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October 8th, 2014, 06:20 PM   #1
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Question Simplify this formula (MIN/MAX)

Hi all, I hope this is the right math section to post this, I'm having trouble simplifying this formula.

$\displaystyle MIN(1,(MIN(1,MIN(x,y)+MAX(0,x-0.9))*MIN(y*100,1))+0.0000001)$

x and y is a percent value from 0% to 100% (0-1).

My brain cannot handle this

I would really appreciate it if someone could help me.

Thanks in advance for your help

Last edited by Apple30; October 8th, 2014 at 06:23 PM.
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October 9th, 2014, 10:59 PM   #2
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Never mind, please delete this thread.
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October 9th, 2014, 11:32 PM   #3
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The problem is it isn't really just a formula, it has decisions in it (min,max).
If this is an Excel formula, you can reduce (I think!) the number of MINs
If that's the case, then this is as simple as I could manage:
0.0000001+MIN(0.9999999,y*100, (MIN(x,y) + MAX(0,x−0.9)) * MIN(y*100,1))
(I think that's right )
Do you have a description of what result you are trying to achieve, there may be a different solution.
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October 10th, 2014, 02:48 AM   #4
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You can use $\displaystyle \min(x, y) = \frac{x + y}{2} - \left|\frac{x - y}{2}\right|$
and $\displaystyle \max(x, y) = \frac{x + y}{2} + \left|\frac{x - y}{2}\right|$
If abs is too much of a decision, you can use $\displaystyle |x| = \sqrt{x^2}$ and the taylorexpansion of $\displaystyle \sqrt{1+u} = \sum_{n=0}^{\infty} \frac{(-1)^n(2n)!}{(1-2n)\cdot (n!)^2\cdot (4^n)}u^n= 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16}- \cdots$ where u = 1-x^2 to get |x| or a good approximation of it.
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October 13th, 2014, 05:31 AM   #5
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Abs should be sufficient, not too sure this is going to simplify much further though
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