" and "MassCoefficients" that can be used to solve diffusion equation as well". Wher is it documented?
user64494
Jan 28, 2024 16:21
@AlexTrounev: I don't see any diffusion equation in the Wiki article. I recall `cdata =
InitializePDECoefficients[vd, sd,
"DiffusionCoefficients" -> {{-IdentityMatrix[2]}},
"MassCoefficients" -> {{1}}];` from your code.
user64494
Jan 28, 2024 16:17
@AlexTrounnev: As I wrote I don't understand the argument from the link. As I understand it, you, following the linked answer by Henrik Schumacher, solve the diffusion equation in reg1. Why is it a conformal map from reg1 to the open unit disk?
user64494
Jan 28, 2024 16:17
@AlexTrounnev: How do we use Riemann mapping theorem to compute f1? Please give us details. No, my request does not concern the corners of the triangle.
user64494
Jan 28, 2024 16:17
@AlexTrounnev: Sorry, why is f1 an approximation of a conformal map from reg1 to the unit disk?
user64494
Jan 28, 2024 16:17
@AlexTrounnev: Thank you. I still don't understand why "we have some numerical approximation to conformal map". Any picture is not a serious argument.
user64494
Jan 28, 2024 16:17
Can you explain why your code does produce a conformal map? I don't understand arguments from the link. TIA.
 

 Discussion on answer by Alex Trounev:…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Jan 30, 2024 05:20
Let us continue this discussion in chat.
user64494
Jan 30, 2024 05:20
@AlexTrounnev: As I wrote I don't understand the argument from the link. As I understand it, you, following the linked answer by Henrik Schumacher, solve the diffusion equation in reg1. Why is it a conformal map from reg1 to the open unit disk?
user64494
Jan 30, 2024 05:20
@AlexTrounnev: How do we use Riemann mapping theorem to compute f1? Please give us details. No, my request does not concern the corners of the triangle.
user64494
Jan 30, 2024 05:20
@AlexTrounnev: Sorry, why is f1 an approximation of a conformal map from reg1 to the unit disk?
user64494
Jan 30, 2024 05:20
@AlexTrounnev: Thank you. I still don't understand why "we have some numerical approximation to conformal map". Any picture is not a serious argument.
user64494
Jan 30, 2024 05:20
Can you explain why your code does produce a conformal map? I don't understand arguments from the link. TIA.
 

 Discussion on answer by Roman: Why do…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Dec 3, 2023 19:52
@drer: An answer to "But Roman proves the convergence only for a special subsequence of the sequence of the partial sum. Hope you feel the difference".
user64494
Dec 3, 2023 19:52
@drer: Sorry, this is off-topic. You avoid of an answer.
user64494
Dec 3, 2023 19:52
@drer: But Roman proves the convergence only for a special subsequence of the sequence of the partial sum. Hope you feel the difference.
user64494
Dec 3, 2023 19:52
I'd like to comment the @Roman's answer. In fact, handling Floors by hand, he proves the convergence of the subsequence of the partial sums $S_{4k}$ to $\frac{1}{4} (\pi +\log (4))$ as $k$ tends to infinity. This does imply the convergence $S_{4k+1},S_{4k+2}$, and $S_{4k+3}$ to $\frac{1}{4} (\pi +\log (4))$ as $k$ tends to infinity. since the differences tend to zero. This argument is not stated by @Roman. That's all.
user64494
Dec 3, 2023 19:52
@MishLavrov: Don't understand you at all. Can you refer to some theorems?
user64494
Dec 3, 2023 19:52
@drer: Can you ground " For convergent series regrouping does not change the result (the absolute convergence is not required)"?
user64494
Dec 3, 2023 19:52
@Roman: More exactly speaking, you use grouping which is not grounded. A simple example: a series 1-1+1-1+1... diverges, but (1-1)+(1-1)+... converges. Hope I am clear.
user64494
Dec 3, 2023 19:52
The series does not absolutely converge. Therefore, such a rearrangement is not grounded (see Wiki for info).
 

 Discussion on question by Tim: How to…

Imported from a comment discussion on mathematica.stackexchang...
user64494
Nov 28, 2023 16:33
@UlrichNeumann: The value of z[1] makes no sense here. I recommend you to read this book.
user64494
Nov 28, 2023 16:33
@UlrichNeumann; Despite your words, NDSolve works well with DiracDeltain many cases. Here is an example:sol = NDSolve[{z'[t] == DiracDelta[t - 1], z[0] == 1}, z[t], {t, 0, 3}];Plot[z[t] /. sol, {t, 0, 3}].