Hmm. It looks like this may not work, unfortunately: apparently though the continuum sum has the property
\( \sum_{n=0}^{-x-1} f(n) = \sum_{n=1}^{x} -f(-n) \)
for integer values of \( x \), it seems to not necessarily hold for non-integer ones. Thus a different way will be needed to generalize
\( \sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x) \)
to ideally, general analytic functions at least.
However, we can apply the formula to those bases who converge to a fixed point at positive infinity, like how the regular formula can. Probably not, however, for the endpoints \( b = e^{-e} \) and \( b = e^{1/e} \) as their convergence rate is too slow (hyperbolic at best) for the above infinite sum to exist.
Base \( \sqrt{2} \)
Consider, for example, \( b = \sqrt{2} \). We can already apply the continuum sum formula above in the "forward" direction in this case due to the convergence toward the fixed point 2 as the tower grows.
We use the summing formula
\( \sum_{n=1}^{x} f(n) = xL + \sum_{n=1}^{\infty} f(n) - f(n + x) \).
and since for tetration/hyper4, \( f(0) = 1 \), we get
\( \sum_{n=0}^{x-1} f(x) = 1 + (x-1)L + \sum_{n=1}^{\infty} f(n) - f(n + x - 1) \).
We can then approximate a solution using the iterating form of Ansus' formula:
\( f(x) = K \int_{-1}^{x} (\log(b))^t \exp_b\left(\sum_{k=0}^{t-1} f(k)\right) dt \).
by sampling an interval of length 1 on the real axis (here, I use [-1, 0]) with nodes for an approximation (this test was done with quadratic spline), and then use the tetration recurrence formulas to fill up the real axis and iterate over the nodes with the continuum sums and the integrals. The constant K is determined at each step by the reciprocal of the value of the function at 0, or the last node in the array.
Using 129 nodes and 8 iterations, starting with a linear initial guess and normalizing the formula at each iteration using the last node's value, and capping the infinite sum at \( n = 100 \) I get
\( ^{1/2} \sqrt{2} \approx 1.243621 \)
which agrees well with the value obtained via the regular iteration. More nodes could be used to get more accuracy. A graph constructed this way is shown here:
(x-axis = tower, y-axis = value of tetration)
Base \( \frac{1}{4} \)
We do this the same way as before. The formula still converges, only this time to complex values instead of real ones. The fixed point is \( L = \frac{1}{2} \). For example, we have
\( ^{\frac{1}{2}} \left(\frac{1}{4}\right) \approx 0.460022 + 0.311787i \)
which, again agrees with the value from the regular formula. Graphs of this function for real height are given here:
Real part:
(x-axis = tower, y-axis = real part of value of tetration)
Imag part:
(x-axis = tower, y-axis = imag part of value of tetration)
\( \sum_{n=0}^{-x-1} f(n) = \sum_{n=1}^{x} -f(-n) \)
for integer values of \( x \), it seems to not necessarily hold for non-integer ones. Thus a different way will be needed to generalize
\( \sum_{n=1}^{x} f(n) = \sum_{n=1}^{\infty} f(n) - f(n + x) \)
to ideally, general analytic functions at least.
However, we can apply the formula to those bases who converge to a fixed point at positive infinity, like how the regular formula can. Probably not, however, for the endpoints \( b = e^{-e} \) and \( b = e^{1/e} \) as their convergence rate is too slow (hyperbolic at best) for the above infinite sum to exist.
Base \( \sqrt{2} \)
Consider, for example, \( b = \sqrt{2} \). We can already apply the continuum sum formula above in the "forward" direction in this case due to the convergence toward the fixed point 2 as the tower grows.
We use the summing formula
\( \sum_{n=1}^{x} f(n) = xL + \sum_{n=1}^{\infty} f(n) - f(n + x) \).
and since for tetration/hyper4, \( f(0) = 1 \), we get
\( \sum_{n=0}^{x-1} f(x) = 1 + (x-1)L + \sum_{n=1}^{\infty} f(n) - f(n + x - 1) \).
We can then approximate a solution using the iterating form of Ansus' formula:
\( f(x) = K \int_{-1}^{x} (\log(b))^t \exp_b\left(\sum_{k=0}^{t-1} f(k)\right) dt \).
by sampling an interval of length 1 on the real axis (here, I use [-1, 0]) with nodes for an approximation (this test was done with quadratic spline), and then use the tetration recurrence formulas to fill up the real axis and iterate over the nodes with the continuum sums and the integrals. The constant K is determined at each step by the reciprocal of the value of the function at 0, or the last node in the array.
Using 129 nodes and 8 iterations, starting with a linear initial guess and normalizing the formula at each iteration using the last node's value, and capping the infinite sum at \( n = 100 \) I get
\( ^{1/2} \sqrt{2} \approx 1.243621 \)
which agrees well with the value obtained via the regular iteration. More nodes could be used to get more accuracy. A graph constructed this way is shown here:
(x-axis = tower, y-axis = value of tetration)
Base \( \frac{1}{4} \)
We do this the same way as before. The formula still converges, only this time to complex values instead of real ones. The fixed point is \( L = \frac{1}{2} \). For example, we have
\( ^{\frac{1}{2}} \left(\frac{1}{4}\right) \approx 0.460022 + 0.311787i \)
which, again agrees with the value from the regular formula. Graphs of this function for real height are given here:
Real part:
(x-axis = tower, y-axis = real part of value of tetration)
Imag part:
(x-axis = tower, y-axis = imag part of value of tetration)
