$f(x) = 0$ when $x$ is $0$, and $1$ otherwise












5














I've been trying to create a function that will return $0$ when $x$ is $0$, and for any other $x$ value it should return $1$. I've searched for a pre-existing function online too and wasn't able to find one.



Do you know of any function that can do this?










share|cite|improve this question




















  • 9




    This will do: $$f(x)=begin{cases}0,&x=0\{}1,&xneq 0end{cases}$$
    – DonAntonio
    Jun 13 '16 at 21:59








  • 1




    Perhaps $f(x)=|text{sgn}(x)|$ or $f(x)=text{sgn}(x)^2$ using the absolute value or square of the sign function
    – Henry
    Jun 13 '16 at 22:03












  • @David_Shmij: That is a combination of elementary function, namely the constant function $0$ and the constant function $1$ (which are both pretty elementary)!
    – Henning Makholm
    Jun 13 '16 at 22:05






  • 1




    @David_Shmij: Which programming language are you talking about here? Most I know will allow you to write either x==0?0:1 or something like if x=0 then 0 else 1, which both seem pretty easy to me.
    – Henning Makholm
    Jun 13 '16 at 22:17










  • @David_Shmij: Wolfram's syntax is slightly bizzarre, but you can write Piecewise[{{0,x=0}},1] and then multiply by $5$ to your heart's content. Example here.
    – Henning Makholm
    Jun 13 '16 at 23:05


















5














I've been trying to create a function that will return $0$ when $x$ is $0$, and for any other $x$ value it should return $1$. I've searched for a pre-existing function online too and wasn't able to find one.



Do you know of any function that can do this?










share|cite|improve this question




















  • 9




    This will do: $$f(x)=begin{cases}0,&x=0\{}1,&xneq 0end{cases}$$
    – DonAntonio
    Jun 13 '16 at 21:59








  • 1




    Perhaps $f(x)=|text{sgn}(x)|$ or $f(x)=text{sgn}(x)^2$ using the absolute value or square of the sign function
    – Henry
    Jun 13 '16 at 22:03












  • @David_Shmij: That is a combination of elementary function, namely the constant function $0$ and the constant function $1$ (which are both pretty elementary)!
    – Henning Makholm
    Jun 13 '16 at 22:05






  • 1




    @David_Shmij: Which programming language are you talking about here? Most I know will allow you to write either x==0?0:1 or something like if x=0 then 0 else 1, which both seem pretty easy to me.
    – Henning Makholm
    Jun 13 '16 at 22:17










  • @David_Shmij: Wolfram's syntax is slightly bizzarre, but you can write Piecewise[{{0,x=0}},1] and then multiply by $5$ to your heart's content. Example here.
    – Henning Makholm
    Jun 13 '16 at 23:05
















5












5








5


3





I've been trying to create a function that will return $0$ when $x$ is $0$, and for any other $x$ value it should return $1$. I've searched for a pre-existing function online too and wasn't able to find one.



Do you know of any function that can do this?










share|cite|improve this question















I've been trying to create a function that will return $0$ when $x$ is $0$, and for any other $x$ value it should return $1$. I've searched for a pre-existing function online too and wasn't able to find one.



Do you know of any function that can do this?







functions boolean-algebra binary






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 13 '16 at 22:15









David_Shmij

368217




368217










asked Jun 13 '16 at 21:58









Albert Renshaw

7271627




7271627








  • 9




    This will do: $$f(x)=begin{cases}0,&x=0\{}1,&xneq 0end{cases}$$
    – DonAntonio
    Jun 13 '16 at 21:59








  • 1




    Perhaps $f(x)=|text{sgn}(x)|$ or $f(x)=text{sgn}(x)^2$ using the absolute value or square of the sign function
    – Henry
    Jun 13 '16 at 22:03












  • @David_Shmij: That is a combination of elementary function, namely the constant function $0$ and the constant function $1$ (which are both pretty elementary)!
    – Henning Makholm
    Jun 13 '16 at 22:05






  • 1




    @David_Shmij: Which programming language are you talking about here? Most I know will allow you to write either x==0?0:1 or something like if x=0 then 0 else 1, which both seem pretty easy to me.
    – Henning Makholm
    Jun 13 '16 at 22:17










  • @David_Shmij: Wolfram's syntax is slightly bizzarre, but you can write Piecewise[{{0,x=0}},1] and then multiply by $5$ to your heart's content. Example here.
    – Henning Makholm
    Jun 13 '16 at 23:05
















  • 9




    This will do: $$f(x)=begin{cases}0,&x=0\{}1,&xneq 0end{cases}$$
    – DonAntonio
    Jun 13 '16 at 21:59








  • 1




    Perhaps $f(x)=|text{sgn}(x)|$ or $f(x)=text{sgn}(x)^2$ using the absolute value or square of the sign function
    – Henry
    Jun 13 '16 at 22:03












  • @David_Shmij: That is a combination of elementary function, namely the constant function $0$ and the constant function $1$ (which are both pretty elementary)!
    – Henning Makholm
    Jun 13 '16 at 22:05






