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Anonymous Sat 28 Dec 08:48:20 2019 No.11258940 View Reply Original Report

Quoted By:

Can anyone explain this math here for PCA, specifically, how to get the part in 2.64? Where c is a vector of an encoding for x, and D is a decoding matrix. Not sure if I need a better understanding of vector calculus or what. Thoughts?

Anonymous

Anonymous Sat 28 Dec 2019 08:58:21 No.11258950 Report

Quoted By: >>11258974

It’s a fucking derivative dude

Anonymous

Anonymous Sat 28 Dec 2019 09:10:21 No.11258968 Report

Quoted By:

See section 4.3 if you do not know how to do this.

Anonymous

Anonymous Sat 28 Dec 2019 09:18:11 No.11258974 Report

Quoted By: >>11258977 >>11258980 >>11258984 >>11258998

>>11258950
It's saying the gradient of
$<br />-2x^TDc<br />$
is
$<br />-2D^Tx<br />$
?
$c^Tc$ being 2c makes sense, but I don't understand that.

Anonymous

Anonymous Sat 28 Dec 2019 09:29:01 No.11258977 Report

Quoted By: >>11258982 >>11258996

>>11258974
x’D = (D’x)’
You take the derivative, it loses c, the resultant vector has to have same dimensions as c. Don’t think about it too much

Anonymous

Anonymous Sat 28 Dec 2019 09:30:55 No.11258980 Report

Quoted By:

>>11258974
giga brainlet

Anonymous

Anonymous Sat 28 Dec 2019 09:31:30 No.11258982 Report

Quoted By: >>11258984 >>11258996

>>11258977
Ie c is 5x1, D 5x5 x 5x1. The form of the left term has to be 5x1 if your right term is 5x1. Don’t get too caught up in transposes the rules are kinda silly

Anonymous

Anonymous Sat 28 Dec 2019 09:33:11 No.11258984 Report

Quoted By:

>>11258982
>>11258974
Also look at trace. Trace properties explain this too

Anonymous

Anonymous Sat 28 Dec 2019 09:42:12 No.11258996 Report

Quoted By: >>11258999

>>11258977
>>11258982
Cool I think I got it. Was just a small thing looks like. Thanks for helping me out.

Anonymous

Anonymous Sat 28 Dec 2019 09:45:26 No.11258998 Report

Quoted By:

>>11258974
$q = tr(x^TDc) \\ = tr( (D^Tx)^Tc) \\ = D^Tx : c \\ dq = D^Tx : dc \\ \pdv{q}{c} = D^Tx [\math]$

Anonymous

Anonymous Sat 28 Dec 2019 09:46:29 No.11258999 Report

Quoted By:

>>11258996
oops, here:

$q = tr(x^TDc) \\ = tr( (D^Tx)^Tc) \\ = D^Tx : c \\ dq = D^Tx : dc \\ \pdv{q}{c} = D^Tx$

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