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From: Narek Saribekyan on 26 Apr 2010 13:16
Hello dear community, please take a minute on this.
Suppose we have a real valued function u(x1,...xn) such that,
u(0)=0
du^k (0) / dxi^k = 0. k <= qi, i=1,..,n
Can we conclude that u = o( x1^q1+1 ) + ... + o( xn^qn+1 ) ?
Thanks
From: Chip Eastham on 26 Apr 2010 14:41
On Apr 26, 1:16 pm, Narek Saribekyan <narek.saribek...(a)gmail.com>
wrote:
> Hello dear community, please take a minute on this.
> Suppose we have a real valued function u(x1,...xn)  such that,
> u(0)=0
> du^k (0) / dxi^k  = 0. k <= qi, i=1,..,n
> Can we conclude that u = o( x1^q1+1 )  + ... + o( xn^qn+1 )  ?
> Thanks

No, I don't think so. Let's take n = 2,
for simplicity, and define u(x,y) to be
a function symmetric about the origin:

u(0,0) = 0

u(x,y) = xy/(x^2 + y^2)

Then u = 0 identically along the x- and y-
axes, so partials du^k/d^k x and du^k/d^k y
of all orders are zero at the origin.

However along a line y = mx through the
origin, u(x,mx) = m/(m^2 + 1), so u is
not even continuous at the origin.

regards, chip
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