acer

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19 years, 310 days
Ontario, Canada

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These are replies submitted by acer

Plot problem reported, regression in Maple 2024.0,

plot(Pi/2-arccos(h/sqrt(h^2+2)),h=0..3),
plot(Pi/2-arccos(h/sqrt(h^2+2)),h=0..3,adaptive=true);

plot(arccos(1/sqrt(x))),
plot(arccos(1/sqrt(x)),adaptive=true);

@minhthien2016 The result is correct.

But you can get rid of the warning with,

   solve(temp = Pi/3, h, useassumptions);

What happens if you replace sum by add?

note: It'd be more useful if you would provide the complete code-to-reproduce.

@nmacsai If there were a way to change the color of the edge of a POLYGONS structure then we'd likely give it in an answer.

I believe that a POLYGONS plotting structure can have either a style where the edge is shown and black, or a style where the edge is not shown.

A POLYGONS plotting structure only has one COLOUR substructure, and that relates only to the color on the interior face.

@AHSAN 

You should,
1) define properly (clearly and in full detail) what you mean by, "check the convergence or validity of the numeric method".

2) define properly (dlearly and in full detail) what you mean by, "streamline pattern" in this context. Simply restating it as, say, "stream functions" is not adequate.

If I don't respond further it's likely because I didn't see adequate details.

@AHSAN I didn't put the "error" in before because you hadn't yet told us what you meant by that.

Now that you've responded that you want the absolute error, here it is:
   HelpAhsan_acc.mw

Do you think that your series solution is accurate? If so, why?

@WA573 You didn't mention D[2](s)(n+1,t) before, and I so didn't consider it because as-is it's not like, "functions with s(n+1,t)" as you originally wrote. But it might first be changed to Diff form, to get that.

Is this correct, as far as D[2](s)(n+1,t) goes? I mean, except for your other query. I'd have to think about your other query, about "(∂)/(∂ (sigma2*t))=1/(sigma2)(∂)/(∂ t), sigma2 is constant".

shift_acc.mw

@WA573 I gave three alternatives for assigning to sourcefolder. I don't know whether all three fail for you.

More importantly, I cannot tell whether you have a file named wkptest_cpc in that folder (wherever it may be).

@MaPal93 Yes, I had already seen (yesterday) that your s2 gave rise to a negative Gamma for your rho=0.5 example (whose 2D plot showed 3 roots.).

And using the s2 formula and your choice rho=0.5 also showed the three roots for lambda when using implicitplot3d -- though you had used that command only for your s3.

However, I looked though your worksheet and it seemed a sparse to me in terms of clear explanation of requirements. I didn't see any mention that Gamma be positive.

You used Gamma=Gamma__1, and described Gamma_1>Gamma_0, and gave both by formulas in terms of rho. But you didn't even plug your rho=0.5 into those formula. I plugged them in an saw that it made Gamma__1 be a negative value, and also used in a range for implicitplot3d.

But I saw no description that Gamma=Gamma__1 ought to be taken as positive.

Indeed you query 1) in your Question at top states,

  1. Why implicitplot3d returns a unique strictly positive root for all rho and Gamma but plot() of the quartic evaluated for a specific pair of value returns three positive roots?

Why does it say "all rho and Gamma" if you're only concerned with Gamma>0 ?

Also, your worksheet starts with the query,
  "Question 1: does the root of the quartic Eq=0 coincide with r, the RootOf the second quartic?"
You don't actually say what "the second quartic" is. Is that phrasing a reference to the quartic inside the RootOf, with _Z=Lambda? If so, then why not,

evala(eval(Eq, Lambda=r));
        0

I adjusted the answer to show how that second plot can be obtained, using either odeplot itself or either of the two other ways to utilize the S(t) in the dsolve results.

@Carl Love Using your code,

periode(1/20);

Error, (in periode) invalid input: Base expects its 3rd argument, d, to be of type posint, but received 0


Also, the your individual routines (as building blocks) might be as fast as their NumberTheory equivalents, but your revised periode is not always as fast as can be had using NumberTheory. (There's quite a bit of garbage collection.)

CodeTools:-Usage(seq(F(1007/(20035+2*i)),i=1..1050)):

memory used=1.09GiB, alloc change=319.50MiB, cpu time=11.04s, real time=10.45s, gc time=2.24s

CodeTools:-Usage(seq(periode(1007/(20035+2*i)),i=1..1050)):

memory used=28.10GiB, alloc change=383.50MiB, cpu time=117.72s, real time=65.73s, gc time=88.87s


RP_CLAC.mw

This reply doesn't address the OP's original question.

@Carl Love It seems to me that there a few interesting bits below.

restart;

ee := F( ((x->x)-rhs)(a=b) );

F((a = b)-b)

%;

F(a-b = 0)

eval(ee,1);

F((a = b)-b)

op(eval(ee,1));

a-b = 0

lprint(op(eval(ee,1)));

(a = b)-b

lprint(eval(ee,1));

F((a = b)-b)

hh:='(a = b) - b';

a-b = 0

lprint(eval(hh,1));

a-b = 0

ff := ''(a = b) - b'';

'(a = b)-b'

lprint(eval(ff,1));

'(a = b)-b'

op(eval(ff,1));

a-b = 0

lprint(op(eval(ff,1)));

(a = b)-b

Download fcn_eval_01.mw

@Paras31 Sorry, I don't know what version or licensing you are entitled to purchase.

You're probably best off contacting Maplesoft customer service, or sales.

Please put followups on this here, instead of in wholly separate new Posts.

@C_R A tweak,

restart;

ineq := a/c < b/c;

a/c < b/c

signum(c)*map(`*`,ineq,abs(c)) assuming c>0;

a < b

signum(c)*map(`*`,ineq,abs(c)) assuming c<0;

b < a

map(`*`,ineq,abs(c)) assuming c>0;

a < b

map(`*`,ineq,abs(c)) assuming c<0;

-a < -b

 

Download ineq_signum.mw

Naturally, a merit of both vv's approach and my tweak is that there's no burden to remember the assumption on c (eg. its signum) and explicitly enter it as part of the formula.

You shouldn't have to type in the extra factor of -1 explicitly, just because you earlier did assume(c<0) or similar.

Either formula directly handles either assumption.

Eg, using assume instead of assuming,

restart;

assume(c<0);

ineq := a/c < b/c:

signum(c)*map(`*`,ineq,abs(c));

b < a

map(`*`,ineq,abs(c));

-a < -b

restart;

assume(c>0);

ineq := a/c < b/c:

signum(c)*map(`*`,ineq,abs(c));

a < b

map(`*`,ineq,abs(c));

a < b

Download ineq_signum2.mw

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