Smooth function that vanishes only on unit cubeCantor Set and ternary expansions.The level set of a smooth functionProve that there is no function $f:BbbRtoBbbR$ with $f(0)>0$ such that $forall x,yinBbbR, f(x+y)geq f(x)+y f(f(x))$Deciding $displaystyle o,omega,Theta$ notationsProve that the function $f(n)=n^3+(fracn2^n)^5$ satisfies some property (where $ninBbbN$)Convergence when the derivative is uniformly continuousIdeas on how to find the number of unit cubes that intersect ball with radius RTrying to prove that this function is continuous for all real numbers.proving limits by ϵ−δ and ϵ−MQuestion on notation of 3rd degree Taylor expansion

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Smooth function that vanishes only on unit cube


Cantor Set and ternary expansions.The level set of a smooth functionProve that there is no function $f:BbbRtoBbbR$ with $f(0)>0$ such that $forall x,yinBbbR, f(x+y)geq f(x)+y f(f(x))$Deciding $displaystyle o,omega,Theta$ notationsProve that the function $f(n)=n^3+(fracn2^n)^5$ satisfies some property (where $ninBbbN$)Convergence when the derivative is uniformly continuousIdeas on how to find the number of unit cubes that intersect ball with radius RTrying to prove that this function is continuous for all real numbers.proving limits by ϵ−δ and ϵ−MQuestion on notation of 3rd degree Taylor expansion













3












$begingroup$


I am having hard time defining a smooth function $f:Bbb R^3 to Bbb R$ such that :



$f(x,y,z) = 0$ if and only if $(x,y,z)$ belongs to the unit cube $[0,1]^3$.



I tried generalizing the case of $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.



I would really appreciate any help,
Thanks in advance!










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    I am having hard time defining a smooth function $f:Bbb R^3 to Bbb R$ such that :



    $f(x,y,z) = 0$ if and only if $(x,y,z)$ belongs to the unit cube $[0,1]^3$.



    I tried generalizing the case of $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.



    I would really appreciate any help,
    Thanks in advance!










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      I am having hard time defining a smooth function $f:Bbb R^3 to Bbb R$ such that :



      $f(x,y,z) = 0$ if and only if $(x,y,z)$ belongs to the unit cube $[0,1]^3$.



      I tried generalizing the case of $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.



      I would really appreciate any help,
      Thanks in advance!










      share|cite|improve this question











      $endgroup$




      I am having hard time defining a smooth function $f:Bbb R^3 to Bbb R$ such that :



      $f(x,y,z) = 0$ if and only if $(x,y,z)$ belongs to the unit cube $[0,1]^3$.



      I tried generalizing the case of $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.



      I would really appreciate any help,
      Thanks in advance!







      calculus differential-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 13 at 17:52









      Adam Chalumeau

      1,617420




      1,617420










      asked May 13 at 17:25









      Oriki77Oriki77

      185




      185




















          1 Answer
          1






          active

          oldest

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          10












          $begingroup$

          If $f:Bbb Rto Bbb R$ is a smooth function such that $f(x)=0iff xin[0,1]$ then the map $g:Bbb R^3to Bbb R$ defined by $$g(x,y,z)=f(x)^2+f(y)^2+f(z)^2$$ is smooth and has the property that $$g(x,y,z)=0iff f(x)=f(y)=f(z)=0iff 0leq x,y,zleq 1,$$
          so you just have to do the case $f:Bbb Rto Bbb R$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
            $endgroup$
            – Dan Uznanski
            May 13 at 21:47











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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          10












          $begingroup$

          If $f:Bbb Rto Bbb R$ is a smooth function such that $f(x)=0iff xin[0,1]$ then the map $g:Bbb R^3to Bbb R$ defined by $$g(x,y,z)=f(x)^2+f(y)^2+f(z)^2$$ is smooth and has the property that $$g(x,y,z)=0iff f(x)=f(y)=f(z)=0iff 0leq x,y,zleq 1,$$
          so you just have to do the case $f:Bbb Rto Bbb R$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
            $endgroup$
            – Dan Uznanski
            May 13 at 21:47















          10












          $begingroup$

          If $f:Bbb Rto Bbb R$ is a smooth function such that $f(x)=0iff xin[0,1]$ then the map $g:Bbb R^3to Bbb R$ defined by $$g(x,y,z)=f(x)^2+f(y)^2+f(z)^2$$ is smooth and has the property that $$g(x,y,z)=0iff f(x)=f(y)=f(z)=0iff 0leq x,y,zleq 1,$$
          so you just have to do the case $f:Bbb Rto Bbb R$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
            $endgroup$
            – Dan Uznanski
            May 13 at 21:47













          10












          10








          10





          $begingroup$

          If $f:Bbb Rto Bbb R$ is a smooth function such that $f(x)=0iff xin[0,1]$ then the map $g:Bbb R^3to Bbb R$ defined by $$g(x,y,z)=f(x)^2+f(y)^2+f(z)^2$$ is smooth and has the property that $$g(x,y,z)=0iff f(x)=f(y)=f(z)=0iff 0leq x,y,zleq 1,$$
          so you just have to do the case $f:Bbb Rto Bbb R$.






          share|cite|improve this answer











          $endgroup$



          If $f:Bbb Rto Bbb R$ is a smooth function such that $f(x)=0iff xin[0,1]$ then the map $g:Bbb R^3to Bbb R$ defined by $$g(x,y,z)=f(x)^2+f(y)^2+f(z)^2$$ is smooth and has the property that $$g(x,y,z)=0iff f(x)=f(y)=f(z)=0iff 0leq x,y,zleq 1,$$
          so you just have to do the case $f:Bbb Rto Bbb R$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited May 14 at 10:25

























          answered May 13 at 17:32









          Adam ChalumeauAdam Chalumeau

          1,617420




          1,617420











          • $begingroup$
            You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
            $endgroup$
            – Dan Uznanski
            May 13 at 21:47
















          • $begingroup$
            You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
            $endgroup$
            – Dan Uznanski
            May 13 at 21:47















          $begingroup$
          You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
          $endgroup$
          – Dan Uznanski
          May 13 at 21:47




          $begingroup$
          You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
          $endgroup$
          – Dan Uznanski
          May 13 at 21:47

















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