Cauchy integral also for b< e^(1/e)? bo198214 Administrator Posts: 1,624 Threads: 103 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 $be^{1/e}$ 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 Fellow   Posts: 151 Threads: 9 Joined: Apr 2008 03/22/2009, 08:29 PM (This post was last modified: 03/22/2009, 08:41 PM by Kouznetsov.) bo198214 Wrote:why not apply a similar Cauchy-integral technique for bases $be^{1/e}$, the Cauchi integral seems to be the only robust way to evaluate the superexponential; but for \( b

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