Previous in Relation and Functions Next in Relation and Functions
Question Number 108711 by mathmax by abdo last updated on 18/Aug/20
![calculate U_n =∫_0 ^∞ (−1)^(2[x]−1) cos(n[x])dx find nature of Σ U_n](Q108711.png)
$$\mathrm{calculate}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{2}\left[\mathrm{x}\right]−\mathrm{1}} \mathrm{cos}\left(\mathrm{n}\left[\mathrm{x}\right]\right)\mathrm{dx} \\ $$$$\mathrm{find}\:\mathrm{nature}\:\mathrm{of}\:\Sigma\:\mathrm{U}_{\mathrm{n}} \\ $$
Answered by mathmax by abdo last updated on 21/Aug/20
![U_n =∫_0 ^∞ (−1)^(2[x]−1) cos(n[x])dx ⇒ U_n =Σ_(p=0) ^∞ ∫_p ^(p+1) (−1)^(2p−1) cos(pn)dx =−Σ_(p=0) ^∞ cos(np) =−Re(Σ_(p=0) ^∞ e^(inp) ) we hsve Σ_(p=0) ^∞ e^(inp) =Σ_(p=0) ^∞ (e^(in) )^p =(1/(1−e^(in) )) =(1/(1−cosn−isinn)) =((1−cos(n)+isin(n))/((1−cosn)^2 +sin^2 n)) =((1−cosn+isin(n))/(1−2cos(n)+1)) ⇒ U_n =((cos(n)−1)/(2−2cos(n)))](Q109150.png)
$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{2}\left[\mathrm{x}\right]−\mathrm{1}} \mathrm{cos}\left(\mathrm{n}\left[\mathrm{x}\right]\right)\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{p}} ^{\mathrm{p}+\mathrm{1}} \:\left(−\mathrm{1}\right)^{\mathrm{2p}−\mathrm{1}} \:\mathrm{cos}\left(\mathrm{pn}\right)\mathrm{dx} \\ $$$$=−\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{np}\right)\:=−\mathrm{Re}\left(\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{inp}} \right)\:\mathrm{we}\:\mathrm{hsve} \\ $$$$\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\mathrm{e}^{\mathrm{inp}} \:=\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\left(\mathrm{e}^{\mathrm{in}} \right)^{\mathrm{p}} \:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{e}^{\mathrm{in}} }\:=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cosn}−\mathrm{isinn}} \\ $$$$=\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{n}\right)+\mathrm{isin}\left(\mathrm{n}\right)}{\left(\mathrm{1}−\mathrm{cosn}\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \mathrm{n}}\:=\frac{\mathrm{1}−\mathrm{cosn}+\mathrm{isin}\left(\mathrm{n}\right)}{\mathrm{1}−\mathrm{2cos}\left(\mathrm{n}\right)+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{cos}\left(\mathrm{n}\right)−\mathrm{1}}{\mathrm{2}−\mathrm{2cos}\left(\mathrm{n}\right)} \\ $$