Is there a way to re-write $\min(a,b)$ in terms of an analytical function?
Also, if not, is there a nice analytic function that is a tight upper bound?
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But absolute value is not analytic. Right? – Boby Sep 18 '14 at 21:33
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What do you mean by a $\textbf{tight}$ upper bound? – D Poole Sep 18 '14 at 21:38
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@Alex The $\min$ function isn't analytic! – azarel Sep 18 '14 at 21:38
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That's what I though. Because it's not differentiable. Is there a nice upper bound that is differential? – Boby Sep 18 '14 at 21:39
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@DPoole I guess, something that upper bounds it well. – Boby Sep 18 '14 at 21:41
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@Alex Upper bound in absolute value or not? There aren't many real-valued analytic functions $f(z=a+bi)$. – D Poole Sep 18 '14 at 21:43
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Is this what you're looking for?
$$\min(a,\,b) \; = \; \frac{a\,+\,b\; - \; |a-b|}{2} $$ $$\max(a,\,b) \; = \; \frac{a\,+\,b\; + \; |a-b|}{2} $$
Dave L. Renfro
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1I was assuming you meant "analytic" in the sense of some kind of explicit expression, not in the "analytic" sense involving power series stuff. – Dave L. Renfro Sep 18 '14 at 21:38
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@robjohn is $\sqrt{a^2-2ab+b^2}$ an upper bound on $|a-b|$ or lower bound? – Boby Sep 18 '14 at 21:45
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@Alex: Note that $\sqrt{u^2} = |u|$ and $a^2 - 2ab + b^2 = (a-b)^{2}.$ – Dave L. Renfro Sep 18 '14 at 21:58
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