# Tetration Forum

Full Version: Cauchy integral also for b< e^(1/e)?
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andydude Wrote:
Ansus Wrote:Which Wiki?

I think he means Citizendium.
I mean

http://en.wikipedia.org/wiki/Cauchy%27s_...al_formula
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 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:

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