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let-f-x-cos-2arctanx-1-calculate-f-n-0-2-developp-f-at-integr-serie-

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Question Number 66791 by mathmax by abdo last updated on 19/Aug/19
let f(x) =cos(2arctanx) 1) calculate f^((n)) (0) 2)developp f at integr serie
$${let}\:\:{f}\left({x}\right)\:={cos}\left(\mathrm{2}{arctanx}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$
Commented by mathmax by abdo last updated on 20/Aug/19
1) we have f(x)=cos(2arctanx) ⇒f(x)=2cos^2 (arctan(x))−1 2((1/( (√(1+x^2 )))))^2 −1 =(2/(1+x^2 ))−1 ⇒f^((n)) (x) =2 ((1/(1+x^2 )))^((n)) if n>0 we have (1/(x^2 +1)) =(1/((x−i)(x+i))) =(1/(2i)){(1/(x−i))−(1/(x+i))} ⇒ 2((1/(x^2 +1)))^((n)) =(1/i){((1/(x−i)))^((n)) −((1/(x+i)))^n }=(1/i){(((−1)^n n!)/((x−i)^(n+1) ))−(((−1)^n n!)/((x+i)^(n+1) ))} =(((−1)^n n!)/i){(((x+i)^(n+1) −(x−i)^(n+1) )/((x^2 +1)^(n+1) ))} ⇒ f^((n)) (0) =(((−1)^n n!)/i){i^(n+1) −(−i)^(n+1) } =(((−1)^n n!)/i)(2i Im(i^(n+1) )) =2(−1)^n n!Im(e^(((i(n+1))/2)π) ) =2(−1)^n n! sin(((n+1)π)/2) =2(−1)^n n!cos(((nπ)/2)) 2)f(x) =Σ_(n=0) ^(∞ ) ((f^((n)) (0))/(n!)) x^n =2Σ_(n=0) ^∞ (((−1)^n n!)/(n!))cos(((nπ)/2))x^n =2 Σ_(n=0) ^∞ (−1)^n cos(((nπ)/2))x^n .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)={cos}\left(\mathrm{2}{arctanx}\right)\:\:\Rightarrow{f}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left({arctan}\left({x}\right)\right)−\mathrm{1} \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)^{\mathrm{2}} −\mathrm{1}\:\:=\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)\:=\mathrm{2}\:\left(\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\left({n}\right)} \:\:{if}\:{n}>\mathrm{0} \\ $$$${we}\:{have}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{1}}{\left({x}−{i}\right)\left({x}+{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\frac{\mathrm{1}}{{x}−{i}}−\frac{\mathrm{1}}{{x}+{i}}\right\}\:\Rightarrow \\ $$$$\mathrm{2}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\right)^{\left({n}\right)} =\frac{\mathrm{1}}{{i}}\left\{\left(\frac{\mathrm{1}}{{x}−{i}}\right)^{\left({n}\right)} −\left(\frac{\mathrm{1}}{{x}+{i}}\right)^{{n}} \right\}=\frac{\mathrm{1}}{{i}}\left\{\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{i}\right)^{{n}+\mathrm{1}} }−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+{i}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}}\left\{\frac{\left({x}+{i}\right)^{{n}+\mathrm{1}} −\left({x}−{i}\right)^{{n}+\mathrm{1}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}+\mathrm{1}} }\right\}\:\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}}\left\{{i}^{{n}+\mathrm{1}} −\left(−{i}\right)^{{n}+\mathrm{1}} \right\}\:=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{i}}\left(\mathrm{2}{i}\:{Im}\left({i}^{{n}+\mathrm{1}} \right)\right) \\ $$$$=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {n}!{Im}\left({e}^{\frac{{i}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\pi} \right)\:=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {n}!\:{sin}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\:=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {n}!{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty\:} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:=\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{{n}!}{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right){x}^{{n}} \\ $$$$=\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{cos}\left(\frac{{n}\pi}{\mathrm{2}}\right){x}^{{n}} \:\:. \\ $$

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