Properties of functions with $0$ second partial derivatives Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does every nonzero polynomial take a nonzero value at one of its multi-indices?Second partial derivatives of harmonic functionsAll second partial derivatives of harmonic function are $0$How many n-th Order Partial Derivatives Exist for a Function of k Variables?(Rudin's) Definition of a harmonic functionWhy can't we have higher order derivative test for functions of two independent variable?If $-Delta u(x)+partial _x_1u=0$ prove that $sup_bar Omega u=sup_partial Omega u$.How to prove this is an harmonic function?L2 Trace of Harmonic FunctionsProduct $u v$ of harmonic functions is harmonic if $langle nabla u, nabla v rangle$ is $equiv 0$ everywhereWays to express sum of $n$th-power partial derivatives?
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Properties of functions with $0$ second partial derivatives
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does every nonzero polynomial take a nonzero value at one of its multi-indices?Second partial derivatives of harmonic functionsAll second partial derivatives of harmonic function are $0$How many n-th Order Partial Derivatives Exist for a Function of k Variables?(Rudin's) Definition of a harmonic functionWhy can't we have higher order derivative test for functions of two independent variable?If $-Delta u(x)+partial _x_1u=0$ prove that $sup_bar Omega u=sup_partial Omega u$.How to prove this is an harmonic function?L2 Trace of Harmonic FunctionsProduct $u v$ of harmonic functions is harmonic if $langle nabla u, nabla v rangle$ is $equiv 0$ everywhereWays to express sum of $n$th-power partial derivatives?
$begingroup$
I have a $n$-dimensional polynomial that I am evaluating on some domain $Omega subset mathbbR^n$
$$
f:Omegarightarrow mathbbR
$$
where I know that all the second partials are zero
$$
dfracpartial^2 fpartial x_k^2 equiv0
$$
however, the mixed partials may be non-zero.
It is easy to see that this function is harmonic as $nabla^2f = 0$. From this we get lots of results, such as the minima $m$ and maxima $M$ is on the boundary: $m,Min partialOmega$.
This condition that all second derivatives seems stronger than being harmonic though, so I was wondering if there was a name / other properties relating to functions like this.
Thank you!
calculus functions polynomials partial-derivative harmonic-functions
asked Mar 31 at 21:14
wjmccannwjmccann
618218
$endgroup$
add a comment |
$begingroup$
I have a $n$-dimensional polynomial that I am evaluating on some domain $Omega subset mathbbR^n$
$$
f:Omegarightarrow mathbbR
$$
where I know that all the second partials are zero
$$
dfracpartial^2 fpartial x_k^2 equiv0
$$
however, the mixed partials may be non-zero.
It is easy to see that this function is harmonic as $nabla^2f = 0$. From this we get lots of results, such as the minima $m$ and maxima $M$ is on the boundary: $m,Min partialOmega$.
This condition that all second derivatives seems stronger than being harmonic though, so I was wondering if there was a name / other properties relating to functions like this.
Thank you!
calculus functions polynomials partial-derivative harmonic-functions
asked Mar 31 at 21:14
wjmccannwjmccann
618218
$endgroup$
$begingroup$
Perhaps $f$ is a polynomial, of degree $;le 1$ in each variable. For example, when $n=2$, $f(x,y) = a + bx + cy + dxy$.
$endgroup$
– GEdgar
Apr 5 at 21:45
$begingroup$
That is the form of solutions I found for higher dimensions too, but I was wondering if that type of polynomial had a special name
$endgroup$
– wjmccann
Apr 5 at 22:32
add a comment |
$begingroup$
I have a $n$-dimensional polynomial that I am evaluating on some domain $Omega subset mathbbR^n$
$$
f:Omegarightarrow mathbbR
$$
where I know that all the second partials are zero
$$
dfracpartial^2 fpartial x_k^2 equiv0
$$
however, the mixed partials may be non-zero.
