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Question Number 53957 by maxmathsup by imad last updated on 05/Feb/19
letf(x)=arctan(1+2x) 1)calculatef(n)(x)thenf(n)(0) 2)developpfatintegrserie. wehavef′(x)=21+(1+2x)2⇒f(n)(x)=2{1(2x+1)2+1}(n−1)withn>0 letW(x)=1(2x+1)2+1⇒W(x)=1(2x+1+i)(2x+1−i) =14(x+1+i2)(x+1−i2)=14(x+12eiπ4)(x+12e−iπ4)but (1x+12e−iπ4−1x+12eiπ4)=12(2isin(π4))(x+12e−iπ4)(x+12eiπ4)⇒ W(x)=14i{1x+12e−iπ4−1x+12eiπ4}⇒ W(n−1)(x)=14i{(−1)n−1(n−1)!(x+12e−iπ4)n−(−1)n−1(n−1)!(x+12eiπ4)n}⇒ f(n)(x)=(−1)n−1(n−1)!2i{1(x+12e−iπ4)n−1(x+12eiπ4)n}and f(n)(0)=(−1)n−1(n−1)!2i{12)−ne−inπ4−1(2)−neinπ4} =(−1)n−1(n−1)!2i{(2)neinπ4−(2)ne−inπ4} =(−1)n−1(n−1)!2i(2)n2isin(nπ4)=(−1)n−1(n−1)!(2)nsin(nπ4)
Commented bymaxmathsup by imad last updated on 05/Feb/19
2)wehavef(x)=∑n=0∞xnn!f(n)(0)=π4+∑n=1∞(−1)n−1(2)nnsin(nπ4)xn.
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