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 Cauchy integral also for b< e^(1/e)? bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 03/03/2009, 07:19 PM (This post was last modified: 03/03/2009, 07:25 PM by bo198214.) Hi Dmitrii, why not apply a similar Cauchy-integral technique for bases $b? 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. « Next Oldest | Next Newest »

 Messages In This Thread Cauchy integral also for b< e^(1/e)? - by bo198214 - 03/03/2009, 07:19 PM RE: Cauchy integral also for b< e^(1/e)? - by Kouznetsov - 03/22/2009, 08:29 PM RE: Cauchy integral also for b< e^(1/e)? - by Kouznetsov - 03/24/2009, 12:25 PM

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