Can this be solved to remove the floor function and simplify the answer?












1












$begingroup$


I've been working through a derivation and have arrived at the following exprssion:



$$E = 1 - frac{x}y left( bigglfloor dfrac{2x+yx-2}{2y} biggrfloor right)^{-1}$$



where $x,y in mathbb{R^+}$.



I would like to know whether this can be reduced further, by removing the floor function and reduce the number of terms to simplify expression?



Kind regards!






floor-function ceiling-function







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  • 1




    $begingroup$
    Could you add further context? Specifically, what derivation were you working on, and what work did you do to get to the equation you posted?
    $endgroup$
    – kkc
    Jan 9 at 3:25
















1












$begingroup$


I've been working through a derivation and have arrived at the following exprssion:



$$E = 1 - frac{x}y left( bigglfloor dfrac{2x+yx-2}{2y} biggrfloor right)^{-1}$$



where $x,y in mathbb{R^+}$.



I would like to know whether this can be reduced further, by removing the floor function and reduce the number of terms to simplify expression?



Kind regards!






floor-function ceiling-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Could you add further context? Specifically, what derivation were you working on, and what work did you do to get to the equation you posted?
    $endgroup$
    – kkc
    Jan 9 at 3:25














1












1








1





$begingroup$


I've been working through a derivation and have arrived at the following exprssion:



$$E = 1 - frac{x}y left( bigglfloor dfrac{2x+yx-2}{2y} biggrfloor right)^{-1}$$



where $x,y in mathbb{R^+}$.



I would like to know whether this can be reduced further, by removing the floor function and reduce the number of terms to simplify expression?



Kind regards!






floor-function ceiling-function







share|cite|improve this question











$endgroup$




I've been working through a derivation and have arrived at the following exprssion:



$$E = 1 - frac{x}y left( bigglfloor dfrac{2x+yx-2}{2y} biggrfloor right)^{-1}$$



where $x,y in mathbb{R^+}$.



I would like to know whether this can be reduced further, by removing the floor function and reduce the number of terms to simplify expression?



Kind regards!






floor-function ceiling-function




floor-function ceiling-function






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share|cite|improve this question













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share|cite|improve this question








edited Jan 9 at 10:06









Lee David Chung Lin

4,38031241




4,38031241










asked Jan 9 at 2:59









AfroeskimoAfroeskimo

114




114








  • 1




    $begingroup$
    Could you add further context? Specifically, what derivation were you working on, and what work did you do to get to the equation you posted?
    $endgroup$
    – kkc
    Jan 9 at 3:25














  • 1




    $begingroup$
    Could you add further context? Specifically, what derivation were you working on, and what work did you do to get to the equation you posted?
    $endgroup$
    – kkc
    Jan 9 at 3:25








1




1




$begingroup$
Could you add further context? Specifically, what derivation were you working on, and what work did you do to get to the equation you posted?
$endgroup$
– kkc
Jan 9 at 3:25




$begingroup$
Could you add further context? Specifically, what derivation were you working on, and what work did you do to get to the equation you posted?
$endgroup$
– kkc
Jan 9 at 3:25










2 Answers
2






active

oldest

votes


















2












$begingroup$

An observation rather than a general answer



Because you haven't said what you mean by "simplify" (fewer operations? Fewer characters? Fewer divisions? ...) this isn't really answerable in its current form. But it's still possible to make at least one observation:



For a moment, fix $y = 2$. Then your formula becomes
$$
E(x) = 1 - frac{x}{2lfloor x-frac{1}{2} rfloor}
$$
At every point $x$ of the form $0.5, 1.5, 2.5, ldots$, $E$ has a discontinuity.



If you're hoping for a simpler expression, it needs to be one that can be discontinuous at a countable number of locations. You could probably generate this using something like the "signum" function and a "mod", but that's really no less complex than "floor". But if you're hoping to generate it with polynomials, trig functions, etc., I don't see an obvious way to make it happen.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    In general, a floor function cannot be simplified away: it creates an often-infinite series of discontinuities that make it impossible to characterize the function without a way to get all the discontinuities in. For this, the floor function is ideal, and so we use it.