It is easy to see that this function is harmonic as $nabla^2f = 0$. From this we get lots of results, such as the minima $m$ and maxima $M$ is on the boundary: $m,Min partialOmega$.
This condition that all second derivatives seems stronger than being harmonic though, so I was wondering if there was a name / other properties relating to functions like this.
Thank you!
calculus functions polynomials partial-derivative harmonic-functions
asked Mar 31 at 21:14
wjmccannwjmccann
618218
$endgroup$
I have a $n$-dimensional polynomial that I am evaluating on some domain $Omega subset mathbbR^n$
$$
f:Omegarightarrow mathbbR
$$
where I know that all the second partials are zero
$$
dfracpartial^2 fpartial x_k^2 equiv0
$$
however, the mixed partials may be non-zero.
It is easy to see that this function is harmonic as $nabla^2f = 0$. From this we get lots of results, such as the minima $m$ and maxima $M$ is on the boundary: $m,Min partialOmega$.
This condition that all second derivatives seems stronger than being harmonic though, so I was wondering if there was a name / other properties relating to functions like this.
Thank you!
calculus functions polynomials partial-derivative harmonic-functions
calculus functions polynomials partial-derivative harmonic-functions
asked Mar 31 at 21:14
wjmccannwjmccann
618218
asked Mar 31 at 21:14
wjmccannwjmccann
618218
asked Mar 31 at 21:14
wjmccannwjmccann
618218
asked Mar 31 at 21:14
wjmccannwjmccann
618218
asked Mar 31 at 21:14
wjmccannwjmccann
618218
618218
$begingroup$
Perhaps $f$ is a polynomial, of degree $;le 1$ in each variable. For example, when $n=2$, $f(x,y) = a + bx + cy + dxy$.
$endgroup$
– GEdgar
Apr 5 at 21:45
$begingroup$
That is the form of solutions I found for higher dimensions too, but I was wondering if that type of polynomial had a special name
$endgroup$
– wjmccann
Apr 5 at 22:32
add a comment |
$begingroup$
Perhaps $f$ is a polynomial, of degree $;le 1$ in each variable. For example, when $n=2$, $f(x,y) = a + bx + cy + dxy$.
$endgroup$
– GEdgar
Apr 5 at 21:45
$begingroup$
That is the form of solutions I found for higher dimensions too, but I was wondering if that type of polynomial had a special name
$endgroup$
– wjmccann
Apr 5 at 22:32
$begingroup$
Perhaps $f$ is a polynomial, of degree $;le 1$ in each variable. For example, when $n=2$, $f(x,y) = a + bx + cy + dxy$.
$endgroup$
– GEdgar
Apr 5 at 21:45
$begingroup$
Perhaps $f$ is a polynomial, of degree $;le 1$ in each variable. For example, when $n=2$, $f(x,y) = a + bx + cy + dxy$.
$endgroup$
– GEdgar
Apr 5 at 21:45
$begingroup$
That is the form of solutions I found for higher dimensions too, but I was wondering if that type of polynomial had a special name
$endgroup$
– wjmccann
Apr 5 at 22:32
$begingroup$
That is the form of solutions I found for higher dimensions too, but I was wondering if that type of polynomial had a special name
$endgroup$
– wjmccann
Apr 5 at 22:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
These polynomials are called multilinear. They are sometimes called polynomials of max degree $1$ (as opposed to total degree $1,$ which would mean affine.)
They come up naturally in Fourier analysis on $-1,1^n$: the Fourier transform passes between the representation of a multilinear polynomial by its values on $-1,1^n$ and the values of its coefficients.
They also come up in things like the polynomial method, Schwartz-Zippel lemma, and related results. The important property here is that if a multilinear polynomial $p$ is zero on a grid of the form $a_1,b_1timesdotstimesa_n,b_n$ with $a_ineq b_i$ for each $i,$ then $p$ is identically zero.
Since you have an algebraic property of polynomials, the restriction to $Omega$ is irrelevant. But it is true that any distribution on a domain $Omega$ satisfying $partial^2 f/partial x_k^2=0$ must be a multilinear polynomial.
