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mike lowry
science forum beginner
Joined: 20 May 2006
Posts: 7
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 Posted: Sat May 20, 2006 5:40 am Post subject:
locally invertible
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Let L: R^2 -> R^2 be an invertible linear map and f(x) = L(x) + g(x), where |g(x) < or = |x|^2. Show that f is locally invertible in some open neighborhood of x = 0. |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Sat May 20, 2006 10:47 am Post subject:
Re: locally invertible
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On Sat, 20 May 2006 01:40:31 EDT, mike lowry <mmehdiza@gmu.edu> wrote:
| Quote: | Let L: R^2 -> R^2 be an invertible linear map and f(x) = L(x) + g(x), where |g(x) < or = |x|^2. Show that f is locally invertible in some open neighborhood of x = 0.
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Probably you've covered a certain big theorem that says if f satisfies
a certain condition or conditions then f is invertible, right?
************************
David C. Ullrich |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Sat May 20, 2006 8:37 pm Post subject:
Re: locally invertible
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In article
<30114433.1148103661533.JavaMail.jakarta@nitrogen.mathforum.org>,
mike lowry <mmehdiza@gmu.edu> wrote:
| Quote: | Let L: R^2 -> R^2 be an invertible linear map and f(x) = L(x) + g(x), where
|g(x) < or = |x|^2. Show that f is locally invertible in some open
neighborhood of x = 0.
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This is false. f need not be 1-1 in any neighborhood of 0. Start
in one dimension: we can find f such that x <= f(x) <= x + x^2,
but such that every open interval about 0 contains a subinterval
where f is constant. The extension to 2 dimensions is not
difficult. |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Sun May 21, 2006 12:04 pm Post subject:
Re: locally invertible
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On Sat, 20 May 2006 13:37:17 -0700, The World Wide Wade
<waderameyxiii@comcast.remove13.net> wrote:
Hmm, yes of course it is. Wonder how I thought it was true
the other day...
Oh. It's true if we assume that g is smooth enough (C^1, I
think). I think.
| Quote: | f need not be 1-1 in any neighborhood of 0. Start
in one dimension: we can find f such that x <= f(x) <= x + x^2,
but such that every open interval about 0 contains a subinterval
where f is constant. The extension to 2 dimensions is not
difficult.
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************************
David C. Ullrich |
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The World Wide Wade
science forum Guru
Joined: 24 Mar 2005
Posts: 790
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| Posted: Sun May 21, 2006 7:03 pm Post subject:
Re: locally invertible
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In article <pol0725q12sllipi4sglspqg9atgf3in9k@4ax.com>,
David C. Ullrich <ullrich@math.okstate.edu> wrote:
| Quote: | On Sat, 20 May 2006 13:37:17 -0700, The World Wide Wade
waderameyxiii@comcast.remove13.net> wrote:
In article
30114433.1148103661533.JavaMail.jakarta@nitrogen.mathforum.org>,
mike lowry <mmehdiza@gmu.edu> wrote:
Let L: R^2 -> R^2 be an invertible linear map and f(x) = L(x) + g(x),
where
|g(x) < or = |x|^2. Show that f is locally invertible in some open
neighborhood of x = 0.
This is false.
Hmm, yes of course it is. Wonder how I thought it was true
the other day...
Oh. It's true if we assume that g is smooth enough (C^1, I
think). I think.
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Sure, then you could use the inverse function theorem. But
there's an elementary direct proof if g is C^1.
| Quote: | f need not be 1-1 in any neighborhood of 0. Start
in one dimension: we can find f such that x <= f(x) <= x + x^2,
but such that every open interval about 0 contains a subinterval
where f is constant. The extension to 2 dimensions is not
difficult.
************************
David C. Ullrich |
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