14
votes
A polynomial with nowhere surjective derivative
As a disclaimer, I will follow an "algebraic geometer" approach. Thus, all the morphisms will essentially be polynomials (or rational functions).
To be short, my answer is "yes". The exaplanation is ...
- 4,554
13
votes
Accepted
If $f���C^1$ and $\{∇f=0\}$ has Lebesgue measure $0$, then $\{f∈B\}$ has Lebesgue measure $0$ for all Borel measurable $B⊆ℝ$ with Lebesgue measure $0$
Non-singular maps. A map $f\,:\,\mathbb{R}^N\rightarrow\mathbb{R}^M$ whose inverse image preserves null-sets, -- i.e., $\mu(f^{-1}(B))=0$ for any null-set $B$, -- is often referred to as a non-...
- 2,160
12
votes
Accepted
A polynomial with nowhere surjective derivative
Here is an elementary proof.
First, some notation and terminology. Capital letters such as $F,G$ denote polynomials in $\mathbb{R}[x,y]$. Also, $f$ (and slight variants of it, such as $\tilde{f},f_1,...
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7
votes
$\frac{\partial F}{\partial y}\neq0\implies$ continuous contour line? (Implicit Function Theorem)
Yes, it's correct. Here's a sketch of the argument. Note first of all that by the implicit function theorem, the portion of a level curve of $F$ in the given region must be a one-dimensional manifold ...
- 128k
5
votes
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Multivariable implicit function theorem proof
In the context of implicit function theorem especially, the Leibniz notation for partial derivatives is absolutely horrible and confusing at best when first learning. One needs to be very careful ...
- 68.3k
5
votes
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(Frechet) Differentiability of Implicit function in Banach spaces
Most of the treatments of the inverse function theorem or the implicit function theorem are based on finding fixed points of a contraction in Banach spaces. Smoothness of the implicit solution $y\...
- 46.8k
4
votes
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Understanding implicit function theorem
I'll make two remarks:
It's true that that $F(x,g(x))=0$ for all $x\in [-1,1]$, but the idea of the implicit function theorem is that it gives sufficient conditions for a point in the level set of $F$ ...
- 453
4
votes
Coordinate change to make function linear on a neighborhood
First of all, I hope you know that both the inverse and implicit functions are equivalent. Having said this, I sometimes find it easier to work with one over the other, and in this case, I find it ...
- 68.3k
4
votes
Hessian matrix of the function defined with Implicit function theorem
For everything which follows, my go-to reference is Loomis and Sternberg's book on Advanced Calculus, and Henri Cartan's book on Differential Calculus. First let me establish some basic notational ...
- 68.3k
4
votes
Accepted
Implicit function Theorem - I have an example which I think is wrong
You have the system of equations of
$$e^{-x} - y = 0 \iff e^{-x} = y \tag{1}\label{eq1A}$$
$$e^{3x} + 10 - y = 0 \tag{2}\label{eq2A}$$
This does have a solution for $x$ and $y$. To see this, ...
- 53.2k
4
votes
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Implicit equation for the image of immersion $(\cosh(t),\sinh(t))$
Observe that $x^2-y^2=1$ so your image is contained in the hyperbola. Also, since $\cosh t>0$ your image is the right branch of the above equilateral hyperbola.
- 18k
4
votes
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Connectedness of a set defined analytically
Write $S_\pm = \{ (x, y, z)\in S: \pm z \ge 0\}$. Then $S = S_+ \cup S_-$ and $S_+ \cap S_- = \{ (x, y, 0): x^2 + y^2 = 7\}$. Thus it suffices to show that each $S_\pm$ is connected. Since $S_- = P S_+...
- 17.2k
4
votes
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Question about showing the equation defines an implicit function and give the Taylor expansion
You are asked to show that $\phi$ exists in the first place - the implicit function theorem comes to your aid here. It is sufficient to show that there exists $(a,b)$ such that $f(a,b)=0$, and that $...
- 48.2k
4
votes
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Coordinate system $(\phi, U, x_1, x_2)$ around $p \in S^1$ such that $S^1 \cap U = \{(x_1, x_2): x_2 = 0\}$.
We consider the continously differentiable function
$$\Phi : \mathbb R^2 \to \mathbb R, f(x_1,x_2) = x_1^2+x_2^2 $$
with $S^1 = \Phi^{-1}(1)$.
The Jacobian matrix of $\Phi$ at $ p = (p_1,p_2)$ is
$$J\...