  • 1




    @David_Shmij: Which programming language are you talking about here? Most I know will allow you to write either x==0?0:1 or something like if x=0 then 0 else 1, which both seem pretty easy to me.
    – Henning Makholm
    Jun 13 '16 at 22:17










  • @David_Shmij: Wolfram's syntax is slightly bizzarre, but you can write Piecewise[{{0,x=0}},1] and then multiply by $5$ to your heart's content. Example here.
    – Henning Makholm
    Jun 13 '16 at 23:05










9




9




This will do: $$f(x)=begin{cases}0,&x=0\{}1,&xneq 0end{cases}$$
– DonAntonio
Jun 13 '16 at 21:59






This will do: $$f(x)=begin{cases}0,&x=0\{}1,&xneq 0end{cases}$$
– DonAntonio
Jun 13 '16 at 21:59






1




1




Perhaps $f(x)=|text{sgn}(x)|$ or $f(x)=text{sgn}(x)^2$ using the absolute value or square of the sign function
– Henry
Jun 13 '16 at 22:03






Perhaps $f(x)=|text{sgn}(x)|$ or $f(x)=text{sgn}(x)^2$ using the absolute value or square of the sign function
– Henry
Jun 13 '16 at 22:03














@David_Shmij: That is a combination of elementary function, namely the constant function $0$ and the constant function $1$ (which are both pretty elementary)!
– Henning Makholm
Jun 13 '16 at 22:05




@David_Shmij: That is a combination of elementary function, namely the constant function $0$ and the constant function $1$ (which are both pretty elementary)!
– Henning Makholm
Jun 13 '16 at 22:05




1




1




@David_Shmij: Which programming language are you talking about here? Most I know will allow you to write either x==0?0:1 or something like if x=0 then 0 else 1, which both seem pretty easy to me.
– Henning Makholm
Jun 13 '16 at 22:17




@David_Shmij: Which programming language are you talking about here? Most I know will allow you to write either x==0?0:1 or something like if x=0 then 0 else 1, which both seem pretty easy to me.
– Henning Makholm
Jun 13 '16 at 22:17












@David_Shmij: Wolfram's syntax is slightly bizzarre, but you can write Piecewise[{{0,x=0}},1] and then multiply by $5$ to your heart's content. Example here.
– Henning Makholm
Jun 13 '16 at 23:05






@David_Shmij: Wolfram's syntax is slightly bizzarre, but you can write Piecewise[{{0,x=0}},1] and then multiply by $5$ to your heart's content. Example here.
– Henning Makholm
Jun 13 '16 at 23:05












8 Answers
8






active

oldest

votes


















4














How about $f(x)=leftlceilfrac{x^2}{x^2+1}rightrceil$



*Works for real numbers, with imaginary numbers you may divide by 0.






share|cite|improve this answer



















  • 3




    That's pretty unreadable compared to Joanpemo's straightforward definition.
    – Henning Makholm
    Jun 13 '16 at 22:07










  • The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
    – The Great Duck
    Jun 13 '16 at 22:14






  • 2




    @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
    – user228113
    Jun 13 '16 at 22:17








  • 1




    @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
    – The Great Duck
    Jun 13 '16 at 22:42










  • @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
    – paw88789
    Jun 13 '16 at 23:16



















10














You've already defined your function (assuming you've also chosen its domain).



One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so.



Typical notation for a function created by the sort of description you give is a definition by cases:



$$ f(x) := begin{cases} 0 & x = 0 \ 1 & x neq 0 end{cases} $$



For many applications — most applications, I expect — this is one of the best descriptions of said function. If need be, name it with a letter, and continue on with whatever you're doing.





The complementary function



$$ g(x) := begin{cases} 1 & x = 0 \ 0 & x neq 0 end{cases} $$



which is related to your function by $f(x) = 1 - g(x)$ comes up often enough in some contexts to have been given a name and notation: e.g.




  • The Kronecker delta. A few different notations exist depending on the setting; e.g. $delta_x$, $delta[x]$, or $delta_{x,0}$.

  • The Iverson bracket. This would be notated $[x = 0]$. This notation is, IMO, indispensable for doing complicated calculations with summations.


  • x == 0 computes this function in C and C++, and many other programming languages allow similar.




Some applications might want to represent such a function in particular ways. For example, if one only cares about the value of $g(x)$ when $x$ is an integer, but strongly prefers to work with analytic functions (e.g. because you're studying a sequence using complex analysis), one has the fact that



$$ g(x) = mathop{mathrm{sinc}}(pi x) $$



holds whenever $x$ is an integer.



(if you're unfamiliar with it, $mathop{mathrm{sinc}}(z)$ is the continuous extension of $sin(z) / z$)






share|cite|improve this answer



















  • 1




    In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
    – Hurkyl
    Jun 13 '16 at 22:45






  • 2




    $0^x$ would be iffy at best if $x$ can be negative ...
    – Henning Makholm
    Jun 13 '16 at 23:07










  • +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
    – Albert Renshaw
    Jun 14 '16 at 4:43










  • Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
    – Albert Renshaw
    Jun 14 '16 at 4:44












  • @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
    – J. M. is not a mathematician
    Jun 15 '16 at 18:04



















3














How about $f(x)= 1-delta_{x,0}$ (using the Kronecker Delta function, in Mathematica/WolframAlpha can write the $delta_{x,0}$ as




kroneckerdelta(x,0)




)






share|cite|improve this answer































    2














    Here's one using $sum$ notation although it only works for the natural numbers:



    $f(x) = sumlimits_{i = 1}^{x}{frac{1}{x}} $



    Due to the empty sum being 0.



