Can You Apply the Implicit Function Theorem Correctly?

[Lambda96]

Homework Statement

see post

Relevant Equations

Implicit function theorem

Hi,

I'm not sure if I've understood the task here correctly

For the Implicit function theorem, $F(x,y)=0$ must hold for all $(x,y)$ for which $f(x,y)=f(x_0,y_0)$ it follows that $f(x,y)-f(x_0,y_0)=0$ so I can apply the Implicit function theorem for these $(x,y)$.

Then I can write $x_0y_0e^{-x_0-y_0}=xye^{-x-y}$ then I would have to solve the equation for $y$ and get the relation y=f(x).


This is exactly where I have my problems, unfortunately, I don't know how to solve the equation for $y$.

 

[fresh_42]

Try to apply the theorem literally. We start with the function $F=f(x,y)-f(x_0,y_0)\, : \,\mathbb{R}^2\rightarrow \mathbb{R}.$ Thus
$$
\dfrac{\partial F}{\partial x} =\dfrac{\partial f}{\partial x} = (1-x)ye^{-x-y}\, , \,\dfrac{\partial F}{\partial y}=(1-y)xe^{-x-y}
$$
Now $F(x_0,y_0)=0$ and $\dfrac{\partial F}{\partial y}(x_0,y_0)=(1-y_0)x_0e^{-x_0-y_0}\neq 0$ if $x_0\neq 0$ and $y_0\neq 1.$ What does the theorem say in this case? And what does it say about points on the $x-$axis, except the origin?

 

[WWGD]

My understanding was that you need to have an invertible Jacobian at a point for F to have a local inverse for F(x1,x2,..,xn)=0. Not aware of this version.

 

[docnet]

My understanding was that you need to have an invertible Jacobian at a point for F to have a local inverse for F(x1,x2,..,xn)=0. Not aware of this version.

I believe that the conditions described in @fresh_42's post belong to the simplified 2-D case of the multivariate theorem. Wiki link

 

[Lambda96]

Thank you fresh_42, WGD and docnet for your help


I think I have understood it now. In order for the Implicit function theorem to be applied, two things must be true

$$1) F(x_0, y_0)=0$$
$$2) \frac{\partial F(x_0, y_0)}{\partial y} \neq 0$$


The first condition already applies, since for all elements (x,y) of the set $X_{x_0,y_0}$ $f(x,y)=f(x_0,y_0) \rightarrow f(x,y)-f(x_0,y_0)=0$ applies according to the task


As you have already written, fresh_42 the following $\frac{\partial F(x_0,y_0)}{\partial y}=(1-y_0)x_0e^{-x_0-y_0}$ applies except for the cases $x_0=0$ and $y_0=1$.


Since the two operations are fulfilled except for the points $(0,0)$ and $(1,1)$, there exists a function $g$ for which all points in the neighbourhood of $(x_0,y_0)$ can be written as follows $y=g(x)$


So the Implicit function theorem tells me that a function $g$ exists, but I can't get the function using the Implicit function theorem.

 

[fresh_42]

Thank you fresh_42, WGD and docnet for your help


I think I have understood it now. In order for the Implicit function theorem to be applied, two things must be true

$$1) F(x_0, y_0)=0$$
$$2) \frac{\partial F(x_0, y_0)}{\partial y} \neq 0$$


The first condition already applies, since for all elements (x,y) of the set $X_{x_0,y_0}$ $f(x,y)=f(x_0,y_0) \rightarrow f(x,y)-f(x_0,y_0)=0$ applies according to the task


As you have already written, fresh_42 the following $\frac{\partial F(x_0,y_0)}{\partial y}=(1-y_0)x_0e^{-x_0-y_0}$ applies except for the cases $x_0=0$ and $y_0=1$.


Since the two operations are fulfilled except for the points $(0,0)$ and $(1,1)$, there exists a function $g$ for which all points in the neighbourhood of $(x_0,y_0)$ can be written as follows $y=g(x)$


So the Implicit function theorem tells me that a function $g$ exists, but I can't get the function using the Implicit function theorem.

The Lambert W function $f(w)=we^w$ is related. It is not really a function, only locally a.e. But you cannot write it as $w=g(f(w)).$ Here we have $ye^{-y}=x_0y_0e^{-x_0-y_0}\cdot \dfrac{e^x}{x}=:g(x)$ and like the Lambert W function, there is no explicit formula $y=\ldots$

 

[docnet]

That is a good attempt! But I the task is not complete or correct as you wrote it yet, and here is a hint: $2)$ should be a statement involving $\frac{\partial F(x_0, y_0)}{\partial y}\neq 0$ and involving $\frac{\partial F(x_0, y_0)}{\partial x}\neq 0$ linked by the proper conjunction. Think through why that is necessary to complete the task.

If I know how nitpicky TAs can be, they would likely rip it apart (I'm sorry to say).