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Question Number 136570 by mnjuly1970 last updated on 23/Mar/21

$$.......nice.....calculus.....$$$$\mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}{cos}^n\left(x\right).cos\left(nx\right)=?$$$$solution::::$$$$\mathrm{\Omega }=\frac{1}{2}\underset{n=0}{\sum ^{\mathrm{\infty }}}{cos}^{n-1}\left(x\right)\{cos(x-nx)+cos(x+nx)$$$$\therefore 2\mathrm{\Omega }=\underset{n=0}{\sum ^{\mathrm{\infty }}}{cos}^{n-1}\left(x\right).cos(n-1)x$$$$+\underset{n=0}{\sum ^{\mathrm{\infty }}}{cos}^{n-1}\left(x\right).cos(n+1)x$$$$=1+\underset{n=1}{\sum ^{\mathrm{\infty }}}{cos}^{n-1}\left(x\right).cos(n-1)x+\frac{1}{{cos}^2\left(x\right)}\underset{n=0}{\sum ^{\mathrm{\infty }}}{cos}^{n+1}\left(x\right).cos(n+1)x$$$$=1+\mathrm{\Omega }+\frac{1}{{cos}^2\left(x\right)}(\mathrm{\Omega }-1)$$$$\mathrm{\Omega }(1-\frac{1}{{cos}^2\left(x\right)})=1-\frac{1}{{cos}^2\left(x\right)}$$$$\therefore \mathrm{\Omega }=1.................$$$$$$

Answered by Dwaipayan Shikari last updated on 23/Mar/21

$$\underset{n=0}{\sum ^{\mathrm{\infty }}}{cos}^n\left(x\right)cos\left(nx\right)g=cosx$$$$=1+\underset{n=1}{\sum ^{\mathrm{\infty }}}cos\left(nx\right)g^n=1+\frac{1}{2}\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(e^{ix}g\right)}^n+{\left(e^{-ix}g\right)}^n$$$$=1+\frac{1}{2}(\frac{1}{(1-e^{ix}g)}+\frac{1}{(1-e^{-ix}g)})=1+\frac{1}{2}\left(\frac{1-e^{-ix}g+1-e^{ix}g}{1-2gcosx+g^2}\right)$$$$=1+\frac{1}{2}\left(\frac{2-2gcosx}{1-{cos}^{2}x}\right)=1+1=2$$

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