Menu Close

let-f-x-arctan-2x-1-calculate-f-n-x-then-f-n-0-2-developp-f-at-integr-serie-3-let-F-t-0-t-arctan-2x-dx-developp-F-at-integr-serie-4-give-F-1-at-form-of-serie-

Tinku Tara 0 Comments
Pin




Question Number 39292 by math khazana by abdo last updated on 04/Jul/18
let f(x)= arctan(2x) 1) calculate f^((n)) (x) then f^((n)) (0) 2) developp f at integr serie 3) let F(t)= ∫_0 ^t arctan(2x)dx developp F at integr serie 4) give F(1) at form of serie.
$${let}\:{f}\left({x}\right)=\:{arctan}\left(\mathrm{2}{x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{then}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{F}\left({t}\right)=\:\int_{\mathrm{0}} ^{{t}} \:\:{arctan}\left(\mathrm{2}{x}\right){dx} \\ $$$${developp}\:{F}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{4}\right)\:{give}\:{F}\left(\mathrm{1}\right)\:{at}\:{form}\:{of}\:{serie}. \\ $$
Commented by math khazana by abdo last updated on 06/Jul/18
1) we have f(x)=arctan(2x) ⇒ f^′ (x)= (2/(1+4x^2 )) ⇒ f^((n)) (x)=2 ((1/(1+4x^2 )))^((n−1)) let decompose inside C(x) w(x)= (1/(4x^2 +1))w w(x) = (1/((2x−i)(2x+i))) = (1/(2i))((1/(2x−i)) −(1/(2x+i))) = (1/(4i)){ ( (1/(x−(i/2))))−( (1/(x +(i/2))))^ } ⇒ w^((n−1)) (x) =(1/(4i)){ ((1/(x−(i/2))))^((n−1)) −((1/(x+(i/2))))^((n−1)) } = (1/(4i)){ (((−1)^(n−1) (n−1)!)/((x−(i/2))^n )) − (((−1)^(n−1) (n−1)!)/((x+(i/2))^n ))}⇒ f^((n)) (x)= (1/(2i))(−1)^(n−1) (n−1)!{ (1/((x−(i/2))^n )) −(1/((x+(i/2))^n ))} =(((−1)^(n−1) (n−1)!)/(2i)){ (((x+(i/2))^n −(x−(i/2))^n )/((x^2 +(1/4))^n ))}. f^((n)) (0) =(((−1)^(n−1) (n−1)!)/(2i)) 4^n {((i/2))^n −(−(i/2))^n } =((4^n (−1)^(n−1) (n−1)!)/(2i)) 2i Im( ((i/2))^n ) =4^n (−1)^(n−1) (n−1)! (1/2^n ) sin(((nπ)/2)) = 2^n (−1)^(n−1) (n−1)! sin(((nπ)/2)) ⇒ f^((2n)) (0) =0 and f^((2n+1)) (0) = 2^(2n+1) (2n)! sin( (((2n+1)π)/2)) =2^(2n+1) (2n)! (−1)^n
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)={arctan}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:\frac{\mathrm{2}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\Rightarrow\:{f}^{\left({n}\right)} \left({x}\right)=\mathrm{2}\:\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\right)^{\left({n}−\mathrm{1}\right)} \\ $$$${let}\:{decompose}\:{inside}\:{C}\left({x}\right)\:{w}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}{w} \\ $$$${w}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{2}{x}−{i}\right)\left(\mathrm{2}{x}+{i}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\frac{\mathrm{1}}{\mathrm{2}{x}−{i}}\:−\frac{\mathrm{1}}{\mathrm{2}{x}+{i}}\right)\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\:\left(\:\:\frac{\mathrm{1}}{{x}−\frac{{i}}{\mathrm{2}}}\right)−\left(\:\frac{\mathrm{1}}{{x}\:+\frac{{i}}{\mathrm{2}}}\right)^{} \right\}\:\Rightarrow \\ $$$${w}^{\left({n}−\mathrm{1}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\left(\frac{\mathrm{1}}{{x}−\frac{{i}}{\mathrm{2}}}\right)^{\left({n}−\mathrm{1}\right)} \:−\left(\frac{\mathrm{1}}{{x}+\frac{{i}}{\mathrm{2}}}\right)^{\left({n}−\mathrm{1}\right)} \right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{n}} }\:−\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\left({n}−\mathrm{1}\right)!