About the fake abs : f(x) = f(-x) - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: About the fake abs : f(x) = f(-x) (/showthread.php?tid=942) About the fake abs : f(x) = f(-x) - tommy1729 - 12/11/2014 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 RE: About the fake abs : f(x) = f(-x) - sheldonison - 12/11/2014 (12/11/2014, 01:22 PM)tommy1729 Wrote: When discussing fake function theory we came across the fake sqrt. ... fake_sqrt(x^2). $f(x)=\sum_{n=0}^{\infty} a_n x^n\;\;\;\;\; a_n = \frac{1}{\Gamma(n+0.5)}$ for large positive numbers, $g(x)=f(x)\exp(-x) \approx \sqrt{x}$ $g(x^2) = g((-x)^2) \approx x\;\;$ this is true if |real(x)| is large enough, and |imag(x)| isn't too large However, at the imaginary axis $g(x^2)$ grows large exponentially, and does not behave like x at all. And g(x^2) never behaves like abs(x), anywhere in the complex plane Here are some example calculations: $g(25)=5 + 1.5\cdot10^{-13}\;\;\;$ 5^2, small error term $g((5+0.1i)^2) = 5+0.1i - k\cdot10^{-13}\;\;\;$also a small error term, but not abs(x^2) $g(-25)=-866955233 \;\;\;$ (5i)^2, huge error term, nowhere near 5i $g(25i) = 3.53768061172 + 3.52450328163i \;\;\;\;\sqrt{25i}\approx 3.535534 +3.535534i$ RE: About the fake abs : f(x) = f(-x) - tommy1729 - 12/11/2014 Im aware that there is no fake abs(z) that is a good approximation on all of C. That is why I used the letter x. So fake in the sense of asymptotic near the real line , or fake of the absolute value of the reals if you like. The focus for fake functions is Always mainly on the real line anyway , as you well know. Im aware of what you said sheldon. You are completely correct. But I see no problem with that. Or did you not claim a problem ? Are you conjecturing a growth rate perhaps ? regards tommy1729