Inverse mapping theorem & local inverses

In summary, for each of the following transformations (u,v) = f(x,y), (i) compute det Df, (ii) find formulas for the local inverses of f when they exist. However, they did not show any of the steps, nor do they show me how they arrive at the answers. Can someone please explain this part?
  • #1
kingwinner
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[Related concepts: Inverse mapping theorem, transformations and coordinate systems]

1) For each of the following transformations (u,v) = f(x,y), (i) compute det Df, (ii) find formulas for the local inverses of f when they exist.
a) u=x^2, v=y/x
b) u=(e^x) cos y, v=(e^x) sin y



I got part (i) easily, but I don't understand part (iii) at all.

Here are the answers from the solutions manual:
For 1a)(iii), the answer is f^-1 (u,v) = (sqrt u, v (sqrt u) )
For 1b)(iii), the answer is
x= ln(u^2 + v^2) / 2
y is given up to multiples of 2pi by arctan(v/u) when u>0, pi/2 - arctan(u/v) when v>0, pi + arctan(v/u) when u<0, 3pi/2 - arctan(u/v) when v<0

But they didn't show any of the steps, nor do they show me how they arrive at the answers. Can someone please explain this part? How exactly can I find the formula for the local inverse? I am terribly confused...

Thanks a million!
 
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  • #2
kingwinner said:
But they didn't show any of the steps, nor do they show me how they arrive at the answers. Can someone please explain this part? How exactly can I find the formula for the local inverse? I am terribly confused...
You have two equations. Solve for x and y.
 
  • #3
The first one should be fairly easy. In the second one what you need to do is eliminate x so that you can solve for y in terms of u and v only and then eliminate y so that you can solve for x in terms of u and v only.

If you're stuck on solving for x and y, then just consider u^2 + v^2 and u/v.
 
  • #4
The crucial point is what Hurkyl said: solve u= x2, v= y/x for x and y in terms of u and v. Try it. If you have difficulty get back to us.
 
  • #5
1a)(iii) Solving for x and y, I get x=+/-sqrt u, y=+/-v (sqrt u), but the answer says that f^-1 (u,v) = (sqrt u, v (sqrt u) ), they only take the positive signs, this is what I don't understand...

In the second one what you need to do is eliminate x so that you can solve for y in terms of u and v only and then eliminate y so that you can solve for x in terms of u and v only.
How can I do so?

Thanks!
 
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  • #6
kingwinner said:
1a)(iii) Solving for x and y, I get x=+/-u, y=+/-vu, but the answer says that f^-1 (u,v) = (sqrt u, v (sqrt u) ), they only take the positive signs, this is what I don't understand...
A function does not have two values! Since you say you got part i easily, I presume you showed that there is a local inverse as long as x is not 0. You can guarantee a local inverse by restricting x to be positive or negative. If you restrict x to be positive, the "local" inverse function is [itex]x= \sqrt{u}[/itex], [itex]y= v\sqrt{u}[/itex]. If you restrict x to be negative, the "local" inverse function is [itex]x= -\sqrt{u}[/itex], [itex]y= -v\sqrt{u}[/itex]. Each part has a unique inverse function- you can't write them together.

How can I do so?

Thanks!
 
  • #7
HallsofIvy said:
A function does not have two values! Since you say you got part i easily, I presume you showed that there is a local inverse as long as x is not 0. You can guarantee a local inverse by restricting x to be positive or negative. If you restrict x to be positive, the "local" inverse function is [itex]x= \sqrt{u}[/itex], [itex]y= v\sqrt{u}[/itex]. If you restrict x to be negative, the "local" inverse function is [itex]x= -\sqrt{u}[/itex], [itex]y= -v\sqrt{u}[/itex]. Each part has a unique inverse function- you can't write them together.
Yes, I showed that there is a local inverse as long as x is not 0


So now there are two answers to (iii) [itex]x= \sqrt{u}[/itex], [itex]y= v\sqrt{u}[/itex] and [itex]x= -\sqrt{u}[/itex], [itex]y= -v\sqrt{u}[/itex].
Now how can I decide which one to take? If I just take one of them, would it still be a complete answer? Which one is the correct answer that the question is actually looking for?
 
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  • #8
kingwinner said:
Yes, I showed that there is a local inverse as long as x is not 0


So now there are two answers to (iii) [itex]x= \sqrt{u}[/itex], [itex]y= v\sqrt{u}[/itex] and [itex]x= -\sqrt{u}[/itex], [itex]y= -v\sqrt{u}[/itex].
Now how can I decide which one to take? If I just take one of them, would it still be a complete answer? Which one is the correct answer that the question is actually looking for?

Can someone please explain this part? I really want to understand it...Thanks!
 

FAQ: Inverse mapping theorem & local inverses

What is the Inverse Mapping Theorem?

The Inverse Mapping Theorem is a mathematical theorem that states a sufficient condition for a function to have a local inverse. It states that if a function is continuous, differentiable, and has a non-zero derivative at a point, then it has a local inverse at that point.

What is the significance of the Inverse Mapping Theorem?

The Inverse Mapping Theorem is significant because it allows us to find local inverses of functions that satisfy its conditions. This can be useful in many applications, such as optimization problems and solving systems of equations.

How is the Inverse Mapping Theorem related to the Implicit Function Theorem?

The Inverse Mapping Theorem is a special case of the Implicit Function Theorem. The Inverse Mapping Theorem applies to functions that map from one-dimensional spaces to multi-dimensional spaces, while the Implicit Function Theorem applies to functions that map from multi-dimensional spaces to multi-dimensional spaces.

Can the Inverse Mapping Theorem be applied to functions with more than one variable?

Yes, the Inverse Mapping Theorem can be applied to functions with any number of variables. However, it only guarantees the existence of a local inverse at a specific point, not for the entire function.

What is the difference between a global and a local inverse?

A global inverse is a function that maps every element in the range of the original function back to its corresponding element in the domain. A local inverse, on the other hand, is only defined for a certain subset of the range of the original function. The Inverse Mapping Theorem only guarantees the existence of a local inverse, not a global one.

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