An equation that is undefined at every value except one (plotting a single point)












1












$begingroup$


Suppose I wanted to plot an equation that would only place a single point at a specific (x,y) value, and nothing anywhere else, that is, the equation, y=g(x), would be undefined for all other x values except the x from that one (x,y) point. What might that look like?



I toyed around with dividing by zero in the equation (to make all x-values undefined and therefor not plotted), but I wasn’t sure how to not divide by zero IFF x is the x-value I want plotted. I know Kronecker Delta function could be used here but afaik it only has a definition as a piecewise function and obviously that defeats the entire purpose of what I’m trying to do. I suspect the answer will involve limits. Any ideas?



I also have an idea of plotting the equation for a circle with its center at the (x,y) point and the radius approaching zero, but I don’t know enough about limits to know if that will suffice?






calculus functions graphing-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By definition, functions are defined on every value of their domains so you would have to make your domain a one point set.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:04










  • $begingroup$
    @john douma is an equation like $y=3/x$ not a function even though it has no value at x=0?
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:10






  • 1




    $begingroup$
    Yes but its domain does not include $0$.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:10










  • $begingroup$
    @john douma thanks, I have no formal training in math so I often get terminology wrong. Editing question now to reflect what I’m really asking
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:13






  • 1




    $begingroup$
    You could still use the Kronecker delta in a division if you really want to make this thing undefined almost everywhere.
    $endgroup$
    – Sean Roberson
    Dec 25 '18 at 2:23
















1












$begingroup$


Suppose I wanted to plot an equation that would only place a single point at a specific (x,y) value, and nothing anywhere else, that is, the equation, y=g(x), would be undefined for all other x values except the x from that one (x,y) point. What might that look like?



I toyed around with dividing by zero in the equation (to make all x-values undefined and therefor not plotted), but I wasn’t sure how to not divide by zero IFF x is the x-value I want plotted. I know Kronecker Delta function could be used here but afaik it only has a definition as a piecewise function and obviously that defeats the entire purpose of what I’m trying to do. I suspect the answer will involve limits. Any ideas?



I also have an idea of plotting the equation for a circle with its center at the (x,y) point and the radius approaching zero, but I don’t know enough about limits to know if that will suffice?






calculus functions graphing-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    By definition, functions are defined on every value of their domains so you would have to make your domain a one point set.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:04










  • $begingroup$
    @john douma is an equation like $y=3/x$ not a function even though it has no value at x=0?
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:10






  • 1




    $begingroup$
    Yes but its domain does not include $0$.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:10










  • $begingroup$
    @john douma thanks, I have no formal training in math so I often get terminology wrong. Editing question now to reflect what I’m really asking
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:13






  • 1




    $begingroup$
    You could still use the Kronecker delta in a division if you really want to make this thing undefined almost everywhere.
    $endgroup$
    – Sean Roberson
    Dec 25 '18 at 2:23














1












1








1


1



$begingroup$


Suppose I wanted to plot an equation that would only place a single point at a specific (x,y) value, and nothing anywhere else, that is, the equation, y=g(x), would be undefined for all other x values except the x from that one (x,y) point. What might that look like?



I toyed around with dividing by zero in the equation (to make all x-values undefined and therefor not plotted), but I wasn’t sure how to not divide by zero IFF x is the x-value I want plotted. I know Kronecker Delta function could be used here but afaik it only has a definition as a piecewise function and obviously that defeats the entire purpose of what I’m trying to do. I suspect the answer will involve limits. Any ideas?



I also have an idea of plotting the equation for a circle with its center at the (x,y) point and the radius approaching zero, but I don’t know enough about limits to know if that will suffice?






calculus functions graphing-functions







share|cite|improve this question











$endgroup$




Suppose I wanted to plot an equation that would only place a single point at a specific (x,y) value, and nothing anywhere else, that is, the equation, y=g(x), would be undefined for all other x values except the x from that one (x,y) point. What might that look like?



I toyed around with dividing by zero in the equation (to make all x-values undefined and therefor not plotted), but I wasn’t sure how to not divide by zero IFF x is the x-value I want plotted. I know Kronecker Delta function could be used here but afaik it only has a definition as a piecewise function and obviously that defeats the entire purpose of what I’m trying to do. I suspect the answer will involve limits. Any ideas?



