Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 13031 by FilupS last updated on 12/May/17

MrW1    Going off of Q12883     How many unique angles angles in Z^3 ?  What about Z^n ?

$$\mathrm{MrW1} \\ $$$$ \\ $$$$\mathrm{Going}\:\mathrm{off}\:\mathrm{of}\:\mathrm{Q12883} \\ $$$$\: \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{unique}\:\mathrm{angles}\:\mathrm{angles}\:\mathrm{in}\:\mathbb{Z}^{\mathrm{3}} ? \\ $$$$\mathrm{What}\:\mathrm{about}\:\mathbb{Z}^{{n}} ? \\ $$

Answered by mrW1 last updated on 12/May/17

For 1≤x_i ≤p_i  with 1≤i≤n and x_i  ∈P^n   the number of unique angles for points   P(x_(1,) x_2 ,∙∙∙,x_i ,∙∙∙,x_n ) is  Φ=Π_(i=1) ^n p_i −Σ_(x_1 =1) ^p_1  Σ_(x_2 =1) ^p_2  ∙∙∙Σ_(x_i =1) ^p_i  ∙∙∙Σ_(x_n =1) ^p_n  sign[GCD(x_1 ,x_2 ,∙∙∙,x_i ,∙∙∙,x_n )−1]

$${For}\:\mathrm{1}\leqslant{x}_{{i}} \leqslant{p}_{{i}} \:{with}\:\mathrm{1}\leqslant{i}\leqslant{n}\:{and}\:{x}_{{i}} \:\in\mathbb{P}^{{n}} \\ $$$${the}\:{number}\:{of}\:{unique}\:{angles}\:{for}\:{points}\: \\ $$$${P}\left({x}_{\mathrm{1},} {x}_{\mathrm{2}} ,\centerdot\centerdot\centerdot,{x}_{{i}} ,\centerdot\centerdot\centerdot,{x}_{{n}} \right)\:{is} \\ $$$$\Phi=\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}{p}_{{i}} −\underset{{x}_{\mathrm{1}} =\mathrm{1}} {\overset{{p}_{\mathrm{1}} } {\sum}}\underset{{x}_{\mathrm{2}} =\mathrm{1}} {\overset{{p}_{\mathrm{2}} } {\sum}}\centerdot\centerdot\centerdot\underset{{x}_{{i}} =\mathrm{1}} {\overset{{p}_{{i}} } {\sum}}\centerdot\centerdot\centerdot\underset{{x}_{{n}} =\mathrm{1}} {\overset{{p}_{{n}} } {\sum}}{sign}\left[{GCD}\left({x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,\centerdot\centerdot\centerdot,{x}_{{i}} ,\centerdot\centerdot\centerdot,{x}_{{n}} \right)−\mathrm{1}\right] \\ $$

Commented by mrW1 last updated on 12/May/17

For −r_i ≤x_i ≤p_i  with 1≤i≤n and x_i  ∈Z^n   the solution is...  in thinking...how to write the formula

$${For}\:−{r}_{{i}} \leqslant{x}_{{i}} \leqslant{p}_{{i}} \:{with}\:\mathrm{1}\leqslant{i}\leqslant{n}\:{and}\:{x}_{{i}} \:\in\mathbb{Z}^{{n}} \\ $$$${the}\:{solution}\:{is}... \\ $$$${in}\:{thinking}...{how}\:{to}\:{write}\:{the}\:{formula} \\ $$

Commented by FilupS last updated on 12/May/17

why is x_i ∈P^n ?

$$\mathrm{why}\:\mathrm{is}\:{x}_{{i}} \in\mathbb{P}^{{n}} ? \\ $$

Commented by mrW1 last updated on 12/May/17

I wanted to solve this simple case at  first: 1≤x_i ≤p_i .  The result for the case −r_i ≤x_i ≤−1 is  the same as  for 1≤x_i ≤r_i .  It means the problem for x_i ∈Z^n  will  be solved as a couple of problems  for x_i ∈P^(n.)

$${I}\:{wanted}\:{to}\:{solve}\:{this}\:{simple}\:{case}\:{at} \\ $$$${first}:\:\mathrm{1}\leqslant{x}_{{i}} \leqslant{p}_{{i}} . \\ $$$${The}\:{result}\:{for}\:{the}\:{case}\:−{r}_{{i}} \leqslant{x}_{{i}} \leqslant−\mathrm{1}\:{is} \\ $$$${the}\:{same}\:{as}\:\:{for}\:\mathrm{1}\leqslant{x}_{{i}} \leqslant{r}_{{i}} . \\ $$$${It}\:{means}\:{the}\:{problem}\:{for}\:{x}_{{i}} \in\mathbb{Z}^{{n}} \:{will} \\ $$$${be}\:{solved}\:{as}\:{a}\:{couple}\:{of}\:{problems} \\ $$$${for}\:{x}_{{i}} \in\mathbb{P}^{{n}.} \\ $$

