## anonymous one year ago (Introductory Real Analysis) I'm trying to transform a given expression with absolute value signs into a version without it by applying the formal definition of the absolute value function (example below), but I'm running into trouble when I get more than one absolute value sign inside another.

1. anonymous

i could see how that would be annoying

2. anonymous

(Sorry, just having trouble putting up the LaTeX, I'll try to put up a pic instead of the math maybe)

3. anonymous

How do I do a piecewise function bracket again? I'm trying to do it, but am somewhere making a mistake.

4. anonymous

let me do one, you can copy it

5. anonymous

I'm trying this: \begin{displaymath} f(x) = \left\{ \begin{array}{lr} 1 & : x \in \mathbb{Q}\\ 0 & : x \notin \mathbb{Q} \end{array} \right. \end{displaymath} to display the abs. value function, and it's not rendering

6. anonymous

Sure, thanks.

7. anonymous

$f(x) = |x+4| = \left\{\begin{array}{rcc} x + 4 & \text{if} & x \geq -4 \\ - x - 4& \text{if} & x < -4 \end{array} \right.$

8. anonymous

What the hell? I just typed it all up in the equation box, and when I entered, nothing was there. OS isn't playing nice with me atm.

9. anonymous

$f(x) = \left\{\begin{array}{rcc} 1 & \text{if} & x \in\mathbb{Q} \\ 0& \text{if} & x\notin\mathbb{Q} \end{array} \right.$

10. anonymous

not really an absolute value equation though

11. anonymous

or function or whatever

12. anonymous

$f(x) = \left\{ \begin{array}{lr} 1 & : x \in \mathbb{Q}\\ 0 & : x \notin \mathbb{Q} \end{array} \right.$

13. anonymous

that is what you wrote i just omitted the "displaymath" in favor of $14. ganeshie8 i smell a continuity/discontinuity proof is on the way 15. zzr0ck3r I think sense they are working with the definition of absolute, they will be doing stuff like triangle inequality, and results of it :) 16. anonymous @Mendicant_Bias you know how to copy this right? 17. anonymous oh i thought it would be something like \[f(x)=||x+3|-2|$

18. anonymous

...And now it is, nevermind, just tremendous lag/delay between it rendering.

19. anonymous

draw it!

20. zzr0ck3r

What are you trying to do with it, most of the time we don't use the definition.

21. anonymous

I'm trying to change the expression to not have the absolute value signs, which I did by using (max) and (min). http://i.imgur.com/l7T5TkD.png

22. anonymous

|dw:1441165013039:dw|

23. anonymous

looks right to me

24. anonymous

What I did was simplify the insides and then use max on this one, e.g. $a+b+|a-b|=2 \ \text{max}(a,b)$

25. anonymous

course the first one is $$2a$$ and the second is $$2b$$

26. anonymous

$a+b+2x+|a-b|+|a+b-2c+|a-b||$ I'd in principle, like to do the same thing, but I don't know what will happen/how the absolute value inside another one will work, really.

27. anonymous

Whoops, no x, that's c.

28. anonymous

yeah was afraid of those nested absolute values

29. ganeshie8

$$a\gt b$$ : $a+b+2x+|a-b|+|a+b-2c+|a-b|| \\= a+b+2x+a-b + |a+b-2c+a-b|\\=2a+2x + |2a-2c|$

30. ganeshie8

you just follow the rules but what exactly is the actual problem

31. anonymous

"Write the following expressions in equivalent forms not involving absolute values."

32. ganeshie8

ahh that really going to require some clever thinking..

33. anonymous

Do you have to do it recursively, then? e.g. are there four possible outcomes or something like that, or do you just take the same condition you first applied to the outside absolute value signs and apply it again? Yeah, and I've never actually seen or used the max/min(value, value, value...) notation, came up unexpectedly.

34. anonymous

But yeah, x is actually c, typo

35. ganeshie8

oh then its easy

36. ganeshie8

$$a\gt b$$ : $a+b+2c+|a-b|+|a+b-2c+|a-b|| \\= a+b+2c+a-b + |a+b-2c+a-b|\\=2a+2c + |2a-2c|\\ = 2max(a,c)$

37. anonymous

Guessing you just apply the same logic of the last problem after the first abs. val sign is gone

38. anonymous

Yeah

39. ganeshie8

Yes looks its similar..

40. anonymous

Alright, cool. Gonna play with this for a second, but I think I got it.

41. ganeshie8

does this work ? $a+b+2c+|a-b|+|a+b-2c+|a-b|| \\~\\ = 2max(a+b+|a-b|, ~2c)\\~\\ = 2max(2max(a,b), 2c)$

42. anonymous

Wow, this is kind of bizarre. I don't know if I just haven't noticed something like this in math yet, but I don't think I've ever noticed something that feels like an operation/changing something (e.g. a>b) and it influencing an expression in layers all at once without having to break them down to get to the "inside" layer. It's not really math, it's an assumption, but I'm so used to order of ops. problems that having something change like that seems strange. The answer is a little different, and a little more aesthetically pleasing, if that's a hint, heh

43. anonymous

Think I got it, one sec. http://i.imgur.com/V1jKffw.png Now turning that into max/min things.

44. ganeshie8

4max(a,b,c) ?

45. anonymous

Yeah, it's just $4 \ \text{max(a,b,c)}$ Yeah, hah

46. anonymous

Has a ring to how it's written. Well, cool! Thanks so much for helping me figure this out! I definitely wouldn't have gotten it had I not asked...lol

47. anonymous

I didn't know that max worked like that, awesome.

48. anonymous

(I mean, I didn't know how max worked at all a couple of hours ago, heh)

49. ganeshie8

they are usually messy the given expression was cooked up to work out smoothly haha

50. ganeshie8

$\color{blue}{a+b+|a-b|}+\color{red}{2c}+|\color{blue}{a+b+|a-b|}+\color{red}{2c}| \\~\\ = 2max(\color{blue}{a+b+|a-b|}, ~\color{red}{2c})\\~\\ = 2max(\color{blue}{2max(a,b)}, \color{red}{2c})\\~\\ =4max(\color{blue}{max(a,b)},c)\\~\\ =4max(a,b,c)$