acer

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19 years, 314 days
Ontario, Canada

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These are replies submitted by acer

@MalakMMK Do you understand than the two images compute two different things, which is one reason why they differ in speed (as well as the main reason why they different in coloring)?

The first image has four colors because it shades according to which root, and your current example has four roots. It takes some extra time to classify the points -- by root.

You wrote, "...in the second image I'm Always getting 3 colors" but that is not true. It has a variety of colors, whose hue corresponds to the number of iterations counted.

In my earlier code I tried to simplify the real and imaginary parts of the iterative formula. That simplification can be too expensive, and even unnecesary as my later code showed for the Hafiz method.

You should try to understand the earlier code to the degree that you could verify that it's working properly. (I might have made mistakes!) And if you cannot edit and adjust the code then how can you adequately understand it? If you cannot adequately understand the code then how can you use it in your coursework?

@MalakMMK In your most recent followup query (ie. the Repley immediately above) you have not made it completely clear what you want.

You have not provided the code you used to generate those images, which doesn't help with any guessing on out part.

You do not have to inline all of any attachments, here. You can upload it and then insert a link (instead of the whole contents, inlined).

I think that you might be trying to ask for something like this:
  - utilize the looped code (as used by my densityplot examples)
  - instead of producing the HUE-based densityplot image (colored by number of iterations to attain tolerance), produce the other kind of image I previously showed where the color is according to which root is closest to the converged value.

If that guess by me is correct then it could be done by extracting the hue data from the densityplot result and then treating it similarly to this line in my earlier procedure:

   # ...set the hue by position in the list of all roots
   G:=Array(zip((a,b)->min[index](map[evalhf](abs,rts-~(a+b*I))),
                  newt[..,..,1],newt[..,..,2]),datatype=float[8],order=C_order):

I may have some time to look at that tomorrow evening, or possibly the day after.

ps. I don't quite understand why you might prefer the manually written looped code. That Escape command provides a much faster (multithreaded, then compiled or evalhf'd) implementation of the loop's same iterate-while-counting-until-tolerance-met idea. The faster way worked fine for the Hafiz 4th order method.

@Stretto You wrote, "printing has nothing to do with mouse hovering".

I can think of one mechanism where a displayed list of point data could cause arbitrary extra information to be revealed upon mouse hover, another for mouse single left-click, another for mouse right-click, and another for mouse left-button-selection. None of those need show positional "graphs" of the data.

@PhearunSeng 

Tom's worksheet has eqs as a list.

Your image shows that you rewrote that so that eqs is a Matrix.

You claimed that your image shows you running Tom's worksheet. But your image shows that the command's that you are using are different from his.

Download Tom's attachment. Open that .mw file, and run it. Don't paste in something else.

Also, it makes little sense to have the equations be some scalars expressions added to Vectors. Be more careful about your syntax.

@PhearunSeng You are still calling restart after you define eqs.

You wrote, "Is there a way to print points that have..." (italics mine). But then you talk of "graphs".

So, are you trying to plot the point data, or print it like output, or...?

And you want hover-over information to popup, for different points?

(You description of how you're reading in the data, and the slowness of that, seems to be a quite different query. But it's difficult to offer best advice when you don't show code-to-reproduce.)

@Carl Love The OP sent me a link to another paper.

https://www.sciencedirect.com/science/article/pii/S0377042708003555

He said, "I used equation 14, Method 1 , and took values for alpha and gamma = 1".

And he sent me a worksheet that had an implementation of the formulas. As a rough test of that I plotted its "escape-time" using densityplot.

restart:

ee := z^3-1:

__FF := proc(z) option inline; z^3-1 end proc:
Wei1Count := subs(g=3, F=__FF, dF=D(__FF),
  proc(x0,y0) local u,i,v,A,p,fpu,fpv,fpuu,ut,W;
  u := evalf(x0+y0*I);
  for i from 1 to 20 do
    v := u - F(u)/dF(u);
    A := F(u)+(g-2)*F(v);
    p := v - 1/A*(F(u)+g*F(v))*F(v)/dF(u);
    fpu := (F(u)-F(p))/(u-p);
    fpv := (F(v)-F(p))/(v-p);
    fpuu := (dF(u)-fpu)/(u-p);
    ut := F(p)/F(u);
    W:= 1+2*ut+ut^2+ut^3;
    u := p - F(p)*W/(fpv +fpuu*(p-v));
    if abs(F(u))<1e-3 then return i; end if;
  end do:
  return i-1;
end proc):

 

plots:-densityplot(Wei1Count, -2..2, -2..2, grid=[201,201],
                   title="Wei's method (1)", size=[300,300],
                   colorstyle=HUE, style=surface);

NULL

Download Wei_method_ac2.mw

I haven't time to figure out why my earlier EscapeTime template runs away on it...