    In this particular case, we have a large division-by-zero area when $2y>2x+yx-2$, equal to when $y > 2frac{x-1}{2-x}, x < 2$, and a series of curves of the form $y=2frac{x-1}{2k-x}, kinmathbb{Z}$ where the function is discontinuous. Without the floor function, or something similar (and frankly, the floor function is way nicer than the alternatives), we cannot describe these discontinuities and thus the behavior of the function in a finite fashion.



    In this image, I have plotted the area where the function is undefined, and many of the curves where the function is discontinuous. As $k$ increases, the discontinuity curves get closer and closer to the $x$ axis; I've only plotted to $k=100$.



    Plots of the discontinuities of E(x,y)






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      An observation rather than a general answer



      Because you haven't said what you mean by "simplify" (fewer operations? Fewer characters? Fewer divisions? ...) this isn't really answerable in its current form. But it's still possible to make at least one observation:



      For a moment, fix $y = 2$. Then your formula becomes
      $$
      E(x) = 1 - frac{x}{2lfloor x-frac{1}{2} rfloor}
      $$
      At every point $x$ of the form $0.5, 1.5, 2.5, ldots$, $E$ has a discontinuity.



      If you're hoping for a simpler expression, it needs to be one that can be discontinuous at a countable number of locations. You could probably generate this using something like the "signum" function and a "mod", but that's really no less complex than "floor". But if you're hoping to generate it with polynomials, trig functions, etc., I don't see an obvious way to make it happen.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        An observation rather than a general answer



        Because you haven't said what you mean by "simplify" (fewer operations? Fewer characters? Fewer divisions? ...) this isn't really answerable in its current form. But it's still possible to make at least one observation:



        For a moment, fix $y = 2$. Then your formula becomes
        $$
        E(x) = 1 - frac{x}{2lfloor x-frac{1}{2} rfloor}
        $$
        At every point $x$ of the form $0.5, 1.5, 2.5, ldots$, $E$ has a discontinuity.



        If you're hoping for a simpler expression, it needs to be one that can be discontinuous at a countable number of locations. You could probably generate this using something like the "signum" function and a "mod", but that's really no less complex than "floor". But if you're hoping to generate it with polynomials, trig functions, etc., I don't see an obvious way to make it happen.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          An observation rather than a general answer



          Because you haven't said what you mean by "simplify" (fewer operations? Fewer characters? Fewer divisions? ...) this isn't really answerable in its current form. But it's still possible to make at least one observation:



          For a moment, fix $y = 2$. Then your formula becomes
          $$
          E(x) = 1 - frac{x}{2lfloor x-frac{1}{2} rfloor}
          $$
          At every point $x$ of the form $0.5, 1.5, 2.5, ldots$, $E$ has a discontinuity.



          If you're hoping for a simpler expression, it needs to be one that can be discontinuous at a countable number of locations. You could probably generate this using something like the "signum" function and a "mod", but that's really no less complex than "floor". But if you're hoping to generate it with polynomials, trig functions, etc., I don't see an obvious way to make it happen.






          share|cite|improve this answer









          $endgroup$



          An observation rather than a general answer



          Because you haven't said what you mean by "simplify" (fewer operations? Fewer characters? Fewer divisions? ...) this isn't really answerable in its current form. But it's still possible to make at least one observation:



          For a moment, fix $y = 2$. Then your formula becomes
          $$
          E(x) = 1 - frac{x}{2lfloor x-frac{1}{2} rfloor}
          $$
          At every point $x$ of the form $0.5, 1.5, 2.5, ldots$, $E$ has a discontinuity.



