# Tetration Forum

Full Version: TPID 4
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2 3 4
here i give a ( nonunique short ) proof of TPID 4.

remember that entire taylor series are coo everywhere. ( infinitely differentiable for all finite complex )

let f(z,1) be an entire periodic function with f(0,1)=f(1,1)=1 and period 1.

and f(z,1) is not identically 1 for all z.

we will prove that for complex b with arg(b) <> 0 , the only solution to the equations is f(z,1) * b^z and hence the proof follows.

let k and n be positive integers.

f(0) = 1
f(z+k) = b^k f(z)
f = entire

then

take the derivative of the equation f(z+k) = b^k f(z) on both sides

f ' (z+k) = b^k f ' (z)
again
f '' (z+k) = b^k f '' (z)

and in general

f^(n) (z+k) = b^k f^(n) (z)

hence because of taylors theorem we must conclude

f(z) = f(0) * f(z,1) * b^z in the neighbourhood of 0.

but since f is entire it must be true everywhere and f(0) = 1 hence

f(z) = f(z,1) b^z

for all z.

if arg(b) <> 0 then the period of b^z does not have Re <> 0 and hence b^z is unbounded on the strip.

if f(z) needs to be bounded and b^z is not bounded , this implies that f(z,1) needs to be bounded.

but this is impossible since f(z,1) has a real period and is entire , it must be unbounded on the strip.

( remember f(z,1) =/= 1 everywhere by definition )

the product of two functions unbounded in the same region must be unbounded in that region.

QED

regards

tommy1729

* post has been edited *
i edited post 1

proof is much better now.

i make a new post to make it clear something changed.

regards

tommy1729
Notice 0 < b^z < oo for finite complex z !

hence since b^z is unbounded to make f(z,1) b^z bounded we need f(z,1) going to 0 near the imaginary limits +/- oo i.

But f(z,1) b^z needs to be entire. And hence (by the above) f(z,1) cannot be entire.

Thus if f(z,1) b^z needs to be entire and f(z,1) goes to oo (f(z,1) goes to oo somewhere because its not entire ) ,we MUST conclude we need points z1 where b^z1 are 0.

But b^z is NEVER 0 in any strip.

This completes and clarfies the proof of post nr. 1.

I see that TPID 4 is still considered unproved in the open questions page.

I hope this post convinces everyone that I did indeed have proven TPID 4.

regards

tommy1729
Im not sure why this is still being ignored.

TPID 4 is imho solved !

regards

tommy1729
(04/26/2014, 12:24 PM)tommy1729 Wrote: [ -> ]Im not sure why this is still being ignored.

TPID 4 is imho solved !

regards

tommy1729
Agreed
Thank you
(03/28/2014, 12:04 AM)tommy1729 Wrote: [ -> ]Notice 0 < b^z < oo for finite complex z !

hence since b^z is unbounded to make f(z,1) b^z bounded we need f(z,1) going to 0 near the imaginary limits +/- oo i.

But f(z,1) b^z needs to be entire. And hence (by the above) f(z,1) cannot be entire.

Thus if f(z,1) b^z needs to be entire and f(z,1) goes to oo (f(z,1) goes to oo somewhere because its not entire ) ,we MUST conclude we need points z1 where b^z1 are 0.

But b^z is NEVER 0 in any strip.

I viewed this proof only applied to b^z; that any other solution of b^(z+theta(z)) would be unbounded on the strip, and that's how I interpreted the proof. I didn't view this as a proof relating to Tetration, and I didn't think the OP viewed this as applying to Kneser's tetraton solution, a non-entire function, since tet(z) and sexp(z) are never mentioned in the post. But for b^z, you can multiply by a 1-cyclic function and still have a solution to b^z, so I interpreted the OP as intending this post to only apply to b^z. You can't multiply tet(z) by a 1-cyclic function and get an alternative solution to tet(z). But maybe I'm missing something.
Indeed it applies to all real-analytic superfunctions !

For clarity its a uniqueness proof , not an existance proof.

regards

tommy1729
(06/15/2014, 06:35 PM)tommy1729 Wrote: [ -> ]Indeed it applies to all real-analytic superfunctions !
How so? Here $\theta(z)$ is an entire 1-cyclic function.
$\text{tet}(z+1)=\exp(\text{tet}(z))$
$\text{tet}(z+1)\times \theta(z) \;<>\; \exp(\theta(z) \times \text{tet}(z))\;\;$ unless theta(z)=1 everywhere

But if you replace tet(z) with b^z, then it works, so that's how I interpreted the Op's proof, given that the proof never mentioned superfunctions or anything like that.

$b^{z+1}\times \theta(z) \;=\; b^{\theta(z) \times b^z}\;\;$ for any entire theta(z) function

(06/15/2014, 06:42 PM)sheldonison Wrote: [ -> ]
(06/15/2014, 06:35 PM)tommy1729 Wrote: [ -> ]Indeed it applies to all real-analytic superfunctions !
How so? Here $\theta(z)$ is an entire 1-cyclic function.
$\text{tet}(z+1)=\exp(\text{tet}(z))$
$\text{tet}(z+1)\times \theta(z) \;<>\; \exp(\theta(z) \times \text{tet}(z))\;\;$ unless theta(z)=1 everywhere

But if you replace tet(z) with b^z, then it works, so that's how I interpreted the Op's proof, given that the proof never mentioned superfunctions or anything like that.

$b^{z+1}\times \theta(z) \;=\; b^{\theta(z) \times b^z}\;\;$ for any entire theta(z) function

z + theta(z) takes on all values in the strip -1=<Re(z)=<1 apart from possibly one value.
This follows from picard's little theorem and the periodicity of theta(z).

So since in the strip we take on all complex values (apart from 1 possible value) it follows that the range of sexp in that strip is the same range as sexp.

since the range of sexp is unbounded , than so is the range of sexp(strip).

Q.e.d.

The similarity with the unboundedness of the theta in the OP is striking.

Hope that clarifies.

regards

tommy1729
Pages: 1 2 3 4