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Question Number 81431 by abdomathmax last updated on 13/Feb/20
let f(x)=((arctan(2x))/(1+x)) 1) calculate f^((n)) (x) and f^((n)) (0) 2) developp f at integr serie
$${let}\:{f}\left({x}\right)=\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{x}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Commented by abdomathmax last updated on 20/Feb/20
1) we have f(x)=((arctan(2x))/(1+x)) ⇒ f^((n)) (x)=Σ_(k=0) ^n C_n ^k (arctan(2x))^((k)) ((1/(1+x)))^((n−k)) =arctan(2x)×(((−1)^n )/((x+1)^(n+1) )) +Σ_(k=1) ^n C_n ^k (arctan(2x))^((k)) ×(((−1)^(n−k) )/((x+1)^(n−k+1) )) we have (arctan(2x))^((1)) =(2/(1+4x^2 )) ⇒ (arctan(2x))^((k)) =((2/(1+4x^2 )))^((k−1)) =(1/2){ (1/(x^2 +(1/4)))}^()k−1)) =(1/2){ (1/((x−(i/2))(x+(i/2))))}^((k−1)) =(1/(2i)){(1/(x−(i/2)))−(1/(x+(i/2)))}^((k−1)) =(1/(2i)){ (((−1)^(k−1) )/((x−(i/2))^k ))−(((−1)^(k−1) )/((x+(i/2))^k ))} =(((−1)^(k−1) )/(2i)){(((x+(i/2))^k −(x−(i/2))^k )/((x^2 +(1/4))^k ))} =(−1)^(k−1) ×((Im(x+(i/2))^k )/((x^2 +(1/4))^k )) ⇒ f^((n)) (x)=(((−1)^n arctan(2x))/((x+1)^n )) +(−1)^(n−1) Σ_(k=1) ^n C_n ^k ×((Im(x+(i/2))^k )/((x^2 +(1/4))^k )) ×(1/((x+1)^(n−k+1) ))
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\frac{{arctan}\left(\mathrm{2}{x}\right)}{\mathrm{1}+{x}}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)^{\left({n}−{k}\right)} \\ $$$$={arctan}\left(\mathrm{2}{x}\right)×\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({x}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:+\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} ×\frac{\left(−\mathrm{1}\right)^{{n}−{k}} }{\left({x}+\mathrm{1}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$${we}\:{have}\:\left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left(\mathrm{1}\right)} =\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\left({arctan}\left(\mathrm{2}{x}\right)\right)^{\left({k}\right)} \:=\left(\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\left({k}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right\}^{\left.\right)\left.{k}−\mathrm{1}\right)} \:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}}{\left({x}−\frac{{i}}{\mathrm{2}}\right)\left({x}+\frac{{i}}{\mathrm{2}}\right)}\right\}^{\left({k}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\frac{\mathrm{1}}{{x}−\frac{{i}}{\mathrm{2}}}−\frac{\mathrm{1}}{{x}+\frac{{i}}{\mathrm{2}}}\right\}^{\left({k}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} }−\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }\right\}\: \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{\mathrm{2}{i}}\left\{\frac{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} −\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\right\} \\ $$$$=\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} \:×\frac{{Im}\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} \:{arctan}\left(\mathrm{2}{x}\right)}{\left({x}+\mathrm{1}\right)^{{n}} } \\ $$$$+\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\:\:×\frac{{Im}\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{k}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} }\:×\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{{n}−{k}+\mathrm{1}} } \\ $$$$ \\ $$$$ \\ $$
Commented by abdomathmax last updated on 20/Feb/20
f^((n)) (0) =(−1)^(n−1) Σ_(k=1) ^n C_n ^k ((sin(((kπ)/2)))/(2^k ((1/4))^k )) =(−1)^(n−1) Σ_(k=1) ^n C_n ^k 2^k sin(((kπ)/2)) 2) f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n =f(0) +Σ_(n=1) ^∞ (−1)^(n−1) ×(1/(n!)){Σ_(k=1) ^n C_n ^k 2^k sin(((kπ)/2))}x^n =Σ_(n=1) ^∞ (((−1)^(n−1) )/(n!)) Σ_(k=1) ^n ((n!)/(k!(n−k)!)) 2^k sin(((kπ)/2)) x^n f(x) =Σ_(n=1) ^∞ (−1)^(n−1) (Σ_(k=1) ^n ((2^k sin(((kπ)/2)))/(k!(n−k)!)))x^n
$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\frac{{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{\mathrm{2}^{{k}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{k}} } \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\mathrm{2}^{{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right)\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$={f}\left(\mathrm{0}\right)\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ×\frac{\mathrm{1}}{{n}!}\left\{\sum_{{k}=\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\mathrm{2}^{{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)\right\}{x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}!}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:\mathrm{2}^{{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)\:{x}^{{n}} \\ $$$${f}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left(\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}^{{k}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}}\right)}{{k}!\left({n}−{k}\right)!}\right){x}^{{n}} \\ $$

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