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Cauchy integral also for b< e^(1/e)?
#11
andydude Wrote:
Ansus Wrote:Which Wiki?

I think he means Citizendium.
I mean

http://en.wikipedia.org/wiki/Cauchy%27s_...al_formula
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#12
Ansus Wrote:

Since , 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:

for on the imaginary axis :

then it can also be used to iteratively compute the superexponential (base ) on the imaginary axis:



Any volunteer to implement this formula?

PS: This formula needs no assumption about the value of convergence of for , the only arbitrarity is the choosen branch of logarithm.
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#13
bo198214 Wrote:

I just see that the formula is not yet usable for implementation, but if we substitute then we have the same range of the imaganiray axis left and right:

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#14
Ansus Wrote:

Since , we derive f(x).

But now I doubt the formula is true.
Setting for example .
Then

But if I compute this numerially I get on the right side something close to 0.
While the left side is .

Where did you get this formula? Is it applicable only to certain functions?
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#15
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 , 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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