Question

(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.

Answer 1

i could see how that would be annoying

Answer 2

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

Answer 3

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

Answer 4

let me do one, you can copy it

Answer 5

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

Answer 6

Sure, thanks.

Answer 7

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

Answer 8

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.

Answer 9

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

Answer 10

not really an absolute value equation though

Answer 11

or function or whatever

Answer 12

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

Answer 13

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

Answer 14

i smell a continuity/discontinuity proof is on the way

Answer 15

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

Answer 16

@Mendicant_Bias you know how to copy this right?

Answer 17

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

Answer 18

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

Answer 19

draw it!

Answer 20

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

Answer 21

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

Answer 22

looks right to me

Answer 23

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

Answer 24

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

Answer 25

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

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.

Answer 26

Whoops, no x, that's c.

Answer 27

yeah was afraid of those nested absolute values

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

ahh that really going to require some clever thinking..

Answer 32

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.

Answer 33

But yeah, x is actually c, typo

Answer 34

oh then its easy

Answer 35

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

Answer 36

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

Answer 37

Yeah

Answer 38

Yes looks its similar..

Answer 39

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

Answer 40

does this work ?

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

Answer 41

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

Answer 42

Think I got it, one sec.

http://i.imgur.com/V1jKffw.png

Now turning that into max/min things.

Answer 43

4max(a,b,c) ?

Answer 44

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

Yeah, hah

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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