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 TPID 4 tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 08/23/2012, 04:26 PM (This post was last modified: 08/24/2012, 03:10 PM by tommy1729.) 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 * tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 08/24/2012, 03:12 PM i edited post 1 proof is much better now. i make a new post to make it clear something changed. regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 03/28/2014, 12:04 AM (This post was last modified: 03/28/2014, 12:06 AM by 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 tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 04/26/2014, 12:24 PM Im not sure why this is still being ignored. TPID 4 is imho solved ! regards tommy1729 sheldonison Long Time Fellow    Posts: 641 Threads: 22 Joined: Oct 2008 04/27/2014, 04:37 AM (04/26/2014, 12:24 PM)tommy1729 Wrote: Im not sure why this is still being ignored. TPID 4 is imho solved ! regards tommy1729Agreed tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 04/27/2014, 01:40 PM Thank you  sheldonison Long Time Fellow    Posts: 641 Threads: 22 Joined: Oct 2008 06/15/2014, 06:22 PM (This post was last modified: 06/15/2014, 06:37 PM by sheldonison.) (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. - Sheldon tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 06/15/2014, 06:35 PM Indeed it applies to all real-analytic superfunctions ! For clarity its a uniqueness proof , not an existance proof. regards tommy1729 sheldonison Long Time Fellow    Posts: 641 Threads: 22 Joined: Oct 2008 06/15/2014, 06:42 PM (This post was last modified: 06/15/2014, 06:51 PM by sheldonison.) (06/15/2014, 06:35 PM)tommy1729 Wrote: Indeed it applies to all real-analytic superfunctions !How so? Here is an entire 1-cyclic function. 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. for any entire theta(z) function - Sheldon tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 06/15/2014, 07:09 PM (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 is an entire 1-cyclic function. 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. for any entire theta(z) function z + theta(z) takes on all values in the strip -1=

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