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 Can we prove these coefficients must be constant? JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 06/03/2012, 09:17 PM (This post was last modified: 06/03/2012, 09:24 PM by JmsNxn.) I have boiled down the recursion of analytic hyper operators into a formula based on their coefficients. If we write these coefficients as follows: $s \in \mathbb{C}\,\,s = \sigma + it$ $a \Delta_0 b = a + b$ $a \Delta_s (a \Delta_{s+1} b) = a \Delta_{s+1} (b+1)$ $a \Delta_s b = \sum_{i=0}^{\infty} \chi_i(a,b) s^i$ We can write the recursive formula; without giving a proof for it (it just requires a few series rearrangement); as: $\chi_n(a,b + 1) = \sum_{i=0}^{\infty} \chi_{n+i}(a, a \Delta_{s} b) \frac{(n+i)!}{n! i!}(-1)^i\$ As you can see; this appears very off. $s$ can vary freely and the result on the L.H.S. doesn't change at all. However, it's being summed across an infinite series so that may compensate. But I wonder if declaring, that since $a \Delta_{s} b$ takes on every value in between $a + b$ and $a^b$; at least; we can say over that interval $c \in [a+b, a^b]$ $\chi_{n+i}(a, c) = \text{Constant}$ Since we can set $n=0$ this implies a strict contradiction: $\sum_{i=0}^{\infty} \chi_{i}(a,c_0)s^i = \sum_{i=0}^{\infty} \chi_{i}(a,c_1)s^i$ This is a contradiction because it implies $a \Delta_s c$ is constant and therefore constant for all b in $a \Delta_s b$. This would imply there is no analytic continuation of hyper operators! At least, not representable by its Taylor series. I didn't write out the proof because I'm stuck and I'm curious if it's justifiable to do that last move, or if there is some other routine I can go about to prove the constancy of these coefficients. If hyper operators aren't analytic; and hopefully I can prove not continuous; I have a separate way of defining them that admit a discrete solution with a more number theoretical algebraic approach. « Next Oldest | Next Newest »

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