    It can't be simplified to $f(x) = x times frac{1}{x}$ because then f(0) would be undefined.






    share|cite|improve this answer



















    • 1




      I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
      – Albert Renshaw
      Oct 21 at 4:12






    • 1




      When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
      – omer
      Oct 21 at 16:23












    • Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
      – omer
      Oct 21 at 16:35





















    0














    Without using floor or ceiling it can be done with limits. Start with a function of this form:



    $$lim_{c to infty} left (frac{y}{sqrt2}-frac{(xcdot c)}{sqrt2} = sqrt[3]{1-left ( frac{y}{sqrt2}+frac{(xcdot c)}{sqrt2} right )^3 } right )$$



    (Note: Using real-valued root not principal root)





    This infinitely squashes (along x-axis) a 45º rotated graph of the form $x^3+y^3=1$ which causes the output values to be $0$ everywhere and $sqrt[6]{2}$ at $0$. This can now easily be worked to achieve the desired affect by dividing by $sqrt[6]{2}$ and subtracting from $1$






    share|cite|improve this answer































      0














      The following equation I’ve made works without using ceiling function or limits or infinite sums when plotting the real values. 1 at x=0 and 0 and a complex pair counterpart at all other values for x



      $(y-1-sqrt{x}-sqrt{-x})cdotfrac{y}{y-x}=0$



      This can then be inverted to a less elegant form to have 0 for x=0 and 1 in all other places.



      I like this best because it’s pure closed form expression and doesn’t involve any elements like ceiling function that are hard to work with when using this in other places. However it only works if you ignore complex answers which has problems of its own, but still is noteworthy and has application






      share|cite|improve this answer































        0














        The following equation I've made is a closed-form expression of KroneckerDelta (j=0 form) which will evaluate y=1 at x=0 and y=0 for x≠0.



        $$(frac{y}{y-x})cdot((y-1)^2+x^2)=0$$



        This function can be subtracted from 1 to achieve the desired result





        The function was formed by taking the function for a horizontal line at 0, divided by a 45º line (y=x) through the origin to create a hole in the function at x=0, then multiplied by a circle with radius 0 whose origin is at (0,1) to create a point at (0,1).






        share|cite|improve this answer





























          -1














          Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all numbers, including imaginary/complex.



          $f(x) = left lceil frac{1}{2Gamma left ( left | x right | right )} right rceil$



          *Note the absolute value inside of Gamma, as it's easy to go unnoticed






          share|cite|improve this answer























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            8 Answers
            8






            active

            oldest

            votes








            8 Answers
            8






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            How about $f(x)=leftlceilfrac{x^2}{x^2+1}rightrceil$



            *Works for real numbers, with imaginary numbers you may divide by 0.






            share|cite|improve this answer



















            • 3




              That's pretty unreadable compared to Joanpemo's straightforward definition.
              – Henning Makholm
              Jun 13 '16 at 22:07










            • The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
              – The Great Duck
              Jun 13 '16 at 22:14






            • 2




              @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
              – user228113
              Jun 13 '16 at 22:17








            • 1




              @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
              – The Great Duck
              Jun 13 '16 at 22:42










            • @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
              – paw88789
              Jun 13 '16 at 23:16
















            4














            How about $f(x)=leftlceilfrac{x^2}{x^2+1}rightrceil$



            *Works for real numbers, with imaginary numbers you may divide by 0.






            share|cite|improve this answer



















            • 3




              That's pretty unreadable compared to Joanpemo's straightforward definition.
              – Henning Makholm
              Jun 13 '16 at 22:07










            • The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
              – The Great Duck
              Jun 13 '16 at 22:14






            • 2




              @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
              – user228113
              Jun 13 '16 at 22:17








            • 1




              @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
              – The Great Duck
              Jun 13 '16 at 22:42










            • @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
              – paw88789
              Jun 13 '16 at 23:16














            4












            4








            4






            How about $f(x)=leftlceilfrac{x^2}{x^2+1}rightrceil$



            *Works for real numbers, with imaginary numbers you may divide by 0.






            share|cite|improve this answer














            How about $f(x)=leftlceilfrac{x^2}{x^2+1}rightrceil$



            *Works for real numbers, with imaginary numbers you may divide by 0.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 14 '16 at 5:00









            Albert Renshaw

            7271627




            7271627










            answered Jun 13 '16 at 22:06









            paw88789

            29k12349




            29k12349








            • 3




              That's pretty unreadable compared to Joanpemo's straightforward definition.
              – Henning Makholm
              Jun 13 '16 at 22:07










            • The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
              – The Great Duck
              Jun 13 '16 at 22:14






            • 2




              @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
              – user228113
              Jun 13 '16 at 22:17








            • 1




              @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
              – The Great Duck
              Jun 13 '16 at 22:42










