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The multiplicative identity of the random function r(x) is any other function f(x) except where f(x)=0.
Originally posted by Loren Booda
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)?
Certainly he does not mean f(r(x)) as you could just set f(x) = 1 for all x.Originally posted by bogdan
What if f(x)=k/r(x) ? Huh ?
Maybe you wanted to say f(r(x))...or I just don't understand...
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.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.