11/06/2009, 04:15 AM
I've also been toying with this, too. It appears, however, that it continues to real values, not complex values, for \( b > e^{1/e} \).
Consider the following “regular iteration” limit formula:
http://math.eretrandre.org/tetrationforu...10#pid1610
\( ^{z} b = \exp_b^z(1) = \lim_{n \rightarrow \infty} \log_b^n \left(F\left(1 - \log(F)^z\right) + \log(F)^z \exp_b^n(1)\right) \)
where \( F = \frac{-W\left(-\log(b)\right)}{\log(b)} \) is the attracting fixed point, i.e. \( ^{\infty} b \).
There's also a series formula, apparently (with respect to "w" in \( \exp_b^z(w) \), so setting w = 1 yields the tetrational), but I haven't yet figured out how one is supposed to evaluate the general coefficient. How can that be done?
To test the analytic continuation, we can use the Cauchy integral: if \( f(z) \) is analytic, then we derive the powerseries coefficients at \( z_0 \) via
\( a_n = \frac{1}{2\pi i} \oint \frac{f(z)}{(z - z_0)^n} dz \)
and
\( f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n \).
This seems to provide a more efficient algorithm for the recovery of the coefficients, than straight numerical differentiation from the difference quotient (which seems to require more rapidly-escalating levels of numerical precision).
We can now choose some \( z_0 \) close to \( e^{1/e} \), set \( f(z) = {}^t z \) for some fractional tower \( t \) where \( ^{t} z \) is obtained from the regular formula, and a path that encircles it, but does not leave the kidneybean ("Shell-Thron" region) of convergence, e.g. a small circle round the point. Then, by increasing n, we obtain the Taylor coefficients. For \( f(z) = {}^{1/2} z \), expanded about \( z_0 = 1.42 \), using a circle of radius 0.01, we get the following estimates for the first 25 coefficients:
For \( z = 1.5 = \frac{3}{2} \), we can use this get \( ^{\frac{1}{2}} \left(\frac{3}{2}\right) \approx 1.28087727794 \), which is real, not complex. How does that agree with other methods of tetration for bases greater than \( e^{1/e} \)? This series should have radius of convergence 0.42, determined by the distance to the nearest singularity/branchpoint, which is at z = 1.
I'm not sure of a formal proof of the "continuability", though one approach may be to try and differentiate the regular iteration formula, then prove that the limit of the derivative as \( b \rightarrow e^{1/e} \) converges -- in order for it to switch to non-real complex values as \( b = e^{1/e} \) is passed, that point would have to be some sort of singularity, like a branch point, and so the function would not be differentiable there, and if it is, then that is not the case.
I'll see if maybe I can get some graphs on the complex plane but calculating the regular iteration is a bear as it requires lots of numerical precision, at least for the limit formula. Maybe that series formula would be better?
Consider the following “regular iteration” limit formula:
http://math.eretrandre.org/tetrationforu...10#pid1610
\( ^{z} b = \exp_b^z(1) = \lim_{n \rightarrow \infty} \log_b^n \left(F\left(1 - \log(F)^z\right) + \log(F)^z \exp_b^n(1)\right) \)
where \( F = \frac{-W\left(-\log(b)\right)}{\log(b)} \) is the attracting fixed point, i.e. \( ^{\infty} b \).
There's also a series formula, apparently (with respect to "w" in \( \exp_b^z(w) \), so setting w = 1 yields the tetrational), but I haven't yet figured out how one is supposed to evaluate the general coefficient. How can that be done?
To test the analytic continuation, we can use the Cauchy integral: if \( f(z) \) is analytic, then we derive the powerseries coefficients at \( z_0 \) via
\( a_n = \frac{1}{2\pi i} \oint \frac{f(z)}{(z - z_0)^n} dz \)
and
\( f(z) = \sum_{n=0}^{\infty} a_n (z - z_0)^n \).
This seems to provide a more efficient algorithm for the recovery of the coefficients, than straight numerical differentiation from the difference quotient (which seems to require more rapidly-escalating levels of numerical precision).
We can now choose some \( z_0 \) close to \( e^{1/e} \), set \( f(z) = {}^t z \) for some fractional tower \( t \) where \( ^{t} z \) is obtained from the regular formula, and a path that encircles it, but does not leave the kidneybean ("Shell-Thron" region) of convergence, e.g. a small circle round the point. Then, by increasing n, we obtain the Taylor coefficients. For \( f(z) = {}^{1/2} z \), expanded about \( z_0 = 1.42 \), using a circle of radius 0.01, we get the following estimates for the first 25 coefficients:
Code:
a_0 ~ 1.24622003310
a_1 ~ 0.447921100148
a_2 ~ -0.194428566238
a_3 ~ 0.143167873861
a_4 ~ -0.144967399774
a_5 ~ 0.182159301224
a_6 ~ -0.263407426426
a_7 ~ 0.417098477762
a_8 ~ -0.702204147888
a_9 ~ 1.23535078363
a_10 ~ -2.24705897139
a_11 ~ 4.19678269601
a_12 ~ -8.00913024091
a_13 ~ 15.5621057277
a_14 ~ -30.7029302261
a_15 ~ 61.3746543453
a_16 ~ -124.093757768
a_17 ~ 253.427621734
a_18 ~ -522.152429631
a_19 ~ 1084.31773542
a_20 ~ -2267.61147731
a_21 ~ 4772.09902388
a_22 ~ -10098.4528601
a_23 ~ 21464.8938685
a_24 ~ -45852.6753827For \( z = 1.5 = \frac{3}{2} \), we can use this get \( ^{\frac{1}{2}} \left(\frac{3}{2}\right) \approx 1.28087727794 \), which is real, not complex. How does that agree with other methods of tetration for bases greater than \( e^{1/e} \)? This series should have radius of convergence 0.42, determined by the distance to the nearest singularity/branchpoint, which is at z = 1.
I'm not sure of a formal proof of the "continuability", though one approach may be to try and differentiate the regular iteration formula, then prove that the limit of the derivative as \( b \rightarrow e^{1/e} \) converges -- in order for it to switch to non-real complex values as \( b = e^{1/e} \) is passed, that point would have to be some sort of singularity, like a branch point, and so the function would not be differentiable there, and if it is, then that is not the case.
I'll see if maybe I can get some graphs on the complex plane but calculating the regular iteration is a bear as it requires lots of numerical precision, at least for the limit formula. Maybe that series formula would be better?
