• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 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: 1,389 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: 1,389 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: 1,389 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: 1,389 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 »

 Possibly Related Threads... Thread Author Replies Views Last Post Where is the proof of a generalized integral for integer heights? Chenjesu 2 1,364 03/03/2019, 08:55 AM Last Post: Chenjesu Kouznetsov-Tommy-Cauchy method tommy1729 0 2,205 02/18/2015, 07:05 PM Last Post: tommy1729 Problem with cauchy method ? tommy1729 0 2,063 02/16/2015, 01:51 AM Last Post: tommy1729 Could be tetration if this integral converges JmsNxn 41 46,800 05/13/2014, 01:58 PM Last Post: JmsNxn [integral] How to integrate a fourier series ? tommy1729 1 2,770 05/04/2014, 03:19 PM Last Post: tommy1729 Some integral transforms related to tetration JmsNxn 0 2,009 05/02/2013, 07:54 PM Last Post: JmsNxn (draft) integral idea tommy1729 0 2,390 06/25/2011, 10:17 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)