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Kainui

  • one year ago

Let's try this! @ikram002p \[\frac{d}{dx}(a\uparrow^x b)\]

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  1. Here_to_Help15
    • one year ago
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    Can i try as well?

  2. Kainui
    • one year ago
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    Yeah anyone can lol

  3. Here_to_Help15
    • one year ago
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    Oki lol let me get paper

  4. Kainui
    • one year ago
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    This might be an impossible question lol.

  5. Here_to_Help15
    • one year ago
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    lol hmm nothing is impossible ;)

  6. ikram002p
    • one year ago
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    i'll write what im thinking of and then lets see :)

  7. anonymous
    • one year ago
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    Woah let me take a stab at this!

  8. Here_to_Help15
    • one year ago
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    Can i give you a question @Kainui :)

  9. Here_to_Help15
    • one year ago
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    \[\frac{ d }{ dx } (2^{x)}\] How do you find ^

  10. Here_to_Help15
    • one year ago
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    You lost @Kainui ? tehee :)

  11. ikram002p
    • one year ago
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    @Kainui lets try on small stuff like |dw:1433362211420:dw|

  12. ganeshie8
    • one year ago
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    * for tomoro

  13. Here_to_Help15
    • one year ago
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    ?

  14. ikram002p
    • one year ago
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    |dw:1433362587173:dw|

  15. Here_to_Help15
    • one year ago
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    Is that

  16. Here_to_Help15
    • one year ago
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    Your writing

  17. ikram002p
    • one year ago
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    thats only special case yes my writing

  18. ikram002p
    • one year ago
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    eh -.- i wish if there is a short cut

  19. Here_to_Help15
    • one year ago
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    lol neat hand writing :)

  20. ikram002p
    • one year ago
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    ty :)

  21. ikram002p
    • one year ago
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    i might end up saying zero since this arrow notation end up defining a finite number hmmm but if we wanna deal with it as exponent we neat ani_arrow notation , right @Kainui ? is there something like this ??

  22. Kainui
    • one year ago
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    Sorry I got distracted right as I posted this orry! Good idea on using the 2 case, that's awesome!

  23. Kainui
    • one year ago
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    One way I was thinking of is seeing if we could weasel our way into a continuous definition of the arrow notation with the definition of derivative: \[\large \lim_{h \to 0}\frac{a \uparrow^{x+h} b - a \uparrow^x b }{h}\]

  24. Kainui
    • one year ago
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    For anyone who doesn't know the arrow notation, which to be honest I barely do either. \[ a \uparrow b = a^b \\ a \uparrow \uparrow b = a^{a^{a^\cdots}} \text{tower b high}\] So the recursive definition is found here, but I'llt ype it out too: \[a \uparrow ^n b = a \uparrow ^{n-1}[a \uparrow^n(b-1)]\] and \[a \uparrow^n 1 = a\] http://mathworld.wolfram.com/PowerTower.html So for example: \[4 \uparrow^2 3 = 4 \uparrow (4 \uparrow^2 2)=4 \uparrow (4 \uparrow (4 \uparrow^2 1)) = 4 \uparrow (4 \uparrow (4 )) = 4^{4^4}\]

  25. geerky42
    • one year ago
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    I don't know, but I think we may need to look at how \(\uparrow^x\) is defined for any real value of x?

  26. Kainui
    • one year ago
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    Yeah it relies on making this definition which is discrete into something continuous which might not really even be possible.

  27. Kainui
    • one year ago
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    Here I should have posted this to begin with for those who aren't into this yet! http://www.numberphile.com/videos/grahamsnumber.html

  28. Kainui
    • one year ago
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    Another example, here are our rules just for reference: \[a \uparrow ^n b = a \uparrow ^{n-1}[a \uparrow^n(b-1)]\] \[a \uparrow^n 1 = a\] Ok so the example: \[4 \uparrow^3 3 \] Just plug and chug that recursion relation:\[ 4 \uparrow^2 (4 \uparrow^3 2) \]Now just looking at that part \((4 \uparrow^3 2)\) I expand that further:\[ 4 \uparrow^2 (4 \uparrow^2(4 \uparrow^3 1)) \] now using the fact that \((4 \uparrow^n 1) = 4 \) also listed above as the other rule we have: \[ 4 \uparrow^2 (4 \uparrow^24) \] Which we then expand further

  29. Kainui
    • one year ago
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    The next step we start expanding: \[ 4 \uparrow^2 (4 \uparrow 4 \uparrow 4 \uparrow 4) = 4 \uparrow^2 4^{4^{4^4}}\] And past this we really can't do anything in terms of writing it cause this is a tower of 4s that is \(4^{4^{4^4}}\) high. lol

  30. geerky42
    • one year ago
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    Here's another problem for you to consider into: Evaluate \(\dfrac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}~~\large x\) :)

  31. geerky42
    • one year ago
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    http://openstudy.com/study#/updates/556f9e4ce4b050a18e84d177

  32. ikram002p
    • one year ago
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    a way not impossible

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