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 About the fake abs : f(x) = f(-x) sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 12/11/2014, 10:58 PM (This post was last modified: 12/11/2014, 11:18 PM by sheldonison.) (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 $g(25i) = 3.53768061172 + 3.52450328163i \;\;\;\;\sqrt{25i}\approx 3.535534 +3.535534i$ - Sheldon « 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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