I also have an idea of plotting the equation for a circle with its center at the (x,y) point and the radius approaching zero, but I don’t know enough about limits to know if that will suffice?






calculus functions graphing-functions




calculus functions graphing-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 25 '18 at 2:14







Albert Renshaw

















asked Dec 25 '18 at 2:00









Albert RenshawAlbert Renshaw

7491728




7491728








  • 1




    $begingroup$
    By definition, functions are defined on every value of their domains so you would have to make your domain a one point set.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:04










  • $begingroup$
    @john douma is an equation like $y=3/x$ not a function even though it has no value at x=0?
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:10






  • 1




    $begingroup$
    Yes but its domain does not include $0$.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:10










  • $begingroup$
    @john douma thanks, I have no formal training in math so I often get terminology wrong. Editing question now to reflect what I’m really asking
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:13






  • 1




    $begingroup$
    You could still use the Kronecker delta in a division if you really want to make this thing undefined almost everywhere.
    $endgroup$
    – Sean Roberson
    Dec 25 '18 at 2:23














  • 1




    $begingroup$
    By definition, functions are defined on every value of their domains so you would have to make your domain a one point set.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:04










  • $begingroup$
    @john douma is an equation like $y=3/x$ not a function even though it has no value at x=0?
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:10






  • 1




    $begingroup$
    Yes but its domain does not include $0$.
    $endgroup$
    – John Douma
    Dec 25 '18 at 2:10










  • $begingroup$
    @john douma thanks, I have no formal training in math so I often get terminology wrong. Editing question now to reflect what I’m really asking
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 2:13






  • 1




    $begingroup$
    You could still use the Kronecker delta in a division if you really want to make this thing undefined almost everywhere.
    $endgroup$
    – Sean Roberson
    Dec 25 '18 at 2:23








1




1




$begingroup$
By definition, functions are defined on every value of their domains so you would have to make your domain a one point set.
$endgroup$
– John Douma
Dec 25 '18 at 2:04




$begingroup$
By definition, functions are defined on every value of their domains so you would have to make your domain a one point set.
$endgroup$
– John Douma
Dec 25 '18 at 2:04












$begingroup$
@john douma is an equation like $y=3/x$ not a function even though it has no value at x=0?
$endgroup$
– Albert Renshaw
Dec 25 '18 at 2:10




$begingroup$
@john douma is an equation like $y=3/x$ not a function even though it has no value at x=0?
$endgroup$
– Albert Renshaw
Dec 25 '18 at 2:10




1




1




$begingroup$
Yes but its domain does not include $0$.
$endgroup$
– John Douma
Dec 25 '18 at 2:10




$begingroup$
Yes but its domain does not include $0$.
$endgroup$
– John Douma
Dec 25 '18 at 2:10












$begingroup$
@john douma thanks, I have no formal training in math so I often get terminology wrong. Editing question now to reflect what I’m really asking
$endgroup$
– Albert Renshaw
Dec 25 '18 at 2:13




$begingroup$
@john douma thanks, I have no formal training in math so I often get terminology wrong. Editing question now to reflect what I’m really asking
$endgroup$
– Albert Renshaw
Dec 25 '18 at 2:13




1




1




$begingroup$
You could still use the Kronecker delta in a division if you really want to make this thing undefined almost everywhere.
$endgroup$
– Sean Roberson
Dec 25 '18 at 2:23




$begingroup$
You could still use the Kronecker delta in a division if you really want to make this thing undefined almost everywhere.
$endgroup$
– Sean Roberson
Dec 25 '18 at 2:23










3 Answers
3






active

oldest

votes


















2












$begingroup$

In the context of real functions
$$f(x) = sqrt x + sqrt{-x}$$



has a graph consisting of only the point $(0,0)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 7:37





















3












$begingroup$

You can just write $(x-x_0)^2+(y-y_0)^2=0$. This equation has only the point $(x_0,y_0)$ as solution.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 7:36



















1












$begingroup$

The graph of the function f = {(x,y)} is a single point.

The constant function f:{0} -> {1}, x -> 1, for example.



You could conjure up a meaningless monster like

f(x) = x/0 if x /= 0; f(0) = 0/0 = 1 or whatever.



