let-f-x-x-x-2-x-1-1-calculate-f-n-0-2-developp-f-at-integr-serie- Tinku Tara June 4, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 36010 by abdo mathsup 649 cc last updated on 27/May/18 letf(x)=xx2+x+11)calculatef(n)(0)2)developpfatintegrserie. Commented by prof Abdo imad last updated on 28/May/18 letdecomposef(x)insideC(x)therootsofp(x)=x2+x+1arearej=ei2π3andj−=e−i2π3sof(x)=x(x−j)(x−j−)=ax−j+bx−j−a=limx→j(x−j)f(x)=jj−j−=j232=j3b=limx→j−(x−j−)f(x)=j−j−−j=−j−3⇒f(x)=j3(x−j)−j−3(x−j−)⇒f(n)(x)=j3{(−1)nn!(x−j)n+1}−j−3{(−1)nn!(x−j−)n+1}f(n)(x)=(−1)nn!3{j(x−j)n+1−j−(x−j−)n+1}=(−1)nn!3{j(x−j−)n+1−j−(x−j)n+1(x2+x+1)n+1} Commented by prof Abdo imad last updated on 28/May/18 sof(n)(0)=(−1)nn!3{j(−j−)n+1−j−(−j)n+11}=(−1)nn!3{(−1)n+1j(j−)n+1−(−1)n+1j−jn+1}=(−1)nn!3(−1)n+1{j−n−jn}=13{2Im(jn))=23sin(2nπ3)f(n)(0)=23sin(2nπ3).2)wehavef(x)=∑n=0∞f(n)(0)n!xn⇒f(x)=∑n=0∞23sin(2nπ3)n!xn.f(x)=23∑n=0∞sin(2nπ3)n!xn. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-167082Next Next post: simplify-interval-number-1-6-3-7- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.