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MrW1-Before-we-concluded-that-x-0-m-y-0-n-1-sgn-x-x-If-you-do-x-0-1-y-0-1-1-sgn-x-x-x-0-1-y-0-1-1-sgn-x-LCM-x-y-y-1-sgn-0-LCM-0-0-0-1-s




Question Number 13034 by FilupS last updated on 12/May/17
MrW1     Before we concluded that:  Φ=Σ_(x=0) ^m Σ_(y=0) ^n (1−sgn(x−x′))     If you do:  Σ_(x=0) ^1 Σ_(y=0) ^1 (1−sgn(x−x′))  =Σ_(x=0) ^1 Σ_(y=0) ^1 (1−sgn(x−((LCM(x,y))/y)))  =(1−sgn(0−((LCM(0,0))/0)))+(1−sgn(1−((LCM(1,0))/0))  +(1−sgn(0−((LCM(0,1))/1)))+(1−sgn(1−((LCM(1,1))/1))     =(1−sgn(−((LCM(0,0))/0)))+(1−sgn(1−(0/0)))  +(1−sgn(−(0/1)))+(1−sgn(1−(1/1)))     =????
$$\mathrm{MrW1} \\ $$$$\: \\ $$$$\mathrm{Before}\:\mathrm{we}\:\mathrm{concluded}\:\mathrm{that}: \\ $$$$\Phi=\underset{{x}=\mathrm{0}} {\overset{{m}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\mathrm{sgn}\left({x}−{x}'\right)\right) \\ $$$$\: \\ $$$$\mathrm{If}\:\mathrm{you}\:\mathrm{do}: \\ $$$$\underset{{x}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\left(\mathrm{1}−\mathrm{sgn}\left({x}−{x}'\right)\right) \\ $$$$=\underset{{x}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{\mathrm{1}} {\sum}}\left(\mathrm{1}−\mathrm{sgn}\left({x}−\frac{\mathrm{LCM}\left({x},{y}\right)}{{y}}\right)\right) \\ $$$$=\left(\mathrm{1}−\mathrm{sgn}\left(\mathrm{0}−\frac{\mathrm{LCM}\left(\mathrm{0},\mathrm{0}\right)}{\mathrm{0}}\right)\right)+\left(\mathrm{1}−\mathrm{sgn}\left(\mathrm{1}−\frac{\mathrm{LCM}\left(\mathrm{1},\mathrm{0}\right)}{\mathrm{0}}\right)\right. \\ $$$$+\left(\mathrm{1}−\mathrm{sgn}\left(\mathrm{0}−\frac{\mathrm{LCM}\left(\mathrm{0},\mathrm{1}\right)}{\mathrm{1}}\right)\right)+\left(\mathrm{1}−\mathrm{sgn}\left(\mathrm{1}−\frac{\mathrm{LCM}\left(\mathrm{1},\mathrm{1}\right)}{\mathrm{1}}\right)\right. \\ $$$$\: \\ $$$$=\left(\mathrm{1}−\mathrm{sgn}\left(−\frac{\mathrm{LCM}\left(\mathrm{0},\mathrm{0}\right)}{\mathrm{0}}\right)\right)+\left(\mathrm{1}−\mathrm{sgn}\left(\mathrm{1}−\frac{\mathrm{0}}{\mathrm{0}}\right)\right) \\ $$$$+\left(\mathrm{1}−\mathrm{sgn}\left(−\frac{\mathrm{0}}{\mathrm{1}}\right)\right)+\left(\mathrm{1}−\mathrm{sgn}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}}\right)\right) \\ $$$$\: \\ $$$$=???? \\ $$
Commented by mrW1 last updated on 12/May/17
the points on the coordinate axes must  be excluded, you can not build fraction  with 0 as denominator.
$${the}\:{points}\:{on}\:{the}\:{coordinate}\:{axes}\:{must} \\ $$$${be}\:{excluded},\:{you}\:{can}\:{not}\:{build}\:{fraction} \\ $$$${with}\:\mathrm{0}\:{as}\:{denominator}. \\ $$
Commented by mrW1 last updated on 12/May/17
for −u≤x≤m and -v≤y≤n  the answer is  Φ=4+mn−Σ_(i=1) ^m Σ_(j=1) ^n sign[GCD(i,j)−1]  +un−Σ_(i=1) ^u Σ_(j=1) ^n sign[GCD(i,j)−1]  +uv−Σ_(i=1) ^u Σ_(j=1) ^v sign[GCD(i,j)−1]  +mv−Σ_(i=1) ^m Σ_(j=1) ^v sign[GCD(i,j)−1]
$${for}\:−{u}\leqslant{x}\leqslant{m}\:{and}\:-{v}\leqslant{y}\leqslant{n} \\ $$$${the}\:{answer}\:{is} \\ $$$$\Phi=\mathrm{4}+{mn}−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{sign}\left[{GCD}\left({i},{j}\right)−\mathrm{1}\right] \\ $$$$+{un}−\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{sign}\left[{GCD}\left({i},{j}\right)−\mathrm{1}\right] \\ $$$$+{uv}−\underset{{i}=\mathrm{1}} {\overset{{u}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}{sign}\left[{GCD}\left({i},{j}\right)−\mathrm{1}\right] \\ $$$$+{mv}−\underset{{i}=\mathrm{1}} {\overset{{m}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{v}} {\sum}}{sign}\left[{GCD}\left({i},{j}\right)−\mathrm{1}\right] \\ $$
Commented by FilupS last updated on 12/May/17
Ahhh so  Σ_(x=0) ^m Σ_(y=0) ^n (1−sgn(x−((LCM(x, y))/y)))  and  Σ_(x=0) ^m Σ_(y=0) ^n (1−sgn(y−((LCM(x, y))/x)))  or  Σ_(x=0) ^m Σ_(y=0) ^n (GCD(x, y)−1)     is only for x∈(0,m) and y∈(0,n)???
$$\mathrm{Ahhh}\:\mathrm{so} \\ $$$$\underset{{x}=\mathrm{0}} {\overset{{m}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\mathrm{sgn}\left({x}−\frac{\mathrm{LCM}\left({x},\:{y}\right)}{{y}}\right)\right) \\ $$$$\mathrm{and} \\ $$$$\underset{{x}=\mathrm{0}} {\overset{{m}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{1}−\mathrm{sgn}\left({y}−\frac{\mathrm{LCM}\left({x},\:{y}\right)}{{x}}\right)\right) \\ $$$$\mathrm{or} \\ $$$$\underset{{x}=\mathrm{0}} {\overset{{m}} {\sum}}\underset{{y}=\mathrm{0}} {\overset{{n}} {\sum}}\left(\mathrm{GCD}\left({x},\:{y}\right)−\mathrm{1}\right) \\ $$$$\: \\ $$$$\mathrm{is}\:\mathrm{only}\:\mathrm{for}\:{x}\in\left(\mathrm{0},{m}\right)\:\mathrm{and}\:{y}\in\left(\mathrm{0},{n}\right)??? \\ $$
Commented by mrW1 last updated on 12/May/17
for x∈(0,m] and y∈(0,n]  Since on the coordinate axes it happens  that tan θ=∞, thus we must exclude  them.
$$\mathrm{for}\:{x}\in\left(\mathrm{0},{m}\right]\:\mathrm{and}\:{y}\in\left(\mathrm{0},{n}\right] \\ $$$${Since}\:{on}\:{the}\:{coordinate}\:{axes}\:{it}\:{happens} \\ $$$${that}\:\mathrm{tan}\:\theta=\infty,\:{thus}\:{we}\:{must}\:{exclude} \\ $$$${them}. \\ $$

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