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Question Number 76190 by abdomathmax last updated on 25/Dec/19
let f(x)=((arctan(1+x))/(2+x))  1) calculate f^((n)) (x) and f^((n)) (0)  2) developp f at integr serie.
$${let}\:{f}\left({x}\right)=\frac{{arctan}\left(\mathrm{1}+{x}\right)}{\mathrm{2}+{x}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
1) f^((n)) (x)=Σ_(k=0) ^n  C_n ^k  (arctan(1+x))^((k)) ×((1/(x+2)))^((n−k))   we have (arctan(1+x))^((1)) =(1/(1+(1+x)^2 )) =(1/(x^2 +2x+2)) ⇒  (arctan(x+1))^((k)) =((1/(x^2  +2x+2)))^((k−1))   Δ^′ =1−2 =−1 ⇒x_1 =−1+i  and x_2 =−1−i  x_1 =(√2)(−(1/( (√2)))+(i/( (√2)))) =(√2)e^((i3π)/4)   and x_2 =(√2)e^(−i((3π)/4))  ⇒  (1/(x^2  +2x+2)) =(1/((x−(√2)e^((i3π)/4) )(x−(√2)e^(−i((3π)/4)) ))) =(1/( (√2)2isin(((3π)/4))))((1/(x−(√2)e^(−((i3π)/4)) ))−(1/(x−(√2)e^(i((3π)/4)) )))  =(1/(2i(√2)(1/( (√2))))){ (1/(x−(√2)e^(−((i3π)/4)) ))−(1/(x−(√2)e^((i3π)/4) ))} ⇒  (d^k /dx^k )(arctan(x+1)) =(1/(2i)){  (((−1)^(k−1) (k−1)!)/((x−(√2)e^(−((i3π)/4)) )^k ))−(((−1)^(k−1) (k−1)!)/((x−(√2)e^((i3π)/4) )^k ))}  =(((−1)^(k−1) (k−1)!)/(2i)){(((x−(√2)e^((i3π)/4) )^k −(x−(√2)e^(−((i3π)/4)) ))/((x^2  +2x+2)^k ))}  f^((n)) (x) =arctan(x+1)(((−1)^n n!)/((x+2)^(n+1) ))  +(1/(2i))Σ_(k=1) ^n  C_n ^k  (−1)^(k−1) (k−1)!×(((x−(√2)e^((i3π)/4) )^k −(x−(√2)e^(−((i3π)/4)) )^k )/((x^2  +2x+2)^k ))×(((−1)^(n−k) (n−k)!)/((x+2)^(n−k+1) ))  =(−1)^n n! ((arctan(x+1))/((x+2)^(n+1) ))  +(−1)^(n−1) Σ_(k=1) ^n  C_n ^k   (k−1)!((Im{(x−(√2)e^(i((3π)/4)) )^k })/((x^2  +2x+2)^k ))×(((n−k)!)/((x+2)^(n−k+1) ))
$$\left.\mathrm{1}\right)\:{f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({arctan}\left(\mathrm{1}+{x}\right)\right)^{\left({k}\right)} ×\left(\frac{\mathrm{1}}{{x}+\mathrm{2}}\right)^{\left({n}−{k}\right)} \\ $$$${we}\:{have}\:\left({arctan}\left(\mathrm{1}+{x}\right)\right)^{\left(\mathrm{1}\right)} =\frac{\mathrm{1}}{\mathrm{1}+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}\:\Rightarrow \\ $$$$\left({arctan}\left({x}+\mathrm{1}\right)\right)^{\left({k}\right)} =\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}\right)^{\left({k}−\mathrm{1}\right)} \\ $$$$\Delta^{'} =\mathrm{1}−\mathrm{2}\:=−\mathrm{1}\:\Rightarrow{x}_{\mathrm{1}} =−\mathrm{1}+{i}\:\:{and}\:{x}_{\mathrm{2}} =−\mathrm{1}−{i} \\ $$$${x}_{\mathrm{1}} =\sqrt{\mathrm{2}}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{{i}}{\:\sqrt{\mathrm{2}}}\right)\:=\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \:\:{and}\:{x}_{\mathrm{2}} =\sqrt{\mathrm{2}}{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \:\Rightarrow \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}}\:=\frac{\mathrm{1}}{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)\left({x}−\sqrt{\mathrm{2}}{e}^{−{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\mathrm{2}{isin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)}\left(\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{\mathrm{2}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\left\{\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} }−\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} }\right\}\:\Rightarrow \\ $$$$\frac{{d}^{{k}} }{{dx}^{{k}} }\left({arctan}\left({x}+\mathrm{1}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} }−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\frac{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} −\left({x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{{k}} }\right\} \\ $$$${f}^{\left({n}\right)} \left({x}\right)\:={arctan}\left({x}+\mathrm{1}\right)\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+\mathrm{2}\right)^{{n}+\mathrm{1}} } \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \left({k}−\mathrm{1}\right)!×\frac{\left({x}−\sqrt{\mathrm{2}}{e}^{\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} −\left({x}−\sqrt{\mathrm{2}}{e}^{−\frac{{i}\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{{k}} }×\frac{\left(−\mathrm{1}\right)^{{n}−{k}} \left({n}−{k}\right)!}{\left({x}+\mathrm{2}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}} {n}!\:\frac{{arctan}\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{2}\right)^{{n}+\mathrm{1}} } \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({k}−\mathrm{1}\right)!\frac{{Im}\left\{\left({x}−\sqrt{\mathrm{2}}{e}^{{i}\frac{\mathrm{3}\pi}{\mathrm{4}}} \right)^{{k}} \right\}}{\left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{2}\right)^{{k}} }×\frac{\left({n}−{k}\right)!}{\left({x}+\mathrm{2}\right)^{{n}−{k}+\mathrm{1}} } \\ $$
Commented by mathmax by abdo last updated on 29/Dec/19
f^((n)) (0) =(π/4)(((−1)^n n!)/2^(n+1) )  +(−1)^(n−1)  Σ_(k=1) ^n  C_n ^k   (((k−1)!(n−k)!)/(2^k ×2^(n−k+1) ))×Im{(−(√2))^k  e^(i((3kπ)/4)) }  =(π/4)(((−1)^n n!)/2^(n+1) ) +(−1)^(n−1)  Σ_(k=1) ^n  ((n!)/(k!(n−k)!))×(((k−1)!(n−k!)/2^(n+1) )×(−(√2))^k sin(((3kπ)/4))  =(π/4)×(((−1)^n n!)/2^(n+1) ) +(((−1)^(n−1) )/2^(n+1) ) Σ_(k=1) ^n  ((n!)/k)(−(√2))^k  sin(((3kπ)/4))
$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\pi}{\mathrm{4}}\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}^{{n}+\mathrm{1}} } \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{\left({k}−\mathrm{1}\right)!\left({n}−{k}\right)!}{\mathrm{2}^{{k}} ×\mathrm{2}^{{n}−{k}+\mathrm{1}} }×{Im}\left\{\left(−\sqrt{\mathrm{2}}\right)^{{k}} \:{e}^{{i}\frac{\mathrm{3}{k}\pi}{\mathrm{4}}} \right\} \\ $$$$=\frac{\pi}{\mathrm{4}}\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}^{{n}+\mathrm{1}} }\:+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}×\frac{\left({k}−\mathrm{1}\right)!\left({n}−{k}!\right.}{\mathrm{2}^{{n}+\mathrm{1}} }×\left(−\sqrt{\mathrm{2}}\right)^{{k}} {sin}\left(\frac{\mathrm{3}{k}\pi}{\mathrm{4}}\right) \\ $$$$=\frac{\pi}{\mathrm{4}}×\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}^{{n}+\mathrm{1}} }\:+\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}^{{n}+\mathrm{1}} }\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}!}{{k}}\left(−\sqrt{\mathrm{2}}\right)^{{k}} \:{sin}\left(\frac{\mathrm{3}{k}\pi}{\mathrm{4}}\right) \\ $$

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