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Subspace of even and odd functions

  1. Sep 24, 2009 #1
    Hey guys. I came upon this problem in the professor's recommended problems, and I have no idea how to solve it. After spending an hour on it, I've got nothing. Any suggestions would be much appreciated!

    A function f : R -> R is called even just in case f(-r) = f(r) for every
    real number r. f : R - > R is called odd just in case f(-r) = -f(r) for
    every real number r. Let V be the real vector space of all functions from
    R to R, U be the set of even functions, and W be the set of odd functions.
    (a) Show that U and W are subspaces of V .
    (b) Show that U + W = V and U [tex]\bigcap[/tex] W = {0}, where here 0 means the
    constant function 0. (In such a situation, as will be pointed out in
    class, we will say that V is the direct sum of U and W and write
    V = U [tex]\otimes[/tex]W.)
     
  2. jcsd
  3. Sep 24, 2009 #2
    Ok, for (a) let's work in U. if [tex]f, g \in U[/tex] is [tex] cf \in U[/tex] where [tex] c \in R[/tex]? is [tex] f + g \in U[/tex]?

    To prove the above, let x be an arbitrary real number and see how cf(-x) and f(-x) + g(-x) behave with respect to being even.

    If you want to be ambitious, you can just try to prove that cf(-x) + dg(-x) are even for c and d both real.
     
    Last edited: Sep 24, 2009
  4. Sep 24, 2009 #3
    Sorry, I messed up my earlier post in that it wasn't as helpful as I meant it to be! I'll edit it.
     
  5. Sep 24, 2009 #4

    Thanks for the rapid reply. I mean that's the definition of a subspace, so it just seemed to simple to me to say "f(-x) + g(-x) = f(x) + g(x) therefore U is a subspace." Is that really all I'd need to say (along with being closed under scalar multiplication?

    And any idea about how to prove part b?

    Regards,

    Hallingrad
     
  6. Sep 24, 2009 #5
    well if you wanted to dress it up some, you could do something like (f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x) but I just wanted to help you get the basic idea. Normally, proving something is a sub-space is pretty easy ;)

    As for part b, suppose that you are given a function f.

    What can you say about the function g(x) = f(x) - f(-x)?
     
  7. Sep 24, 2009 #6
    That depends, is f(x) even or odd? If even, then g(x) = 0, if odd, then g(x) = 2f(x). What's the next logical step that you're hinting at?
     
  8. Sep 24, 2009 #7
    Sorry, f is just a general function on R. What happens when you plug -x in to g?
     
  9. Sep 24, 2009 #8
    It gives the sum of both even and odd functions, and hence all of V? I sense that's where you're leading me, but I still don't fully see it.
     
  10. Sep 24, 2009 #9
    Ok, is g an even function or an odd function?
     
  11. Sep 24, 2009 #10
    It's neither since it's composed of the sum of both an even and odd function?
     
  12. Sep 24, 2009 #11
    I think we got confused. f(x) is neither even nor odd because we assumed that it was a general function on R, right?

    So then f(-x) is neither even nor odd either because it's just f(x) reflected about the y axis when you graph it, correct?

    Substitute -x for x in the definition of g(x) and see what happens.
     
  13. Sep 24, 2009 #12
    g(-x) = f(-x) - f(x), right? So g(x) + g(-x) = 0; therefore, g encompases all of the vector space? Couldn't we just say since even functions map all of R to positive R's, and odd functions map all of R to negative R's, the sum of both functions will allow us to produce any function?

    I still don't fully get it :(. I feel like I'm missing something about the entire question, and that's preventing me from making the next step. Thanks for your continual help though :)
     
  14. Sep 24, 2009 #13
    What do you mean g encompasses all the vector space? g is just one vector in the space!

    you have g(x) = f(x) - f(-x) and g(-x) = f(-x) - f(x).

    Now stop thinking and answer my question! Is g even or odd? ;)

    Stick with it. This stuff is fun.
     
  15. Sep 24, 2009 #14

    Ah, right, it's an odd function! While h(x) = f(-x) is an even function, correct? so h(x) + g(x) = f(x)... So does that mean that any even function added to any odd function can give any function in the space? Or did I skip a step?
     
  16. Sep 24, 2009 #15
    Correct, it's an odd function. Can you prove that h(x) = f(-x) is even? (hint: no because it's not true!)

    Try to cook up an even function from f though. It's pretty similar to g. think about what makes g an odd function and then do the opposite of that. Just as a hint, i'll tell you that when you find it, h + g does not equal f. You have to take a linear combination of them.

    Then we can get on with the second part of part (b)
     
  17. Sep 24, 2009 #16
    okay, so then h(x) = f(x) + f(-x).

    In order to map all the functions in V, we need to be able to go from any x to any y, i.e. f(x) - f(x) + f(-x) - f(-x). Is this the right way of thinking about it? What I'm not sure about is whether functions can be manipulated in this way? i.e. would f(x) - f(x) just cancel out, or can I use it the way I intended?
     
  18. Sep 25, 2009 #17
    Good job finding h!

    No, you don't need to be able to go to any y from any x! Some functions won't do that. think of [tex]\frac{x^{2}}{x^{2} + 1}[/tex] for instance. x and y really have nothing to do with this. See below.

    In response to your question about f(x) being canceled by -f(x), you can do that. Think of sin(x) and -sin(x) for instance. But I don't see what you are getting at with that sum. Remember, we are dealing with vectors here. That's the name of the game. We are only dealing with x because these vectors happen to be functions of x but all the rules for vectors apply. If you can do it with a vector, you can do it with a function!

    So what is h + g? You can probably think of two scalars c and d (not necessarily different) so that cg + dh = f.
     
  19. Sep 25, 2009 #18
    Yup, 1/2 = c = d. I just wasn't sure if we wanted f(x), or if we wanted every possible quadrant. I'm just not used to working with functions as vectors in this way, so my mind was a bit mangled up for a bit :P. But now I think I got it. Thanks so much for all your help!
     
  20. Sep 25, 2009 #19
    No problem. It's a conceptual leap to be certain. It might be obvious to you but just so you know, g and h will be different for each function f.

    Can you get the last part by yourself? After the first part, it's easy.
     
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