}{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{n}} }\right\}\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\left\{\:\:\frac{\mathrm{1}}{\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{n}} }\:−\frac{\mathrm{1}}{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{n}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\left\{\:\frac{\left({x}+\frac{{i}}{\mathrm{2}}\right)^{{n}} \:−\left({x}−\frac{{i}}{\mathrm{2}}\right)^{{n}} }{\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{4}}\right)^{{n}} }\right\}. \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\mathrm{4}^{{n}} \left\{\left(\frac{{i}}{\mathrm{2}}\right)^{{n}} \:−\left(−\frac{{i}}{\mathrm{2}}\right)^{{n}} \right\} \\ $$$$=\frac{\mathrm{4}^{{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}{\mathrm{2}{i}}\:\:\mathrm{2}{i}\:{Im}\left(\:\left(\frac{{i}}{\mathrm{2}}\right)^{{n}} \right) \\ $$$$=\mathrm{4}^{{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$=\:\mathrm{2}^{{n}} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:{sin}\left(\frac{{n}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}^{\left(\mathrm{2}{n}\right)} \left(\mathrm{0}\right)\:=\mathrm{0}\:\:{and} \\ $$$${f}^{\left(\mathrm{2}{n}+\mathrm{1}\right)} \left(\mathrm{0}\right)\:=\:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}{n}\right)!\:{sin}\left(\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}{n}\right)!\:\left(−\mathrm{1}\right)^{{n}} \\ $$
Commented by math khazana by abdo last updated on 06/Jul/18
2) we have f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n = Σ_(n=0) ^∞ ((f^((2n+1)) (0))/((2n+1)!)) x^(2n+1) =Σ_(n=0) ^(∞ ) ((2^(2n+1) (2n)!(−1)^n )/((2n+1)!)) x^(2n+1) = Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) 2^(2n+1) x^(2n+1) .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{{f}^{\left(\mathrm{2}{n}+\mathrm{1}\right)} \left(\mathrm{0}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty\:} \:\:\:\:\:\frac{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \left(\mathrm{2}{n}\right)!\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} \:.\: \\ $$
Commented by math khazana by abdo last updated on 06/Jul/18
3) we have F(t) = ∫_0 ^t arctan(2x)dx due to uniform convergencewe have F(t) = ∫_0 ^t f(x)dx =∫_0 ^t (Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) 2^(2n+1) x^(2n+1) )dx =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) 2^(2n+1) ∫_0 ^t x^(2n+1) dx =Σ_(n=0) ^∞ (((−1)^n 2^(2n+1) )/((2n+1)(2n+2))) t^(2n+2) 4) F(1) =Σ_(n=0) ^∞ (((−1)^n 2^(2n+1) )/((2n+1)(2n+2))).
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{F}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{{t}} \:{arctan}\left(\mathrm{2}{x}\right){dx} \\ $$$${due}\:{to}\:{uniform}\:{convergencewe}\:{have} \\ $$$${F}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{{t}} \:{f}\left({x}\right){dx}\:=\int_{\mathrm{0}} ^{{t}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \:{x}^{\mathrm{2}{n}+\mathrm{1}} \right){dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}+\mathrm{1}}\:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} \:\:\int_{\mathrm{0}} ^{{t}} \:\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:{dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:{t}^{\mathrm{2}{n}+\mathrm{2}} \\ $$$$\left.\mathrm{4}\right)\:{F}\left(\mathrm{1}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *