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Linear copying

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Let's say that [tex]f: \mathbb {R}^3 \rightarrow \mathbb {R}^2[/tex], [tex]g: \mathbb {R}^3 \rightarrow \mathbb {R}^2[/tex], and [tex]h: \mathbb {R}^2 \rightarrow \mathbb {R}^2[/tex] are linear copying defined with [tex]f(x,y,z)=(y,x+z)[/tex], [tex]g(x,y,z)=(2z,x-y)[/tex] and [tex]h(x,y)=(y,2x)[/tex]. Find the linear copying:

    h o f and h o g

    2. Relevant equations



    3. The attempt at a solution

    Sorry, but again, I don't know the principle of solving this task. My book is not good at all. Thanks from the start.
     
  2. jcsd
  3. Feb 29, 2008 #2

    CompuChip

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    What does [itex]h \circ f[/itex] mean?
    What is its domain and co-domain (e.g. what are A and B in [itex]h: A \to B[/itex])? How is it defined ([itex]h(\cdots)= \cdots[/itex] )?
     
  4. Feb 29, 2008 #3
    probably h, f, g are the linear copying
     
  5. Feb 29, 2008 #4

    CompuChip

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    I don't know what language you translated it from, but I think you mean "linear map" instead of "linear copying".
    Also I think you are a little confused as to what [itex]h \circ f[/itex] means. f is a function that takes something in R3 and gives you something in R2. Into h you put something in R2 and get something (else) in R2 back. Now what you can do, is use the output of f as input for h. So
    [tex]h \circ f: \mathbb{R}^3 \to \mathbb{R}^2[/tex]
    is a function defined by
    [tex](h \circ f)(x, y, z) = h(f(x, y, z))[/tex].
    So to calculate (h o f)(x, y, z) you first calculate (u, v) = f(x, y, z) and then h(u, v).
    Does that clarify ?
     
  6. Feb 29, 2008 #5
    Yes, sir, that's what it means. Sorry for my mistranslation.
     
  7. Feb 29, 2008 #6

    HallsofIvy

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    Okay, so do it! If h(u,v)= (v, 2u), and (u,v)= f(x,y,z)= (2z, x- y), what is [itex]h\circle f(x,y,z)[/itex]. What is [itex]f\circle g(u,v)[/itex]?
     
  8. Feb 29, 2008 #7
    How do you know that (x,y)=f(x,y,z)=(2z,x-y)? I truly don't know what is [itex]h \circ f(x,y,z)[/tex] and [itex]f \circ g(u,v)[/itex]
     
    Last edited: Feb 29, 2008
  9. Feb 29, 2008 #8

    HallsofIvy

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    Because that was what YOU told us:
    [itex]h\circ f(x,y,z)= h(f(x,y,z))= h(y, x+ z)[/itex]. If h is defined by h(a,b)= (b, 2a), what is h(y, x+z).
     
  10. Mar 1, 2008 #9
    Can you tell me just this, please, there is one line more, I guess. Thanks.
    Probably [itex]h(y, x+ z)=(b,2a)[/itex]
    But, what next?
     
    Last edited: Mar 1, 2008
  11. Mar 1, 2008 #10

    CompuChip

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    I think you are having some troubles with basic calculus concepts.
    Suppose we have a function f(x) = 2 x + 3.
    Then can you tell me what are f(3), f(a) and f(y + 1) ?
     
  12. Mar 1, 2008 #11
    f(3)=2*3+3
    f(a)=2a + 3
    f(y+1)=2(y+1)+3

    No, I don't have troubles with basic calculations.
     
  13. Mar 1, 2008 #12

    CompuChip

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    Well, this is exactly the same, but there are just two values to substitute. So
    h(y, x + z) is just (b, 2a) with a = y, b = x + z, i.e. h(y, x + z) = (x + z, 2y).
     
  14. Mar 1, 2008 #13
    But it should be h(y, x+z) and (y,2x), so y=y and 2x=x+z. Hmm...
     
  15. Mar 1, 2008 #14

    HallsofIvy

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    No, you should have learned long ago that the letters used as variables in defining functions are just "place holders"- you can replace them with anything you want. If f is defined by f(x, y)= 3x+ y then f(a, b)= 3a+ b, f(y, x)= 3y+ b, etc. [itex]h\circ f(x,y)= h(f(x,y,z))= h(y, x+z)= (x+ y, 2y) as I said.
     
  16. Mar 1, 2008 #15
    I know that, they are just 'place holders'. But can't understand the whole process.

    [tex]h \circ f[/tex] is what we need to find.

    [tex](h \circ f)(x,y,z)=h(f(x,y,z))=h(y,x+z)[/tex]

    Now if [tex]h(x,y)=(y,2x)[/tex] then [tex]h(y,x+z)=(x+z,2y)[/tex], right?

    I think I understand now. Now for [tex]h \circ g[/tex] is the 2nd think that we need to find.

    [tex](h \circ g)(x,y,z)=h(g(x,y,z))=h(2z,x-y)[/tex]

    Now if [tex]h(x,y)=(y,2x)[/tex] then [tex]h(2z,x-y)=(x-y,4z)[/tex], right?

    I think, I understand now.
     
  17. Mar 1, 2008 #16
    And if we get [tex]f \circ h[/tex]?

    [tex](f \circ h)(x,y)=f(h(x,y))=f(y,2x)[/tex]

    Now if [tex]f(x,y,z)=(y,x+z)[/tex] then [tex]f(y,2x)=??[/tex]

    In my book, it says that it is impossible, why?
     
  18. Mar 1, 2008 #17

    CompuChip

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    What could it be? [itex](y, 2x)[/itex] is an element of [itex]\mathbb{R}^2[/itex], whereas [itex]f[/itex] is a function on [itex]\mathbb{R}^3[/itex]. In other words, h produces a pair of numbers, whereas f expects a triple as input. So you cannot plug the output of h into f, in other words: you can't compose them in that order.
     
  19. Mar 1, 2008 #18
    Ok, and what for [tex]h \circ (f+g)[/tex] and [tex]h \circ f + h \circ g[/tex].
    First, are they same?
    Other thing, how will I solve it?
     
  20. Mar 1, 2008 #19

    CompuChip

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    Can you write down the definitions of (f + g) and h o (f + g) and (h o f) + (h o g)?
    In particular, write down from what space into what space they map, like I did in my last post. Then you can already see if there will be a problem in the composition (like: if the space (f + g) maps to is different than the space h maps from, you cannot compose them, as you have just seen).
     
    Last edited: Mar 1, 2008
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