Cauchy integral also for b< e^(1/e)?
#1
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.
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Messages In This Thread
Cauchy integral also for b< e^(1/e)? - by bo198214 - 03/03/2009, 07:19 PM

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