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Question Number 60502 by prof Abdo imad last updated on 21/May/19
let f(x) =arctan(2x) ln (1−x^2 ) 1) calculate f^′ (x) 2) determine f^((n)) (x) and f^((n)) (0) 3) developp f at integr serie .
letf(x)=arctan(2x)ln(1−x2)1)calculatef′(x)2)determinef(n)(x)andf(n)(0)3)developpfatintegrserie.
Commented by maxmathsup by imad last updated on 31/May/19
1) we have f(x)=arctan(2x)ln(1−x^2 ) ⇒ f^′ (x) =(2/(1+4x^2 ))ln(1−x^2 ) +arctan(2x)((−2x)/(1−x^2 )) =((2ln(1−x^2 ))/(1+4x^2 )) −((2x arctan(2x))/(1−x^2 )) 2) leibniz formulae give f^((n)) (x) =Σ_(k=0) ^n C_n ^k (arctan(2x))^((k)) (ln(1−x^2 ))^((n−k)) =arctan(x)(ln(1−x^2 ))^((n)) +Σ_(k=1) ^n C_n ^k (arctan(2x))^((k)) {ln(1−x^2 )}^((n−k)) let w(x) =arctan(2x) ⇒w^′ (x) =(2/(1+4x^2 )) ⇒ w^((k)) (x) =2 ((1/(4x^2 +1)))^((k−1)) but (1/(4x^2 +1)) =(1/(4(x^2 +(1/4)))) =(1/(4(x−(i/2))(x+(i/2)))) =(1/(4i)){(1/(x−(i/2))) −(1/(x+(i/2)))} ⇒w^((k)) (x) =(1/(2i)){ ((1/(x−(i/2))))^((k−1)) −((1/(x+(i/2))))^((k−1)) } =(1/(2i)){ (((−1)^(k−1) (k−1)!)/((x−(i/2))^k )) −(((−1)^(k−1) (k−1)!)/((x+(i/2))^k ))} =(((−1)^(k−1) (k−1)!)/(2i)){(((x+(i/2))^k −(x−(i/2))^k )/((x^2 +(1/4))^k ))} =(((−1)^(k−1) (k−1)!)/((x^2 +(1/(4 )))^k )) Im( (x+(i/2))^k ) (x+(i/2))^k =Σ_(p=0) ^k C_k ^p ((i/2))^p x^(k−p) =Σ_(p=2q) (....)+Σ_(p=2q+1) =Σ_(q=0) ^([(k/2)]) C_k ^(2q) (((−1)^q )/2^(2q) ) x^(k−2q) +Σ_(q=0) ^([((k−1)/2)]) C_k ^(2q+1) ((i(−1)^q )/2^(2q+1) ) x^(k−2q−1) ⇒ Im{(x+(i/2))^k } =Σ_(q=0) ^([((k−1)/2)]) C_k ^(2q+1) (((−1)^q )/2^(2q+1) ) x^(k−2q−1) ⇒ w^((k)) (x) =(((−1)^(k−1) (k−1)!)/((x^2 +(1/4))^k )){ Σ_(q=0) ^([((k−1)/2)]) C_k ^(2q+1) (((−1)^q )/2^(2q+1) ) x^(k−2q−1) }
