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The random function's universality of multiplicative identity

  1. Apr 27, 2003 #1
    The multiplicative identity of the random function r(x) is any other function f(x) except where f(x)=0.
  2. jcsd
  3. May 1, 2003 #2
    In other words, a field in x of random values r(x) each multiplied by a corresponding value for a (non-)random mapping f(x), always retains a random mapping r(x), except where f(x)=0.

    Can you imagine this situation?

    Also, may this be a great way (in theory) to obtain 0's of f(x)?
  4. May 1, 2003 #3
    What if f(x)=k/r(x) ? Huh ?
    Maybe you wanted to say f(r(x))...or I just don't understand...:frown:
  5. May 1, 2003 #4

    Does it always maintain the properties of the random function r(x). For example you might have specified that if the domain is the real numbers, the function r(x) returns values in the interval [0, 1], multiplying this by 2 does not preserve this property.

    Do you need to make your definition of randomness more tight, or am I missing something.

    As to Bogdan's point, f(x)=k/r(x) will be a random function.......and thus not allowed by the Loren Booda's initial statement.
    Last edited: May 1, 2003
  6. May 1, 2003 #5
    In a set X equipped with a binary operation called a product, the multiplicative identity is an element e such that e * x = x * e = x, for all x in X...got it...

    So e = f(x)...hmmm... r(x)*f(x)=r(x)...hmmm...still don't get it...
  7. May 1, 2003 #6
    Certainly he does not mean f(r(x)) as you could just set f(x) = 1 for all x.

    What he appears to be saying that there if r contains no information, i.e. it returns a value on the range with equal probability, then multiplying it by any conventional (not including random) and not equal to zero at x ,function returns a result which contains no information as defined above.

    Perhaps his statement is more general?
  8. May 1, 2003 #7
    I don't know for sure...maybe he'll explain better...
  9. May 1, 2003 #8
    What is returned is the property of randomness, apparently not the function itself as first defined. Indeed I need to "loosen" my definition of random r(x). Perhaps I should require its returned interval to be [-[oo],[oo]]. Thanks for your feedback.

    I hope this helps your understanding of the problem also, bogdan?
  10. May 1, 2003 #9
    Yeap...if (non-)random means not random... ...those brackets...
  11. May 1, 2003 #10

    A random number times a random number, or times a non-random number, is a random number.
  12. May 1, 2003 #11
    But why ?
    if f(x)=1/r(x), then both are random...
    Or am I just stupid ? Or worse ?
  13. May 2, 2003 #12
    What is the probability distribution of your random function on the real line going to look like?
  14. May 2, 2003 #13
    Actually, plus, I don't know if r(x) can be "seen," as its average magnitude may be infinite. Can you think of a finite random function that returns the property of randomness to similar bounds (or better yet, as I initiated, the random function itself?) Perhaps I should require normalization of r(x) and r(x)f(x)?

    bogdan, I appreciate your interest too. Think of a random function with magnitude r(x) returning values from 0 to [oo], multiply each by a nonzero arbitrary number f(x), and one returns the random function along x. The cardinality of "randomness" apparently is greater than that of the real number line.
  15. May 2, 2003 #14
    The only way I could think of to get this in a finite range would be to have modular multiplication. so if it was mod 10,

    8.18* 10 = 0.9 (mod 10)
  16. May 2, 2003 #15
    Beautiful idea. The credit is yours. Do your see any application for the infinite range r(x) (like finding zeroes of a function)?
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