Another property, perhaps only interesting to me: they are "nonzero at one of their multi-indices". By this I mean that if $p$ is a non-zero multilinear polynomial, there exist $alpha_1,dots,alpha_nin0,1$ such that $p(alpha_1,dots,alpha_n)neq 0$ and the $x_1^alpha_1dots x_n^alpha_n$ coefficient of $p$ is non-zero. Specifically, we can take $(alpha_1,dots,alpha_n)$ to be minimal in the product order on $0,1^n,$ such that the coefficient $c$ of $x_1^alpha_1dots x_n^alpha_n$ is non-zero. This gives $p(x_1^alpha_1,dots,x_n^alpha_n)=c.$
answered Apr 9 at 2:54
DapDap
20k842
$endgroup$
add a comment |
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$begingroup$
These polynomials are called multilinear. They are sometimes called polynomials of max degree $1$ (as opposed to total degree $1,$ which would mean affine.)
They come up naturally in Fourier analysis on $-1,1^n$: the Fourier transform passes between the representation of a multilinear polynomial by its values on $-1,1^n$ and the values of its coefficients.
They also come up in things like the polynomial method, Schwartz-Zippel lemma, and related results. The important property here is that if a multilinear polynomial $p$ is zero on a grid of the form $a_1,b_1timesdotstimesa_n,b_n$ with $a_ineq b_i$ for each $i,$ then $p$ is identically zero.
Since you have an algebraic property of polynomials, the restriction to $Omega$ is irrelevant. But it is true that any distribution on a domain $Omega$ satisfying $partial^2 f/partial x_k^2=0$ must be a multilinear polynomial.
Another property, perhaps only interesting to me: they are "nonzero at one of their multi-indices". By this I mean that if $p$ is a non-zero multilinear polynomial, there exist $alpha_1,dots,alpha_nin0,1$ such that $p(alpha_1,dots,alpha_n)neq 0$ and the $x_1^alpha_1dots x_n^alpha_n$ coefficient of $p$ is non-zero. Specifically, we can take $(alpha_1,dots,alpha_n)$ to be minimal in the product order on $0,1^n,$ such that the coefficient $c$ of $x_1^alpha_1dots x_n^alpha_n$ is non-zero. This gives $p(x_1^alpha_1,dots,x_n^alpha_n)=c.$
answered Apr 9 at 2:54
DapDap
20k842
$endgroup$
add a comment |
$begingroup$
These polynomials are called multilinear. They are sometimes called polynomials of max degree $1$ (as opposed to total degree $1,$ which would mean affine.)
They come up naturally in Fourier analysis on $-1,1^n$: the Fourier transform passes between the representation of a multilinear polynomial by its values on $-1,1^n$ and the values of its coefficients.
They also come up in things like the polynomial method, Schwartz-Zippel lemma, and related results. The important property here is that if a multilinear polynomial $p$ is zero on a grid of the form $a_1,b_1timesdotstimesa_n,b_n$ with $a_ineq b_i$ for each $i,$ then $p$ is identically zero.
Since you have an algebraic property of polynomials, the restriction to $Omega$ is irrelevant. But it is true that any distribution on a domain $Omega$ satisfying $partial^2 f/partial x_k^2=0$ must be a multilinear polynomial.
Another property, perhaps only interesting to me: they are "nonzero at one of their multi-indices". By this I mean that if $p$ is a non-zero multilinear polynomial, there exist $alpha_1,dots,alpha_nin0,1$ such that $p(alpha_1,dots,alpha_n)neq 0$ and the $x_1^alpha_1dots x_n^alpha_n$ coefficient of $p$ is non-zero. Specifically, we can take $(alpha_1,dots,alpha_n)$ to be minimal in the product order on $0,1^n,$ such that the coefficient $c$ of $x_1^alpha_1dots x_n^alpha_n$ is non-zero. This gives $p(x_1^alpha_1,dots,x_n^alpha_n)=c.$
answered Apr 9 at 2:54
DapDap
20k842
$endgroup$
add a comment |
$begingroup$
These polynomials are called multilinear. They are sometimes called polynomials of max degree $1$ (as opposed to total degree $1,$ which would mean affine.)