1)wehavef(x)=arctan(2x)ln(1−x2)⇒f′(x)=21+4x2ln(1−x2)+arctan(2x)−2x1−x2=2ln(1−x2)1+4x2−2xarctan(2x)1−x22)leibnizformulaegivef(n)(x)=∑k=0nCnk(arctan(2x))(k)(ln(1−x2))(n−k)=arctan(x)(ln(1−x2))(n)+∑k=1nCnk(arctan(2x))(k){ln(1−x2)}(n−k)letw(x)=arctan(2x)⇒w′(x)=21+4x2⇒w(k)(x)=2(14x2+1)(k−1)but14x2+1=14(x2+14)=14(x−i2)(x+i2)=14i{1x−i2−1x+i2}⇒w(k)(x)=12i{(1x−i2)(k−1)−(1x+i2)(k−1)}=12i{(−1)k−1(k−1)!(x−i2)k−(−1)k−1(k−1)!(x+i2)k}=(−1)k−1(k−1)!2i{(x+i2)k−(x−i2)k(x2+14)k}=(−1)k−1(k−1)!(x2+14)kIm((x+i2)k)(x+i2)k=∑p=0kCkp(i2)pxk−p=∑p=2q(….)+∑p=2q+1=∑q=0[k2]Ck2q(−1)q22qxk−2q+∑q=0[k−12]Ck2q+1i(−1)q22q+1xk−2q−1⇒Im{(x+i2)k}=∑q=0[k−12]Ck2q+1(−1)q22q+1xk−2q−1⇒w(k)(x)=(−1)k−1(k−1)!(x2+14)k{∑q=0[k−12]Ck2q+1(−1)q22q+1xk−2q−1}
Commented by maxmathsup by imad last updated on 31/May/19
let v(x) =ln(1−x^2 ) let determine v^((n)) (x) we have v^′ (x) =((−2x)/(1−x^2 )) =((1/(1+x)) −(1/(1−x))) ⇒ v^((n)) (x) =((1/(x+1)))^((n−1)) +((1/(x−1)))^((n−1)) =(((−1)^(n−1) (n−1)!)/((x+1)^n )) +(((−1)^(n−1) (n−1)!)/((x−1)^n )) =(((−1)^(n−1) (n−1)!)/((x^2 −1)^n )){ (x+1)^n +(x−1)^n } but (x+1)^n +(x−1)^n =Σ_(k=0) ^n C_n ^k 1^k x^(n−k) +Σ_(k=0) ^n C_n ^k (−1)^k x^(n−k) =Σ_(k=0) ^n C_n ^k (1+(−1)^k )x^(n−k) =Σ_(p=0) ^([(n/2)]) C_n ^(2p) 2 x^(n−2p) =2 Σ_(p=0) ^([(n/2)]) C_n ^(2p) x^(n−2p) ⇒ v^((n)) (x) =((2(−1)^(n−1) (n−1)!)/((x^2 −1)^n )) Σ_(p=0) ^([(n/2)]) C_n ^(2p) x^(n−2p) ⇒ v^((n−k)) =((2(−1)^(n−k−1) (n−k−1)!)/((x^2 −1)^(n−k) )) Σ_(p=0) ^([((n−k)/2)]) C_(n−k) ^(2p) x^(n−k−2p) ⇒f^((n)) (x)=arctan(2x)w^((n)) (x) +Σ_(k=1) ^n C_n ^k w^((k)) (x) v^((n−k)) (x)
letv(x)=ln(1−x2)letdeterminev(n)(x)wehavev′(x)=−2x1−x2=(11+x−11−x)⇒v(n)(x)=(1x+1)(n−1)+(1x−1)(n−1)=(−1)n−1(n−1)!(x+1)n+(−1)n−1(n−1)!(x−1)n=(−1)n−1(n−1)!(x2−1)n{(x+1)n+(x−1)n}but(x+1)n+(x−1)n=∑k=0nCnk1kxn−k+∑k=0nCnk(−1)kxn−k=∑k=0nCnk(1+(−1)k)xn−k=∑p=0[n2]Cn2p2xn−2p=2∑p=0[n2]Cn2pxn−2p⇒v(n)(x)=2(−1)n−1(n−1)!(x2−1)n∑p=0[n2]Cn2pxn−2p⇒v(n−k)=2(−1)n−k−1(n−k−1)!(x2−1)n−k∑p=0[n−k2]Cn−k2pxn−k−2p⇒f(n)(x)=arctan(2x)w(n)(x)+∑k=1nCnkw(k)(x)v(n−k)(x)

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