            • @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
              – paw88789
              Jun 13 '16 at 23:16














            • 3




              That's pretty unreadable compared to Joanpemo's straightforward definition.
              – Henning Makholm
              Jun 13 '16 at 22:07










            • The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
              – The Great Duck
              Jun 13 '16 at 22:14






            • 2




              @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
              – user228113
              Jun 13 '16 at 22:17








            • 1




              @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
              – The Great Duck
              Jun 13 '16 at 22:42










            • @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
              – paw88789
              Jun 13 '16 at 23:16








            3




            3




            That's pretty unreadable compared to Joanpemo's straightforward definition.
            – Henning Makholm
            Jun 13 '16 at 22:07




            That's pretty unreadable compared to Joanpemo's straightforward definition.
            – Henning Makholm
            Jun 13 '16 at 22:07












            The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
            – The Great Duck
            Jun 13 '16 at 22:14




            The function you gave is the case of when x <= 0! You need to multiply it with another term. Just multiply it with f(-x). In terms of Boolean manipulation, this would be the same as ANDing the set of x <= 0 with the set -x <= 0. That should fix your problem.
            – The Great Duck
            Jun 13 '16 at 22:14




            2




            2




            @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
            – user228113
            Jun 13 '16 at 22:17






            @TheGreatDuck The function is even. So there is no difference between the case $xle 0$ and the case $xge 0$.
            – user228113
            Jun 13 '16 at 22:17






            1




            1




            @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
            – The Great Duck
            Jun 13 '16 at 22:42




            @G. Sassatelli You are correct. I am used to using x on top when defining iverson brackets with floor. I didn't look carefully enough. Oops. :p
            – The Great Duck
            Jun 13 '16 at 22:42












            @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
            – paw88789
            Jun 13 '16 at 23:16




            @Barry Cipra, Thanks. Yes, you're right. I made the function to be $1$ at $0$ and $0$ elsewhere. I will fix it.
            – paw88789
            Jun 13 '16 at 23:16











            10














            You've already defined your function (assuming you've also chosen its domain).



            One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so.



            Typical notation for a function created by the sort of description you give is a definition by cases:



            $$ f(x) := begin{cases} 0 & x = 0 \ 1 & x neq 0 end{cases} $$



            For many applications — most applications, I expect — this is one of the best descriptions of said function. If need be, name it with a letter, and continue on with whatever you're doing.





            The complementary function



            $$ g(x) := begin{cases} 1 & x = 0 \ 0 & x neq 0 end{cases} $$



            which is related to your function by $f(x) = 1 - g(x)$ comes up often enough in some contexts to have been given a name and notation: e.g.




            • The Kronecker delta. A few different notations exist depending on the setting; e.g. $delta_x$, $delta[x]$, or $delta_{x,0}$.

            • The Iverson bracket. This would be notated $[x = 0]$. This notation is, IMO, indispensable for doing complicated calculations with summations.


            • x == 0 computes this function in C and C++, and many other programming languages allow similar.




            Some applications might want to represent such a function in particular ways. For example, if one only cares about the value of $g(x)$ when $x$ is an integer, but strongly prefers to work with analytic functions (e.g. because you're studying a sequence using complex analysis), one has the fact that



            $$ g(x) = mathop{mathrm{sinc}}(pi x) $$



            holds whenever $x$ is an integer.



            (if you're unfamiliar with it, $mathop{mathrm{sinc}}(z)$ is the continuous extension of $sin(z) / z$)






            share|cite|improve this answer



















            • 1




              In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
              – Hurkyl
              Jun 13 '16 at 22:45






            • 2




              $0^x$ would be iffy at best if $x$ can be negative ...
              – Henning Makholm
              Jun 13 '16 at 23:07










            • +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
              – Albert Renshaw
              Jun 14 '16 at 4:43










            • Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
              – Albert Renshaw
              Jun 14 '16 at 4:44












            • @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
              – J. M. is not a mathematician
              Jun 15 '16 at 18:04
















            10














            You've already defined your function (assuming you've also chosen its domain).



            One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so.



            Typical notation for a function created by the sort of description you give is a definition by cases:



            $$ f(x) := begin{cases} 0 & x = 0 \ 1 & x neq 0 end{cases} $$



            For many applications — most applications, I expect — this is one of the best descriptions of said function. If need be, name it with a letter, and continue on with whatever you're doing.





            The complementary function



            $$ g(x) := begin{cases} 1 & x = 0 \ 0 & x neq 0 end{cases} $$



            which is related to your function by $f(x) = 1 - g(x)$ comes up often enough in some contexts to have been given a name and notation: e.g.




            • The Kronecker delta. A few different notations exist depending on the setting; e.g. $delta_x$, $delta[x]$, or $delta_{x,0}$.

            • The Iverson bracket. This would be notated $[x = 0]$. This notation is, IMO, indispensable for doing complicated calculations with summations.


            • x == 0 computes this function in C and C++, and many other programming languages allow similar.




