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MaplePrimes Activity

These are replies submitted by vv

@Markiyan Hirnyk 

But you have spent a rather long time with this problem as I can see. So, selectively busy.

@Markiyan Hirnyk
It is obvious (as I have already said) that the procedure cannot print an infinity of points.
And I do not have enough spare time to analyse more deeply the problem (for which there are many articles in the literature).
But you are free (and invited) to write a better procedure. I will vote it up!


Using distributions (such as Dirac) is a delicate task. They are not actually functions but Maple allows algebraic operations with them. It is the user's responsibility to ckeck that these operations have a mathematical sense (for example, usually it is not possible to multiply a non-regular distribution by a function which is not C^oo).

For t=7Pi/6 AND 11Pi/6

@Markiyan Hirnyk 

@Markiyan Hirnyk 

Why don't you try to understand the problem first?
t runs in the interval [0,2Pi).
For example, the point [2,0] is double (in this approach): it is obtained for t=0 AND for t=Pi.

@Markiyan Hirnyk 

In a Lissajous curve m,n should be relatively prime i.e. igcd(m,n)=1. Otherwise all the points are double/multiple and the procedure shows only the "extreme" ones. So, use instead dblpoints(1,2,2,3).

For a non-trivial procedure see the chapter "Advanced examples" in

Char B.W. et al - First Leaves, A Tutorial Introduction to Maple V, Springer 1992

It is very fast and works for huge numbers.

@Carl Love 

Hello Carl, glad to "see" you here again.
I think that OP's confusion is due to the fact that t was omitted in 'maximize'.
So, probably it is better to write:

f:= cos(2*t/m) + cos(2*(t+5)/m):
plot('maximize'(f,t), m= 1..10);


It is not the loop. If you recompute several type evalf(add(...)) ==> random results.
And it is not the add, the same thing happens for `+`(seq(...)).
The problem is the randomness, not necessarily the 3 digits!
This is not nice.


In the first procedure (=J) the condition 1<n<k  is not correct in 1D math and not needed in 2D; it should be simply n<k  as in J.
In the last return you have too many ")".


Then it would be nice to post your corrected procedure.


Of course, k>1. I have not included a check for this.
Or, one may insert
if k=1 then return n-1 fi;
but I prefer to keep it as it is.

@Markiyan Hirnyk 

It is possible to compute by hand stating from

sum(k*z^k/(k^2+p^2+k), k = 1 .. infinity)

for |z|<1, and expanding k/(k^2+p^2+k) in partial fractions. Then LerchPhi appears.
But it's very time consumming.

For example,

sum( (a^2*k+sin(a))^3/(1+a^2*k+a^4*k^3), k=1..infinity) assuming real;
returns the generic result infinity, which is valid for a<>0.
With option parametric I would expect the result:
piecewise(a=0, 0, infinity);
Unfortunately it is infinity.

@Markiyan Hirnyk 


A generic answer should be valid for most cases (usually except a finite set).
'parametric' should take care of those exceptions.

My impression is that 'parametric' works only for simple situations.

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