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Question Number 53961 by maxmathsup by imad last updated on 27/Jan/19

let ϕ(x) =((arctan(2x))/(1−x^2 )) 1) calculate ϕ^((n)) (x) 2) calculate ϕ^((n)) (0) anddevelpp ϕ at integr serie

$${let}\:\varphi\left({x}\right)\:=\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\varphi^{\left({n}\right)} \left({x}\right)\: \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\varphi^{\left({n}\right)} \left(\mathrm{0}\right)\:{anddevelpp}\:\varphi\:{at}\:{integr}\:{serie} \\ $$

Commented by maxmathsup by imad last updated on 06/Feb/19

1) we have ϕ(x)=(1/2)arctan(2x){(1/(1−x)) +(1/(1+x))) =(1/2) ((arctan(2x))/(1−x)) +(1/2) ((arctan(2x))/(1+x)) =W(x) −H(x) with W(x)=(1/2) ((arctan(2x))/(x+1)) and H(x)=(1/2) ((arctan(2x))/(x−1)) ⇒ ϕ^((n)) (x)=W^((n)) (x)−H^((n)) (x) leibniz formula give W^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan(2x))^((k)) ((1/(x+1)))^((n−k)) but ((1/(x+1)))^((n−k)) =(((−1)^(n−k) (n−k)!)/((x+1)^(n−k+1) )) we have (arctan(2x))^′ =(2/(1+4x^2 )) ⇒ (arctan(2x))^((k)) =2((1/(4x^2 +1)))^((k−1)) =−i { (1/(2x−i)) −(1/(2x+i))}^((k−1)) =i{(1/(2(x+(i/2)))) −(1/(2(x−(i/2))))}^((k−1)) =(i/2){ (((−1)^(k−1) (k−1)!)/((x+(i/2))^k )) −(((−1)^(k−1) (k−1)!)/((x−(i/2))^k ))} =(i/2)(−1)^(k−1) (k−1)!{ (((x−(i/2))^k −(x+(i/2))^k )/((x^2 +(1/4))^k ))} ⇒ W^((n)) (x) = arctan(2x)(((−1)^n n!)/((x+1)^(n+1) )) +Σ_(k=1) ^n C_n ^k (i/2)(−1)^(k−1) (k−1)!{(((x−(i/2))^k −(x+(i/2))^k )/((x^2 +(1/4))^k ))}(((−1)^(n−k) (n−k)!)/((x+1)^(n−k+1) )) . =(((−1)^n n!)/((x+1)^(n+1) )) arctan(2x) −Σ_(k=1) ^n ((n!)/k) (−1)^k ( (( Im(x+(i/2))^k )/((x^2 +(1/4))^k ))) (1/((x+1)^(n−k+1) )) .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:\varphi\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{arctan}\left(\mathrm{2}{x}\right)\left\{\frac{\mathrm{1}}{\mathrm{1}−{x}}\:+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}−{x}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{x}} \\ $$$$={W}\left({x}\right)\:−{H}\left({x}\right)\:{with}\:{W}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}+\mathrm{1}}\:{and}\:{H}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}−\mathrm{1}} \\ $$$$\Rightarrow\:\varphi^{\left({n}\right)} \left({x}\right)={W}^{\left({n}\right)} \left({x}\right)−{H}^{\left({n}\right)} \left({x}\right)\:\:\:{leibniz}\:{formula}\:{give} \\ $$$${W}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} \left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({n}−{k}\right)} \:\:{but}\: \\ $$$$\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)^{\left({n}−{k}\right)} =\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{\left({x}+\mathrm{1}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$${we}\:{have}\:\:\left({arctan}\left(\mathrm{2}{x}\right)\right)^{'} \:=\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\Rightarrow\:\left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} \:=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\left({k}−\mathrm{1}\right)} \\ $$$$=−{i}\:\left\{\:\frac{\mathrm{1}}{\mathrm{2}{x}−{i}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}+{i}}\right\}^{\left({k}−\mathrm{1}\right)} \:={i}\left\{\frac{\mathrm{1}}{\mathrm{2}\left({x}+\frac{{i}}{\mathrm{2}}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}\left({x}−\frac{{i}}{\mathrm{2}}\right)}\right\}^{\left({k}−\mathrm{1}\right)} \\ $$$$=\frac{{i}}{\mathrm{2}}\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }\:−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} }\right\} \\ $$$$=\frac{{i}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left\{\:\frac{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} −\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\right\}\:\Rightarrow \\ $$$${W}^{\left({n}\right)} \left({x}\right)\:=\:{arctan}\left(\mathrm{2}{x}\right)\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \frac{{i}}{\mathrm{2}}\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!\left\{\frac{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} −\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\right\}\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{\left({x}+\mathrm{1}\right)^{{n}−{k}+\mathrm{1}} }\:\:. \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:{arctan}\left(\mathrm{2}{x}\right)\:−\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{{n}!}{{k}}\:\left(−\mathrm{1}\right)^{{k}} \left(\:\frac{\:{Im}\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\right)\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{{n}−{k}+\mathrm{1}} }\:. \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 06/Feb/19

also we have H^((n)) (x) =(((−1)^n n!)/((x−1)!)) arctan(2x)−Σ_(k=1) ^n n!(((−1)^k )/k) ((Im(x+(i/2))^k )/((x^2 +(1/4))^k (x−1)^(n−k +1) )) . ϕ^((n)) (0) =W^((n)) (0)−H^((n)) (0) =n! Σ_(k=1) ^n (((−1)^(k−1) )/k) ((Im((i^k /2^k )))/(((1/4))^k )) −(n! Σ_(k=1) ^n (((−1)^(k−1) )/k) ((Im((i^k /2^k )))/(((1/4))^k ))(−1)^(n−k +1) =n!{Σ_(k=1) ^n (((−1)^(k−1) )/k) 4^(−k) (1/2^k )sin(((kπ)/2))−Σ_(k=1) ^n (−1)^n 4^(−k) (1/2^k )sin(((kπ)/2))} .

$${also}\:{we}\:{have}\: \\ $$$${H}^{\left({n}\right)} \left({x}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−\mathrm{1}\right)!}\:{arctan}\left(\mathrm{2}{x}\right)−\sum_{{k}=\mathrm{1}} ^{{n}} \:{n}!\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\:\frac{{Im}\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} \left({x}−\mathrm{1}\right)^{{n}−{k}\:+\mathrm{1}} }\:. \\ $$$$\varphi^{\left({n}\right)} \left(\mathrm{0}\right)\:={W}^{\left({n}\right)} \left(\mathrm{0}\right)−{H}^{\left({n}\right)} \left(\mathrm{0}\right)\: \\ $$$$={n}!\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:\:\:\frac{{Im}\left(\frac{{i}^{{k}} }{\mathrm{2}^{{k}} }\right)}{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\:−\left({n}!\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:\frac{{Im}\left(\frac{{i}^{{k}} }{\mathrm{2}^{{k}} }\right)}{\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\left(−\mathrm{1}\right)^{{n}−{k}\:+\mathrm{1}} \right. \\ $$$$={n}!\left\{\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{k}}\:\mathrm{4}^{−{k}} \:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)−\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(−\mathrm{1}\right)^{{n}} \:\mathrm{4}^{−{k}} \:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)\right\}\:. \\ $$

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