Menu Close

nice-calculus-n-0-cos-n-x-cos-nx-solution-1-2-n-0-cos-n-1-x-cos-x-nx-cos-x-nx-2-n-0-cos-n-1-x-cos-n-1-x-

Tinku Tara 0 Comments
Pin




Question Number 136570 by mnjuly1970 last updated on 23/Mar/21
.......nice ..... calculus..... Ω=Σ_(n=0) ^∞ cos^n (x).cos(nx)=? solution:::: Ω=(1/2)Σ_(n=0) ^∞ cos^(n−1) (x){cos(x−nx)+cos(x+nx) ∴ 2Ω=Σ_(n=0) ^∞ cos^(n−1) (x).cos(n−1)x +Σ_(n=0) ^∞ cos^(n−1) (x).cos(n+1)x =1+Σ_(n=1) ^∞ cos^(n−1) (x).cos(n−1)x+(1/(cos^2 (x)))Σ_(n=0) ^∞ cos^(n+1) (x).cos(n+1)x =1+Ω+(1/(cos^2 (x)))(Ω−1) Ω(1−(1/(cos^2 (x))))=1−(1/(cos^2 (x))) ∴ Ω=1 .................
$$\:\:\:\:\:\:\:\:\:…….{nice}\:\:…..\:\:\:{calculus}….. \\ $$$$\:\:\:\:\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}} \left({x}\right).{cos}\left({nx}\right)=? \\ $$$$\:\:\:{solution}:::: \\ $$$$\:\:\:\:\Omega=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}−\mathrm{1}} \left({x}\right)\left\{{cos}\left({x}−{nx}\right)+{cos}\left({x}+{nx}\right)\right. \\ $$$$\:\:\therefore\:\mathrm{2}\Omega=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}−\mathrm{1}} \left({x}\right).{cos}\left({n}−\mathrm{1}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}−\mathrm{1}} \left({x}\right).{cos}\left({n}+\mathrm{1}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{cos}^{{n}−\mathrm{1}} \left({x}\right).{cos}\left({n}−\mathrm{1}\right){x}+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}+\mathrm{1}} \left({x}\right).{cos}\left({n}+\mathrm{1}\right){x} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}+\Omega+\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}\left(\Omega−\mathrm{1}\right) \\ $$$$\:\:\:\Omega\left(\mathrm{1}−\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)}\right)=\mathrm{1}−\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left({x}\right)} \\ $$$$\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\Omega=\mathrm{1}\:…………….. \\ $$$$\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 23/Mar/21
Σ_(n=0) ^∞ cos^n (x)cos(nx) g=cosx =1+Σ_(n=1) ^∞ cos(nx)g^n =1+(1/2)Σ_(n=1) ^∞ (e^(ix) g)^n +(e^(−ix) g)^n =1+(1/2)((1/((1−e^(ix) g)))+(1/((1−e^(−ix) g))))=1+(1/2)(((1−e^(−ix) g+1−e^(ix) g)/(1−2gcosx+g^2 ))) =1+(1/2)(((2−2gcosx)/(1−cos^2 x)))=1+1=2
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}^{{n}} \left({x}\right){cos}\left({nx}\right)\:\:\:\:{g}={cosx} \\ $$$$=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{cos}\left({nx}\right){g}^{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left({e}^{{ix}} {g}\right)^{{n}} +\left({e}^{−{ix}} {g}\right)^{{n}} \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\left(\mathrm{1}−{e}^{{ix}} {g}\right)}+\frac{\mathrm{1}}{\left(\mathrm{1}−{e}^{−{ix}} {g}\right)}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}−{e}^{−{ix}} {g}+\mathrm{1}−{e}^{{ix}} {g}}{\mathrm{1}−\mathrm{2}{gcosx}+{g}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}−\mathrm{2}{gcosx}}{\mathrm{1}−{cos}^{\mathrm{2}} {x}}\right)=\mathrm{1}+\mathrm{1}=\mathrm{2} \\ $$

Pin

Leave a Reply

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