What are the applications of inverses of vector functions?

[askmathquestions]

As an example, consider a vector-valued function of the form $f(x,y) = (g_1(x,y),g_2(x,y))$.
I typed up one example on wolfram to see if this could be visualized

https://www.wolframalpha.com/input?i=plot+f(x,y)+=+(x+y,xy)

which was inspired by this question

https://math.stackexchange.com/questions/350963/inverse-of-a-vector-valued-function

and it returned a result that looks a lot like the kinds of trajectories that arise in systems of ordinary differential equations, though instead of lines, they're "vectors".

What are the applications of vector-valued functions of this form and their inverses functions to physics (or general science)? What is the relationship of such vectors to differential equations?

If I did the math correctly, the inverse function of this, assuming $(u,v) = f(x,y)$ then we solve for $x$ and $y$ in terms of $u$ and $v$ to obtain
$y = u/2 \pm \frac{1}{2} \sqrt{u^2 - 4v}$ and $x = u - u/2 \pm \frac{1}{2} \sqrt{u^2 - 4v}$ though I'm confused as to what the final vector is now.

What is the correct inverse vector function, and why might someone take interest in it?

 

[jedishrfu]

I'm sure someone can conjure up an application that uses inverse vector functions, but there doesn't seem to be anything online about it.

Do you have some use for them?

 

[FactChecker]

Your definition of a vector function would also define an inverse vector function. Either one is from $\mathbb{R}^2$ to $\mathbb{R}^2$, and it is hard to say which is the original function and which is the inverse.
Consider aerodynamics and airflow around a flying airplane. At every point of $\mathbb{R}^3$ there is a direction and magnitude of airflow in $\mathbb{R}^3$. In designing an airplane, it is important to model the aerodynamics of the airplane. One problem to study is that a certain orientation of a flying airplane leads to a local flow at the sensor for Angle Of Attack, AOA. There is another problem that involves the inverse function. Given a reading from the AOA sensor, one would like to know what that implies about the orientation of the airplane relative to the overall wind axis.

 

[askmathquestions]

I'm still confused as to what I actually state the inverse vector is. I can see how there might be applications though. I've solved for $x$ and $y$ in terms of $u$ and $v$, now what?

 

[FactChecker]

I'm still confused as to what I actually state the inverse vector is. I can see how there might be applications though. I've solved for $x$ and $y$ in terms of $u$ and $v$, now what?

The vector has the coordinates (x,y). Why might someone take interest in it depends on the application that the original equations came from (which is not given in the problem statement). It is often the case that there is a good theory and model of how the variables x and y lead to u and v, but in practice, only u and v are known and measurable. Then it is often necessary to determine the associated x and y.

 

[askmathquestions]

The vector has the coordinates (x,y). Why might someone take interest in it depends on the application that the original equations came from (which is not given in the problem statement). It is often the case that there is a good theory and model of how the variables x and y lead to u and v, but in practice, only u and v are known and measurable. Then it is often necessary to determine the associated x and y.

Okay, so this is very context dependent then. I'm still confused though as to what the final inverse vector is. I solved for $x$ and $y$ in terms of $u$ and $v$, now what? The inverse vector is...?

 

[FactChecker]

Okay, so this is very context dependent then. I'm still confused though as to what the final inverse vector is. I solved for $x$ and $y$ in terms of $u$ and $v$, now what? The inverse vector is...?

(x,y)

 

[askmathquestions]

(x,y)

So, $(x,y)$ is the inverse vector field of $(x+y,xy)$?

 

[FactChecker]

I probably should have said $(x,y)=(u - u/2 \pm \frac{1}{2} \sqrt{u^2 - 4v}, u/2 \pm \frac{1}{2} \sqrt{u^2 - 4v})$ (assuming you did these calculations correctly), where the choice of the signs are determined by the problem being solved and by requirements of continuity, etc.

 

[askmathquestions]

I probably should have said $(x,y)=(u - u/2 \pm \frac{1}{2} \sqrt{u^2 - 4v}, u/2 \pm \frac{1}{2} \sqrt{u^2 - 4v})$ (assuming you did these calculations correctly), where the choice of the signs are determined by the problem being solved and by requirements of continuity, etc.

Let's say we want to derive $f(x) = y$ for the inverse function of $f^{-1}(x).$ Well, we say the inverse of $f^{-1}(x)$ is of $x$, not necessarily $f^{-1}(y)$. So I'm still confused as to the $x,y$ coordinates. Should $u = x$ and $v = y$? How do you know?