Commented by mrW1 last updated on 12/May/17

Example: the solution for  −127≤x≤128  −128≤y≤127  is:  Φ=128×127−Σ_(i=1) ^(128) Σ_(j=1) ^(127) sign(GCD(i,j)−1)  +127×127−Σ_(i=1) ^(127) Σ_(j=1) ^(127) sign(GCD(i,j)−1)  +127×128−Σ_(i=1) ^(127) Σ_(j=1) ^(128) sign(GCD(i,j)−1)  +128×128−Σ_(i=1) ^(128) Σ_(j=1) ^(128) sign(GCD(i,j)−1)  +4    with line 1 for quadrant 1 and  line 2 for quadrant 2 and  line 3 for quadrant 3 and  line 4 for quadrant 4 and  line 5 for points on axes

$${Example}:\:{the}\:{solution}\:{for} \\ $$$$−\mathrm{127}\leqslant{x}\leqslant\mathrm{128} \\ $$$$−\mathrm{128}\leqslant{y}\leqslant\mathrm{127} \\ $$$${is}: \\ $$$$\Phi=\mathrm{128}×\mathrm{127}−\underset{{i}=\mathrm{1}} {\overset{\mathrm{128}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{127}} {\sum}}{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right) \\ $$$$+\mathrm{127}×\mathrm{127}−\underset{{i}=\mathrm{1}} {\overset{\mathrm{127}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{127}} {\sum}}{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right) \\ $$$$+\mathrm{127}×\mathrm{128}−\underset{{i}=\mathrm{1}} {\overset{\mathrm{127}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{128}} {\sum}}{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right) \\ $$$$+\mathrm{128}×\mathrm{128}−\underset{{i}=\mathrm{1}} {\overset{\mathrm{128}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{\mathrm{128}} {\sum}}{sign}\left({GCD}\left({i},{j}\right)−\mathrm{1}\right) \\ $$$$+\mathrm{4} \\ $$$$ \\ $$$${with}\:{line}\:\mathrm{1}\:{for}\:{quadrant}\:\mathrm{1}\:{and} \\ $$$${line}\:\mathrm{2}\:{for}\:{quadrant}\:\mathrm{2}\:{and} \\ $$$${line}\:\mathrm{3}\:{for}\:{quadrant}\:\mathrm{3}\:{and} \\ $$$${line}\:\mathrm{4}\:{for}\:{quadrant}\:\mathrm{4}\:{and} \\ $$$${line}\:\mathrm{5}\:{for}\:{points}\:{on}\:{axes} \\ $$

Commented by FilupS last updated on 12/May/17

This is all so fascinating and gives  me so many more questions to explore.     For example:  for  u≤x≤m  and v≤y≤n,  x, y∈Z  What is the average angle?

$$\mathrm{This}\:\mathrm{is}\:\mathrm{all}\:\mathrm{so}\:\mathrm{fascinating}\:\mathrm{and}\:\mathrm{gives} \\ $$$$\mathrm{me}\:\mathrm{so}\:\mathrm{many}\:\mathrm{more}\:\mathrm{questions}\:\mathrm{to}\:\mathrm{explore}. \\ $$$$\: \\ $$$$\mathrm{For}\:\mathrm{example}: \\ $$$$\mathrm{for}\:\:{u}\leqslant{x}\leqslant{m}\:\:\mathrm{and}\:{v}\leqslant{y}\leqslant{n},\:\:{x},\:{y}\in\mathbb{Z} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{average}\:\mathrm{angle}? \\ $$

