Finding inverse of non-linear transformation

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Homework Help Overview

The discussion revolves around finding the inverse of a nonlinear transformation from R² to R², defined by the equations u=3y and v=3x^7-6y. Participants are exploring the implications and methods for determining the inverse of this transformation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Some participants attempt to express y in terms of u and substitute it into the equation for v to isolate x. Others question the nature of the inverse and seek clarification on the reasoning behind the transformations.

Discussion Status

The discussion includes attempts to manipulate the equations to find x and y, with some participants expressing uncertainty about the process. There is an exploration of the concept of an inverse transformation, with one participant reflecting on the definition and implications of the inverse in this context.

Contextual Notes

Participants note that they have not encountered similar problems in class, which may contribute to their uncertainty. There is also a focus on understanding the definitions and properties of inverse transformations.

snoggerT
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Find the inverse of the (nonlinear) transformation from R^2 to R^2 given by

u=3y
v=3x^7-6y

x=?
y=?




The Attempt at a Solution



- I'm really not sure what to do on this problem. We haven't seen any problems even similar to it in class, so I'm looking for help on it.
 
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snoggerT said:
Find the inverse of the (nonlinear) transformation from R^2 to R^2 given by

u=3y
v=3x^7-6y

So y= u/3. Put that into the second equation: v= 3x^7- 6(u/3)= 3x^7- 2u.
Solve that for x.

x=?
y=?

The Attempt at a Solution



- I'm really not sure what to do on this problem. We haven't seen any problems even similar to it in class, so I'm looking for help on it.
 
well, that was much easier than I thought it would be. Can you explain to me why that is the inverse?
 
What do you think an inverse is? You had u= 3y, v= 3x^7- 6y and you said the answer must be in the form x=, y= . I reduce the two equation to that form, solving for x and y.

Perhaps more specifically, if you start with (x, y) and apply the original tranform, you get (3y, 3x^7 - 6y). Now what happens if you apply the tranformation x= ((v+2u)/3)^(1/7, y= u/3? Since u= 3y, the second gives y= (3y)/3= y immediately. Since u= 3y and v= 3x^7- 6y, the x= ((3x^7- 6y+ 2(3y))/3)^(1/7)= ((3x^7/3)^(1/7)= (x^7)^(1/7)= x.
That's what an inverse is supposed to do.
 

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