Taylor series of upx function
#5
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 \( z\le 2 \).
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 \( \text{upx}(S) \). 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.
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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 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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