[note: in my original answer here I preprocessed the Newton's method (compounded) formula and hit its Im and Re components with evalc and simplify. I happened to do the same for my King's method implementation in that same Answer. But I did not do that in my followup implementation of the "Haf1" (Hafiz paper) 4th order method. There I left the Im and Re calls alone, since the simplify/evalc time is large, it's of little merit, and these days evalhf and the Compiler can handle those better.]

@Dkunb Here are a couple of ways to get that first image -- the contour N(beta,alpha)=0.

I used Axel's idea of integrating only out to 6 instead of infinity, and using oscillating method _d01akc instead of semi-infinite method _d01amc.

I used a list of values by rootfinding (in two ways), instead of contourplot or implicitplot.

I also show a simple (simplistic) idea of using Interpolation and using a coarser list of points. But the final curve appears quite smooth.

I also tried Grid:-Seq, and you might try tweaking that.

I may not have any more spare time for looking at this interesting problem. But this might illustrate a few alternate approaches.

restart;

F:=kappa->kappa:
f:=(alpha,delta)->exp(-abs(F(kappa))^2*(1+delta^2)/2-abs(F(kappa))*alpha)/abs(F(kappa)):

L:=(alpha,delta,Lambda) ->
  (lambda^2*exp(-alpha^2/2)/4)*(Int(f(alpha,delta),kappa= -infinity..-Lambda)+Int(f(alpha,delta),kappa= Lambda..infinity)):

CodeTools:-Usage(evalf(L(4,1,0.001)));

memory used=1.88MiB, alloc change=0 bytes, cpu time=65.00ms, real time=65.00ms, gc time=0ns

0.8209373770e-3*lambda^2

X:=Int(exp(-kappa^2*(delta^2 + 1)/2)*cos(kappa*beta)*cos(kappa*gg)/(kappa),
       kappa = Lambda .. infinity):

Y:='Int(-exp(-kappa^2*(delta^2 + 1)/2)*cos(kappa*beta)*Im(exp(kappa*gg*I)*erf( (gg+kappa*I)/sqrt(2)))/(kappa), kappa = Lambda .. infinity)':

theJ:='sqrt(X^2+Y^2)/2*exp(-alpha^2/2)*lambda^2':

J:=unapply(subs(infinity=6, theJ), [alpha,delta,Lambda,beta,gg]): # cut off

thisJ:=unapply(subsindets(J(alpha,1,0.001,beta,3),
                          'specfunc(anything,Int)',
                          'q -> Int(op(q), digits=15, epsilon = 1e-4, method = _d01akc)'),[alpha,beta]):

evalf(thisJ(4,8));

0.7304273940e-3*lambda^2

N:=proc(beta,alpha) option numeric; local res;
     Digits:=15;
     if not [beta,alpha]::[numeric,numeric] then
     return 'procname'(args); end if;
     res:=thisJ(alpha,beta) - L(alpha,1,0.001);
     res:=evalf[15](res);
     res:=evalf[10](res);
     if type(res/lambda^2,numeric) then
     res/lambda^2; else undefined; end if;
end proc:

forget(evalf):
CodeTools:-Usage( evalf(N(4,8)) );

memory used=13.14MiB, alloc change=4.00MiB, cpu time=89.00ms, real time=90.00ms, gc time=0ns

0.7385644022e-14

#forget(evalf);
#CodeTools:-Usage( plots:-contourplot(N(beta,alpha), beta=0..10, alpha=0..2, grid=[7,7]) );

#forget(evalf);
#CodeTools:-Usage( plots:-contourplot(N(beta,alpha), beta=0..10, alpha=0..10,
#                                     contours=[-1e-8, 1e-8], grid=[17,17]) );

N:=subsop(4=NULL,eval(N)): # clear N's remember table
forget(evalf);
time[real](assign('PP',[seq([beta,fsolve(alpha->N(beta,alpha),1..10)],
                            beta=0..10,1.0)]))*`seconds real time`;

