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plz explain kouznetsov slowly
#1
plz explain the kouznetsov method very slowly to me.

step by step , without C++ code.

is it complicated or am i not smart enough ? :p

i believe it relates to some contour integrals , but i dont understand how.

i have the experience that most misunderstanding lie at the beginning so plz explain how to start.

bo did a great job explaining andrews slog , maybe he can explain this too and become the " ultimate tetration forum webmaster of all time " Smile Smile
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#2
(07/21/2010, 10:47 PM)tommy1729 Wrote: plz explain the kouznetsov method very slowly to me.

Tommy, didnt I refer you several times already to the overview paper? Kouznetsov's method is also explained there. The overview paper is announced in the FAQ thread (which is on top of the threads in the General section).
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#3
http://cdn.bitbucket.org/bo198214/bunch/...s/main.pdf

assuming you mean that , and kouznetsov is equal to " the cauchy integral approach "

i have to say that the last limit formula ( recurrence ) on the bottom of page 20 is not clear to me.

i mean , how does one solve this ?

the tetration is on both sides of the equation , the integral is related to values of the tetration itself ??

i understand very well how you arrive at that last equation.

but not how it helps or needs to solved ?

i want to remark that - though maybe i dont understand well - i do think this 'cauchy' formula is very well suited to compare if 2 functions that are both analytic solutions of tetration are in fact equal or not.

more specificly i intended - and independently of this forum or any papers - to test if my solution is equal to kouznetsov and ' equal to it self ' ( needed to show existance and Coo ) with those integrals.

more specificly , use f1 and f2 ( the 2 solutions to tetration ) on both sides of the equation and see if the equation still holds.

thus f1 on LHS and f2 on RHS for clarity ...

furthermore , im dubious if that equation requires uniqueness ...

i dont see why sexp(x + 1periodic(x) ) wont satisfy the same equation , though maybe there is no uniqueness claim to the cauchy integral approach ? ( but kouznetsov claims uniqueness ... based upon a non-unique cauchy integral equation ?? )

sorry if im slow , but the equation seems as nonconstructive as

log(f(x+1)) = f(x)

adding just a limit and an integral and some extra terms and requiring values of f(x) for certain x ...

i dont want to be harsh , but this is how i feel and understand it , unless explained better ... without C+ or C++ or other computer language plz !!!

regards

tommy1729
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#4
(07/24/2010, 10:58 PM)tommy1729 Wrote: i have to say that the last limit formula ( recurrence ) on the bottom of page 20 is not clear to me.

i mean , how does one solve this ?

the tetration is on both sides of the equation , the integral is related to values of the tetration itself ??

Like in every recurrence you have the searched thing at both sides of the equation, you iterate. and if the recurrence is convergent, it gets closer and closer to searched for solution.

In our case we compute at the left side f with arguments i*y (let x_0=0) and on the right side there is an integral involving f applied on i*t or on i*(-t). And that is the way you iterate:
You compute values of f along suitably dense grid on the y-Axis by using these values on the y-Axis with a numeric integration.

And - amazingly enough - the values on the y-Axis converge.
I never verified it myself yet, but the convergence seems not very robust in certain cases (from some comments on this forum).

Quote:i dont see why sexp(x + 1periodic(x) ) wont satisfy the same equation , though maybe there is no uniqueness claim to the cauchy integral approach ? ( but kouznetsov claims uniqueness ... based upon a non-unique cauchy integral equation ?? )

Of course the equation is just a reformulation of the superfunction equation, however it converges to one specific solution.

(Btw see also this thread)
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#5
ok in a recurrence equation we iterate.

but how ?

you said x_0 = 0 ok.

again how ? and does that really converge ?? why ?

to quote :

"You compute values of f along suitably dense grid on the y-Axis by using these values on the y-Axis with a numeric integration."

that sounds vague and mysterious to me.


so lets say we use dummy variables : f(i) = a , f(2i) = b , f(3i) = c

but approaching an integral from -oo to + oo with a few integer points of dummy variables ?

i even doubt if numeric integration is a good approximation of the whole integral !? afterall it goes from -oo to + oo.

am i correct in assuming you use dummy variables for the values of f(i) , f(2i) etc and then try to solve it by replacing the integral on the RHS with " rectangles " and the dummy variables ?

and then further try to control those dummy variables by placing upper and lower bounds on their values ?

is f(i) , f(2i) , f(3i) , ... a good choice or do you mean suitable dense rather as in " the rationals are dense in the reals " ?

is f(a/b i) for all a and b rel prime up to value 100 a good choice for dummy variables ?

is the recurrence equation suppose to give exact values or upper and lower boundaries ?

assuming convergence - hence existance - , how is this necc unique ?

can you proof the recurrence to have a single unique solution ?

( maybe that one is easy , i havent really tried that yet )

sorry if im slow , on the other hand i would be amazed to be alone with my doubts or misunderstandings.

regards

tommy1729
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#6
(07/25/2010, 11:25 PM)tommy1729 Wrote: and does that really converge ?? why ?

It looks so, nobody knows yet why.

Quote:so lets say we use dummy variables : f(i) = a , f(2i) = b , f(3i) = c

but approaching an integral from -oo to + oo with a few integer points of dummy variables ?
...
is f(i) , f(2i) , f(3i) , ... a good choice or do you mean suitable dense rather as in " the rationals are dense in the reals " ?

"Dense" in the normal man's understanding, = tight, close, etc.
I mean we want to numerically approximate a function (here the function f on ), so we place enough supporting points and calculate the function values there.
The more exact we want to be the more points we place, the denser the grid.

Quote:i even doubt if numeric integration is a good approximation of the whole integral !? afterall it goes from -oo to + oo.

Kouznetsov does it by integrating to a bound of instead of .

Quote:am i correct in assuming you use dummy variables for the values of f(i) , f(2i) etc and then try to solve it by replacing the integral on the RHS with " rectangles " and the dummy variables ?

Ya not at integer points, ususally more dense. Kouznetsov uses Gauss legendre integration which places the supporting points as zero's of the legendre polynomials.

Quote:and then further try to control those dummy variables by placing upper and lower bounds on their values ?

No its just assumed that at "i*infinity", i.e. at the value using for approximating i*infinity, the value of f is the fixed point (and at "-i*infinity" the conjugated fixed point).

Quote:assuming convergence - hence existance - , how is this necc unique ?
Well if it satisfies a certain uniqueness criterion, which it seems to do.
Which is basically injectivity on the vertical strip .

Quote:can you proof the recurrence to have a single unique solution ?

A recurrence is a recurrence. Either it converges then this is a solution or not, then its not a solution.
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#7
are there similar recursions investigated earlier ?

any hope of similar recursions with closed form solutions ?

can the recursion be reduced to differential equations ?
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#8
(07/26/2010, 08:47 PM)tommy1729 Wrote: are there similar recursions investigated earlier ?
I dont know about any, but perhaps you have to ask Dmitrii (personal mail) how he developed that idea, by what knowledge he was incluenced or so.

Quote:any hope of similar recursions with closed form solutions ?
no.
Quote:can the recursion be reduced to differential equations ?
Dunno.
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