            Some applications might want to represent such a function in particular ways. For example, if one only cares about the value of $g(x)$ when $x$ is an integer, but strongly prefers to work with analytic functions (e.g. because you're studying a sequence using complex analysis), one has the fact that



            $$ g(x) = mathop{mathrm{sinc}}(pi x) $$



            holds whenever $x$ is an integer.



            (if you're unfamiliar with it, $mathop{mathrm{sinc}}(z)$ is the continuous extension of $sin(z) / z$)






            share|cite|improve this answer



















            • 1




              In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
              – Hurkyl
              Jun 13 '16 at 22:45






            • 2




              $0^x$ would be iffy at best if $x$ can be negative ...
              – Henning Makholm
              Jun 13 '16 at 23:07










            • +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
              – Albert Renshaw
              Jun 14 '16 at 4:43










            • Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
              – Albert Renshaw
              Jun 14 '16 at 4:44












            • @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
              – J. M. is not a mathematician
              Jun 15 '16 at 18:04














            10












            10








            10






            You've already defined your function (assuming you've also chosen its domain).



            One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so.



            Typical notation for a function created by the sort of description you give is a definition by cases:



            $$ f(x) := begin{cases} 0 & x = 0 \ 1 & x neq 0 end{cases} $$



            For many applications — most applications, I expect — this is one of the best descriptions of said function. If need be, name it with a letter, and continue on with whatever you're doing.





            The complementary function



            $$ g(x) := begin{cases} 1 & x = 0 \ 0 & x neq 0 end{cases} $$



            which is related to your function by $f(x) = 1 - g(x)$ comes up often enough in some contexts to have been given a name and notation: e.g.




            • The Kronecker delta. A few different notations exist depending on the setting; e.g. $delta_x$, $delta[x]$, or $delta_{x,0}$.

            • The Iverson bracket. This would be notated $[x = 0]$. This notation is, IMO, indispensable for doing complicated calculations with summations.


            • x == 0 computes this function in C and C++, and many other programming languages allow similar.




            Some applications might want to represent such a function in particular ways. For example, if one only cares about the value of $g(x)$ when $x$ is an integer, but strongly prefers to work with analytic functions (e.g. because you're studying a sequence using complex analysis), one has the fact that



            $$ g(x) = mathop{mathrm{sinc}}(pi x) $$



            holds whenever $x$ is an integer.



            (if you're unfamiliar with it, $mathop{mathrm{sinc}}(z)$ is the continuous extension of $sin(z) / z$)






            share|cite|improve this answer














            You've already defined your function (assuming you've also chosen its domain).



            One of the main ways to "create" a function is simply by specifying its values at all points, and your description has done so.



            Typical notation for a function created by the sort of description you give is a definition by cases:



            $$ f(x) := begin{cases} 0 & x = 0 \ 1 & x neq 0 end{cases} $$



            For many applications — most applications, I expect — this is one of the best descriptions of said function. If need be, name it with a letter, and continue on with whatever you're doing.





            The complementary function



            $$ g(x) := begin{cases} 1 & x = 0 \ 0 & x neq 0 end{cases} $$



            which is related to your function by $f(x) = 1 - g(x)$ comes up often enough in some contexts to have been given a name and notation: e.g.




            • The Kronecker delta. A few different notations exist depending on the setting; e.g. $delta_x$, $delta[x]$, or $delta_{x,0}$.

            • The Iverson bracket. This would be notated $[x = 0]$. This notation is, IMO, indispensable for doing complicated calculations with summations.


            • x == 0 computes this function in C and C++, and many other programming languages allow similar.




            Some applications might want to represent such a function in particular ways. For example, if one only cares about the value of $g(x)$ when $x$ is an integer, but strongly prefers to work with analytic functions (e.g. because you're studying a sequence using complex analysis), one has the fact that



            $$ g(x) = mathop{mathrm{sinc}}(pi x) $$



            holds whenever $x$ is an integer.



            (if you're unfamiliar with it, $mathop{mathrm{sinc}}(z)$ is the continuous extension of $sin(z) / z$)







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jun 13 '16 at 23:57

























            answered Jun 13 '16 at 22:38









            Hurkyl

            111k9117259




            111k9117259








            • 1




              In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
              – Hurkyl
              Jun 13 '16 at 22:45






            • 2




              $0^x$ would be iffy at best if $x$ can be negative ...
              – Henning Makholm
              Jun 13 '16 at 23:07










            • +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
              – Albert Renshaw
              Jun 14 '16 at 4:43










            • Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
              – Albert Renshaw
              Jun 14 '16 at 4:44












            • @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
              – J. M. is not a mathematician
              Jun 15 '16 at 18:04














            • 1




              In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
              – Hurkyl
              Jun 13 '16 at 22:45






            • 2




              $0^x$ would be iffy at best if $x$ can be negative ...
              – Henning Makholm
              Jun 13 '16 at 23:07










            • +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
              – Albert Renshaw
              Jun 14 '16 at 4:43










            • Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
              – Albert Renshaw
              Jun 14 '16 at 4:44












            • @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
              – J. M. is not a mathematician
              Jun 15 '16 at 18:04








            1




            1




            In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
            – Hurkyl
            Jun 13 '16 at 22:45




            In some settings, $0^x$ could be used as another notation for this, but would likely look out of place.
            – Hurkyl
            Jun 13 '16 at 22:45




