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 TPID 4 tommy1729 Ultimate Fellow Posts: 1,614 Threads: 364 Joined: Feb 2009 06/18/2022, 10:40 PM (08/23/2012, 04:26 PM)tommy1729 Wrote: 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. ... ...unless they cancel each other. but notice the general solution is (also) b^(z + theta(z)) = b^z f(z,1) now since theta is entire and 1-periodic , this implies that b^theta(z) =  f(z,1) grows double exp. and therefore b^z f(z,1) grows double exponential ( double exponential dominates exponential ). So going up or down the strip we must get bigger and bigger values ; unbounded. QED tommy1729 « Next Oldest | Next Newest »

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