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contour of what ?
#1
maybe i didnt pay attention but i have a big question.

many methods to compute a certain tetration depend on contour integrals.

for instance kouznetsov defines his coefficients by contour integrals.

but it is not clear to me ?!?

contour of what ? of the function we are defining with the contour ?

selfreference ? or did i miss something ?

i know contour integrals can give the coefficients of taylor or laurent series.

but i dont understand how integral _ path A _ f(z) * g(z)

can be computed when f(z) is not known and only defined by these integrals.

maybe this belongs in the section " computation " better.

explaining this would help me alot.

regards

tommy1729
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#2
(11/18/2009, 04:46 PM)tommy1729 Wrote: contour of what ? of the function we are defining with the contour ?

The gradient theorem and residue theorem both show that in many cases the value of a contour integral/line integral is independent of the contour of integration.

So the simple answer to your question is: in many cases, it doesn't matter what the contour of integration is, you get the same result.
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#3
(11/18/2009, 04:46 PM)tommy1729 Wrote: many methods to compute a certain tetration depend on contour integrals.
I only know one method, i.e. the one of Dmitrii.

Quote:for instance kouznetsov defines his coefficients by contour integrals.

You can always get coefficients by contour integrals, but this is not his method. He obtains an integral equation with the function on the left side and the function is contained in the integral on the right side.
This can be used as a recurrence, to repeatedly compute the function on the left side, like you would do for example to find a fixed point, but this time with a function instead of a point.

For details just consult his paper.
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#4
(11/19/2009, 08:46 AM)bo198214 Wrote:
(11/18/2009, 04:46 PM)tommy1729 Wrote: many methods to compute a certain tetration depend on contour integrals.
I only know one method, i.e. the one of Dmitrii.

Quote:for instance kouznetsov defines his coefficients by contour integrals.

You can always get coefficients by contour integrals, but this is not his method. He obtains an integral equation with the function on the left side and the function is contained in the integral on the right side.
This can be used as a recurrence, to repeatedly compute the function on the left side, like you would do for example to find a fixed point, but this time with a function instead of a point.

For details just consult his paper.

i didnt get his free papers.

his graphics are nice but how he computes his coef is a mystery to me.

bo , you have explained andydude ( andrew robbins ) method here on the forum very well. ( beter than andrew himself imho actually )

i request you do the same for kouznetsov.
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#5
(11/19/2009, 01:07 AM)andydude Wrote:
(11/18/2009, 04:46 PM)tommy1729 Wrote: contour of what ? of the function we are defining with the contour ?

The gradient theorem and residue theorem both show that in many cases the value of a contour integral/line integral is independent of the contour of integration.

So the simple answer to your question is: in many cases, it doesn't matter what the contour of integration is, you get the same result.

gradient theorem and residue theorem are very well known to me.

but andy , i think you got my question wrong.

i didnt ask contour over which ' path ' , i asked over ' what function '.
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#6
This is an old thread but still I felt the need to comment.

The paper of Kouznetsov did not get as popular as some here expected , but I - as a skeptic - did.

One of the reasons might be this :

INTEGRAL TRANSFORMS HAVE CONDITIONS !

A good example is the Mellin inversion theorem.

Those conditions can be quite complex.

If an equation involved unknown functions AND integral transforms , it is a bit handwaving to assume all the conditions are met + uniqueness AND existance.

From my experience in number theory , many of my " unpublished wrong crank ideas " were mistakes by assuming conditions of integral transforms holds.

One could easily prove RH , collatz and Hardy-Littlewood I or II with those silly integrals , that turn out not to be valid in some sense ( conditions not met , some formula's not valid , divergence , summability method fail/dubious , using assumptions equivalent to the whole conjecture , assuming differentiability , ... )

Most pro mathematicians see this immediately.

I am unaware of any paper that adresses these issues , and I assume nobody is willing to pay for it before being convinced.

You see , tetration is not mainstraim , so the access should be easy and the burden of proof is on the claimer side.

Don't get me wrong , I like Kouznetsov and some of his ideas.

But those are my viewpoints , and by the lack of popularity of his paper ( in retrospect years later ) I assume it is the view of most.

regards

tommy1729
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