Cauchy integral also for b< e^(1/e)?

[bo198214]

Hi Dmitrii,

why not apply a similar Cauchy-integral technique for bases \( b<e^{1/e} \)?
Instead of letting the parameter \( A \) go to infinity, we just choose \( A \) to be the period of the tetration, i.e.
\( A=\frac{2\pi}{\ln(\ln(a))} \) where \( a^{1/a}=b \).

In the version \( b>e^{1/e} \) you compute the values of \( f \) along the vertical line \( \gamma_V=[-iA,+iA] \) and compute the values to the left \( \gamma_L=[-1-iA,-1+iA] \) via \( \log(f(z)) \) and to the right on \( \gamma_R=[+1-iA,+1-iA] \) via \( \exp(f(z)) \).

But similarly we can compute the values on the horizontal line \( \gamma_H=[-1,+1] \) and conclude the values on the top line \( \gamma_T = [-1+iA,+1+iA] \) and on the bottom line \( \gamma_B=[-1-iA,+1-iA] \) equal to the values on \( \gamma_H \) via periodicity.

So summarized we had the recursion formula:
\(
f(z)=\frac{1}{2\pi i}\int_{\gamma_L+\gamma_R+\gamma_T+\gamma_B} \frac{f(w)}{w-z} dw=\frac{1}{2\pi i}\int_{\gamma_V} -\frac{\log(f(w))}{w-1-z} + \frac{\exp(f(w))}{w+1-z}dw + \frac{1}{2\pi i}\int_{\gamma_H} -\frac{f(w)}{w+iA-z} + \frac{f(w)}{w-iA-z}dw
\)

Though the question is whether this is faster than the direct limit formula.

[Kouznetsov]

why not apply a similar Cauchy-integral technique for bases \( b<e^{1/e} \)?
...

Because, in the case \( b>e^{1/e} \), the Cauchi integral seems to be the only robust way to evaluate the superexponential; but for \( b<e^{1/e} \), the direct expansion is much faster, giving the same 14 decimal digits.
If you mention, even for b=e, in the final implementation, I use the truncated series instead of the numerical approximation of the integrals. It is at least 2 orders of magnitude faster. There is difference, to get the figure within few seconds or tomorrow, ah?

Now I feel that I can evaluate any tetration with 14 digits, and I work making the evaluation fast; and the algorithm - simple and portable (the same pieces of code run with C++, Mathematica and Maple).
If you implement the integral formulas, we may compare the results with my expansion. I expect, of order of 14 digits will coincide.

In the recursion formula, I would specify the base, to avoid confusions:
\(
f(z)=\frac{1}{2\pi i}\oint \frac{f(w)}{w-z} \mathrm{d}w=\frac{1}{2\pi i}\int_{-iA}^{iA}\Big(-\frac{\log_b(f(w))}{w-1-z} + \frac{\exp_b(f(w))}{w+1-z}\Big) \mathrm{d}w + \frac{1}{2\pi i}\int_{-1}^1 \Big(-\frac{f(w)}{w+iA-z} + \frac{f(w)}{w-iA-z} \Big) \mathrm{d}w
\)
If you want your implementation to converge, in the last integral, \( f(w) \) should be replaced to \( \log_b(w+1) \) at the left hand tip of the interval of integration, and to the \( \exp_b(w-1) \) at the right hand tip, in order to avoid the evaluation of the function in a point close to the contour of integration.
Be careful with the order of updates of the estimates of the values of the function:
the speed of convergence (if at all) may depend on this order.

[Kouznetsov]

why not apply a similar Cauchy-integral technique for bases \( b<e^{1/e} \)?
..

Yes. Please, apply it, reproduce the fugure
http://en.citizendium.org/wiki/Image%3AH...rt2v01.jpg

and compare the values to those by the algorithm (updated version) posted at
http://en.citizendium.org/wiki/Tetration...seSqrt2v01

I expect we can get the agreement with 14 decimal digits..
Who wants to write a paper about it?

[Kouznetsov]

How did you evaluate Taylor series coefficients with Cauchy integral?

I have algorithm \( A \) that returns at least 14 decimal digits of function \( f \).
I express \( f(z) \approx \frac{1}{2\pi i} \oint \frac{ A(t) \mathrm{d}t}{t-z} \).
I differentiate this expression with respect to \( z \) so many times I need.

Usually, I use the circular contour of integration, centered at the point of expansion,
with radius slightly smaller than distance to the closest singularity.
This allows to work with good precision, close to the limit due to the rounding errors of the complex double variables.

[bo198214]

See also Wikipedia about Cauchy's formula.

There is the explicit formula for the \( n \)-th derivative given.

[bo198214]

What about this equation?

\(
f(x)=\ln f(x)-\frac{1}{2\pi i}
\int_{-i\infty}^{+i\infty} \frac{f(z+x)}{z(z-1)}\, dz
\)

Can it be useful?

Looks damn good. Simpler than Dmitrii's default formula.

[bo198214]

Yes, but that formula is recursive.

But that does not matter as long we have once computed f (on the imaginary axis).

[Kouznetsov]

But how to compute it on the imaginary axis?

for b< e^(1/e), the algorithm of the evaluation on the imaginary axis through the integral Cauchi equation is not yet ready.
While, you may recover the function from the asymptotic behavior.
I believe, it is even better than Cauchi.

For b=sqrt(2), the expansion is
F(z)=2- u - v_2 u^2 -v_3 u^3...
where
u=u(z)=exp((z-t)Q)
Substitute the expansion into equation F(z+1)=exp_b(F(z)).
Calculate the Taylor expansion of F(z+1)-exp_b(F(z)) with respect to u,
Taking into account that u(z+1)=Qu(z).
Set to zero the coefficients in this expansion.
These equations determine Q, v_2, v_3, ...
Recover t from the condition F(0)=1.
Then, using F(z-1)=log_b(F(z)) ,
you can evaluate F everywhere, not only along the imaginary axis.

You may use the representation through the asymptotic for evaluation of few first derivatives. If you want a preise evaluation of many coefficients of the Taylor expansion, better, use the Cauchi; you may evaluate a hundred coefficients within a second.

[Kouznetsov]

r b< e^(1/e), the algorithm of the evaluation on the imaginary axis through the integral Cauchi equation is not yet ready.
While, you may recover the function from the asymptotic behavior.

There is a plenty of formulas for tetration on this forum. Why not to use them?

Because they did not allow to plot the map of tetration.
Because they do not give 14 decimal digits.
Because they are slow in the evaluation.

better, use the Cauchi; you may evaluate a hundred coefficients within a second.

How?

Use formulas for the derivatives through the Cauchi integrals from Wiki.
Chose the circular contour of integration.
Distribute some thoudand points uniformly at this contour.
Spend halfsecond of CPU time to evaluate function in these points.
Spend another halfsecond to approximate the Cauchi integrals with the finite sums
and get the coefficients in the Taylor expansion.
Compare the truncated Taylor expansion to the values by some of algorithms you have mentioned.

[andydude]

Which Wiki?

I think he means Citizendium.