let-f-x-x-n-arctan-x-2-with-n-integr-natural-1-calculate-f-n-x-and-f-n-0-2-developp-f-at-integr-serie- Tinku Tara June 4, 2023 Relation and Functions 0 Comments FacebookTweetPin Question Number 55270 by maxmathsup by imad last updated on 20/Feb/19 letf(x)=xnarctan(x2)withnintegrnatural1)calculatef(n)(x)andf(n)(0)2)developpfatintegrserie. Commented by maxmathsup by imad last updated on 02/Mar/19 1)leibnizformulagivef(n)(x)=∑k=0nCnk(xn)(k)(arctan(x2))(n−k)but(xn)(k)=n(n−1)…(n−k+1)xn−k=n!(n−k)!xn−kifk⩽n⇒f(n)(x)=∑k=0nn!k!(n−k)!n!(n−k)!xn−k{arctan(x2)})n−k)=∑k=0n(n!)2k!{(n−k)!}2{arctan(x2)}(n−k)letfind(arctan(x2))(m)wehave{arctan(x2)}(1)=2x1+x4=2x(x2−i)(x2+i)=2x(x−i)(x+i)(x−−i)(x+−i)=2x(x−eiπ4)(x+eiπ4)(x−e−iπ4)(x+e−iπ4)=F(x)=ax−eiπ4+bx+eiπ4+cx−e−iπ4+dx+e−iπ4a=2eiπ42eiπ4(2isin(π4))(2cos(π4))=14i12=12ib=−2eiπ4(−2eiπ4)(−2cos(π4))(−2isin(π4))=12ic=2e−iπ4(−2isin(π4))(2cos(π4))(2e−iπ4)=−12id=−2e−iπ4(−2cos(π4))(2isin(π4))(−2e−iπ4)=−12i⇒(arctan(x2))(1)=12i{1x−eiπ4+1x+eiπ4−1x−e−iπ4−1x+e−iπ4}⇒(arctan(x2))(m)=12i(−1)m−1(m−1)!{1(x−eiπ4)m+1(x+eiπ4)m−1(x−e−iπ4)m−1(x+e−iπ4)}⇒f(n)(x)=12i∑k=0n(n!)2k!{(n−k)!}2(−1)n−k−1(n−k−1)!{1(x−eiπ4)n−k+1(x+eiπ4)n−k−1(x−e−iπ4)n−k−1(x+e−iπ4)n−k} Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: let-f-x-2x-1-ln-1-x-2-1-calculate-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-3-calculate-0-1-f-x-dx-Next Next post: Find-the-largest-number-of-positive-integers-that-can-be-found-in-such-a-way-that-any-two-of-them-a-and-b-a-b-satisfy-the-next-inequality-a-b-ab-100- Leave a Reply Cancel replyYour email address will not be published.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.