            2




            2




            $0^x$ would be iffy at best if $x$ can be negative ...
            – Henning Makholm
            Jun 13 '16 at 23:07




            $0^x$ would be iffy at best if $x$ can be negative ...
            – Henning Makholm
            Jun 13 '16 at 23:07












            +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
            – Albert Renshaw
            Jun 14 '16 at 4:43




            +1 Thanks, this is a resourceful answer because you referenced Kronecker Delta which was great to read about and see some of the sigma-notated and integral-notated versions of it. As for the piece-wise function, unfortunately I'm working on applying this function into a larger function to simplify the super-function, I can't simplify using a piece-wise function because identities don't exist for custom piece-wise functions and therefor it's hard to cancel stuff out.
            – Albert Renshaw
            Jun 14 '16 at 4:43












            Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
            – Albert Renshaw
            Jun 14 '16 at 4:44






            Also for the sinc(πx) only wokring when x is an integer, we can just use floor or ceiling function on x to get it working for all values of x, that's a great answer too for what I'm trying to do except when I look at the definition of sinc, it seems to be piece-wise too, doh
            – Albert Renshaw
            Jun 14 '16 at 4:44














            @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
            – J. M. is not a mathematician
            Jun 15 '16 at 18:04




            @Henning, since you mention it: have you already seen Knuth's article on notations? There is a mention of the interesting history of $0^x$ in there...
            – J. M. is not a mathematician
            Jun 15 '16 at 18:04











            3














            How about $f(x)= 1-delta_{x,0}$ (using the Kronecker Delta function, in Mathematica/WolframAlpha can write the $delta_{x,0}$ as




            kroneckerdelta(x,0)




            )






            share|cite|improve this answer




























              3














              How about $f(x)= 1-delta_{x,0}$ (using the Kronecker Delta function, in Mathematica/WolframAlpha can write the $delta_{x,0}$ as




              kroneckerdelta(x,0)




              )






              share|cite|improve this answer


























                3












                3








                3






                How about $f(x)= 1-delta_{x,0}$ (using the Kronecker Delta function, in Mathematica/WolframAlpha can write the $delta_{x,0}$ as




                kroneckerdelta(x,0)




                )






                share|cite|improve this answer














                How about $f(x)= 1-delta_{x,0}$ (using the Kronecker Delta function, in Mathematica/WolframAlpha can write the $delta_{x,0}$ as




                kroneckerdelta(x,0)




                )







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jun 13 '16 at 22:20

























                answered Jun 13 '16 at 22:07









                David_Shmij

                368217




                368217























                    2














                    Here's one using $sum$ notation although it only works for the natural numbers:



                    $f(x) = sumlimits_{i = 1}^{x}{frac{1}{x}} $



                    Due to the empty sum being 0.



                    It can't be simplified to $f(x) = x times frac{1}{x}$ because then f(0) would be undefined.






                    share|cite|improve this answer



















                    • 1




                      I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
                      – Albert Renshaw
                      Oct 21 at 4:12






                    • 1




                      When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
                      – omer
                      Oct 21 at 16:23












                    • Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
                      – omer
                      Oct 21 at 16:35


















                    2














                    Here's one using $sum$ notation although it only works for the natural numbers:



                    $f(x) = sumlimits_{i = 1}^{x}{frac{1}{x}} $



                    Due to the empty sum being 0.



                    It can't be simplified to $f(x) = x times frac{1}{x}$ because then f(0) would be undefined.






                    share|cite|improve this answer



















                    • 1




                      I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
                      – Albert Renshaw
                      Oct 21 at 4:12






                    • 1




                      When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
                      – omer
                      Oct 21 at 16:23












                    • Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
                      – omer
                      Oct 21 at 16:35
















                    2












                    2








                    2






                    Here's one using $sum$ notation although it only works for the natural numbers:



                    $f(x) = sumlimits_{i = 1}^{x}{frac{1}{x}} $



                    Due to the empty sum being 0.



                    It can't be simplified to $f(x) = x times frac{1}{x}$ because then f(0) would be undefined.






                    share|cite|improve this answer














                    Here's one using $sum$ notation although it only works for the natural numbers:



                    $f(x) = sumlimits_{i = 1}^{x}{frac{1}{x}} $



                    Due to the empty sum being 0.



                    It can't be simplified to $f(x) = x times frac{1}{x}$ because then f(0) would be undefined.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Oct 21 at 16:35

























                    answered Oct 20 at 17:32









                    omer

                    1495




                    1495








                    • 1




                      I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
                      – Albert Renshaw
                      Oct 21 at 4:12






                    • 1




                      When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
                      – omer
                      Oct 21 at 16:23












                    • Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
                      – omer
                      Oct 21 at 16:35
















                    • 1




                      I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
                      – Albert Renshaw
                      Oct 21 at 4:12






                    • 1




                      When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
                      – omer
                      Oct 21 at 16:23












                    • Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
                      – omer
                      Oct 21 at 16:35










                    1




                    1




                    I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
                    – Albert Renshaw
                    Oct 21 at 4:12




