11/12/2011, 04:24 PM

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.

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.