Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 57405 by Abdo msup. last updated on 03/Apr/19

calculate lim_(x→0) ((1−cosx.cos(2x)....cos(nx))/x^2 ) with n integr natural not 0.

$${calculate}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\mathrm{1}−{cosx}.{cos}\left(\mathrm{2}{x}\right)....{cos}\left({nx}\right)}{{x}^{\mathrm{2}} } \\ $$$${with}\:{n}\:{integr}\:{natural}\:{not}\:\mathrm{0}. \\ $$

Answered by Smail last updated on 05/Apr/19

cosx=1−(x^2 /2)+o(x^2 ) cos2x=1−(((2x)^2 )/2)+o(x^2 ) cos3x=1−(((3x)^2 )/2)+o(x^2 ) cosx.cos(2x).cos(3x)...cos(nx)=(1−(x^2 /2))(1−(((2x)^2 )/2))...(1−(((nx)^2 )/2))+o(x^2 ) =1−((x^2 /2)+(((2x)^2 )/2)+(((3x)^2 )/2)+...(((nx)^2 )/2))+o(x^2 ) =1−(x^2 /2)(1+2^2 +3^2 +...+n^2 )+o(x^2 ) 1−cosx.cos2x...cos(nx)∼_0 (x^2 /2)(((n(n+1)(2n+1))/6)) ((1−cosx.cos(2x)...cos(nx))/x^2 )∼_0 ((n(n+1)(2n+1))/(12)) lim_(x→0) ((1−cosx.cos(2x)...cos(nx))/x^2 )=((n(n+1)(2n+1))/(12))

$${cosx}=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${cos}\mathrm{2}{x}=\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${cos}\mathrm{3}{x}=\mathrm{1}−\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right) \\ $$$${cosx}.{cos}\left(\mathrm{2}{x}\right).{cos}\left(\mathrm{3}{x}\right)...{cos}\left({nx}\right)=\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right)...\left(\mathrm{1}−\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{3}{x}\right)^{\mathrm{2}} }{\mathrm{2}}+...\frac{\left({nx}\right)^{\mathrm{2}} }{\mathrm{2}}\right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+{n}^{\mathrm{2}} \right)+{o}\left({x}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}−{cosx}.{cos}\mathrm{2}{x}...{cos}\left({nx}\right)\underset{\mathrm{0}} {\sim}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right) \\ $$$$\frac{\mathrm{1}−{cosx}.{cos}\left(\mathrm{2}{x}\right)...{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }\underset{\mathrm{0}} {\sim}\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\mathrm{1}−{cosx}.{cos}\left(\mathrm{2}{x}\right)...{cos}\left({nx}\right)}{{x}^{\mathrm{2}} }=\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com