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Finding image under phi

  1. Jul 9, 2008 #1
    Let [phi](u,v)=(u^2,v). Is phi one-to-one? If not, determine a domain on which phi is one-to-one. Find the image under phi of:

    - The rectangle R=[-1,1]X[-1,1]





    3. The attempt at a solution

    - I'm not sure at all how to determine whether phi is one-to-one or not, so if somebody can explain that, that would be of great help.

    I thought I knew how to find the image under phi (because I got the right answers the way I did it on the previous problem), but I'm not getting the right answers on this problem. Please help.
     
  2. jcsd
  3. Jul 9, 2008 #2

    Dick

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    That's a pretty weak attempt. Can't you think of two (u,v) points that map to the same point under (u,v)->(u^2,v)??
     
  4. Jul 9, 2008 #3
    - I guess I just don't get it or something. The book just doesn't explain mapping in a way I really understand at all.
     
  5. Jul 9, 2008 #4

    Dick

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    Ok. What are the values of phi(-1,0) and phi(1,0)? What does that say about the possibility of phi being one-to-one.
     
  6. Jul 10, 2008 #5
    - those values would give you (1,0) and (1,0), and I don't think that would be one-to-one because you can only have each value once. So it would only be one-to-one when u>=0 or u<=0.

    If that is the right way of looking at the one-to-one part. What exactly does the mapping part of the question mean?
     
  7. Jul 10, 2008 #6

    Dick

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    Yes, that's the right way of looking at the 1-1 part. If by 'mapping' part you mean the image part, the question is to describe the region that [-1,1]x[-1,1] maps to. Hint: u->u^2 maps [-1,1] to [0,1], doesn't it?
     
  8. Jul 10, 2008 #7
    - I'm sure this is going to sound like a dumb question, but how do you get the zero in the [0,1]. I knew that from looking at the answer in the back of the book, but I'm not completely sure on how you get to that. So I guess I'm just not completely sure on the process involved here. The first region is in the uv plane, right? So is the problem asking to find the image in the xy plane?
     
  9. Jul 10, 2008 #8

    Dick

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    I got the 0 because u->u^2 for u in [-1,1] has a max at 1 and a min at 0 and it's continuous. Draw the graph. So the range is [0,1]. I won't say it's a dumb question, but it's hard to figure out what you aren't getting. Plug a lot of numbers into (u,v)->(u^2,v) and draw arrows connecting them from the uv plane to the xy plane, if that's what you want to call the image. Keep doing that until you get some grasp of what's happening. Then step back and think about what's happening.
     
  10. Jul 10, 2008 #9
    - I'll try what you said and see if I can get a better grasp on it. I think I'm pretty close to understanding it though.
     
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