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Good day everyone. I have very simple question about behaviour of the functions FunctionPoles, FunctionMeromorphic e.t.c. Can these functions work properly with undefined functions? For example when I wrote:

FunctionPoles[1/(k^2 + Subscript[\[Omega], 1]), Subscript[\[Omega], 1]]

I obtained

{{-k^2, 1}}

But If I try to insert an undefined function $B(k)$ like that:

FunctionPoles[1/(k^2 + B[k] + Subscript[\[Omega], 
   1]), Subscript[\[Omega], 1]]

Then I did not obtain expected result. I also tried use Assumptions option but alas have no success. Is there a way to make it work properly? Thanks in advance!

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  • $\begingroup$ What do you expect as the output? $\endgroup$
    – Domen
    Apr 1, 2023 at 17:20
  • $\begingroup$ {{-k^2-B[k], 1}} $\endgroup$
    – George
    Apr 1, 2023 at 17:23
  • $\begingroup$ If you write FunctionPoles[1/(B + k^2 + ω1), ω1], it returns {{-B-k^2,1}}. $\endgroup$
    – Domen
    Apr 1, 2023 at 17:40
  • $\begingroup$ Alas I need it to be of form B[k] because I have a lot of such expressions with B[k],B[q] and B[k-q] $\endgroup$
    – George
    Apr 1, 2023 at 18:44

1 Answer 1

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There is a workaround

FunctionPoles[1/(k^2 + B + Subscript[\[Omega],1]), Subscript[\[Omega], 1]]/.B->B[k]

{{-k^2 - B[k], 1}}

BTW, FunctionSingularities[1/(k^2 + B[k] + Subscript[\[Omega], 1]), Subscript[\[Omega], 1]] works well.

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  • $\begingroup$ Seems like a proper solution. However, it is very sad that FunctionPoles behaves itself in such a strange manner. Anyway, thanks for the answer. $\endgroup$
    – George
    Apr 2, 2023 at 19:00

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