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October 8th, 2014, 06:20 PM  #1 
Member Joined: Oct 2014 From: Mars Posts: 35 Thanks: 0  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. $\displaystyle MIN(1,(MIN(1,MIN(x,y)+MAX(0,x0.9))*MIN(y*100,1))+0.0000001)$ x and y is a percent value from 0% to 100% (01). 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. 
October 9th, 2014, 10:59 PM  #2 
Member Joined: Oct 2014 From: Mars Posts: 35 Thanks: 0 
Never mind, please delete this thread.

October 9th, 2014, 11:32 PM  #3 
Senior Member Joined: Apr 2014 From: UK Posts: 917 Thanks: 331 
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. 
October 10th, 2014, 02:48 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361 
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 $\displaystyle \sqrt{1+u} = \sum_{n=0}^{\infty} \frac{(1)^n(2n)!}{(12n)\cdot (n!)^2\cdot (4^n)}u^n= 1 + \frac{x}{2}  \frac{x^2}{8} + \frac{x^3}{16} \cdots$ where u = 1x^2 to get x or a good approximation of it. 
October 13th, 2014, 05:31 AM  #5 
Senior Member Joined: Apr 2014 From: UK Posts: 917 Thanks: 331 
Abs should be sufficient, not too sure this is going to simplify much further though 

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