792 Reputation

9 years, 296 days

thanks...

@acer   thank you acer

thank you, it was not at all evident...

@Carl Love

Great thanks to you Carl, I think I could have been waste a lot of time on this problem !!!

I am going to look to the many commands you mentionned and try to understand all this,

thanks again

Ok, I understand...

@Preben Alsholm

This is a complete explanation, thank you Preben

Perfect...

It suits me very well.

Smarter than Matrix(N,N, Re~(convert(DC, list))) !!!

Nevertheless : do you observe the same behaviours I did myself ?

@Preben AlsholmThank you Preben, but I h...

@Preben Alsholm

Thank you Preben, but I have this answer.

I would not offend Kitonum (who posted a  pretty solution) but the "other" was smarter.

In fact I think I rediscoverd it by myself (thanks to my anonymous inspirer) :

1. Let A a fixed real interval
2. let B another real interval and B(x) a translation of B of value x
3. Question : "Do A and B(X) overlap ?"

(A) solution :

coverage := proc(A, B, x)
coulditbe('_u', `intersect`(SetOf(A), SetOf(map(y -> x+y, B)) ) )
end proc:

# example :

A := RealRange(0, 2);
B := RealRange(-3, -1);
coverage(A, B, 0.999):
coverage(A, B, 1.001):

No offence meant to Kitonum, and without regard of efficiency, this solution is slightly more aesthetic than the one he proposed to me  :-)

@John Fredsted I obtained the same ...

@John Fredsted I obtained the same conclusion yesterday evening by setting k=r*exp(theta*I).
It seemed a little bit simpler to analyze because r is always positive and sin(theta) is bounded

I will send you the .mw file asap. For the moment the uploading doesn't work correctly (I'm in the office and it could be related to the firewall ?)

@Kitonum Thanks. It is a little bit...

@Kitonum Thanks.

It is a little bit clever than what I coded myself.

I just hoped that the equivalent of `intersect`(set1, set2) for countable sets existed for no countable ones

(Maple is sometimes so "elegant" ... it could have been the case).

Thanks again

@John Fredsted I am really sorry : ...

@John Fredsted I am really sorry : I realized my mistake as I was pushing the "submit" button !!!

In addition to your previous reply  "Whether this is the complete solution for k remains to be investigated" I obtained these results (I hope there is no mistake this time) :

phi2  := (x,k) -> expand(phi(x,k)*conjugate(phi(x,k))) :
PHI2 := phi2(x, a+b*I) assuming x::real, a::real b::real :

limit(PHI2, x=+infinity) assuming b::positive ;    # returns 0
limit(PHI2, x=+infinity) assuming b::negative;    # returns infinity
limit(subs(b=0, PHI2), x=+infinity) ;                  # returns 1

It seems that the boundedness only occur for pure imaginary k ?

(sorry again)

In your last reply you study phi(x,k) for k complex of the form k=a+b*I (a and b real).
You draw phi(x,(a,b)) for the particular case a=0, b=1 and find thie last bell-shaped curve.

But if a=0 and b=1 then k = I and k^2 = -1

and then phi(x,k) is undefined ...

Did I miss something ?

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