Commented by mrW1 last updated on 12/May/17

Let′s say −u≤x≤m and −v≤y≤n.    In quadrant 1 we have  θ_(ij) =tan^(−1) ((y/x))=tan^(−1) ((j/i))=tan^(−1) (((j′)/(i′)))  S_1 =Σ_(i=1) ^m Σ_(j=1) ^n θ_(ij) =Σ_(i=1) ^m Σ_(j=1) ^n tan^(−1) ((j/i))    In quadrant 2 we have  θ_(ij) =tan^(−1) ((y/x))=π−tan^(−1) ((j/i))=π−tan^(−1) (((j′)/(i′)))  S_2 =Σ_(i=1) ^u Σ_(j=1) ^n θ_(ij) =unπ−Σ_(i=1) ^u Σ_(j=1) ^n tan^(−1) ((j/i))    In quadrant 3 we have  θ_(ij) =tan^(−1) ((y/x))=π+tan^(−1) ((j/i))=π+tan^(−1) (((j′)/(i′)))  S_3 =Σ_(i=1) ^u Σ_(j=1) ^v θ_(ij) =uvπ+Σ_(i=1) ^u Σ_(j=1) ^v tan^(−1) ((j/i))    In quadrant 4 we have  θ_(ij) =tan^(−1) ((y/x))=2π−tan^(−1) ((j/i))=2π−tan^(−1) (((j′)/(i′)))  S_4 =Σ_(i=1) ^m Σ_(j=1) ^v θ_(ij) =2mvπ−Σ_(i=1) ^m Σ_(j=1) ^v tan^(−1) ((j/i))    On +x axis: m points with angle 0,  On +y axis: n points with angle (π/2),  On −x axis: u points with angle π,  On −y axis: v points with angle ((3π)/2),  S_5 =m×0+n×(π/2)+uπ+v×((3π)/2)=(((n+2u+3v)π)/2)    Sum of all angles:  S=S_1 +S_2 +S_3 +S_4 +S_5   =Σ_(i=1) ^m Σ_(j=1) ^n tan^(−1) ((j/i))  +unπ−Σ_(i=1) ^u Σ_(j=1) ^n tan^(−1) ((j/i))  +uvπ+Σ_(i=1) ^u Σ_(j=1) ^v tan^(−1) ((j/i))  +2mvπ−Σ_(i=1) ^m Σ_(j=1) ^v tan^(−1) ((j/i))  +(((n+2u+3v)π)/2)  =(((2un+2uv+4mv+n+2u+3v)π)/2)  +[Σ_(i=1) ^m −Σ_(i=1) ^u ][Σ_(j=1) ^n tan^(−1) ((j/i))−Σ_(j=1) ^v tan^(−1) ((j/i))]    The average angle is  θ^− =(S/((m+u+1)(n+v+1)))

$${Let}'{s}\:{say}\:−{u}\leqslant{x}\leqslant{m}\:{and}\:−{v}\leqslant{y}\leqslant{n}. \\ $$$$ \\ $$$${In}\:{quadrant}\:\mathrm{1}\:{we}\:{have} \\ $$$$\theta_{{ij}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}'}{{i}'}\right) \\ $$$${S}_{\mathrm{1}} =\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\theta_{{ij}} =\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$ \\ $$$${In}\:{quadrant}\:\mathrm{2}\:{we}\:{have} \\ $$$$\theta_{{ij}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)=\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)=\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}'}{{i}'}\right) \\ $$$${S}_{\mathrm{2}} =\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\theta_{{ij}} ={un}\pi−\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$ \\ $$$${In}\:{quadrant}\:\mathrm{3}\:{we}\:{have} \\ $$$$\theta_{{ij}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)=\pi+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)=\pi+\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}'}{{i}'}\right) \\ $$$${S}_{\mathrm{3}} =\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\theta_{{ij}} ={uv}\pi+\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$ \\ $$$${In}\:{quadrant}\:\mathrm{4}\:{we}\:{have} \\ $$$$\theta_{{ij}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)=\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)=\mathrm{2}\pi−\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}'}{{i}'}\right) \\ $$$${S}_{\mathrm{4}} =\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\theta_{{ij}} =\mathrm{2}{mv}\pi−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$ \\ $$$${On}\:+{x}\:{axis}:\:{m}\:{points}\:{with}\:{angle}\:\mathrm{0}, \\ $$$${On}\:+{y}\:{axis}:\:{n}\:{points}\:{with}\:{angle}\:\frac{\pi}{\mathrm{2}}, \\ $$$${On}\:−{x}\:{axis}:\:{u}\:{points}\:{with}\:{angle}\:\pi, \\ $$$${On}\:−{y}\:{axis}:\:{v}\:{points}\:{with}\:{angle}\:\frac{\mathrm{3}\pi}{\mathrm{2}}, \\ $$$${S}_{\mathrm{5}} ={m}×\mathrm{0}+{n}×\frac{\pi}{\mathrm{2}}+{u}\pi+{v}×\frac{\mathrm{3}\pi}{\mathrm{2}}=\frac{\left({n}+\mathrm{2}{u}+\mathrm{3}{v}\right)\pi}{\mathrm{2}} \\ $$$$ \\ $$$${Sum}\:{of}\:{all}\:{angles}: \\ $$$${S}={S}_{\mathrm{1}} +{S}_{\mathrm{2}} +{S}_{\mathrm{3}} +{S}_{\mathrm{4}} +{S}_{\mathrm{5}} \\ $$$$=\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$+{un}\pi−\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$+{uv}\pi+\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$+\mathrm{2}{mv}\pi−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right) \\ $$$$+\frac{\left({n}+\mathrm{2}{u}+\mathrm{3}{v}\right)\pi}{\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}{un}+\mathrm{2}{uv}+\mathrm{4}{mv}+{n}+\mathrm{2}{u}+\mathrm{3}{v}\right)\pi}{\mathrm{2}} \\ $$$$+\left[\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}−\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\right]\left[\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)−\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{j}}{{i}}\right)\right] \\ $$$$ \\ $$$${The}\:{average}\:{angle}\:{is} \\ $$$$\overset{−} {\theta}=\frac{{S}}{\left({m}+{u}+\mathrm{1}\right)\left({n}+{v}+\mathrm{1}\right)} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com