Timeline for Some proof that linear translations and rotations of a bound-constrained function are equivalent
Current License: CC BY-SA 4.0
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| Apr 7, 2020 at 17:13 | comment | added | Brian Borchers | It's likely that the steps taken by your optimization method aren't invariant under this transformation, which explains why you might get different results. There are methods (e.g. Newton's method) that are invariant under invertible linear transformations. | |
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| May 16, 2018 at 3:32 | answer | added | Aniruddha Acharya | timeline score: 1 | |
| May 9, 2018 at 16:34 | comment | added | HBR | The minimum of $f(x) = (x-x_0)^T(x-x_0)$ is obviously $x_0$. If this minimum depends on the translation, then it is not translational invariant. | |
| May 9, 2018 at 15:50 | comment | added | aleksv | Then how do I know if a function has a rotational-translational invariance? Isn't all functions rotational-translational invariant from optimization standpoint? | |
| May 9, 2018 at 15:34 | comment | added | HBR | If they are rotational-tranlational invariance... they will be equivalent. i mean, the function $f(x) = x^Tx$ is invariant w.r.t. rotations $x'=Rx$ because $f(x') = (Rx)^TRx = x^TR^TRx=x^Tx$, since $R$ is an orthogonal matrix. | |
| May 9, 2018 at 14:59 | history | edited | aleksv | CC BY-SA 4.0 |
added 53 characters in body
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| May 9, 2018 at 14:38 | history | asked | aleksv | CC BY-SA 4.0 |