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+--- Thread: Can we prove these coefficients must be constant? (/showthread.php?tid=739)
Can we prove these coefficients must be constant? - JmsNxn - 06/03/2012
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.