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Question Number 53957 by maxmathsup by imad last updated on 05/Feb/19

let f(x) =arctan(1+2x)  1) calculate f^((n)) (x) then f^((n)) (0)  2) developp f at integr serie .  we have f^′ (x)=(2/(1+(1+2x)^2 )) ⇒ f^((n)) (x) =2 {(1/((2x+1)^2 +1))}^((n−1))   with n>0  let W(x)=(1/((2x+1)^2  +1)) ⇒W(x) =(1/((2x+1+i)(2x+1−i)))  =(1/(4(x+((1+i)/2))(x+((1−i)/2)))) =(1/(4(x +(1/(√2)) e^((iπ)/4) )( x +(1/(√2)) e^(−((iπ)/4)) ))) but  ((1/(x +(1/(√2))e^(−((iπ)/4)) )) −(1/(x+(1/(√2))e^((iπ)/4) ))) =(1/(√2)) (((2isin((π/4))))/((x+(1/(√2))e^(−((iπ)/4)) )(x+(1/(√2))e^((iπ)/4) ))) ⇒  W(x) =(1/(4i)){  (1/(x+(1/(√2))e^(−((iπ)/4)) )) −(1/(x+(1/(√2))e^((iπ)/4) ))} ⇒  W^((n−1)) (x)=(1/(4i)){   (((−1)^(n−1) (n−1)!)/((x+(1/(√2))e^(−((iπ)/4)) )^n )) −(((−1)^(n−1) (n−1)!)/((x+(1/(√2))e^((iπ)/4) )^n ))} ⇒  f^((n)) (x)=(((−1)^(n−1) (n−1)!)/(2i)){   (1/((x+(1/(√2))e^(−((iπ)/4)) )^n )) −(1/((x+(1/(√2))e^((iπ)/4) )^n ))} and  f^((n)) (0) =(((−1)^(n−1) (n−1)!)/(2i)){  (1/((√2))^(−n)  e^(−i((nπ)/4)) )) −(1/(((√2))^(−n)  e^(i((nπ)/4)) ))}  =(((−1)^(n−1) (n−1)!)/(2i)){ ((√2))^n  e^((inπ)/4)  − ((√2))^n  e^(−((inπ)/4)) }  =(((−1)^(n−1) (n−1)!)/(2i)) ((√2))^n  2i sin(((nπ)/4)) =(−1)^(n−1) (n−1)! ((√2))^n sin(((nπ)/4))

letf(x)=arctan(1+2x) 1)calculatef(n)(x)thenf(n)(0) 2)developpfatintegrserie. wehavef(x)=21+(1+2x)2f(n)(x)=2{1(2x+1)2+1}(n1)withn>0 letW(x)=1(2x+1)2+1W(x)=1(2x+1+i)(2x+1i) =14(x+1+i2)(x+1i2)=14(x+12eiπ4)(x+12eiπ4)but (1x+12eiπ41x+12eiπ4)=12(2isin(π4))(x+12eiπ4)(x+12eiπ4) W(x)=14i{1x+12eiπ41x+12eiπ4} W(n1)(x)=14i{(1)n1(n1)!(x+12eiπ4)n(1)n1(n1)!(x+12eiπ4)n} f(n)(x)=(1)n1(n1)!2i{1(x+12eiπ4)n1(x+12eiπ4)n}and f(n)(0)=(1)n1(n1)!2i{12)neinπ41(2)neinπ4} =(1)n1(n1)!2i{(2)neinπ4(2)neinπ4} =(1)n1(n1)!2i(2)n2isin(nπ4)=(1)n1(n1)!(2)nsin(nπ4)

Commented bymaxmathsup by imad last updated on 05/Feb/19

2) we have f(x) =Σ_(n=0) ^∞  (x^n /(n!)) f^((n)) (0) = (π/4) +Σ_(n=1) ^∞ (−1)^(n−1)  ((((√2))^n )/n) sin(((nπ)/4))x^n   .

2)wehavef(x)=n=0xnn!f(n)(0)=π4+n=1(1)n1(2)nnsin(nπ4)xn.

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