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Question-210786




Question Number 210786 by zhou0429 last updated on 19/Aug/24
Answered by Berbere last updated on 19/Aug/24
x^9 +1=Π_(k=0) ^8 (x−e^(iπ(((1+2k)/9))) )=p(x)  =Σ∫(1/((x−e^(iπ(((1+2k)/9))) )p′(e^(iπ(((1+2k)/9))) )))dx  =Σ_(k=0) ^∞ ((ln(x−e^(iπ(((1+2k)/9))) ))/(P′(e^(iπ(((1+2k)/9))) )))+c
x9+1=8k=0(xeiπ(1+2k9))=p(x)=Σ1(xeiπ(1+2k9))p(eiπ(1+2k9))dx=k=0ln(xeiπ(1+2k9))P(eiπ(1+2k9))+c

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