 

[FactChecker]

Let's say we want to derive $f(x) = y$ for the inverse function of $f^{-1}(x).$ Well, we say the inverse of $f^{-1}(x)$ is of $x$, not necessarily $f^{-1}(y)$.

No. You might see something like that, but it is careless. That would be using the variable $x$ for two completely different things. It is more precise to say that if $y=f(x)$, then $x=f^{-1}(y)$

So I'm still confused as to the $x,y$ coordinates. Should $u = x$ and $v = y$? How do you know?

Back to the original problem: $(x,y)$ and $(u,v)$are vectors in the domain and range, respectively, of the function f. If $(u,v) = f( (x,y) )$, then $(x,y) = f^{-1}( (u,v) )$.

 

[askmathquestions]

No. You might see something like that, but it is careless. That would be using the variable $x$ for two completely different things. It is more precise to say that if $y=f(x)$, then $x=f^{-1}(y)$

Back to the original problem: $(x,y)$ and $(u,v)$are vectors in the domain and range, respectively, of the function f. If $(u,v) = f( (x,y) )$, then $(x,y) = f^{-1}( (u,v) )$.

What I proposed is highly conventional and extends to non-function objects (operators), so I'm confused by what you're saying and still don't see a concise answer.

Consider an invertible linear transform $T$, like say a square matrix. Then we may calculate $T^{-1}\cdot T[x] = I x.$ Notice how $T^{-1}$ is irrespective of the variable itself, yet still cancels out the original transform $T$.

Now apply this to function composition: $f(f^{-1}(x)) = x.$ Observe here that both $f$ and $f^{-1}$ are a function in a single variable. It's not a function of $y$, $f^{-1}$ is a function of $x$. If we can calculate $f(x)$ over some domain of real numbers and its output is another set of real numbers, and it is invertible, we can also calculate $f^{-1}(x)$ over some domain of real numbers.

Apply this principle to the $(u,v)$ coordinate system, and what do we get? My guess is we replace $u$ with $x$ and $v$ with $y$ to calculate the action of the inverse vector field in $(x,y)$ coordinates.

 

[FactChecker]

Now apply this to function composition: $f(f^{-1}(x)) = x.$ Observe here that both $f$ and $f^{-1}$ are a function in a single variable.

No. In $f(f^{-1}(x))$, $f^{-1}(x)$ is a function of $x$ in the range of $f$ and $f$ is a function of $f^{-1}(x)$ in the domain of $f$ which is not necessarily even in the same space as $x$.
Anyway, that is my two cents, take it or leave it.

 

[askmathquestions]

No. In $f(f^{-1}(x))$, $f^{-1}(x)$ is a function of $x$ in the range of $f$ and $f$ is a function of $f^{-1}(x)$ in the domain of $f$ which is not necessarily even in the same space as $x$.
Anyway, that is my two cents, take it or leave it.

$f(f^{-1}(x)) = x$, is that correct or incorrect?

 

[FactChecker]

$f(f^{-1}(x)) = x$, is that correct or incorrect?

That is correct for $x$ in the range of $f$ if $f$ is invertible. But it is wrong to say that $f^{-1}(x))$ and $x$ are the same. They may not even be in the same space. $x$ is in the range of $f$ and $f^{-1}(x))$ is in the domain of $f$.

 

[askmathquestions]

That is correct for $x$ in the range of $f$ if $f$ is invertible. But it is wrong to say that $f^{-1}(x))$ and $x$ are the same. They may not even be in the same space. $x$ is in the range of $f$ and $f^{-1}(x))$ is in the domain of $f$.

I don't think there's any implication of $x = f^{-1}(x)$ unless $x$ is identically the identity function. Both $f$ and $f^{-1}$ may have overlapping domains.

 

[FactChecker]

I don't think there's any implication of $x = f^{-1}(x)$ unless $x$ is identically the identity function. Both $f$ and $f^{-1}$ may have overlapping domains.

Ok. I must have misinterpreted something.

 

[PeterDonis]

$f(f^{-1}(x)) = x$, is that correct or incorrect?

It depends. This would normally be written $f^{-1} ( f(x) ) = x$. That's because we cannot assume that the domains of $f$ and $f^{-1}$ are the same, because we cannot assume that the domain and range of $f$ are the same. The domain of $f^{-1}$ is the range of $f$.

The way I have written the formula, it is correct even if the range of $f$ (and hence the domain of $f^{-1}$) is different from the domain of $f$. The way you have written it, it is only correct if the domain and range of $f$ are the same.