Mendicant_Bias
  • Mendicant_Bias
(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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i could see how that would be annoying
Mendicant_Bias
  • 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
  • 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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anonymous
  • anonymous
let me do one, you can copy it
Mendicant_Bias
  • 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
  • Mendicant_Bias
Sure, thanks.
anonymous
  • 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
  • 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
  • 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
  • anonymous
not really an absolute value equation though
anonymous
  • anonymous
or function or whatever
anonymous
  • anonymous
\[ f(x) = \left\{ \begin{array}{lr} 1 & : x \in \mathbb{Q}\\ 0 & : x \notin \mathbb{Q} \end{array} \right.\]
anonymous
  • anonymous
that is what you wrote i just omitted the "displaymath" in favor of \[
ganeshie8
  • ganeshie8
i smell a continuity/discontinuity proof is on the way
zzr0ck3r
  • 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
  • anonymous
@Mendicant_Bias you know how to copy this right?
anonymous
  • anonymous
oh i thought it would be something like \[f(x)=||x+3|-2|\]
Mendicant_Bias
  • Mendicant_Bias
...And now it is, nevermind, just tremendous lag/delay between it rendering.
anonymous
  • anonymous
draw it!
zzr0ck3r
  • zzr0ck3r
What are you trying to do with it, most of the time we don't use the definition.
Mendicant_Bias
  • 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
  • anonymous
|dw:1441165013039:dw|
anonymous
  • anonymous
looks right to me
Mendicant_Bias
  • 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
  • anonymous
course the first one is \(2a\) and the second is \(2b\)
Mendicant_Bias
  • 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
  • Mendicant_Bias
Whoops, no x, that's c.
anonymous
  • anonymous
yeah was afraid of those nested absolute values
ganeshie8
  • 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
  • ganeshie8
you just follow the rules but what exactly is the actual problem
Mendicant_Bias
  • Mendicant_Bias
"Write the following expressions in equivalent forms not involving absolute values."
ganeshie8
  • ganeshie8
ahh that really going to require some clever thinking..
Mendicant_Bias
  • 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
  • Mendicant_Bias
But yeah, x is actually c, typo
ganeshie8
  • ganeshie8
oh then its easy
ganeshie8
  • 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
  • Mendicant_Bias
Guessing you just apply the same logic of the last problem after the first abs. val sign is gone
Mendicant_Bias
  • Mendicant_Bias
Yeah
ganeshie8
  • ganeshie8
Yes looks its similar..
Mendicant_Bias
  • Mendicant_Bias
Alright, cool. Gonna play with this for a second, but I think I got it.
ganeshie8
  • 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
  • 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
  • Mendicant_Bias
Think I got it, one sec. http://i.imgur.com/V1jKffw.png Now turning that into max/min things.
ganeshie8
  • ganeshie8
4max(a,b,c) ?
Mendicant_Bias
  • Mendicant_Bias
Yeah, it's just \[4 \ \text{max(a,b,c)}\] Yeah, hah
Mendicant_Bias
  • 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
  • Mendicant_Bias
I didn't know that max worked like that, awesome.
Mendicant_Bias
  • Mendicant_Bias
(I mean, I didn't know how max worked at all a couple of hours ago, heh)
ganeshie8
  • ganeshie8
they are usually messy the given expression was cooked up to work out smoothly haha
ganeshie8
  • 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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