The implication of zero mixed partial derivatives for multivariate function's minimization The...
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The implication of zero mixed partial derivatives for multivariate function's minimization
The Next CEO of Stack OverflowIf the second mixed partial is identically zero, then the function can be written as a sum $f(x,y) = f_1(x) + f_2(y)$$dfrac{partial^2 f}{partial x partial y} = 0 nRightarrow f(x,y) = g(x) + h(y)$If a nonnegative function of $x_1,dots,x_n$ can be written as $sum g_k(x_k)$, then the summands can be taken nonnegative$f(x,y) = f_1(x) + f_2(x)$ with continuous differential real-valued functionsShow that both mixed partial derivatives exist at the origin but are not equalFunction on $mathbb{R}^{2}-{0}$.Let $F_1,F_2:bf R^2 to R$ be functions such that…How many n-th Order Partial Derivatives Exist for a Function of k Variables?Maximize $f(textbf{x},textbf{y}) = f_1(textbf{x}) + f_2(textbf{x},textbf{y})$Symmetry of higher order mixed partial derivatives under weaker assumptionsWhy does the partial of $f: Delta to mathbb{R}^2$ fail to exist?To find function satisfying given partial derivatives2 variable, 2 valued function $f(x_1,,x_2)=(x_1,,x_2)$What will be $frac {partial F_i} {partial x_j}$?
$begingroup$
Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_{12}=f''_{21}=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_{textbf x} f(textbf x)equiv min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$
A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.
multivariable-calculus optimization partial-derivative
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
$endgroup$
add a comment |
$begingroup$
Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_{12}=f''_{21}=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_{textbf x} f(textbf x)equiv min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$
A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.
multivariable-calculus optimization partial-derivative
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
$endgroup$
$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_{textbf x} f(textbf x) = min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24
add a comment |
$begingroup$
Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_{12}=f''_{21}=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_{textbf x} f(textbf x)equiv min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$
A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.
multivariable-calculus optimization partial-derivative
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
$endgroup$
Suppose $f(textbf x)=f(x_1,x_2) $ has mixed partial derivatives $f''_{12}=f''_{21}=0$, so can I say: there exist $f_1(x_1)$ and $f_2(x_2)$ such that $min_{textbf x} f(textbf x)equiv min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$? Or even further, as follows:
$$f(textbf x)equiv f_1(x_1)+ f_2(x_2)$$
A positive simple case is $f(x_1,x_2)=x_1^2+x_2^3$. I can not think of any opposite cases, but I am not so sure about it and may need a proof.
multivariable-calculus optimization partial-derivative
multivariable-calculus optimization partial-derivative
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
edited Nov 1 '15 at 20:04
user147263
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
asked Jun 24 '13 at 22:43
jorter.jijorter.ji
258
258
$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_{textbf x} f(textbf x) = min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24
add a comment |
$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_{textbf x} f(textbf x) = min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24
$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_{textbf x} f(textbf x) = min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24
$begingroup$
Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_{textbf x} f(textbf x) = min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$, see this question.
$endgroup$
– user147263
Dec 29 '15 at 0:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For a mixed derivative $f_{xy} = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_{xy} ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_{xy} ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_{yx}$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
$endgroup$
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
|
show 5 more comments
$begingroup$
The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.
Here I try to provide a proof without that assumption.
Restatement
I can restate the conjecture with little weaker conditions:
If $f(x, y)$ has $f_{yx} = 0$, then $z(x, y) = f(x) + g(y)$.
Proof
From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_{yx}(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.
$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$
Annotation
The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.
answered Mar 16 at 8:52
TA123TA123
958
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a mixed derivative $f_{xy} = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_{xy} ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_{xy} ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_{yx}$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
$endgroup$
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
|
show 5 more comments
$begingroup$
For a mixed derivative $f_{xy} = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_{xy} ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_{xy} ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_{yx}$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
$endgroup$
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
|
show 5 more comments
$begingroup$
For a mixed derivative $f_{xy} = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_{xy} ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_{xy} ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_{yx}$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
$endgroup$
For a mixed derivative $f_{xy} = 0$, integrating with respect to $y$ gives:
$$
f_x(x,y) = int f_{xy} ,dy + h(x).
$$
Integrating with respect to $x$:
$$
f(x,y) = iint f_{xy} ,dydx + int h(x)dx + g(y).
$$
Similar result yields if we start from $f_{yx}$, now this implies
$$
f(x,y) = f_1(x) + f_2(y),
$$
and there goes your conclusion in the question.
