Transformation not reproducible
$begingroup$
Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
For
$$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
$mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.
The only advice given is to introduce the function
$$
F(X):=frac{X}{sqrt{1+|X|^2}}
$$
and use a green formula for
$$
F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
$$
Attempt:
When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get
$$
mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
= 0
= mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
$$
Here I am stuck as I don't find a suiting Green formula.
real-analysis integration pde
edited Jan 12 at 14:09
Viktor Glombik
790527
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
For
$$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
$mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.
The only advice given is to introduce the function
$$
F(X):=frac{X}{sqrt{1+|X|^2}}
$$
and use a green formula for
$$
F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
$$
Attempt:
When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get
$$
mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
= 0
= mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
$$
Here I am stuck as I don't find a suiting Green formula.
real-analysis integration pde
edited Jan 12 at 14:09
Viktor Glombik
790527
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
For
$$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
$mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.
The only advice given is to introduce the function
$$
F(X):=frac{X}{sqrt{1+|X|^2}}
$$
and use a green formula for
$$
F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
$$
Attempt:
When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get
$$
mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
= 0
= mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
$$
Here I am stuck as I don't find a suiting Green formula.
real-analysis integration pde
edited Jan 12 at 14:09
Viktor Glombik
790527
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
$endgroup$
Let $f: mathbb{R}^n supset Omega to mathbb{R}$ and $u:Omega rightarrow mathbb{R}$.
For
$$mathrm{div} Bigg (frac{nabla u}{sqrt{1+|nabla u|^2}}Bigg) =0$$
let $u_1,u_2$ be two solutions. I want to show that $v := u_1 - u_2$ is a solution of the form
$mathrm{div} (Anabla v)=0$ for a positive definite and symmetric matrix $A$.
The only advice given is to introduce the function
$$
F(X):=frac{X}{sqrt{1+|X|^2}}
$$
and use a green formula for
$$
F(b)-F(a)=int_0^1partial_tF(ta+(1-t)b),mathrm{d}t.
$$
Attempt:
When choosing $X=nabla u_{1,2}$ and using linearity of the divergence I get
$$
mathrm{div}left (F(nabla u_1)-F(nabla u_2)right )
= 0
= mathrm{div}int_0^1partial_tF(tnabla u_2+(1-t)nabla u_1),mathrm{d}t.
$$
Here I am stuck as I don't find a suiting Green formula.
real-analysis integration pde
real-analysis integration pde
edited Jan 12 at 14:09
Viktor Glombik
790527
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
edited Jan 12 at 14:09
Viktor Glombik
790527
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
edited Jan 12 at 14:09
Viktor Glombik
790527
edited Jan 12 at 14:09
Viktor Glombik
790527
edited Jan 12 at 14:09
Viktor Glombik
790527
790527
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
asked Jan 12 at 14:03
EpsilonDeltaEpsilonDelta
6611615
6611615
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Now you need to compute. First note that
$$
partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
$$
which in particular means that
$$
partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
$$
Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
$$
0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
$$
and we get the desired result by setting
$$
A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
$$
To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
$$
partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
$$
which we can rewrite as
$$
DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
$$
We can then plug in above to get more information about $A$.
EDIT: As requested, here is some more information about $A$.
It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
$$
DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
$$
where the last inequality follows from Cauchy-Schwarz. Next note that
$$
|t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
$$
which means that
$$
(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
$$
Then
$$
A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
$$
and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
$endgroup$
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070929%2ftransformation-not-reproducible%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Now you need to compute. First note that
$$
partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
$$
which in particular means that
$$
partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
$$
Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
$$
0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
$$
and we get the desired result by setting
$$
A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
$$
To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
$$
partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
$$
which we can rewrite as
$$
DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
$$
We can then plug in above to get more information about $A$.
EDIT: As requested, here is some more information about $A$.
It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
$$
DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
$$
where the last inequality follows from Cauchy-Schwarz. Next note that
$$
|t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
$$
which means that
$$
(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
$$
Then
$$
A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
$$
and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
$endgroup$
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
add a comment |
$begingroup$
Now you need to compute. First note that
$$
partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
$$
which in particular means that
$$
partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
$$
Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
$$
0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
$$
and we get the desired result by setting
$$
A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
$$
To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
$$
partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
$$
which we can rewrite as
$$
DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
$$
We can then plug in above to get more information about $A$.
