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+--- 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