I don't see any slick answer for a dubious problem.

Use the set theory definition of function

instead of a naive calculus notion.






share|cite|improve this answer









$endgroup$














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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    In the context of real functions
    $$f(x) = sqrt x + sqrt{-x}$$



    has a graph consisting of only the point $(0,0)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:37


















    2












    $begingroup$

    In the context of real functions
    $$f(x) = sqrt x + sqrt{-x}$$



    has a graph consisting of only the point $(0,0)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:37
















    2












    2








    2





    $begingroup$

    In the context of real functions
    $$f(x) = sqrt x + sqrt{-x}$$



    has a graph consisting of only the point $(0,0)$






    share|cite|improve this answer









    $endgroup$



    In the context of real functions
    $$f(x) = sqrt x + sqrt{-x}$$



    has a graph consisting of only the point $(0,0)$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 25 '18 at 3:06









    WW1WW1

    7,3701712




    7,3701712












    • $begingroup$
      This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:37




















    • $begingroup$
      This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:37


















    $begingroup$
    This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 7:37






    $begingroup$
    This is really interesting because it seems to still be continuous, right? Just everything else is hidden in the complex plane. I like this answer best because it behaves most like a real function and can be manipulated (translated) to any point I desire. Thank you!
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 7:37













    3












    $begingroup$

    You can just write $(x-x_0)^2+(y-y_0)^2=0$. This equation has only the point $(x_0,y_0)$ as solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:36
















    3












    $begingroup$

    You can just write $(x-x_0)^2+(y-y_0)^2=0$. This equation has only the point $(x_0,y_0)$ as solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:36














    3












    3








    3





    $begingroup$

    You can just write $(x-x_0)^2+(y-y_0)^2=0$. This equation has only the point $(x_0,y_0)$ as solution.






    share|cite|improve this answer









    $endgroup$



    You can just write $(x-x_0)^2+(y-y_0)^2=0$. This equation has only the point $(x_0,y_0)$ as solution.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 25 '18 at 3:10









    Ross MillikanRoss Millikan

    302k24201375




    302k24201375












    • $begingroup$
      Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:36


















    • $begingroup$
      Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
      $endgroup$
      – Albert Renshaw
      Dec 25 '18 at 7:36
















    $begingroup$
    Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 7:36




    $begingroup$
    Yes this is what I was looking into doing with my circle attempt I had referenced. For some reason I was under the impression it would have to be the limit as the radius approaches zero, hadn't realized it could work directly at zero, but upon seeing it written out this is obvious, I was just thinking visually in my head, thanks!
    $endgroup$
    – Albert Renshaw
    Dec 25 '18 at 7:36











    1












    $begingroup$

    The graph of the function f = {(x,y)} is a single point.

    The constant function f:{0} -> {1}, x -> 1, for example.



    You could conjure up a meaningless monster like

    f(x) = x/0 if x /= 0; f(0) = 0/0 = 1 or whatever.



    I don't see any slick answer for a dubious problem.

    Use the set theory definition of function

    instead of a naive calculus notion.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The graph of the function f = {(x,y)} is a single point.

      The constant function f:{0} -> {1}, x -> 1, for example.



      You could conjure up a meaningless monster like

      f(x) = x/0 if x /= 0; f(0) = 0/0 = 1 or whatever.



      I don't see any slick answer for a dubious problem.

      Use the set theory definition of function

      instead of a naive calculus notion.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The graph of the function f = {(x,y)} is a single point.

        The constant function f:{0} -> {1}, x -> 1, for example.



        You could conjure up a meaningless monster like

        f(x) = x/0 if x /= 0; f(0) = 0/0 = 1 or whatever.



        I don't see any slick answer for a dubious problem.

        Use the set theory definition of function

        instead of a naive calculus notion.






        share|cite|improve this answer









        $endgroup$



        The graph of the function f = {(x,y)} is a single point.

        The constant function f:{0} -> {1}, x -> 1, for example.



        You could conjure up a meaningless monster like

        f(x) = x/0 if x /= 0; f(0) = 0/0 = 1 or whatever.



        I don't see any slick answer for a dubious problem.

        Use the set theory definition of function

        instead of a naive calculus notion.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 2:38









        William ElliotWilliam Elliot

        9,3162820




        9,3162820






























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