108.788*`seconds real time`

#eval(PP,1);

PPSPL:=Interpolation:-Interpolate(PP[..,1],PP[..,2]):

plot(PPSPL, 0..10, view=[0..10,0..10],
     filled=true, color=gray, labels=[beta,alpha], title=(gamma=3),
     transparency=0, background="aa0000");

N:=subsop(4=NULL,eval(N)): # clear N's remember table
forget(evalf);
time[real](assign('PNZ',
                  [seq([bval,
                        RootFinding:-NextZero(subs(beta=bval,
                                                   alpha->N(beta,alpha)),
                                              1.0)],bval=0..10,1.0)]) )*`seconds real time`;

106.904*`seconds real time`

#eval(PNZ,1);

PNZSPL:=Interpolation:-Interpolate(PNZ[..,1],PNZ[..,2]):

plot(PNZSPL, 0..10, view=[0..10,0..10],
     filled=true, color=gray, labels=[beta,alpha], title=(gamma=3),
     transparency=0, background="aa0000");

N:=subsop(4=NULL,eval(N)):
forget(evalf);
Grid:-Set(N);Grid:-Set(L);Grid:-Set(thisJ);Grid:-Set(J);Grid:-Set(f);Grid:-Set(F);
time[real](assign('PGR',
                  [Grid:-Seq['tasksize'=max(1,floor(10/kernelopts('numcpus')))]([beta,
                                                                                 fsolve(N(beta,alpha),
                                                                                        alpha=1..10)],
                                                       beta=0..10,1.0)]) )*`seconds real time`;

57.881*`seconds real time`

#eval(PGR,1);

PGRSPL:=Interpolation:-Interpolate(PGR[..,1],PGR[..,2]):

plot(PGRSPL,0..10, view=[0..10,0..10],
     filled=true, color=gray, labels=[beta,alpha], title=(gamma=3),
     transparency=0, background="aa0000");

 

Download Negativity_v1_acU0.mw

Please don't create duplicate Question threads of your followup (below) about an 8th order method.

(I hope to find time to look at your formulation. But the fact that nobody's yet done so here is not justification to submit duplicates.)

You could try renaming your .mw worksheet with a file-name that doesn't include special characters (`+`, "(), etc). Try it with only alpha-numeric characters, underscores, and a period.

Some special characters can confuse the backend of this forum, preventing downloads.

@Carl Love Does your Windows Task Manager show a similar CPU load if the option engine=triade is used instead?

@MaPal93 It is counter-productive and unnecessarily difficult for the readership if a derivative followup query is separated from its parent, since that obscures connections amongst the information and details given by the original poster as well as the responders.

@Carl Love Despite the wording of the error message I don't see that the sign command is necessarily the problem here.

I'd guess that signum is being used at some internal juncture, and the calling routine is not properly prepared to deal with its result...

@MaPal93 Please do not put that followup in a (duplicate) separate Question thread.

How would you feel about taking limits instead of using eval? Or do you want to know precisely how dsolve's internals are handling it differently?

note: in Maple 2017.3 (and 2022.1) a single call to limit can do the same.

restart

kernelopts(version);

`Maple 2015.2, X86 64 LINUX, Dec 20 2015, Build ID 1097895`

sys_1 := {diff(x(t), t$2)=sin(t)-x(t), x(0)=0, D(x)(0)=0}:

sol_1 := dsolve(sys_1);

x(t) = (1/2)*sin(t)-(1/2)*cos(t)*t

sys_2 := {(M__p+M__a)*diff(x(t), t$2)=M__p*sin(t)-x(t), x(0)=0, D(x)(0)=0}:
sol_2 := dsolve(sys_2)

x(t) = sin(t/(M__p+M__a)^(1/2))*M__p*(M__p+M__a)^(1/2)/(M__p+M__a-1)-M__p*sin(t)/(M__p+M__a-1)

limit(eval(sol_2, M__p=1), M__a=0);

x(t) = (1/2)*sin(t)-(1/2)*cos(t)*t

limit(eval(sol_2, [M__a=0]), M__p=1);

x(t) = (1/2)*sin(t)-(1/2)*cos(t)*t

Download SomethingWrong_ac.mw

Shout if you have reservations.

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