                    I think this satisfies the 0 when x=0 part but what about the 1 otherwise part? Is there something I do, with x, to f(x), after to get to 1 every time?
                    – Albert Renshaw
                    Oct 21 at 4:12




                    1




                    1




                    When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
                    – omer
                    Oct 21 at 16:23






                    When $x = 1$ it's just $frac{1}{1}$, when $x = 2$ it's $frac{1}{2} + frac{1}{2}$ which is $frac{2}{2}$ which is 1, in general it's $frac{1}{n} + frac{1}{n} + ... + frac{1}{n}$ repeated $n$ times, which is simply $frac{n}{n}$ which is $1$
                    – omer
                    Oct 21 at 16:23














                    Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
                    – omer
                    Oct 21 at 16:35






                    Oh I'm sorry! I wrote $n$ instead of $x$. I'll edit it. Should make more sense now
                    – omer
                    Oct 21 at 16:35













                    0














                    Without using floor or ceiling it can be done with limits. Start with a function of this form:



                    $$lim_{c to infty} left (frac{y}{sqrt2}-frac{(xcdot c)}{sqrt2} = sqrt[3]{1-left ( frac{y}{sqrt2}+frac{(xcdot c)}{sqrt2} right )^3 } right )$$



                    (Note: Using real-valued root not principal root)





                    This infinitely squashes (along x-axis) a 45º rotated graph of the form $x^3+y^3=1$ which causes the output values to be $0$ everywhere and $sqrt[6]{2}$ at $0$. This can now easily be worked to achieve the desired affect by dividing by $sqrt[6]{2}$ and subtracting from $1$






                    share|cite|improve this answer




























                      0














                      Without using floor or ceiling it can be done with limits. Start with a function of this form:



                      $$lim_{c to infty} left (frac{y}{sqrt2}-frac{(xcdot c)}{sqrt2} = sqrt[3]{1-left ( frac{y}{sqrt2}+frac{(xcdot c)}{sqrt2} right )^3 } right )$$



                      (Note: Using real-valued root not principal root)





                      This infinitely squashes (along x-axis) a 45º rotated graph of the form $x^3+y^3=1$ which causes the output values to be $0$ everywhere and $sqrt[6]{2}$ at $0$. This can now easily be worked to achieve the desired affect by dividing by $sqrt[6]{2}$ and subtracting from $1$






                      share|cite|improve this answer


























                        0












                        0








                        0






                        Without using floor or ceiling it can be done with limits. Start with a function of this form:



                        $$lim_{c to infty} left (frac{y}{sqrt2}-frac{(xcdot c)}{sqrt2} = sqrt[3]{1-left ( frac{y}{sqrt2}+frac{(xcdot c)}{sqrt2} right )^3 } right )$$



                        (Note: Using real-valued root not principal root)





                        This infinitely squashes (along x-axis) a 45º rotated graph of the form $x^3+y^3=1$ which causes the output values to be $0$ everywhere and $sqrt[6]{2}$ at $0$. This can now easily be worked to achieve the desired affect by dividing by $sqrt[6]{2}$ and subtracting from $1$






                        share|cite|improve this answer














                        Without using floor or ceiling it can be done with limits. Start with a function of this form:



                        $$lim_{c to infty} left (frac{y}{sqrt2}-frac{(xcdot c)}{sqrt2} = sqrt[3]{1-left ( frac{y}{sqrt2}+frac{(xcdot c)}{sqrt2} right )^3 } right )$$



                        (Note: Using real-valued root not principal root)





                        This infinitely squashes (along x-axis) a 45º rotated graph of the form $x^3+y^3=1$ which causes the output values to be $0$ everywhere and $sqrt[6]{2}$ at $0$. This can now easily be worked to achieve the desired affect by dividing by $sqrt[6]{2}$ and subtracting from $1$







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 17 at 18:12

























                        answered Dec 17 at 10:27









                        Albert Renshaw

                        7271627




                        7271627























                            0














                            The following equation I’ve made works without using ceiling function or limits or infinite sums when plotting the real values. 1 at x=0 and 0 and a complex pair counterpart at all other values for x



                            $(y-1-sqrt{x}-sqrt{-x})cdotfrac{y}{y-x}=0$



                            This can then be inverted to a less elegant form to have 0 for x=0 and 1 in all other places.



                            I like this best because it’s pure closed form expression and doesn’t involve any elements like ceiling function that are hard to work with when using this in other places. However it only works if you ignore complex answers which has problems of its own, but still is noteworthy and has application






                            share|cite|improve this answer




























                              0














                              The following equation I’ve made works without using ceiling function or limits or infinite sums when plotting the real values. 1 at x=0 and 0 and a complex pair counterpart at all other values for x



                              $(y-1-sqrt{x}-sqrt{-x})cdotfrac{y}{y-x}=0$



                              This can then be inverted to a less elegant form to have 0 for x=0 and 1 in all other places.



