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Question Number 140638 by Mathspace last updated on 10/May/21
let f(x)=arctan((2/x)) developp f at integr serie
$${let}\:{f}\left({x}\right)={arctan}\left(\frac{\mathrm{2}}{{x}}\right) \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Answered by Dwaipayan Shikari last updated on 10/May/21
f(x)=f(0)+((f′(0))/(1!))x+((f′′(0))/(2!))x^2 +.. f(x)=tan^(−1) ((2/x)) f′(x)=−(2/x^2 ).(1/(1+(4/x^2 )))=((−2)/(x^2 +4))=(1/(2i))((1/((x+2i)))−(1/((x−2i)))) f′′(x)=−(1/(2i)).(1/((x+2i)^2 ))+(1/(2i)).(1/((x−2i)^2 )) f^n (x)=(1/(2i))(−1)^(n+1) (n−1)!((1/((x+2i)^n ))−(1/((x−2i)^n ))) f^n (0)=(1/(2i))(−1)^(n+1) (n−1)!(1/2^n )(e^(−((πin)/2)) −e^((πin)/2) )=(((−1)^n sin(((πn)/2)))/2^n )n(n−1)! tan^(−1) ((2/x))=(π/2)+Σ_(n=1) ^∞ (((−1)^n sin(((πn)/2))(n−1)!)/(2^n n!))x^n =(π/2)+Σ_(n=1) ^∞ (((−1)^n sin(((πn)/2)))/(2^n n))x^n
$${f}\left({x}\right)={f}\left(\mathrm{0}\right)+\frac{{f}'\left(\mathrm{0}\right)}{\mathrm{1}!}{x}+\frac{{f}''\left(\mathrm{0}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +.. \\ $$$${f}\left({x}\right)={tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{x}}\right) \\ $$$${f}'\left({x}\right)=−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{4}}{{x}^{\mathrm{2}} }}=\frac{−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\frac{\mathrm{1}}{\left({x}+\mathrm{2}{i}\right)}−\frac{\mathrm{1}}{\left({x}−\mathrm{2}{i}\right)}\right) \\ $$$${f}''\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}{i}}.\frac{\mathrm{1}}{\left({x}+\mathrm{2}{i}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}{i}}.\frac{\mathrm{1}}{\left({x}−\mathrm{2}{i}\right)^{\mathrm{2}} } \\ $$$${f}^{{n}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({n}−\mathrm{1}\right)!\left(\frac{\mathrm{1}}{\left({x}+\mathrm{2}{i}\right)^{{n}} }−\frac{\mathrm{1}}{\left({x}−\mathrm{2}{i}\right)^{{n}} }\right) \\ $$$${f}^{{n}} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({n}−\mathrm{1}\right)!\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left({e}^{−\frac{\pi{in}}{\mathrm{2}}} −{e}^{\frac{\pi{in}}{\mathrm{2}}} \right)=\frac{\left(−\mathrm{1}\right)^{{n}} {sin}\left(\frac{\pi{n}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} }{n}\left({n}−\mathrm{1}\right)! \\ $$$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{x}}\right)=\frac{\pi}{\mathrm{2}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {sin}\left(\frac{\pi{n}}{\mathrm{2}}\right)\left({n}−\mathrm{1}\right)!}{\mathrm{2}^{{n}} {n}!}{x}^{{n}} \\ $$$$=\frac{\pi}{\mathrm{2}}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} {sin}\left(\frac{\pi{n}}{\mathrm{2}}\right)}{\mathrm{2}^{{n}} {n}}{x}^{{n}} \\ $$
Answered by Dwaipayan Shikari last updated on 10/May/21
tan^(−1) ((2/x))=(π/2)−tan^(−1) ((x/2))=(π/2)−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)))((x/2))^(2n+1)
$${tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{x}}\right)=\frac{\pi}{\mathrm{2}}−{tan}^{−\mathrm{1}} \left(\frac{{x}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)}\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}{n}+\mathrm{1}} \\ $$
Answered by mathmax by abdo last updated on 11/May/21
if x>0 ⇒f(x)=(π/2) −arctan((x/2)) we know arctan^′ u =(1/(1+u^2 )) =Σ_(n=0) ^∞ (−1)^n u^(2n) for ∣u∣<1 ⇒arctan(u) =Σ_(n=0) ^∞ (−1)^n (u^(2n+1) /(2n+1)) ⇒ f(x)=(π/2)−Σ_(n=0) ^∞ (−1)^n ((((x/2))^(2n+1) )/(2n+1))=(π/2)−Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)2^(2n+1) )) if x<0 ⇒f(x)=−(π/2)−arctan((x/2)) =−(π/2) −Σ_(n=0) ^∞ (((−1)^n x^(2n+1) )/((2n+1)2^(2n+1) )) with ∣x∣<2
$$\mathrm{if}\:\mathrm{x}>\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\pi}{\mathrm{2}}\:−\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{arctan}^{'} \mathrm{u}\:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{u}^{\mathrm{2n}} \:\:\mathrm{for}\:\mid\mathrm{u}\mid<\mathrm{1} \\ $$$$\Rightarrow\mathrm{arctan}\left(\mathrm{u}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\frac{\mathrm{u}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\pi}{\mathrm{2}}−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\frac{\left(\frac{\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}=\frac{\pi}{\mathrm{2}}−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2n}+\mathrm{1}} } \\ $$$$\mathrm{if}\:\mathrm{x}<\mathrm{0}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=−\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$=−\frac{\pi}{\mathrm{2}}\:−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{2}^{\mathrm{2n}+\mathrm{1}} }\:\:\:\mathrm{with}\:\mid\mathrm{x}\mid<\mathrm{2} \\ $$

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