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 About the fake abs : f(x) = f(-x) tommy1729 Ultimate Fellow     Posts: 1,493 Threads: 356 Joined: Feb 2009 12/11/2014, 01:22 PM When discussing fake function theory we came across the fake sqrt. the fake abs function is then fake_abs(x) = fake_sqrt(x^2). Clearly when we want a real-entire function f(x) to satisfy f(x) = f(-x) then we consider f(fake_abs(x)). But what if we have a real-entire function f(x) and we want to remove the property f(x) = f(-x). We can do many things like : g(x) = f(x) + exp(-x). (*) But lets consider the context of fake functions : Find g such that f(x) = g(fake_abs(x)) => g(x) = f(fake_abs^[-1](x)) g(x) = f( sqrt ( inv_fake_sqrt(x) ) ) Lets call inv_fake_sqrt(x) := fakesquare(x). g(x) = f ( sqrt ( fakesquare(x) ) ) Now since f(x) = f(-x) we have that F(x) = f(sqrt(x)) is also an entire function. g(x) = F( fakesquare(x) ) Now if we want g(x) to be entire then F ( fakesquare(x) ) needs to be entire. Since fakesquare is a multivalued function ( an inverse of an entire ) its not entire. SO when is F( fakesquare(x) ) entire ? And how does that look like ? Also of interest ( when its not entire ) : fake ( F ( fakesquare(x) ) ) From (*) one then also wonders about F( fakesquare(x) ) - f(x) and how that looks like. I have some ideas and guesses but no evidence or plots. Seems like chapter 2 in fake function theory. The analogue questions exist for exp(x) instead of x^2 ; removing the periodic property. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread About the fake abs : f(x) = f(-x) - by tommy1729 - 12/11/2014, 01:22 PM RE: About the fake abs : f(x) = f(-x) - by sheldonison - 12/11/2014, 10:58 PM RE: About the fake abs : f(x) = f(-x) - by tommy1729 - 12/11/2014, 11:59 PM

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