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
answered Jun 24 '13 at 23:14
Shuhao CaoShuhao Cao
16.1k34292
16.1k34292
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
|
show 5 more comments
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
Hi again. Can I say $f_1$ and $f_2$ are unique? Excluding constants.
$endgroup$
– jorter.ji
Jun 26 '13 at 6:13
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
@jorter.ji If you only know that the mixed derivative is zero, then any differentiable function $f_1$ and $f_2$ will do. So not unique unless you have other conditions.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 12:32
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
Like what kind of conditions? Or can you think of any instances such that $f=f_1+f_2=f_3+f_4$?
$endgroup$
– jorter.ji
Jun 26 '13 at 16:49
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
@jorter.ji I meant to say that provided only this differential relation $f_{xy} = f_{yx} = 0$ is known, then there are infinitely choices for $g$ and $h$ in $f = g(x) + h(y)$, $f$ can be $f = x+y$, or $f= x^2+y^2$. Not that they are equal or something.
$endgroup$
– Shuhao Cao
Jun 26 '13 at 16:52
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
$begingroup$
OIC your meaning. Actually, I already know that $f(x,y)=f_1(x)+f_2(y)+V(x,y)$, where $V(x,y)$ is a implicit but determined function, i.e., $f(x,y)$ is somehow implicitly determined, then I wonder if $f=g(x)+h(y)$ is unique.
$endgroup$
– jorter.ji
Jun 26 '13 at 17:41
|
show 5 more comments
$begingroup$
The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.
Here I try to provide a proof without that assumption.
Restatement
I can restate the conjecture with little weaker conditions:
If $f(x, y)$ has $f_{yx} = 0$, then $z(x, y) = f(x) + g(y)$.
Proof
From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_{yx}(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.
$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$
Annotation
The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.
answered Mar 16 at 8:52
TA123TA123
958
$endgroup$
add a comment |
$begingroup$
The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.
Here I try to provide a proof without that assumption.
Restatement
I can restate the conjecture with little weaker conditions:
If $f(x, y)$ has $f_{yx} = 0$, then $z(x, y) = f(x) + g(y)$.
Proof
From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_{yx}(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.
$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$
Annotation
The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.
answered Mar 16 at 8:52
TA123TA123
958
$endgroup$
add a comment |
$begingroup$
The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.
Here I try to provide a proof without that assumption.
Restatement
I can restate the conjecture with little weaker conditions:
If $f(x, y)$ has $f_{yx} = 0$, then $z(x, y) = f(x) + g(y)$.
Proof
From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_{yx}(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.
$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$
Annotation
The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.
answered Mar 16 at 8:52
TA123TA123
958
$endgroup$
The answer of @Shuhao Cao needs an assumption that the first partial derivative is integrable.
Here I try to provide a proof without that assumption.
Restatement
I can restate the conjecture with little weaker conditions:
If $f(x, y)$ has $f_{yx} = 0$, then $z(x, y) = f(x) + g(y)$.
Proof
From Mean Value Theorem, $$f_y(x, y) = f_y(0, y) + f_{yx}(xi , y) x, xi in (0, x)$$This implies $f_y = f_y(0, y)$.
$forall x_0$, $f(x_0, y)$ is an antiderivative of $f_y(0, y)$. Any two antiderivatives differ by constant. So, we can write that $$f(x, y) = f(0, y) + c(x) = f(0, y) + f(x, 0) - f(0, 0) = f_1(x) + f_2(y)$$
Annotation
The key is that "any two antiderivatives differ by constant" can be proved only based on Mean Value Theorem, but nothing wih Reimann Integral.
answered Mar 16 at 8:52
TA123TA123
958
answered Mar 16 at 8:52
TA123TA123
958
answered Mar 16 at 8:52
TA123TA123
958
answered Mar 16 at 8:52
TA123TA123
958
958
add a comment |
add a comment |
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Concerning the passage from $f(textbf x)equiv f_1(x_1)+ f_2(x_2)$ to $min_{textbf x} f(textbf x) = min_{x_1}f_1(x_1)+ min_{x_2}f_2(x_2)$, see this question.
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– user147263
Dec 29 '15 at 0:24