09/20/2007, 05:46 AM
Start with the two primary fixed points of base e: \( 0.318131505204764\ \pm\ 1.3372357014306894 i \)
Now, let's perform hyperbolic iteration from both of them. To see this better, we have to look at the slog function.
If hyperbolic iteration at a fixed point looks like exponentiation, then the slog function in the immediate vicinity of the fixed point will look like a logarithm. The base of the logarithm is the multiplicative constant of the distance from the fixed point of two consecutive integer iterates.
For base e, this distance, of course, is the fixed point itself.
Therefore, to a first approximation, the slog function should behave like \( \log_{a_0}\left(z-a_0\right) + \log_{\overline{a_0}}\left(z-\overline{a_0}\right) \), where a_0 is the primary fixed point, and the overline indicates the complex conjugate.
Notice that these two functions are complex conjugates of each other for real z. Therefore, the imaginary parts will cancel, giving a real-valued function for real inputs. This is the key to understanding how continuous iteration from complex fixed points will nonetheless yield a real-valued function for real inputs.
Notice that far from the fixed points, this function will behave very little like the true slog function. But in the vicinity of either, it's should be a very good approximation.
Adding in the other fixed points seems logical, and yields a terribly surprising result: rational coefficients of the power series! This is dependent on how one calculates the logarithms at the other fixed points (i.e., each logarithm has to be calculated in the branch of the particular fixed point).
The power series using the sum of the logarithmic functions at all the fixed points has the following first few coefficients:
\(
\sum_{k=0}^{\infty} \log_{a_k}\left(z-a_k\right) + \log_{\overline{a_k}}\left(z-\overline{a_k}\right)\ =\ C_0 + \\
\\[8pt]
\\
\frac{1}{1^2 \times 1}z^{1} + \\
\\[8pt]
\\
\frac{1}{2^2 \times 1}z^{2}\ + \\
\\[8pt]
\\
\frac{-1}{3^2 \times 1}z^{3}\ + \\
\\[8pt]
\\
\frac{-3}{4^2 \times 2}z^{4}\ + \\
\\[8pt]
\\
\frac{1}{5^2 \times 4}z^{5}\ + \\
\\[8pt]
\\
\frac{31}{6^2 \times 24}z^{6}\ + \\
\\[8pt]
\\
\frac{29}{7^2 \times 90}z^{7}\ + \\
\\[8pt]
\\
\frac{-63}{8^2 \times 80}z^{8}\ + \\
\\[8pt]
\\
\frac{-2087}{9^2 \times 4032}z^{9}\ + \\
\\[8pt]
\\
\frac{35953}{10^2 \times 120960}z^{10}\ + \\
\\[8pt]
\\
\frac{137918}{11^2 \times 302400}z^{11}\ + \\
\\[8pt]
\\
\frac{-146387}{12^2 \times 3628800}z^{12}\ + \\
\\[8pt]
\\
\frac{10264123}{13^2 \times 34214400}z^{13}
\)
Here, C_0 is rather arbitrary, and could just as well be -1.
Note that besides the meagre pattern I've already extracted, the denominators are all fairly composite. I suspect there is in fact a very tidy pattern to the denominators, which is obscured because these fractions are reduced to lowest terms.
At any rate, that the sums of powers of reciprocals of complex irrational fixed points would lead to rational coefficients just totally blew me away, since any particular logarithm in the sum has irrational coefficients. It's like magic to me that the all of them put together lead to rational coefficients.
I'm not entirely sure that this series of rational coefficients will be particular useful in "the" solution, because I'm not sure whether each logarithm should be computed relative to its own branch. By this I mean:
\( \log_b(z) = \frac{\ln(z)}{\ln(b)} \)
Notice the division by ln(b). Should this ln(b) be equal to b? For example, should we consider ln(2.06227773+7.58863118i) to be 2.06227773+7.58863118i or 2.06227773+1.30544587i?
If the former, I suspect this power series will play an integral role in the "correct" solution, though it's obviously not correct alone. There would appear to be other singularities, other functions embedded within the slog. Perhaps they too are logarithms, but I'm not sure yet.
