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 Cauchy integral also for b< e^(1/e)? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 03/29/2009, 10:37 AM andydude Wrote:Ansus Wrote:Which Wiki? I think he means Citizendium.I mean http://en.wikipedia.org/wiki/Cauchy%27s_...al_formula bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 04/09/2009, 10:35 AM (This post was last modified: 04/09/2009, 10:47 AM by bo198214.) Ansus Wrote:$\Delta[f]=\exp f - f$ Since $\Delta[f] = -\frac{1}{2\pi i} \int_{-1-i\infty}^{-1+i\infty} \frac{f(z+x)}{z(z-1)}\, dz$, we derive f(x). Though I dont know where from you got this formula, if I assume that the formula is correct and slightly reformulate it: $\exp(f(z_0)) - f(z_0) = -\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{f(it+z_0-1)}{(it-1)(it-2)} dt$ for $z_0$ on the imaginary axis $z_0=is$: $f(i s)-\exp(f(i s))= - \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{f(it+is-1)}{(it-1)(it-2)} dt$ then it can also be used to iteratively compute the superexponential (base $e$) on the imaginary axis: $\fbox{f(is)=\exp(f(i s)) + \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\log(f(i(t+s)))}{(it-1)(it-2)} dt}$ Any volunteer to implement this formula? PS: This formula needs no assumption about the value of convergence of $f$ for $z\to i\infty$, the only arbitrarity is the choosen branch of logarithm. bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 04/09/2009, 04:57 PM bo198214 Wrote:$\fbox{f(is)=\exp(f(i s)) + \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\log(f(i(t+s)))}{(it-1)(it-2)} dt}$ I just see that the formula is not yet usable for implementation, but if we substitute $t=t-s$ then we have the same range of the imaganiray axis left and right: $f(is)=\exp(f(i s)) + \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\log(f(it))}{(it-is-1)(it-is-2)} dt$ bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 04/09/2009, 06:10 PM Ansus Wrote:$\Delta[f]=\exp f - f$ Since $\Delta[f] = -\frac{1}{2\pi i} \int_{-1-i\infty}^{-1+i\infty} \frac{f(z+x)}{z(z-1)}\, dz$, we derive f(x). But now I doubt the formula is true. Setting for example $f=\exp$. Then $\exp(\exp(0)) - \exp(0) = -\frac{1}{2\pi i} \int_{-1-i\infty}^{-1+i\infty} \frac{\exp(z)}{z(z-1)}\, dz$ But if I compute this numerially I get on the right side something close to 0. While the left side is $e - 1$. Where did you get this formula? Is it applicable only to certain functions? bo198214 Administrator Posts: 1,391 Threads: 90 Joined: Aug 2007 04/24/2009, 05:29 PM Ansus Wrote:Quote:I mean without references I can not conclude that myself. It is Nörlund–Rice integral (http://en.wikipedia.org/wiki/N%C3%B6rlun...e_integral). bo198214 Wrote:Is it applicable only to certain functions? See Ansus, your formula is only applicable to (in the right halfplane) polynomially bounded functions $f$, you can read it in your reference. So it is not applicable here, and I dont need to wonder why the formula doesnt work. « Next Oldest | Next Newest »

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