They come up naturally in Fourier analysis on $-1,1^n$: the Fourier transform passes between the representation of a multilinear polynomial by its values on $-1,1^n$ and the values of its coefficients.
They also come up in things like the polynomial method, Schwartz-Zippel lemma, and related results. The important property here is that if a multilinear polynomial $p$ is zero on a grid of the form $a_1,b_1timesdotstimesa_n,b_n$ with $a_ineq b_i$ for each $i,$ then $p$ is identically zero.
Since you have an algebraic property of polynomials, the restriction to $Omega$ is irrelevant. But it is true that any distribution on a domain $Omega$ satisfying $partial^2 f/partial x_k^2=0$ must be a multilinear polynomial.
Another property, perhaps only interesting to me: they are "nonzero at one of their multi-indices". By this I mean that if $p$ is a non-zero multilinear polynomial, there exist $alpha_1,dots,alpha_nin0,1$ such that $p(alpha_1,dots,alpha_n)neq 0$ and the $x_1^alpha_1dots x_n^alpha_n$ coefficient of $p$ is non-zero. Specifically, we can take $(alpha_1,dots,alpha_n)$ to be minimal in the product order on $0,1^n,$ such that the coefficient $c$ of $x_1^alpha_1dots x_n^alpha_n$ is non-zero. This gives $p(x_1^alpha_1,dots,x_n^alpha_n)=c.$
answered Apr 9 at 2:54
DapDap
20k842
$endgroup$
These polynomials are called multilinear. They are sometimes called polynomials of max degree $1$ (as opposed to total degree $1,$ which would mean affine.)
They come up naturally in Fourier analysis on $-1,1^n$: the Fourier transform passes between the representation of a multilinear polynomial by its values on $-1,1^n$ and the values of its coefficients.
They also come up in things like the polynomial method, Schwartz-Zippel lemma, and related results. The important property here is that if a multilinear polynomial $p$ is zero on a grid of the form $a_1,b_1timesdotstimesa_n,b_n$ with $a_ineq b_i$ for each $i,$ then $p$ is identically zero.
Since you have an algebraic property of polynomials, the restriction to $Omega$ is irrelevant. But it is true that any distribution on a domain $Omega$ satisfying $partial^2 f/partial x_k^2=0$ must be a multilinear polynomial.
Another property, perhaps only interesting to me: they are "nonzero at one of their multi-indices". By this I mean that if $p$ is a non-zero multilinear polynomial, there exist $alpha_1,dots,alpha_nin0,1$ such that $p(alpha_1,dots,alpha_n)neq 0$ and the $x_1^alpha_1dots x_n^alpha_n$ coefficient of $p$ is non-zero. Specifically, we can take $(alpha_1,dots,alpha_n)$ to be minimal in the product order on $0,1^n,$ such that the coefficient $c$ of $x_1^alpha_1dots x_n^alpha_n$ is non-zero. This gives $p(x_1^alpha_1,dots,x_n^alpha_n)=c.$
answered Apr 9 at 2:54
DapDap
20k842
answered Apr 9 at 2:54
DapDap
20k842
answered Apr 9 at 2:54
DapDap
20k842
answered Apr 9 at 2:54
DapDap
20k842
20k842
add a comment |
add a comment |
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$begingroup$
Perhaps $f$ is a polynomial, of degree $;le 1$ in each variable. For example, when $n=2$, $f(x,y) = a + bx + cy + dxy$.
$endgroup$
– GEdgar
Apr 5 at 21:45
$begingroup$
That is the form of solutions I found for higher dimensions too, but I was wondering if that type of polynomial had a special name
$endgroup$
– wjmccann
Apr 5 at 22:32