Question Number 60502 by prof Abdo imad last updated on 21/May/19
$${let}\:{f}\left({x}\right)\:={arctan}\left(\mathrm{2}{x}\right)\:{ln}\:\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{determine}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right)\:{developp}\:\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
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) }](https://www.tinkutara.com/question/Q61309.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)={arctan}\left(\mathrm{2}{x}\right){ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:+{arctan}\left(\mathrm{2}{x}\right)\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:−\frac{\mathrm{2}{x}\:{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{leibniz}\:{formulae}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} \left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right)^{\left({n}−{k}\right)} \\ $$$$={arctan}\left({x}\right)\left({ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right)^{\left({n}\right)} \:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} \left\{{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\right\}^{\left({n}−{k}\right)} \\ $$$${let}\:{w}\left({x}\right)\:={arctan}\left(\mathrm{2}{x}\right)\:\Rightarrow{w}^{'} \left({x}\right)\:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$${w}^{\left({k}\right)} \left({x}\right)\:=\mathrm{2}\:\left(\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\left({k}−\mathrm{1}\right)} \:{but} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{4}\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)}\:=\frac{\mathrm{1}}{\mathrm{4}\left({x}−\frac{{i}}{\mathrm{2}}\right)\left({x}+\frac{{i}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\frac{\mathrm{1}}{{x}−\frac{{i}}{\mathrm{2}}}\:−\frac{\mathrm{1}}{{x}+\frac{{i}}{\mathrm{2}}}\right\}\:\Rightarrow{w}^{\left({k}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\left(\frac{\mathrm{1}}{{x}−\frac{{i}}{\mathrm{2}}}\right)^{\left({k}−\mathrm{1}\right)} −\left(\frac{\mathrm{1}}{{x}+\frac{{i}}{\mathrm{2}}}\right)^{\left({k}−\mathrm{1}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} }\:−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\frac{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} −\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}\:}\right)^{{k}} }\:{Im}\left(\:\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} \right) \\ $$$$\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} \:=\sum_{{p}=\mathrm{0}} ^{{k}} \:\:{C}_{{k}} ^{{p}} \left(\frac{{i}}{\mathrm{2}}\right)^{{p}} \:{x}^{{k}−{p}} \:\:=\sum_{{p}=\mathrm{2}{q}} \left(….\right)+\sum_{{p}=\mathrm{2}{q}+\mathrm{1}} \\ $$$$=\sum_{{q}=\mathrm{0}} ^{\left[\frac{{k}}{\mathrm{2}}\right]} \:\:\:\:\:{C}_{{k}} ^{\mathrm{2}{q}} \:\:\frac{\left(−\mathrm{1}\right)^{{q}} }{\mathrm{2}^{\mathrm{2}{q}} }\:{x}^{{k}−\mathrm{2}{q}} \:\:+\sum_{{q}=\mathrm{0}} ^{\left[\frac{{k}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{k}} ^{\mathrm{2}{q}+\mathrm{1}} \:\frac{{i}\left(−\mathrm{1}\right)^{{q}} }{\mathrm{2}^{\mathrm{2}{q}+\mathrm{1}} }\:{x}^{{k}−\mathrm{2}{q}−\mathrm{1}} \:\Rightarrow \\ $$$${Im}\left\{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} \right\}\:=\sum_{{q}=\mathrm{0}} ^{\left[\frac{{k}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{k}} ^{\mathrm{2}{q}+\mathrm{1}} \frac{\left(−\mathrm{1}\right)^{{q}} }{\mathrm{2}^{\mathrm{2}{q}+\mathrm{1}} }\:{x}^{{k}−\mathrm{2}{q}−\mathrm{1}} \:\Rightarrow \\ $$$${w}^{\left({k}\right)} \left({x}\right)\:=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\left\{\:\sum_{{q}=\mathrm{0}} ^{\left[\frac{{k}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{k}} ^{\mathrm{2}{q}+\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{q}} }{\mathrm{2}^{\mathrm{2}{q}+\mathrm{1}} }\:{x}^{{k}−\mathrm{2}{q}−\mathrm{1}} \right\} \\ $$$$ \\ $$
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)](https://www.tinkutara.com/question/Q61310.png)
$${let}\:{v}\left({x}\right)\:={ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\:\:{let}\:{determine}\:{v}^{\left({n}\right)} \left({x}\right) \\ $$$${we}\:{have}\:{v}^{'} \left({x}\right)\:=\frac{−\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\:=\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\:−\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\:\Rightarrow \\ $$$${v}^{\left({n}\right)} \left({x}\right)\:=\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({n}−\mathrm{1}\right)} +\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}\right)^{\left({n}−\mathrm{1}\right)} \:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}+\mathrm{1}\right)^{{n}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−\mathrm{1}\right)^{{n}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} }\left\{\:\left({x}+\mathrm{1}\right)^{{n}} \:+\left({x}−\mathrm{1}\right)^{{n}} \right\}\:\:{but} \\ $$$$\left({x}+\mathrm{1}\right)^{{n}} \:+\left({x}−\mathrm{1}\right)^{{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \mathrm{1}^{{k}} \:{x}^{{n}−{k}} \:\:+\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} \:{x}^{{n}−{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \left(\mathrm{1}+\left(−\mathrm{1}\right)^{{k}} \right){x}^{{n}−{k}} \:\:=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{p}} \:\mathrm{2}\:{x}^{{n}−\mathrm{2}{p}} \\ $$$$=\mathrm{2}\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}} \:{x}^{{n}−\mathrm{2}{p}} \:\Rightarrow \\ $$$${v}^{\left({n}\right)} \left({x}\right)\:=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} }\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{p}} \:{x}^{{n}−\mathrm{2}{p}} \:\Rightarrow \\ $$$${v}^{\left({n}−{k}\right)} \:\:=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}−{k}} }\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−{k}}{\mathrm{2}}\right]} \:\:{C}_{{n}−{k}} ^{\mathrm{2}{p}} \:{x}^{{n}−{k}−\mathrm{2}{p}} \\ $$$$\Rightarrow{f}^{\left({n}\right)} \left({x}\right)={arctan}\left(\mathrm{2}{x}\right){w}^{\left({n}\right)} \left({x}\right)\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{w}^{\left({k}\right)} \left({x}\right)\:{v}^{\left({n}−{k}\right)} \left({x}\right) \\ $$