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Hello! First off, I have to thank everybody who contributed to the thread at http://math.eretrandre.org/tetrationforu...php?tid=47, as that was what started me on this line of reasoning. In the attached pdf (I don't know how to use TeX yet, so I did it in Mathematica), I derived what I thought was a logical derivative of e[4]x, but upon actually running the numbers, it turned out not to work. I can't figure out why, but hopefully somebody here can. Thanks in advance for reading through this post (and hopefully the pdf as well). :-)
The attachment is broken. Somehow firefox shows it also as *.php instead of *.pdf. What's going on? (When I downloaded the file explicitely, the usual beginning of a *.pdf-file is missing, perhaps this helps to find the source of the problem...).
That's strange. It displayed fine for me in Chrome but not in Firefox and not when downloaded again. In that case, I'll just post the Mathematica code that was there originally.

The Derivative of E tetra x

First off, a few notes on the notation used by this paper. Also, as this is my first time trying anything like this, I apologize for any formatting errors or obvious math mistakes made here.

T[x] = n tetra x
TE[x] = E tetra x
D[f[x], x] = f'[x]

Ok, to business. To find the derivative, let's start with a basic identity.

TE[x] == TE[x]

Taking the natural log of both sides gives

Log[TE[x]] == Log[TE[x]]

One of the tetration identies is

Log[T[x]] == T[x - 1]*Log[n]

Or, using E as the base:

Log[TE[x]] == TE[x - 1]

As a result,

D[Log[TE[x]], x] == TE'[x - 1]

TE'[x]/TE[x] == TE'[x - 1]

TE'[x] == TE[x]*TE[x - 1]

Thus we have a recurrence relation for the derivative. This can be continued further.

TE'[x] == TE[x]*TE[x - 1]*TE'[x - 2]

TE'[x] == TE[x]*TE[x - 1]*TE[x - 2]*TE'[x - 3]

This is, so far, based entirely off of http://math.eretrandre.org/tetrationforu...php?tid=47. However, I took it a bit further, hoping that (for whole numbers, anyway), you could find the product of the TE terms:

TE'[x] == TE'[0] Product[TE[x - k], {k, 0, x}]

I don't know enough about partial products to be able to know what to do in the case of non-integers here, but I figured that figuring out a general formula even only for integer values of x would be useful, so I tried solving the product the same way (more or less) you would solve a sum of powers:

P == Product[TE[x - k], {k, 0, x}]

P == TE[x]*TE[x - 1]*TE[x - 2]*TE[x - 3] ...*TE[x - k]

E^P == TE[x + 1]*TE[x]*TE[x - 1]*TE[x - 2] ...*TE[x - k + 1]

E^P*TE[x - k] == TE[x + 1]*TE[x]*TE[x - 1]*TE[x - 2] ...*TE[x - k + 1]*TE[x - k]

E^P*TE[x - k] == TE[x + 1]*P

Since

x - k == 0,

TE[x - k] = 1

As a result,

E^P == TE[x + 1]*P

Now, we rearrange the equation a bit.

Log[E^P] == Log[TE[x + 1]]*Log[P]

P == TE[x]*Log[P]

P == Log[P^TE[x]]

E^P == P^TE[x]

Substituting into the above equation gives

TE[x + 1]*P == P^TE[x]

TE[x + 1] == P^(TE[x] - 1)

P == TE[x + 1]^(1/(TE[x] - 1))

Now that there is a formula for the product:

TE'[x] == TE'[0]*TE[x + 1]^(1/(TE[x] - 1))

Sadly, this can be easily proven not to work. If x=2, and with the derivative recurrence equation listed above,

TE'[x] = TE[x]*TE'[x - 1]

TE'[0]*TE[x + 1]^(1/(TE[x] - 1)) == TE[x]*TE'[0]*TE[x]^(1/(TE[x - 1] - 1))

TE[x + 1]^(1/(TE[x] - 1)) == TE[x]*TE[x]^(1/(TE[x - 1] - 1))

TE[3]^(1/(TE[2] - 1)) == TE[2]^(TE[1]/(TE[1] - 1))

2.917275 != 73.71885

So after all that, it turns out not to be true. What I can't figure out is why. I'm hoping you guys could show me what's wrong with this derivation. Thanks in advance for your help.
(11/12/2011, 04:24 PM)Forehead Wrote: [ -> ]That's strange. It displayed fine for me in Chrome but not in Firefox and not when downloaded again. In that case, I'll just post the Mathematica code that was there originally.

Hmm, just for the record, here is a bitmap of the binary beginning of the file. Surprisingly- before the "official" beginning of the pdf-file (in the picture framed by a orange frame containing the "%pdf-1.4"-"magic stamp") there is some code which I cannot assign to something (it looks like some pdf-content-code). Usually a correct *.pdf-file begins with the "%pdf-1.4"-stamp, so the pdf-producer-software seems to have scrambled the content of the file, or appended something unwanted...
Gottfried

Excerpt from the vicinity of the entry, which refers the "producing" software:
Code:
`<</Producer (Mathematica PDF Export)/Creator (Mathematica)/CreationDate (D:20111112005354-05'00')/ModDate (D:20111112005354-05'00')/Title (Derivative of E tetra x.nb)>>`

(11/12/2011, 04:24 PM)Forehead Wrote: [ -> ]As a result,

E^P == TE[x + 1]*P

Now, we rearrange the equation a bit.

Log[E^P] == Log[TE[x + 1]]*Log[P]

I think this is the error. In going from the first to the second equation, it looks like you took, on the right hand side, $\log(ab) = \log(a) \log(b)$. But that is wrong. Instead, $\log(ab) = \log(a) + \log(b)$ and so your second equation should be

Log[E^P] == Log[TE[x + 1]] + Log[P]

If we continue your steps with this corrected equation, we get

P == TE[x] + Log[P]
E^P == E^(TE[x] + Log[P])
E^P == E^TE[x] E^Log[P]
E^P == TE[x+1] P
TE[x+1] P == TE[x+1] P

a tautological equation. Though perhaps you could solve for P in the first equation via the Lambert function?
(11/12/2011, 04:24 PM)Forehead Wrote: [ -> ]P == TE[x]*TE[x - 1]*TE[x - 2]*TE[x - 3] ...*TE[x - k]

E^P == TE[x + 1]*TE[x]*TE[x - 1]*TE[x - 2] ...*TE[x - k + 1]

I think there is a fatal problem here. In general, $\exp(ab) \ne \exp(a) \exp(b)$.
(11/12/2011, 10:02 PM)mike3 Wrote: [ -> ]I think there is a fatal problem here. In general, $\exp(ab) \ne \exp(a) \exp(b)$.

That would do it. The question arises, then: how do you compute E raised to the power of each term in a product individually?

EDIT: I just realized that there can't be a function $f$ such that $f(a*b) = \exp(a)\exp(b)$ and $f(x) = \exp(x)$. Back to the drawing board...
I think the first error is on page 2, where you make the transition from Product to E^Product, the second equation is false. You can't distribute exponentiation that way. You can with the identity:

$a = b + c$ to
$e^a = e^b e^c$

but that requires that you are starting with addition, but you start with multiplication.