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Apr 14, 2018 at 3:38 vote accept James
Apr 11, 2018 at 1:44 comment added James Many thanks! Well given this, it seems that the maximum number of roots is 2K. Do you agree?
Apr 10, 2018 at 22:49 comment added Robert Israel In this case $$R(s) = \left( {A_{{{\it cc}}}}^{2}-2\,A_{{{\it cc}}}A_{{{\it ss}}}+{A_{{{ \it sc}}}}^{2}+{A_{{{\it ss}}}}^{2} \right) {s}^{4}+ \left( 2\,A_{{c}} A_{{{\it sc}}}-2\,A_{{{\it cc}}}A_{{s}}+2\,A_{{s}}A_{{{\it ss}}} \right) {s}^{3}+ \left( {A_{{c}}}^{2}-2\,{A_{{{\it cc}}}}^{2}+2\,A_{{ {\it cc}}}A_{{{\it ss}}}+{A_{{s}}}^{2}-{A_{{{\it sc}}}}^{2} \right) {s }^{2}+ \left( -2\,A_{{c}}A_{{{\it sc}}}+2\,A_{{{\it cc}}}A_{{s}} \right) s-{A_{{c}}}^{2}+{A_{{{\it cc}}}}^{2}$$ Most computer algebra systems have a resultant function.
Apr 10, 2018 at 19:15 comment added James Thanks. I am trying to digest your suggestion. Let's say K=2, hence we have $P(s,c)=A_s s+A_c c+A_{sc} sc+ A_{ss} s^2+ A_{cc} c^2$. Would you please tell me how I can find $R(s)$ now? I did not find a way to calculate the resultant.
Apr 10, 2018 at 18:06 history answered Robert Israel CC BY-SA 3.0