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Self-Containing Functions

  1. Sep 29, 2005 #1
    First off, I'd like to say that I'm new here and not very advanced. I'm really here in the hopes of learning something, so that my mathematical tools might match up with the mathematical level of my ideas ><.

    So... First off: Is it possible for a function to contain itself? For instance, here's something I've been messing around with today:

    [tex]f(x) = a ( f(x) ) +x[/tex]

    Now... If I change the notation from [tex]f(x)[/tex] to [tex]y[/tex] for clarity, I get this equation, which I can rearrange in the following way:

    [tex]y = ay + x[/tex]

    [tex]y - ay = x[/tex]

    [tex]y(1-a) = x[/tex]

    [tex]y = \frac{x}{1-a}[/tex]

    And, returning the notation to its original form:

    [tex]f(x) = \frac{x}{1-a}[/tex]


    In which case, my result (without all of the work I posted above) would be that

    [tex]f(x) = a( f(x) ) + x[/tex]

    is identical to

    [tex]f(x) = \frac{x}{1-a}[/tex]


    Is this correct, or incorrect; is there some property of functions that makes what I've done wrong? If it is correct, is there a name for this sort of "self-containing" function?

    Thanks in advance for your replies... ^^
     
  2. jcsd
  3. Sep 29, 2005 #2
    Yes, that is valid. I would call it recursive, but I am a programmer, not a mathematician. Replacing a function with a variable in that case, though, is fine.
     
  4. Sep 29, 2005 #3
    Alright, that's good then. Thank you. ^^

    So... This is as different from what I just asked about as multiplication is from addition. But I guess it's related somehow ><.

    Is it possible to solve this equation excusively for the function P(w)? What I mean is, isolate P(w) completely to one side, defining the function, without having it appear anywhere within itself. No D(P(w)) or anything like that.

    And, if so, how, if you know? Or do you not know at all ><? My problem:

    [tex]P(w) = \frac{N(w)}{D(P(w))}[/tex]
     
  5. Sep 29, 2005 #4
    I do not know of a way to take a function outside of another unknown function. If you know what the function “D” is, though, just replace it with the formula inside that function, and simplify.

    The problem is that, if you do not know what the function D looks like, you do not know how it is transforming P(w), and therefore, you do not know how to reverse that to get it out of it… though I am not very advanced in mathematics, do not trust me that it is impossible with an unknown function! :)
     
  6. Sep 29, 2005 #5
    In general, the answer is no. For example, let [tex] N(w)=1 [/tex] , let [tex]D(u)=sin(u) [/tex], and call [tex]P(w)=y [/tex],then your problem is: [tex]y*Sin(y)=1 [/tex], in this case P(w) will just equal a constant which is about as simple as a function can get, but it can't be found in closed form, which I would call unsolvable.
    (p.s. y can be found approximately using root finding methods, like newtons method)
     
  7. Sep 29, 2005 #6
    I know what function D and function N are. P is the unknown. I can't use the inverse of D, because if I do, it looks like this:

    [tex]P(w) = \frac{N(w)}{D(P(w))}[/tex]

    [tex]P(w) * D(P(w)) = N(w)[/tex]

    [tex]D^{-1}(P(w)) * P(w) = D^{-1}(N(w))[/tex]

    As you can see, that doesn't help at all. Which is why I ask if it can be done at all...? It might, I don't know. And if it can, I really need to know how ><.

    Thanks again for your help so far!
     
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