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Mendicant_Bias

  • 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.

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  1. anonymous
    • one year ago
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    i could see how that would be annoying

  2. Mendicant_Bias
    • one year ago
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    (Sorry, just having trouble putting up the LaTeX, I'll try to put up a pic instead of the math maybe)

  3. Mendicant_Bias
    • one year ago
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    How do I do a piecewise function bracket again? I'm trying to do it, but am somewhere making a mistake.

  4. anonymous
    • one year ago
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    let me do one, you can copy it

  5. Mendicant_Bias
    • one year ago
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    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. Mendicant_Bias
    • one year ago
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    Sure, thanks.

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

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

  10. anonymous
    • one year ago
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    not really an absolute value equation though

  11. anonymous
    • one year ago
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    or function or whatever

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

  13. anonymous
    • one year ago
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    that is what you wrote i just omitted the "displaymath" in favor of \[

  14. ganeshie8
    • one year ago
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    i smell a continuity/discontinuity proof is on the way

  15. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    @Mendicant_Bias you know how to copy this right?

  17. anonymous
    • one year ago
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    oh i thought it would be something like \[f(x)=||x+3|-2|\]

  18. Mendicant_Bias
    • one year ago
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    ...And now it is, nevermind, just tremendous lag/delay between it rendering.

  19. anonymous
    • one year ago
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    draw it!

  20. zzr0ck3r
    • one year ago
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    What are you trying to do with it, most of the time we don't use the definition.

  21. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    |dw:1441165013039:dw|

  23. anonymous
    • one year ago
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    looks right to me

  24. Mendicant_Bias
    • one year ago
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    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
    • one year ago
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    course the first one is \(2a\) and the second is \(2b\)

  26. Mendicant_Bias
    • one year ago
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    \[a+b+2x+|a-b|+|a+b-2c+|a-b||\] 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. Mendicant_Bias
    • one year ago
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    Whoops, no x, that's c.

  28. anonymous
    • one year ago
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    yeah was afraid of those nested absolute values

  29. ganeshie8
    • one year ago
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    \(a\gt b\) : \[a+b+2x+|a-b|+|a+b-2c+|a-b|| \\= a+b+2x+a-b + |a+b-2c+a-b|\\=2a+2x + |2a-2c|\]

  30. ganeshie8
    • one year ago
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    you just follow the rules but what exactly is the actual problem

  31. Mendicant_Bias
    • one year ago
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    "Write the following expressions in equivalent forms not involving absolute values."

  32. ganeshie8
    • one year ago
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    ahh that really going to require some clever thinking..

  33. Mendicant_Bias
    • one year ago
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    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. Mendicant_Bias
    • one year ago
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    But yeah, x is actually c, typo

  35. ganeshie8
    • one year ago
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    oh then its easy

  36. ganeshie8
    • one year ago
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    \(a\gt b\) : \[a+b+2c+|a-b|+|a+b-2c+|a-b|| \\= a+b+2c+a-b + |a+b-2c+a-b|\\=2a+2c + |2a-2c|\\ = 2max(a,c)\]

  37. Mendicant_Bias
    • one year ago
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    Guessing you just apply the same logic of the last problem after the first abs. val sign is gone

  38. Mendicant_Bias
    • one year ago
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    Yeah

  39. ganeshie8
    • one year ago
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    Yes looks its similar..

  40. Mendicant_Bias
    • one year ago
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    Alright, cool. Gonna play with this for a second, but I think I got it.

  41. ganeshie8
    • one year ago
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    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. Mendicant_Bias
    • one year ago
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    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. Mendicant_Bias
    • one year ago
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    Think I got it, one sec. http://i.imgur.com/V1jKffw.png Now turning that into max/min things.

  44. ganeshie8
    • one year ago
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    4max(a,b,c) ?

  45. Mendicant_Bias
    • one year ago
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    Yeah, it's just \[4 \ \text{max(a,b,c)}\] Yeah, hah

  46. Mendicant_Bias
    • one year ago
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    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. Mendicant_Bias
    • one year ago
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    I didn't know that max worked like that, awesome.

  48. Mendicant_Bias
    • one year ago
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    (I mean, I didn't know how max worked at all a couple of hours ago, heh)

  49. ganeshie8
    • one year ago
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    they are usually messy the given expression was cooked up to work out smoothly haha

  50. ganeshie8
    • one year ago
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    \[\color{blue}{a+b+|a-b|}+\color{red}{2c}+|\color{blue}{a+b+|a-b|}+\color{red}{2c}| \\~\\ = 2max(\color{blue}{a+b+|a-b|}, ~\color{red}{2c})\\~\\ = 2max(\color{blue}{2max(a,b)}, \color{red}{2c})\\~\\ =4max(\color{blue}{max(a,b)},c)\\~\\ =4max(a,b,c) \]

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