Breaking New Ground In The Quest For The "Analytical" Formula For Tetration.
#2
(03/20/2011, 03:45 AM)mike3 Wrote: But beyond here we run out of luck. The next step would be to set the sum for \( N_n \) in this and try to solve for a recurrence for the \( \mathrm{smag}_{n,j} \) and then try to come up with an explicit non-recursive formula for that, but it gets hairy and we don't have an expression for the expanded-out product there. So the first order of business here would be to find the explicit, non-recursive formula for the coefficients of

\( \prod_{j=m}^{n-2} (1 - u^j) \).

Any ideas? The degree of the resulting polynomial is \( \sum_{j=m}^{n-2} j = \frac{(n-1)(n-2) - m(m-1)}{2} \).

Also, \( N_1 = -1 \). We don't set it to 1, since if you look at the denominator formula, you'll notice it is equal to -1 there and we have \( \chi_1 = 1 \).

Add:
For this product, we can write

\( \prod_{j=m}^{n-2} (1 - u^j) = \sum_{j=m}^{\frac{(n-1)(n-2) - m(m-1)}{2}} R_{m,n-2,j} u^j \)

where \( R_{m,n,j} \) is another sequence of magic numbers defined for \( n \ge m \). To find the equations for \( R_{m,n,j} \), we use

\( P_{m,n} = \prod_{j=m}^{n} (1 - u^j) \)

then

\( P_{m,n} = (1 - u^n) P_{m,n-1} \).

Now, letting \( P_{m,n} = \sum_{j=0}^{\frac{n(n+1) - m(m-1)}{2}} R_{m,n,j} u^j \), we have

\(
\begin{align}
\sum_{j=0}^{\frac{n(n+1) - m(m-1)}{2}} R_{m,n,j} u^j &= (1 - u^n) \sum_{j=0}^{\frac{n(n-1) - m(m-1)}{2}} R_{m,n-1,j} u^j\\
&= \left(\sum_{j=0}^{\frac{n(n-1) - m(m-1)}{2}} R_{m,n-1,j} u^j\right) - \left(\sum_{j=0}^{\frac{n(n-1) - m(m-1)}{2}} R_{m,n-1,j} u^{j+n}\right)\\
&= \left(\sum_{j=0}^{\frac{n(n-1) - m(m-1)}{2}} R_{m,n-1,j} u^j\right) - \left(\sum_{j=n}^{\frac{n(n+1) - m(m-1)}{2}} R_{m,n-1,j-n} u^j\right)
\end{align}
\)

The last formula allows for equating coefficients, yielding a three-piece recurrence:

\( R_{m,n,j} = R_{m,n-1,j} \), for \( 0 \le j < n \)
\( R_{m,n,j} = R_{m,n-1,j} - R_{m,n-1,j-n} \), for \( n \le j \le \frac{n(n-1) - m(m-1)}{2} \)
\( R_{m,n,j} = -R_{m,n-1,j-n} \), for \( \frac{n(n-1) - m(m-1)}{2} < j \le \frac{n(n+1) - m(m-1)}{2} \)

with the appropriate initial conditions, e.g.

\( R_{m,m,0} = 1 \),
\( R_{m,m,j} = 0 \), for \( 0 < j < m \),
\( R_{m,m,m} = -1 \).

Not sure if this is of any use or helps in getting explicit coefficient formulas, though.
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Messages In This Thread
RE: Breaking New Ground In The Quest For The "Analytical" Formula For Tetration. - by mike3 - 03/28/2011, 11:51 AM

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