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log-2-x-1-2015-y-1-2015-1-e-2-ixy-2015-

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Question Number 7159 by Master Moon last updated on 14/Aug/16
log_2 Π_(x=1) ^(2015) Π_(y=1) ^(2015) (1+e^((2𝛑ixy)/(2015)) ) = ?
$$\boldsymbol{{log}}_{\mathrm{2}} \underset{\boldsymbol{{x}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\:\underset{\boldsymbol{{y}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\mathrm{1}+\boldsymbol{{e}}^{\frac{\mathrm{2}\boldsymbol{\pi{ixy}}}{\mathrm{2015}}} \right)\:=\:? \\ $$
Commented by FilupSmith last updated on 14/Aug/16
S=log_2 (Π_(x=1) ^(2015) Π_(y=1) ^(2015) (1+e^((2𝛑ixy)/(2015)) ))=log_2 (P) P =Π_(x=1) ^(2015) (Π_(y=1) ^(2015) (1+e^((2𝛑ixy)/(2015)) )) (−1)^k =e^(iπk) =(e^(iπ) )^k P =Π_(x=1) ^(2015) (Π_(y=1) ^(2015) (1+(−1)^((2xy)/(2015)) )) S∈C⇔P∉Z trying to think how to continue
$${S}=\mathrm{log}_{\mathrm{2}} \left(\underset{\boldsymbol{{x}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\:\underset{\boldsymbol{{y}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\mathrm{1}+\boldsymbol{{e}}^{\frac{\mathrm{2}\boldsymbol{\pi{ixy}}}{\mathrm{2015}}} \right)\right)=\mathrm{log}_{\mathrm{2}} \left({P}\right) \\ $$$${P}\:=\underset{\boldsymbol{{x}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\:\left(\underset{\boldsymbol{{y}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\mathrm{1}+\boldsymbol{{e}}^{\frac{\mathrm{2}\boldsymbol{\pi{ixy}}}{\mathrm{2015}}} \right)\right) \\ $$$$\left(−\mathrm{1}\right)^{{k}} ={e}^{{i}\pi{k}} =\left({e}^{{i}\pi} \right)^{{k}} \\ $$$${P}\:=\underset{\boldsymbol{{x}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\:\left(\underset{\boldsymbol{{y}}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\mathrm{1}+\left(−\mathrm{1}\right)^{\frac{\mathrm{2}{xy}}{\mathrm{2015}}} \right)\right) \\ $$$${S}\in\mathbb{C}\Leftrightarrow{P}\notin\mathbb{Z} \\ $$$$\mathrm{trying}\:\mathrm{to}\:\mathrm{think}\:\mathrm{how}\:\mathrm{to}\:\mathrm{continue} \\ $$
Commented by Yozzia last updated on 14/Aug/16
−−−−−−−−−−−−−−−−−−−− Let θ=((xyπ)/(2015))⇒u(x,y)=1+e^((2πixy)/(2015)) =1+cos2θ+isin2θ u(x,y)=2cos^2 θ+2isinθcosθ u(x,y)=2cosθ(cosθ+isinθ) u(x,y)=(2cosθ)e^(iθ) ={2cos((xyπ)/(2015))}e^((πxyi)/(2015)) ⇒Π_(y=1) ^(2015) u(x,y)=2^(2015) (Π_(y=1) ^(2015) cos((xyπ)/(2015)))e^(((xπi)/(2015))(1+2+3+4+...+2015)) Π_(y=1) ^(2015) u(x,y)=2^(2015) [Π_(y=1) ^(2015) cos((xyπ)/(2015))]e^(((xπi)/(2015))×((2015×2016)/2)) Π_(y=1) ^(2015) u(x,y)=2^(2015) [Π_(y=1) ^(2015) cos((xyπ)/(2015))]e^(1008πix) x∈N⇒e^(1008πix) =cos1008πx+0=cos1008πx=1 ⇒Π_(y=1) ^(2015) u(x,y)=2^(2015) [Π_(y=1) ^(2015) cos((xyπ)/(2015))] Π_(x=1) ^(2015) (Π_(y=1) ^(2015) u(x,y))=Π_(x=1) ^(2015) [2^(2015) Π_(y=1) ^(2015) cos((πxy)/(2015))] Π_(x=1) ^(2015) (Π_(y=1) ^(2015) u(x,y))=2^(2015^2 ) Π_(x=1) ^(2015) [Π_(y=1) ^(2015) cos((πxy)/(2015))] Π_(x=1) ^(2015) (Π_(y=1) ^(2015) u(x,y))=2^(2015^2 ) Π_(y=1) ^(2015) [Π_(x=1) ^(2015) cos((πxy)/(2015))]
$$−−−−−−−−−−−−−−−−−−−− \\ $$$${Let}\:\theta=\frac{{xy}\pi}{\mathrm{2015}}\Rightarrow{u}\left({x},{y}\right)=\mathrm{1}+{e}^{\frac{\mathrm{2}\pi{ixy}}{\mathrm{2015}}} =\mathrm{1}+{cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta \\ $$$${u}\left({x},{y}\right)=\mathrm{2}{cos}^{\mathrm{2}} \theta+\mathrm{2}{isin}\theta{cos}\theta \\ $$$${u}\left({x},{y}\right)=\mathrm{2}{cos}\theta\left({cos}\theta+{isin}\theta\right) \\ $$$${u}\left({x},{y}\right)=\left(\mathrm{2}{cos}\theta\right){e}^{{i}\theta} =\left\{\mathrm{2}{cos}\frac{{xy}\pi}{\mathrm{2015}}\right\}{e}^{\frac{\pi{xyi}}{\mathrm{2015}}} \: \\ $$$$\Rightarrow\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)=\mathrm{2}^{\mathrm{2015}} \left(\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{{xy}\pi}{\mathrm{2015}}\right){e}^{\frac{{x}\pi{i}}{\mathrm{2015}}\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+…+\mathrm{2015}\right)} \\ $$$$\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)=\mathrm{2}^{\mathrm{2015}} \left[\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{{xy}\pi}{\mathrm{2015}}\right]{e}^{\frac{{x}\pi{i}}{\mathrm{2015}}×\frac{\mathrm{2015}×\mathrm{2016}}{\mathrm{2}}} \\ $$$$\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)=\mathrm{2}^{\mathrm{2015}} \left[\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{{xy}\pi}{\mathrm{2015}}\right]{e}^{\mathrm{1008}\pi{ix}} \\ $$$${x}\in\mathbb{N}\Rightarrow{e}^{\mathrm{1008}\pi{ix}} ={cos}\mathrm{1008}\pi{x}+\mathrm{0}={cos}\mathrm{1008}\pi{x}=\mathrm{1} \\ $$$$\Rightarrow\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)=\mathrm{2}^{\mathrm{2015}} \left[\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{{xy}\pi}{\mathrm{2015}}\right] \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)\right)=\underset{{x}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left[\mathrm{2}^{\mathrm{2015}} \underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{\pi{xy}}{\mathrm{2015}}\right] \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)\right)=\mathrm{2}^{\mathrm{2015}^{\mathrm{2}} } \underset{{x}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left[\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{\pi{xy}}{\mathrm{2015}}\right] \\ $$$$\underset{{x}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left(\underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{u}\left({x},{y}\right)\right)=\mathrm{2}^{\mathrm{2015}^{\mathrm{2}} } \underset{{y}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}\left[\underset{{x}=\mathrm{1}} {\overset{\mathrm{2015}} {\prod}}{cos}\frac{\pi{xy}}{\mathrm{2015}}\right] \\ $$$$ \\ $$$$ \\ $$

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