EDIT: As requested, here is some more information about $A$.
It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
$$
DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
$$
where the last inequality follows from Cauchy-Schwarz. Next note that
$$
|t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
$$
which means that
$$
(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
$$
Then
$$
A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
$$
and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
$endgroup$
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
add a comment |
$begingroup$
Now you need to compute. First note that
$$
partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
$$
which in particular means that
$$
partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
$$
Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
$$
0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
$$
and we get the desired result by setting
$$
A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
$$
To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
$$
partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
$$
which we can rewrite as
$$
DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
$$
We can then plug in above to get more information about $A$.
EDIT: As requested, here is some more information about $A$.
It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
$$
DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
$$
where the last inequality follows from Cauchy-Schwarz. Next note that
$$
|t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
$$
which means that
$$
(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
$$
Then
$$
A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
$$
and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
$endgroup$
Now you need to compute. First note that
$$
partial_t F(t X + (1-t) Y) = DF(tX + (1-t)Y)(X-Y),
$$
which in particular means that
$$
partial_t F(t nabla u_2 + (1-t) nabla u_1) = DF(tnabla u_2 + (1-t)nabla u_1)(nabla u_2-nabla u_1).
$$
Returning to your formula, we note that $nabla u_2 - nabla u_1$ is independent of $t$, so
$$
0 = text{div} left( left[int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dtright] (nabla u_2-nabla u_1) right)
$$
and we get the desired result by setting
$$
A = int_0^1 DF(tnabla u_2 + (1-t)nabla u_1) dt.
$$
To get more specific information about $A$ (ellipticity, etc), we need to compute $DF$. First note
$$
partial_j F(X)_k = frac{ sqrt{1+ |X|^2} delta_{jk} - X_k X_j/ sqrt{1 + |X|^2} }{1+ |X|^2},
$$
which we can rewrite as
$$
DF(X) = frac{I}{(1+|X|^2)^{1/2}} - frac{X otimes X}{(1+ |X|^2)^{3/2}}.
$$
We can then plug in above to get more information about $A$.
EDIT: As requested, here is some more information about $A$.
It's obvious that $A$ is symmetric since $DF(cdot)$ is. Let $z in mathbb{R}^n$ and note that
$$
DF(X) z cdot z = frac{|z|^2}{(1+|X|^2)^{1/2}} - frac{|X cdot z|^2}{(1+ |X|^2)^{3/2}} = frac{|z|^2 + |z|^2|X|^2 - |zcdot X|^2 }{(1+ |X|^2)^{3/2}} ge frac{|z|^2}{(1+ |X|^2)^{3/2}},
$$
where the last inequality follows from Cauchy-Schwarz. Next note that
$$
|t nabla u_2 + (1-t) nabla u_1| le Vert nabla u_2Vert_{L^infty} + Vert nabla u_1Vert_{L^infty} =: C_0,
$$
which means that
$$
(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2} le (1+ C_0^2)^{3/2}.
$$
Then
$$
A z cdot z ge |z|^2 int_0^1 frac{dt}{(1+ |t nabla u_2 + (1-t) nabla u_1|^2)^{3/2}} ge frac{|z|^2}{(1+ C_0^2)^{3/2}},
$$
and we have that $A$ is elliptic with the ellipticity constant depending on $nabla u_i$ via the constant $C_0$.
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
edited Jan 12 at 17:15
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
answered Jan 12 at 15:40
GlitchGlitch
5,5881030
5,5881030
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
add a comment |
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
Very elegant! If I wanted to prove ellipticity then I would have to show that this matrix is positiv definite and symmetric. Symmetry follows immediately from commutativity of the tensor product for commutative rings. Do you have an idea for showing the pos. def. property?
$endgroup$
– EpsilonDelta
Jan 12 at 16:30
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
$begingroup$
@EpsilonDelta I'm glad you like it. See the edit for ellipticity.
$endgroup$
– Glitch
Jan 12 at 17:15
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070929%2ftransformation-not-reproducible%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