          If you're hoping for a simpler expression, it needs to be one that can be discontinuous at a countable number of locations. You could probably generate this using something like the "signum" function and a "mod", but that's really no less complex than "floor". But if you're hoping to generate it with polynomials, trig functions, etc., I don't see an obvious way to make it happen.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 10:48









          John HughesJohn Hughes

          64.3k24191




          64.3k24191























              1












              $begingroup$

              In general, a floor function cannot be simplified away: it creates an often-infinite series of discontinuities that make it impossible to characterize the function without a way to get all the discontinuities in. For this, the floor function is ideal, and so we use it.



              In this particular case, we have a large division-by-zero area when $2y>2x+yx-2$, equal to when $y > 2frac{x-1}{2-x}, x < 2$, and a series of curves of the form $y=2frac{x-1}{2k-x}, kinmathbb{Z}$ where the function is discontinuous. Without the floor function, or something similar (and frankly, the floor function is way nicer than the alternatives), we cannot describe these discontinuities and thus the behavior of the function in a finite fashion.



              In this image, I have plotted the area where the function is undefined, and many of the curves where the function is discontinuous. As $k$ increases, the discontinuity curves get closer and closer to the $x$ axis; I've only plotted to $k=100$.



              Plots of the discontinuities of E(x,y)






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                In general, a floor function cannot be simplified away: it creates an often-infinite series of discontinuities that make it impossible to characterize the function without a way to get all the discontinuities in. For this, the floor function is ideal, and so we use it.



                In this particular case, we have a large division-by-zero area when $2y>2x+yx-2$, equal to when $y > 2frac{x-1}{2-x}, x < 2$, and a series of curves of the form $y=2frac{x-1}{2k-x}, kinmathbb{Z}$ where the function is discontinuous. Without the floor function, or something similar (and frankly, the floor function is way nicer than the alternatives), we cannot describe these discontinuities and thus the behavior of the function in a finite fashion.



                In this image, I have plotted the area where the function is undefined, and many of the curves where the function is discontinuous. As $k$ increases, the discontinuity curves get closer and closer to the $x$ axis; I've only plotted to $k=100$.



                Plots of the discontinuities of E(x,y)






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  In general, a floor function cannot be simplified away: it creates an often-infinite series of discontinuities that make it impossible to characterize the function without a way to get all the discontinuities in. For this, the floor function is ideal, and so we use it.



                  In this particular case, we have a large division-by-zero area when $2y>2x+yx-2$, equal to when $y > 2frac{x-1}{2-x}, x < 2$, and a series of curves of the form $y=2frac{x-1}{2k-x}, kinmathbb{Z}$ where the function is discontinuous. Without the floor function, or something similar (and frankly, the floor function is way nicer than the alternatives), we cannot describe these discontinuities and thus the behavior of the function in a finite fashion.



                  In this image, I have plotted the area where the function is undefined, and many of the curves where the function is discontinuous. As $k$ increases, the discontinuity curves get closer and closer to the $x$ axis; I've only plotted to $k=100$.



                  Plots of the discontinuities of E(x,y)






                  share|cite|improve this answer









                  $endgroup$



                  In general, a floor function cannot be simplified away: it creates an often-infinite series of discontinuities that make it impossible to characterize the function without a way to get all the discontinuities in. For this, the floor function is ideal, and so we use it.



                  In this particular case, we have a large division-by-zero area when $2y>2x+yx-2$, equal to when $y > 2frac{x-1}{2-x}, x < 2$, and a series of curves of the form $y=2frac{x-1}{2k-x}, kinmathbb{Z}$ where the function is discontinuous. Without the floor function, or something similar (and frankly, the floor function is way nicer than the alternatives), we cannot describe these discontinuities and thus the behavior of the function in a finite fashion.



                  In this image, I have plotted the area where the function is undefined, and many of the curves where the function is discontinuous. As $k$ increases, the discontinuity curves get closer and closer to the $x$ axis; I've only plotted to $k=100$.



                  Plots of the discontinuities of E(x,y)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 9 at 11:06









                  Dan UznanskiDan Uznanski

                  6,86521528




                  6,86521528






























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