(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.

- Mendicant_Bias

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- schrodinger

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- anonymous

i could see how that would be annoying

- Mendicant_Bias

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

- Mendicant_Bias

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

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## More answers

- anonymous

let me do one, you can copy it

- Mendicant_Bias

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

- Mendicant_Bias

Sure, thanks.

- anonymous

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

- Mendicant_Bias

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.

- anonymous

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

- anonymous

not really an absolute value equation though

- anonymous

or function or whatever

- anonymous

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

- anonymous

that is what you wrote
i just omitted the "displaymath" in favor of \[

- ganeshie8

i smell a continuity/discontinuity proof is on the way

- zzr0ck3r

I think sense they are working with the definition of absolute, they will be doing stuff like triangle inequality, and results of it :)

- anonymous

@Mendicant_Bias you know how to copy this right?

- anonymous

oh i thought it would be something like
\[f(x)=||x+3|-2|\]

- Mendicant_Bias

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

- anonymous

draw it!

- zzr0ck3r

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

- Mendicant_Bias

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

- anonymous

|dw:1441165013039:dw|

- anonymous

looks right to me

- Mendicant_Bias

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

- anonymous

course the first one is \(2a\) and the second is \(2b\)

- Mendicant_Bias

\[a+b+2x+|a-b|+|a+b-2c+|a-b||\]
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.

- Mendicant_Bias

Whoops, no x, that's c.

- anonymous

yeah was afraid of those nested absolute values

- ganeshie8

\(a\gt b\) :
\[a+b+2x+|a-b|+|a+b-2c+|a-b|| \\= a+b+2x+a-b + |a+b-2c+a-b|\\=2a+2x + |2a-2c|\]

- ganeshie8

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

- Mendicant_Bias

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

- ganeshie8

ahh that really going to require some clever thinking..

- Mendicant_Bias

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.

- Mendicant_Bias

But yeah, x is actually c, typo

- ganeshie8

oh then its easy

- ganeshie8

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

- Mendicant_Bias

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

- Mendicant_Bias

Yeah

- ganeshie8

Yes looks its similar..

- Mendicant_Bias

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

- ganeshie8

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

- Mendicant_Bias

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

- Mendicant_Bias

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

- ganeshie8

4max(a,b,c) ?

- Mendicant_Bias

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

- Mendicant_Bias

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

- Mendicant_Bias

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

- Mendicant_Bias

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

- ganeshie8

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

- ganeshie8

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

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