- 92.3k
4
votes
Accepted
Implicit function theorem: from local to global
Yes this argument works quite well provided that you insert one minor edit: you should state explicitly that the open set $U$ is an open interval (a connected open subset of the real line.) Your ...
- 6,070
4
votes
Implicit function theorem: when a level set intersects itself
At the intersection point there are two independent directions a particle on the curve could go. If the derivative of $f$ at this point was not zero there could only be one such direction because $\...
- 7,056
3
votes
Is it possible to find an explicit form of the solution to $y'=\frac{1-x+y}{x-y}$
It's possible to express $y$ as an explicit function of $x$ using the Lambert W function:
$$
\frac{1}{2}z+\frac{1}{4}\ln(2z-1)=x+c \implies (2z-1)e^{2z}=e^{4x+4c}
$$
$$
\implies (2z-1)e^{2z-1}=e^{4x+...
- 16.5k
3
votes
Implicit function theorem exercise with higher derivatives
You can try these formulas
\begin{cases} \begin{split} \dfrac{\partial^2z}{\,\partial\,\!x^2}&=\dfrac{1}{{F_z}^3} \begin{vmatrix} F_{xx}&F_{xz}&F_{x}\\ F_{zx}&F_{zz}&F_{z}\\ F_{x}&...
- 1,259
3
votes
Accepted
Higher order partial derivatives of implicit function?
By the Inverse Function Theorem, the equations $x=u+v^2$ and $y=u^2-v^3$ locally define $u$ and $v$ as $C^1$ functions of $x$ and $y$ near the point in question, since $x(u,v)=u+v^2$ and $y(u,v)=u^2-v^...
- 55.3k
3
votes
Global Implicit Function Theorem
The way you've set things up, it looks like you've set $n = m = 1$. In that case, the additional condition that there exists a $c > 0$ such that $\frac{\delta f}{\delta y}(x, y) \ge c$ for every $(...
- 31
3
votes
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Inverse function theorem regarding solutions to a cubic equation
We know that if $x_1, x_2, x_3 \in \mathbb{R}$ are the three roots of $X^3 - y_1 X^2 + y_2 X - y_3 = 0$, then $y_1 = x_1 + x_2 + x_3$, $y_2 = x_1 x_2 + x_1 x_3 + x_2 x_3$, and $y_3 = x_1 x_2 x_3$. ...
- 23.2k
3
votes
Accepted
Application of Implicit Function Theorem: Problem 2-40 from Spivak's Calculus on Manifolds
What you did looks good to me. On my side I would have started from the beginning using matrix/vector notations in order to have a more synthetic view of the question. Which means that you can define ...
- 72.1k
3
votes
Accepted
Use the implicit function theorem to prove that $f=f^{-1}$.
Note that by the chain rule, $Df^2 \equiv I_n$. Thus, by Lagrange's mean value theorem,
$$
f^2(b) - f^2(a) = Df^2(b-a) = b-a \quad (\forall a,b \in \mathbb{R}^n)
$$
In particular, since $f^2(x_0) = ...
- 17.7k
3
votes
Accepted
When does $F(x,y) = 0$ imply an implicit function?
To apply the implicit function theorem, you have to fix a point $(a, b) \in \mathbb{R^2}$, such that $f(a, b) = 0$. But there is no such point in $\mathbb{R}^2$: Assume the validity of (1) for some $x,...
- 777
3
votes
Implicit Function Theorem Application.
For $y=y(x)$ to exist it is necessary for the Jacobian to be non-singular. In this case, the Jacobian is just a $1\times 1$ matrix. It is just a number with the value
$$
\frac{d}{dy}(x^3 + xy^2 + y^3)...
- 7,726
3
votes
If $f∈C^1$ and $\{∇f=0\}$ has Lebesgue measure $0$, then $\{f∈B\}$ has Lebesgue measure $0$ for all Borel measurable $B⊆ℝ$ with Lebesgue measure $0$
I don't know about the implicit function theorem, but you can a one related theorem, the Local Submersion Theorem. With your notations, locally around $a$, $f$ looks like a projections onto the first ...
- 3,343
3
votes
Accepted
How will fixed point change if the mapping changes?
If $c$ is a small perturbation, then the fixed point will be a small perturbation of the original fixed point, whose size is in fact quite explicitly controlled by $c$. This is a feature known as ...
- 19.6k
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