Kainui
  • Kainui
Let's try this! @ikram002p \[\frac{d}{dx}(a\uparrow^x b)\]
Calculus1
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Here_to_Help15
  • Here_to_Help15
Can i try as well?
Kainui
  • Kainui
Yeah anyone can lol
Here_to_Help15
  • Here_to_Help15
Oki lol let me get paper

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More answers

Kainui
  • Kainui
This might be an impossible question lol.
Here_to_Help15
  • Here_to_Help15
lol hmm nothing is impossible ;)
ikram002p
  • ikram002p
i'll write what im thinking of and then lets see :)
anonymous
  • anonymous
Woah let me take a stab at this!
Here_to_Help15
  • Here_to_Help15
Can i give you a question @Kainui :)
Here_to_Help15
  • Here_to_Help15
\[\frac{ d }{ dx } (2^{x)}\] How do you find ^
Here_to_Help15
  • Here_to_Help15
You lost @Kainui ? tehee :)
ikram002p
  • ikram002p
@Kainui lets try on small stuff like |dw:1433362211420:dw|
ganeshie8
  • ganeshie8
* for tomoro
Here_to_Help15
  • Here_to_Help15
?
ikram002p
  • ikram002p
|dw:1433362587173:dw|
Here_to_Help15
  • Here_to_Help15
Is that
Here_to_Help15
  • Here_to_Help15
Your writing
ikram002p
  • ikram002p
thats only special case yes my writing
ikram002p
  • ikram002p
eh -.- i wish if there is a short cut
Here_to_Help15
  • Here_to_Help15
lol neat hand writing :)
ikram002p
  • ikram002p
ty :)
ikram002p
  • ikram002p
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 ??
Kainui
  • Kainui
Sorry I got distracted right as I posted this orry! Good idea on using the 2 case, that's awesome!
Kainui
  • Kainui
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: \[\large \lim_{h \to 0}\frac{a \uparrow^{x+h} b - a \uparrow^x b }{h}\]
Kainui
  • Kainui
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: \[4 \uparrow^2 3 = 4 \uparrow (4 \uparrow^2 2)=4 \uparrow (4 \uparrow (4 \uparrow^2 1)) = 4 \uparrow (4 \uparrow (4 )) = 4^{4^4}\]
geerky42
  • geerky42
I don't know, but I think we may need to look at how \(\uparrow^x\) is defined for any real value of x?
Kainui
  • Kainui
Yeah it relies on making this definition which is discrete into something continuous which might not really even be possible.
Kainui
  • Kainui
Here I should have posted this to begin with for those who aren't into this yet! http://www.numberphile.com/videos/grahamsnumber.html
Kainui
  • Kainui
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
Kainui
  • Kainui
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
geerky42
  • geerky42
Here's another problem for you to consider into: Evaluate \(\dfrac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}~~\large x\) :)
ikram002p
  • ikram002p
a way not impossible

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