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
$begingroup$
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!
calculus differential-geometry
edited 5 hours ago
Adam Chalumeau
1,289217
asked 5 hours ago
Oriki77Oriki77
113
$endgroup$
add a comment |
$begingroup$
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!
calculus differential-geometry
edited 5 hours ago
Adam Chalumeau
1,289217
asked 5 hours ago
Oriki77Oriki77
113
$endgroup$
add a comment |
$begingroup$
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!
calculus differential-geometry
edited 5 hours ago
Adam Chalumeau
1,289217
asked 5 hours ago
Oriki77Oriki77
113
$endgroup$
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
calculus differential-geometry
edited 5 hours ago
Adam Chalumeau
1,289217
asked 5 hours ago
Oriki77Oriki77
113
edited 5 hours ago
Adam Chalumeau
1,289217
asked 5 hours ago
Oriki77Oriki77
113
edited 5 hours ago
Adam Chalumeau
1,289217
edited 5 hours ago
Adam Chalumeau
1,289217
edited 5 hours ago
Adam Chalumeau
1,289217
1,289217
asked 5 hours ago
Oriki77Oriki77
113
asked 5 hours ago
Oriki77Oriki77
113
asked 5 hours ago
Oriki77Oriki77
113
113
add a comment |
add a comment |
1 Answer
1
active
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votes
$begingroup$
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 $f:Bbb Rto Bbb R$.
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
$endgroup$
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 $f:Bbb Rto Bbb R$.
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
$endgroup$
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
add a comment |
$begingroup$
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 $f:Bbb Rto Bbb R$.
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
$endgroup$
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
add a comment |
$begingroup$
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 $f:Bbb Rto Bbb R$.
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
$endgroup$
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 $f:Bbb Rto Bbb R$.
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
answered 5 hours ago
Adam ChalumeauAdam Chalumeau
1,289217
1,289217
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
add a comment |
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
$begingroup$
You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
$endgroup$
– Dan Uznanski
1 hour ago
add a comment |
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