                              I like this best because it’s pure closed form expression and doesn’t involve any elements like ceiling function that are hard to work with when using this in other places. However it only works if you ignore complex answers which has problems of its own, but still is noteworthy and has application






                              share|cite|improve this answer


























                                0












                                0








                                0






                                The following equation I’ve made works without using ceiling function or limits or infinite sums when plotting the real values. 1 at x=0 and 0 and a complex pair counterpart at all other values for x



                                $(y-1-sqrt{x}-sqrt{-x})cdotfrac{y}{y-x}=0$



                                This can then be inverted to a less elegant form to have 0 for x=0 and 1 in all other places.



                                I like this best because it’s pure closed form expression and doesn’t involve any elements like ceiling function that are hard to work with when using this in other places. However it only works if you ignore complex answers which has problems of its own, but still is noteworthy and has application






                                share|cite|improve this answer














                                The following equation I’ve made works without using ceiling function or limits or infinite sums when plotting the real values. 1 at x=0 and 0 and a complex pair counterpart at all other values for x



                                $(y-1-sqrt{x}-sqrt{-x})cdotfrac{y}{y-x}=0$



                                This can then be inverted to a less elegant form to have 0 for x=0 and 1 in all other places.



                                I like this best because it’s pure closed form expression and doesn’t involve any elements like ceiling function that are hard to work with when using this in other places. However it only works if you ignore complex answers which has problems of its own, but still is noteworthy and has application







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 2 days ago

























                                answered 2 days ago









                                Albert Renshaw

                                7271627




                                7271627























                                    0














                                    The following equation I've made is a closed-form expression of KroneckerDelta (j=0 form) which will evaluate y=1 at x=0 and y=0 for x≠0.



                                    $$(frac{y}{y-x})cdot((y-1)^2+x^2)=0$$



                                    This function can be subtracted from 1 to achieve the desired result





                                    The function was formed by taking the function for a horizontal line at 0, divided by a 45º line (y=x) through the origin to create a hole in the function at x=0, then multiplied by a circle with radius 0 whose origin is at (0,1) to create a point at (0,1).






                                    share|cite|improve this answer


























                                      0














                                      The following equation I've made is a closed-form expression of KroneckerDelta (j=0 form) which will evaluate y=1 at x=0 and y=0 for x≠0.



                                      $$(frac{y}{y-x})cdot((y-1)^2+x^2)=0$$



                                      This function can be subtracted from 1 to achieve the desired result





                                      The function was formed by taking the function for a horizontal line at 0, divided by a 45º line (y=x) through the origin to create a hole in the function at x=0, then multiplied by a circle with radius 0 whose origin is at (0,1) to create a point at (0,1).






                                      share|cite|improve this answer
























                                        0












                                        0








                                        0






                                        The following equation I've made is a closed-form expression of KroneckerDelta (j=0 form) which will evaluate y=1 at x=0 and y=0 for x≠0.



                                        $$(frac{y}{y-x})cdot((y-1)^2+x^2)=0$$



                                        This function can be subtracted from 1 to achieve the desired result





                                        The function was formed by taking the function for a horizontal line at 0, divided by a 45º line (y=x) through the origin to create a hole in the function at x=0, then multiplied by a circle with radius 0 whose origin is at (0,1) to create a point at (0,1).






                                        share|cite|improve this answer












                                        The following equation I've made is a closed-form expression of KroneckerDelta (j=0 form) which will evaluate y=1 at x=0 and y=0 for x≠0.



                                        $$(frac{y}{y-x})cdot((y-1)^2+x^2)=0$$



                                        This function can be subtracted from 1 to achieve the desired result





                                        The function was formed by taking the function for a horizontal line at 0, divided by a 45º line (y=x) through the origin to create a hole in the function at x=0, then multiplied by a circle with radius 0 whose origin is at (0,1) to create a point at (0,1).







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 2 days ago









                                        Albert Renshaw

                                        7271627




                                        7271627























                                            -1














                                            Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all numbers, including imaginary/complex.



                                            $f(x) = left lceil frac{1}{2Gamma left ( left | x right | right )} right rceil$



                                            *Note the absolute value inside of Gamma, as it's easy to go unnoticed






                                            share|cite|improve this answer




























                                              -1














                                              Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all numbers, including imaginary/complex.



                                              $f(x) = left lceil frac{1}{2Gamma left ( left | x right | right )} right rceil$



                                              *Note the absolute value inside of Gamma, as it's easy to go unnoticed






                                              share|cite|improve this answer


























                                                -1












                                                -1








                                                -1






                                                Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all numbers, including imaginary/complex.



                                                $f(x) = left lceil frac{1}{2Gamma left ( left | x right | right )} right rceil$



                                                *Note the absolute value inside of Gamma, as it's easy to go unnoticed






                                                share|cite|improve this answer














                                                Paw88789's answer worked great for what I'm trying to do; the only issue was that it didn't work for all "numbers"; some imaginary numbers would cause division by 0. Luckily tonight I was able to create a function that produced the desired result for all numbers, including imaginary/complex.



                                                $f(x) = left lceil frac{1}{2Gamma left ( left | x right | right )} right rceil$



                                                *Note the absolute value inside of Gamma, as it's easy to go unnoticed







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited Apr 13 '17 at 12:20









                                                Community

                                                1




                                                1










                                                answered Jun 14 '16 at 4:54









                                                Albert Renshaw

                                                7271627




                                                7271627






























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