If the latter, then this curiosity will have to remain a curiosity, exquisitely interesting and unfortunately not of much use.
Now, let's perform hyperbolic iteration from both of them. To see this better, we have to look at the slog function.
If hyperbolic iteration at a fixed point looks like exponentiation, then the slog function in the immediate vicinity of the fixed point will look like a logarithm. The base of the logarithm is the multiplicative constant of the distance from the fixed point of two consecutive integer iterates.
For base e, this distance, of course, is the fixed point itself.
Therefore, to a first approximation, the slog function should behave like \( \log_{a_0}\left(z-a_0\right) + \log_{\overline{a_0}}\left(z-\overline{a_0}\right) \), where a_0 is the primary fixed point, and the overline indicates the complex conjugate.
Notice that these two functions are complex conjugates of each other for real z. Therefore, the imaginary parts will cancel, giving a real-valued function for real inputs. This is the key to understanding how continuous iteration from complex fixed points will nonetheless yield a real-valued function for real inputs.
Notice that far from the fixed points, this function will behave very little like the true slog function. But in the vicinity of either, it's should be a very good approximation.
Adding in the other fixed points seems logical, and yields a terribly surprising result: rational coefficients of the power series! This is dependent on how one calculates the logarithms at the other fixed points (i.e., each logarithm has to be calculated in the branch of the particular fixed point).
The power series using the sum of the logarithmic functions at all the fixed points has the following first few coefficients:
\(
\sum_{k=0}^{\infty} \log_{a_k}\left(z-a_k\right) + \log_{\overline{a_k}}\left(z-\overline{a_k}\right)\ =\ C_0 + \\
\\[8pt]
\\
\frac{1}{1^2 \times 1}z^{1} + \\
\\[8pt]
\\
\frac{1}{2^2 \times 1}z^{2}\ + \\
\\[8pt]
\\
\frac{-1}{3^2 \times 1}z^{3}\ + \\
\\[8pt]
\\
\frac{-3}{4^2 \times 2}z^{4}\ + \\
\\[8pt]
\\
\frac{1}{5^2 \times 4}z^{5}\ + \\
\\[8pt]
\\
\frac{31}{6^2 \times 24}z^{6}\ + \\
\\[8pt]
\\
\frac{29}{7^2 \times 90}z^{7}\ + \\
\\[8pt]
\\
\frac{-63}{8^2 \times 80}z^{8}\ + \\
\\[8pt]
\\
\frac{-2087}{9^2 \times 4032}z^{9}\ + \\
\\[8pt]
\\
\frac{35953}{10^2 \times 120960}z^{10}\ + \\
\\[8pt]
\\
\frac{137918}{11^2 \times 302400}z^{11}\ + \\
\\[8pt]
\\
\frac{-146387}{12^2 \times 3628800}z^{12}\ + \\
\\[8pt]
\\
\frac{10264123}{13^2 \times 34214400}z^{13}
\)
Here, C_0 is rather arbitrary, and could just as well be -1.
Note that besides the meagre pattern I've already extracted, the denominators are all fairly composite. I suspect there is in fact a very tidy pattern to the denominators, which is obscured because these fractions are reduced to lowest terms.
At any rate, that the sums of powers of reciprocals of complex irrational fixed points would lead to rational coefficients just totally blew me away, since any particular logarithm in the sum has irrational coefficients. It's like magic to me that the all of them put together lead to rational coefficients.
I'm not entirely sure that this series of rational coefficients will be particular useful in "the" solution, because I'm not sure whether each logarithm should be computed relative to its own branch. By this I mean:
\( \log_b(z) = \frac{\ln(z)}{\ln(b)} \)
Notice the division by ln(b). Should this ln(b) be equal to b? For example, should we consider ln(2.06227773+7.58863118i) to be 2.06227773+7.58863118i or 2.06227773+1.30544587i?
If the former, I suspect this power series will play an integral role in the "correct" solution, though it's obviously not correct alone. There would appear to be other singularities, other functions embedded within the slog. Perhaps they too are logarithms, but I'm not sure yet.
If the latter, then this curiosity will have to remain a curiosity, exquisitely interesting and unfortunately not of much use.
~ Jay Daniel Fox
