Least Squares

[@drpeyam]

Note: At 4:30, part of the video was accidentally cut out. That part is simply the calculation of b^, which is on the right-hand side of the whiteboard. Also at around 5:00, I switched b and b^ but I correct that mistake a couple of minutes later in the video.

[@thomasjefferson6225]

Thank you for this. After rewinding two times i Said lemme read comments. Dr peyam youve inspired my love of math. BTW my text book, lay, uses the peyam method as the alternative method.... how dare they don't give you Credit!!!

[@gastonsolaril.237]

You've just clarified so many things about a subject that I've never been able to totally understand throughout my 9 years of graduate and post-graduate engineering courses... Hail and long live Dr Peyam, First of His Name, Ruler of the Seven Kingdoms of Mathematos, Lord Paramount of Integrals, Series, Transforms, Eigenvectors and Stochastics!

[@taubone9257]

The awesome Peyam method is the best method. Also, bless you (14:11).

[@drpeyam]

Thanks :)

[@blackpenredpen]

Bless u

[@JLConawayII]

Vastly superior to the most-squares method.

[@loganreina2290]

For my final calc II project, I gave a derivation and a few examples of the least squares method in use as both an introduction to partials and a practical application of them. I was pretty proud of the presentation.

[@tomatrix7525]

Wow great video Peyam! Happy New years!

[@shareefsyed2968]

Thank you so much for making this! It really helps!

[@joebaldwin9005]

18:55 How is, for Q orthogonal, QQ^T not the identity but Q^TQ is? Im confused since Q^-1 = Q^T and therefore Q^Q-1 = I and Q^-1Q = I

[@joebaldwin9005]

oh becuase Q is not necessarily square?

[@codebits4461]

Am watching one of your old vids😂😂😂 You've really upgraded🙌🔥 Good teach though✔💪

[@AndDiracisHisProphet]

I have never seen any of those methods.

[@estuardodiaz2720]

Awesome video!!

[@JorgetePanete]

So, do you mean that it's like **I WANT A FUCKING SOLUTION, YOU FUCKING FUNCTION!!** ?

[@drpeyam]

Exactly 😂 And I’ll do whatever it takes!

[@dhunt6618]

sneezing from all the white board dust?

[@drpeyam]

Hahaha, yeah, all the dust from the marker 😂 lol, I’ve also been having a cold, but I’ve recovered fortunately 😊

[@justcarcrazy]

Why is (Q_t)(Q)=(I)? Isn't (Q^-1)(Q)=(I)? Is this a definition, a happy coincidence, or a necessary condition?

[@drpeyam]

IF Q is orthogonal and square, then yes, indeed Q^-1 = Q^T, which makes inverses easy to calculate. In general, if Q is not square, Q^-1 isn’t defined

[@mitchkovacs1396]

Was this cut incorrectly or am I missing something? There seems to be a jump at 4:30

[@drpeyam]

Mitch Kovacs You’re right!!! But fortunately you didn’t miss much: all I cut out was the calculation of b^ which is on the right side of the whiteboard!

[@mitchkovacs1396]

Thanks!

[@arthurribaudet559]

Your first method only works if the matrix is already orthogonal?...

[@arthurribaudet559]

Else, after obtaining x hat through your method, we need to multiply by Q-transpose A (or R in the QR factorisation.) so after applying gram-schmidt we actually need to multiply by the inverse of R... x hat = R^-1Q^Tb

[@chdfnh]

Genial!!!

[@weinihao3632]

14:10 Gesundheit!

[@michalbotor]

frankly, i think that if Ax=b doesn't have a solution, then calling x-hat its least squares solution is, forgive me, preposterous; even if it's already established in the literature, i believe, that we should insist on calling it what it actually is -- least squares approximated solution.

[@thatdude_93]

can you prove that the integral from -inf to inf of exp(-x²) = sqrt(pi), but using the wallis product and the gamma function (gauss' representation of the gamma function specifically)? i haven't seen a video about this here on youtube, and i think it's a lot more interesting that the ole "polar coordinate substitution" method XD

[@drpeyam]

Huh, that’s a really interesting idea! But polar coordinates are great too!

[@thatdude_93]

yeah i know, but there a few hundred videos on youtube with that approach :D. i tried it myself with the wallis product and the gauss-gamma function, and it's pretty amazing :) maybe you can give it a try sometime aswell

[@Bbdu75yg]

Amazinggggg,!!!

[@harikishan5690]

nice!!

[@mipmip4575]

second

[@mainakdeb9188]

first