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Taylor series of upx function
#6
bo198214 Wrote:
Kouznetsov Wrote:There is EXACT Tailor series, if we insist that upx(z^*) = upx(z)^*
and upx(z) is holomorphic outside the part of the negative part of axis,
id est, holomorphic everywhere except .
Then the solution is unique.

Dmitrii, that is not yet proven. Even with what you call lemma about almost identical functions, we still need conditions for the slog, to make something unique here. I do not yet see a uniqueness criterion that does not depend on the shape of . We have to make a waterproof proof before announcing something wrong! See my thread Universal Uniqueness Criterion II about what we really have in the moment.
Can this solution be verified to be different from the approximation that Andrew came up with using linear equations for slog? The values for Dmitri's this solution appear very close to Andrew's approximation.

To me, it would seem as if the first couple of derivatives at x=-1,0 were the same for both solutions, than it is likely that they're the same solution, though the two could still diverge in the higher derivatives.
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Messages In This Thread
Taylor series of upx function - by Zagreus - 09/07/2008, 10:56 AM
RE: Taylor series of upx function - by bo198214 - 09/10/2008, 01:56 PM
RE: Taylor series of upx function - by andydude - 10/23/2008, 10:16 PM
RE: Taylor series of upx function - by Kouznetsov - 11/16/2008, 01:40 PM
RE: Taylor series of upx function - by bo198214 - 11/16/2008, 07:17 PM
RE: Taylor series of upx function - by sheldonison - 11/17/2008, 12:21 AM
RE: Taylor series of upx function - by Kouznetsov - 11/17/2008, 04:11 AM
RE: Taylor series of upx function - by bo198214 - 11/18/2008, 11:49 AM
RE: Taylor series of upx function - by Kouznetsov - 12/02/2008, 12:03 PM
RE: Taylor series of upx function - by Kouznetsov - 12/05/2008, 12:30 AM

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