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From: greg@endor.harvard.edu (Greg)
Newsgroups: sci.math
Subject: Re: Geometry/Algebraic Geometry query.
Message-ID: <494@husc6.HARVARD.EDU>
Date: Tue, 21-Oct-86 11:15:08 EDT
Article-I.D.: husc6.494
Posted: Tue Oct 21 11:15:08 1986
Date-Received: Wed, 22-Oct-86 04:29:11 EDT
References: <471@husc6.HARVARD.EDU>
Sender: news@husc6.HARVARD.EDU
Reply-To: greg@endor.UUCP (Greg)
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Organization: Harvard
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In article <471@husc6.HARVARD.EDU> greg@endor.UUCP (Greg) writes:
>Let T be a triangulation of the
>interior of a polyhedron P in R^n.  Let g be a function from the vertices of T
>to the reals, and let k be a non-negative integer.  Is there always a function
>f from P to R such that:
>
>1) The k'th derivative of f is defined and continuous.
>2) For any n-dimensional simplex S of T, f restricted to S is a polynomial of
>degree k+1.
>3) For any vertex v of T, f(v) = y(v)?

Here are the essentials of an article that I stupidly posted to net.math rather
than sci.math.  I also wish to make a minor correction:

I'm convinced that if f has degree k+1, f usually doesn't have enough degrees
of freedom to satisfy all of the above conditions.  So I'll change the
question some:  For what integer d (which may depend on k, n, and T) can we
always find an f with the above properties such that f restricted to S is a
polynomial of degree d?
----
Greg