## Kainui one year ago Let's try this! @ikram002p $\frac{d}{dx}(a\uparrow^x b)$

1. Here_to_Help15

Can i try as well?

2. Kainui

Yeah anyone can lol

3. Here_to_Help15

Oki lol let me get paper

4. Kainui

This might be an impossible question lol.

5. Here_to_Help15

lol hmm nothing is impossible ;)

6. ikram002p

i'll write what im thinking of and then lets see :)

7. anonymous

Woah let me take a stab at this!

8. Here_to_Help15

Can i give you a question @Kainui :)

9. Here_to_Help15

$\frac{ d }{ dx } (2^{x)}$ How do you find ^

10. Here_to_Help15

You lost @Kainui ? tehee :)

11. ikram002p

@Kainui lets try on small stuff like |dw:1433362211420:dw|

12. ganeshie8

* for tomoro

13. Here_to_Help15

?

14. ikram002p

|dw:1433362587173:dw|

15. Here_to_Help15

Is that

16. Here_to_Help15

17. ikram002p

thats only special case yes my writing

18. ikram002p

eh -.- i wish if there is a short cut

19. Here_to_Help15

lol neat hand writing :)

20. ikram002p

ty :)

21. 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 ??

22. Kainui

Sorry I got distracted right as I posted this orry! Good idea on using the 2 case, that's awesome!

23. 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: $\large \lim_{h \to 0}\frac{a \uparrow^{x+h} b - a \uparrow^x b }{h}$

24. 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: $4 \uparrow^2 3 = 4 \uparrow (4 \uparrow^2 2)=4 \uparrow (4 \uparrow (4 \uparrow^2 1)) = 4 \uparrow (4 \uparrow (4 )) = 4^{4^4}$

25. 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?

26. Kainui

Yeah it relies on making this definition which is discrete into something continuous which might not really even be possible.

27. 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

28. 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

29. 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

30. geerky42

Here's another problem for you to consider into: Evaluate $$\dfrac{\mathrm d^{1/2}}{\mathrm dx^{1/2}}~~\large x$$ :)

31. geerky42
32. ikram002p

a way not impossible