Power Tower Derivative

[Shadow]

Hey there,

 

First of all, I'm completely new to calculus in any form, so please excuse any completely obvious errors if there are any.

 

I was just wondering the other day about the derivative of a Power Tower. I tried to treat it as [math]ax^n[/math], and came up with

 

[math]f(a, b) = a \uparrow \uparrow b[/math]

[math]f'(a, b) = (a \uparrow \uparrow b) a^{(a \uparrow \uparrow [b-1]) -1}[/math]

 

Is this correct?

 

Cheers,

 

Gabe

Edited December 16, 2008 by Shadow

[ajb]

I do not understand your notation, but lets do a simple exercise.

 

Let [math]f(x) = a(x)^{b(x)}[/math] and lets suppose we wish to calculate the derivative of [math]f[/math].

 

To do this, lets take the [math]\log[/math] thus

 

[math]\log[f](x) = b(x) \log[a[x]][/math].

 

Now take the derivative,

 

[math]\frac{f'(x)}{f(x)} = b'(x)\log[a(x)] + b(x) \frac{a'(x)}{a(x)}[/math],

 

where hence

 

[math]f'(x) = a(x)^{b(x)} \left( b'(x) \log[a(x)] + b(x) \frac{a'(x)}{a(x)} \right)[/math].

 

You can now use this formula iteratively to calculate higher order towers.

[Shadow]

Yeah, I believe Maple gave me something similar. Anyway, the notation I'm using is Knuth's up-arrow notation. And why can't I treat a Power Tower like I would [math]a^n[/math]? I mean, if we have a power tower [math]a \uparrow \uparrow b[/math] and treat it as [math]a^n[/math], then [math]a=a[/math] and [math]n= a \uparrow \uparrow (b-1)[/math]...doesn't it?

[ajb]

Isn't the reason simply that [math]n \neq n(x)[/math] which allows you to simplify things?

 

The formula I have given works for [math]a(x)^{b}[/math] with [math]b \neq b(x)[/math].

[Bignose]

The power tower is different from an exponent -- that's why there is a different notation. Your question, to me at least, isn't terribly clear, because you use a prime, but there are two variables given f(a,b). Is the prime meant to represent differentiation with respect to a or b?

[Shadow]

Yeah, I wasn't entirely sure I could get away with that But I guess with respect to a, although I'm not completely sure; as I said, I'm as new to calculus as one can get, and I think I've already jumped a little too far for my level.

 

And while I'm aware that tetration and exponentiation is different, you can do the same thing with multiplication;

 

[math]\underbrace{a*a*a*a*a...a*a}_{b}[/math] can be represented as either [math]a^b[/math] or as [math]a*c[/math], where [math]c=\underbrace{a*a*a*a*a...a*a}_{b-1}[/math]. So why can't we treat tetration the same way? If we do [math]\underbrace{a^{a^{a^{.^{.^{.}}}}}}_{b}[/math], then it should be equal to both [math]a \uparrow \uparrow b[/math] and to [math]a^c[/math], where [math]c= \underbrace{a^{a^{a^{.^{.^{.}}}}}}_{b-1}[/math]

 

Cheers,

 

Gabe

[ajb]

If I am reading you right, you want to take a derivative with respect to the function a(x)? (or b(x)).

 

If so you can do such things it is called the functional derivative and is very important in physics.