acer

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These are replies submitted by acer

@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

@sand15 To respond to your own query: I don't care. That would be shallow.

I care about it little enough that I sometimes don't pay it enough attention. I try to base it upon whether the content might lead to an actual solution. (But sometimes I Reply instead of Answer. And sometimes I err the other way. To err is human.)

I'll change this one, since it's moved you to comment. If you feel I've made a similar mistake in future then please consider changing it yourself.

Try this code for such labels on the 2D contours, by Kitonum.

Please put your followup queries about this problem as Reply/Comment here, rather than spawning a wholly separate new thread for it.

Or you could use the "Branch" item at the bottom of this Question's body, to started a linked new Question.

And also please make any new posting as a Question, not a Post.

@janhardo You could use the "Branch" item at the bottom of this Question, or of the relevant Answer or Reply? Then the cross-reference links will be automatically inserted in both threads.

(I asked just that you not spawn the related thread as wholly separate.)

@janhardo If you have followup comments/queries about this Answer (expressing Zeta(2) as an infinite product, etc) then please add such here rather than spawn a wholly separate Question thread.

@Hullzie16 You might first decide how accurate you need the results, then figure out how tight the precision and tolerances need to be to achieve that for a single m value, optimize that, and then figure out how many m values you need and optionally reuse a procedure to attain such, etc.

@Prakash J You could import using either ImportMatrix or ExcelTools:-Import , getting a Matrix say. Eg, (using Maple 18.02 as you have),

   M := Matrix(ImportMatrix("mapleprimes/foo1.xls")):

Then you could plot pairs of columns. Eg, you could plot the first column against the second,

   plot(M[..,[1,2]]);

You wrote of multiple plots, for different parameter values, but you're only showing two columns of Excel data.

ps. You didn't respond to my Anwer to your earlier differential equations Question (Blasius solution). Did it serve?

@Prakash J I cannot make sense of what you're trying to say.

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