kiraMou

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These are replies submitted by kiraMou

No worries @mmcdara ! my message was not that clear either,  and thank you very  much for the further details which I am really interested in. 

Thank you @Carl Love for the clarification. It is more clear now than before. 

@mmcdara

The range 0.5..1 for x[1] and x[2] and  -1..-0.5 for x[3] and x[4] was just an experiment for me to restrict the range of the solution in fsolve. However, as I mentioned before, I am interested in those that lie on the lines Im x_i = ±π/2.

@Carl Love 

Yes, you are right. I should mention that I was talking about one solution among many… next time I will be more careful. However, as I mentioned before, I am interested in those that lie on the lines Im x_i = ±π/2.

@mmcdara 

Thank you very much for your detailed answer. I guess that these solutions (for a=0.8) :

x[1] = -.43334870250217230+1.5707963267948966*I, x[2] = .43334870250217230+1.5707963267948966*I, x[3] = -.43334870250217230-1.5707963267948966*I, x[4] = .43334870250217230-1.5707963267948966*I

verified the criteria. Any comments will be very appreciated.
if the solutions are valid, how can I guess the solutions for N=8 ?

@tomleslie 

I am sorry! I did not insert the link. Now, I edited my questions. Thanks for letting me know!

@Mariusz Iwaniuk 

for my point of view (numerically), It is better to keep  k(x) in the momentum space because you will need to deal with the convolution numerically, then you will have just to multiply both functions (k and log(1+f_1)) instead of dealing with convolution' integrals. 

@Mariusz Iwaniuk  it works, thank you !

 

Thank you very much  @vv  for your help. when I tried to run it, it gives me the following  error "Error, invalid sum/difference" . The cursor is pointing on 

[crt++"here"]

 


Thank you @vv . Yes, Indeed. There are too many solutions and some of them are identical. I just learnt that the physical solutions are expected to sit on the two lines Im(x_j) ± Pi/2. Therefore, how to avoid physically meaningless solutions and identical solutions?

@Thomas Richard  it is done.

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