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Question Number 158186 by mkam last updated on 31/Oct/21

Answered by puissant last updated on 31/Oct/21

sinx=x−(x^3 /(3!))+(x^5 /(5!))−(x^7 /(7!))+......  ⇒ ((sinx)/x) = 1−(x^2 /(3!))+(x^4 /(5!))−(x^6 /(7!))+.....  the solutons of this polynomial are   the solution of sinx (x=nπ) ; (n∈Z)  ⇒ ((sinx)/x)=(1+(x/π))(1−(x/π))(1+(x/(2π)))(1−(x/(2π)))..  ⇒ ((sinx)/x)=(1−(x^2 /π^2 ))(1−(x^2 /(4π^2 )))(1−(x^2 /(9π^2 )))...  ⇒ ((sinx)/x) = Π_(k=1) ^∞ (1−(x^2 /((nπ)^2 )))...               ...........Le puissant..........

$${sinx}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}!}+...... \\ $$$$\Rightarrow\:\frac{{sinx}}{{x}}\:=\:\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{7}!}+..... \\ $$$${the}\:{solutons}\:{of}\:{this}\:{polynomial}\:{are}\: \\ $$$${the}\:{solution}\:{of}\:{sinx}\:\left({x}={n}\pi\right)\:;\:\left({n}\in\mathbb{Z}\right) \\ $$$$\Rightarrow\:\frac{{sinx}}{{x}}=\left(\mathrm{1}+\frac{{x}}{\pi}\right)\left(\mathrm{1}−\frac{{x}}{\pi}\right)\left(\mathrm{1}+\frac{{x}}{\mathrm{2}\pi}\right)\left(\mathrm{1}−\frac{{x}}{\mathrm{2}\pi}\right).. \\ $$$$\Rightarrow\:\frac{{sinx}}{{x}}=\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{9}\pi^{\mathrm{2}} }\right)... \\ $$$$\Rightarrow\:\frac{{sinx}}{{x}}\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\left({n}\pi\right)^{\mathrm{2}} }\right)... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:...........\mathscr{L}{e}\:{puissant}.......... \\ $$

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