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# Jacobians help Watch

1. Calculate the Jacobian of the function .
For which values of (x; y) does this function have an inverse in a (small) neighbourhood of (x; y)? Justify your answer. Do not calculate the inverse.

I have calculated the Jacobian as . I just honestly don't know what it would mean for the function to have an inverse in a small neighbour hood of (x;y).
2. (Original post by Happy2Guys1Hammer)
Calculate the Jacobian of the function .
For which values of (x; y) does this function have an inverse in a (small) neighbourhood of (x; y)? Justify your answer. Do not calculate the inverse.

I have calculated the Jacobian as . I just honestly don't know what it would mean for the function to have an inverse in a small neighbour hood of (x;y).
THe function have an inverse in a small neigbourhood of that point
where the Jacobian determinant is nonzero. Even the determinant is nonzero
everywhere we can not say that F is invertible over its entire image because maybe F is not injective.
f.e F(x,y)=F(-x,-y) here
3. (Original post by Happy2Guys1Hammer)
Calculate the Jacobian of the function .
For which values of (x; y) does this function have an inverse in a (small) neighbourhood of (x; y)? Justify your answer. Do not calculate the inverse.

I have calculated the Jacobian as . I just honestly don't know what it would mean for the function to have an inverse in a small neighbour hood of (x;y).
f(x)=sin(x) is a good example of what it means for a function to invertible about a "small neighbourhood" of a point. We know that f isn't invertible, because the points {2n pi| n an integer} all map to 0, for example. However, if we restrict f to a neighbourhood of a point, it is possible for this restricted version of f to have an inverse: for example, f restricted to the interval is invertible for any , so f has an inverse in a neighbourhood of 0, despite not having an inverse when considered as having all real numbers in its domain.

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Updated: April 9, 2013
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