Menu Close

Advanced-Calculus-prove-that-determinant-i-n-0-1-x-2-2n-1-2-cos-pix-2-ii-n-0-1-x-2-2n-1-2-cosh

Tinku Tara 0 Comments
Pin




Question Number 135610 by mnjuly1970 last updated on 14/Mar/21
.... Advanced ...... Calculus.... prove that : determinant (((i :: Π_(n=0) ^∞ (1−(x^2 /((2n+1)^2 ))) =cos(((πx)/2)) ✓ )),((ii :: Π_(n=0) ^∞ (1+(x^2 /((2n+1)^2 )))= cosh(((πx)/2)) ✓✓))) .............
$$\:\:\:\:\:\:\:\:\:\:….\:\mathscr{A}{dvanced}\:\:……\:\:\mathscr{C}{alculus}…. \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:{that}\:: \\ $$$$\:\:\:\begin{array}{|c|c|}{{i}\:::\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:={cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\:\:\:\checkmark\:\:}\\{{ii}\:::\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)=\:{cosh}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\checkmark\checkmark}\\\hline\end{array}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…………. \\ $$
Answered by Dwaipayan Shikari last updated on 14/Mar/21
cosx=C((π/2)−x)((π/2)+x)(((3π)/2)−x)(((3π)/2)+x)... cos(x) has zeros at ={(π/2),−(π/2),((3π)/2),−((3π)/2)....} x=0 ⇒1=C(π/2).(π/2).((3π)/2).((3π)/2)...⇒C=(2/π).(2/π).(2/(3π)).(2/(3π))... cosx=(1−((2x)/π))(1+((2x)/π))(1−((2x)/(3π)))(1+((2x)/(3π))).... cosx=Π_(n=0) ^∞ (1−((2x^2 )/(π^2 (2n+1)^2 ))) cos((π/2)x)=Π_(n=0) ^∞ (1−(x^2 /((2n+1)^2 ))) cosh((π/2)x)=Π_(n=0) ^∞ (1+(x^2 /((2n+1)^2 ))) x→xi
$${cosx}={C}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\left(\frac{\pi}{\mathrm{2}}+{x}\right)\left(\frac{\mathrm{3}\pi}{\mathrm{2}}−{x}\right)\left(\frac{\mathrm{3}\pi}{\mathrm{2}}+{x}\right)… \\ $$$${cos}\left({x}\right)\:{has}\:{zeros}\:{at}\:=\left\{\frac{\pi}{\mathrm{2}},−\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}},−\frac{\mathrm{3}\pi}{\mathrm{2}}….\right\} \\ $$$${x}=\mathrm{0}\:\:\:\Rightarrow\mathrm{1}={C}\frac{\pi}{\mathrm{2}}.\frac{\pi}{\mathrm{2}}.\frac{\mathrm{3}\pi}{\mathrm{2}}.\frac{\mathrm{3}\pi}{\mathrm{2}}…\Rightarrow{C}=\frac{\mathrm{2}}{\pi}.\frac{\mathrm{2}}{\pi}.\frac{\mathrm{2}}{\mathrm{3}\pi}.\frac{\mathrm{2}}{\mathrm{3}\pi}… \\ $$$${cosx}=\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\pi}\right)\left(\mathrm{1}−\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)\left(\mathrm{1}+\frac{\mathrm{2}{x}}{\mathrm{3}\pi}\right)…. \\ $$$${cosx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$${cos}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:\:\:\:\:\: \\ $$$${cosh}\left(\frac{\pi}{\mathrm{2}}{x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right)\:\:\:{x}\rightarrow{xi} \\ $$$$ \\ $$

Pin

Leave a Reply

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