What does the function $f(x,y)$ reduce to?











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What does the function $f(x,y)$ reduce to?



$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$










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  • When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
    – Semiclassical
    Nov 24 at 3:52












  • @Semiclassical no it is correct as written.
    – The Great Duck
    Nov 24 at 5:36










  • @TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
    – metamorphy
    Nov 25 at 13:02












  • @metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
    – The Great Duck
    Nov 25 at 20:38















up vote
0
down vote

favorite












What does the function $f(x,y)$ reduce to?



$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$










share|cite|improve this question
























  • When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
    – Semiclassical
    Nov 24 at 3:52












  • @Semiclassical no it is correct as written.
    – The Great Duck
    Nov 24 at 5:36










  • @TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
    – metamorphy
    Nov 25 at 13:02












  • @metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
    – The Great Duck
    Nov 25 at 20:38













up vote
0
down vote

favorite









up vote
0
down vote

favorite











What does the function $f(x,y)$ reduce to?



$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$










share|cite|improve this question















What does the function $f(x,y)$ reduce to?



$$f(x,y) = lim_{n to infty} sum_{i = 0}^{n} (((x bmod 2^{i-1})-(x bmod 2^i)) cdot ((y bmod 2^{i-1})-(y bmod 2^i)) cdot frac {1}{2^i})$$







limits summation






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 3:22









metamorphy

2,9971518




2,9971518










asked Nov 23 at 3:17









The Great Duck

11732047




11732047












  • When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
    – Semiclassical
    Nov 24 at 3:52












  • @Semiclassical no it is correct as written.
    – The Great Duck
    Nov 24 at 5:36










  • @TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
    – metamorphy
    Nov 25 at 13:02












  • @metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
    – The Great Duck
    Nov 25 at 20:38


















  • When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
    – Semiclassical
    Nov 24 at 3:52












  • @Semiclassical no it is correct as written.
    – The Great Duck
    Nov 24 at 5:36










  • @TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
    – metamorphy
    Nov 25 at 13:02












  • @metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
    – The Great Duck
    Nov 25 at 20:38
















When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52






When $i=0$, you've got $xbmod 2^{i-1}=xbmod 2^{-1}$. Should the summation start at $i=1$ instead, so that you only take mod with respect to integers?
– Semiclassical
Nov 24 at 3:52














@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36




@Semiclassical no it is correct as written.
– The Great Duck
Nov 24 at 5:36












@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02






@TheGreatDuck, I wouldn't say so. No one understands what do you mean by $xbmod 2^{-1}$ (including me; and even - to start with - what are $x$ and $y$). Looks like a flimsy conundrum with $x&y$ buried in (where $&$ stands for bitwise "and"; I'm not saying here that it is your $f(x,y)$ - it would be wrong).
– metamorphy
Nov 25 at 13:02














@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38




@metamorphy it’s a function. The inputs are named $x$ and $y$. The question is correct as written. Of course modulo is defined for $frac 12$. I’m confused why you’d think otherwise.
– The Great Duck
Nov 25 at 20:38















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