not sure if this equation is right?? Watch

hey_venus
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I need to find the set of points X \subset R^n where F is not a local diffeomorphism. I'm told it's not a LD when J_F = 0, therefore

x_1...x_n(2^n - (-3)^nx_1...x_n)^2 = 0

But if you look at the end of the question, it states what J_F is- which no matter how I re-arrange it, is not the same to what I have just written above. Am I getting it wrong?




This is the question if it's needed:
Consider the smooth map F: R_X^n \rightarrow R_y^n given by the equations:
y_1 = x_1^2 + x_2^3

y_2 = x_2^2 + x_3^3

...

y_i = x_i^2 + x_{i+1}^3

...

y_n = x_n^2 + x_1^3

The Jacobi matrix is:
\begin{pmatrix} 2x_1 & 3x_2^2 & 0 & 0 & 0 & ... & 0 \\0 & 2x_2 & 3x_3^2 & 0 & 0 & ... & 0\\0 & 0 & 2x_3 & 3x_4^2 & 0 & ... & 0\\ & & & & & ... &\\3x_1^2 & 0 & 0 & 0 & 0 & ... & 2x_n\end{pmatrix}

and the determinant (Jacobian) is:
J_F = 2^n(x_1x_2...x_n) - (-3)^n(x_1...x_n)^2

any help really appreciated!!
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ghostwalker
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(Original post by hey_venus)
any help really appreciated!!
If it's just a case of evaluating the determinant then a quick check with n=2 say will tell you, if it's anything else, you need someone more up to speed with this.
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hey_venus
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(Original post by ghostwalker)
If it's just a case of evaluating the determinant then a quick check with n=2 say will tell you, if it's anything else, you need someone more up to speed with this.
thanks i've just substituted 2 for n in both versions of the J_F but it still comes out different:

the first version i get:
x_1...x_n(2^2-(-3)^2x_1...x_n)^2

= x_1...x_n(-5x_1...x_n)^2

=x_1...x_n25(x_1...x_n)^2

but if i do it to the original i get
2^2(x_1...x_n) - (-3)^2(x_1...x_n)^2

4(x_1...x_n) - 9(x_1...x_n)^2.

i'm really not sure how the lecturer got the first version from the original :s this is annoying!
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ghostwalker
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(Original post by hey_venus)
thanks i've just substituted 2 for n in both versions of the J_F but it still comes out different:

the first version i get:
x_1...x_n(2^2-(-3)^2x_1...x_n)^2

= x_1...x_n(-5x_1...x_n)^2

=x_1...x_n25(x_1...x_n)^2

but if i do it to the original i get
2^2(x_1...x_n) - (-3)^2(x_1...x_n)^2

4(x_1...x_n) - 9(x_1...x_n)^2.

i'm really not sure how the lecturer got the first version from the original :s this is annoying!
Your evaluation of the first one's not correct; should be:

x_1...x_n(2^2-(-3)^2x_1...x_n)^2

= x_1...x_n(4 -9x_1...x_n)^2

Which when you compare it with the original, makes me think the final squaring shouldn't be there, in which case your first version should be x_1...x_n(2^2-(-3)^2x_1...x_n) and they are equal.
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