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has a root. I think it is possible to use something like "mean value theorem". But i can not find any mean value theorem for R^n -> R^n.

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- Thread starter steffka
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- #1

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has a root. I think it is possible to use something like "mean value theorem". But i can not find any mean value theorem for R^n -> R^n.

- #2

matt grime

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x^2+y^2=1

irrespective of the second equation over R, so are you talking about C?

- #3

VietDao29

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Hmmm, ... Yes, it does... x = 0, y = 1 is one example.matt grime said:the mean value theorem is not true in multiple dimensions, and unless you can be more specific your problem as stated is false, there is no real solution to

x^2+y^2=1

The one does not have real root is x^2 + y^2 = -1.

Viet Dao,

- #4

matt grime

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sorry, meant it to be -1 not 1.

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I have two equations (nonlinear) with two variles. With some aproximative methods I get some potential roots (but not exact), so i know (or hope) there are some.

I need to show, that there exists at least one root.

I thought that somethinq like mena value theorem could help.

In attachment are equations and also the potential roots.

(the root x=0 and y=0 is trivial, but i need some other)

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