Cauchy integral also for b< e^(1/e)?

Cauchy integral also for b< e^(1/e)? - Printable Version

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RE: Cauchy integral also for b< e^(1/e)? - Kouznetsov - 03/29/2009

andydude Wrote:
Ansus Wrote:Which Wiki?

I think he means Citizendium.

I mean

http://en.wikipedia.org/wiki/Cauchy%27s_integral_formula

RE: Cauchy integral also for b< e^(1/e)? - bo198214 - 04/09/2009

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.

RE: Cauchy integral also for b< e^(1/e)? - bo198214 - 04/09/2009

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:

\( f(is)=\exp(f(i s)) + \frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{\log(f(it))}{(it-is-1)(it-is-2)} dt \)

RE: Cauchy integral also for b< e^(1/e)? - bo198214 - 04/09/2009

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?

RE: Cauchy integral also for b< e^(1/e)? - bo198214 - 04/24/2009

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%B6rlund%E2%80%93Rice_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.