Normed Linear Spaces | Introduction, L1 and L2 Norms

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  • Опубліковано 26 вер 2024

КОМЕНТАРІ • 43

  • @maniam5460
    @maniam5460 3 роки тому +13

    These videos are very helpful and they deserve more recognition from the UA-cam algorithm

    • @DrWillWood
      @DrWillWood  3 роки тому +4

      It honestly makes me so happy to hear that people find these videos useful! Thank you!

  • @stevenschilizzi4104
    @stevenschilizzi4104 2 роки тому +11

    These presentations are wonderfully clear and pedagogical. I will get my students to use them to get a better understanding of regressions than just blindly pushing buttons, as they are too often taught. I must say I also love that (northern?) British accent! Thanks so much for putting it all together.

  • @hugeturnip3520
    @hugeturnip3520 Рік тому +2

    you deserve wayyyyyy more attention these videos are insanely good

  • @navjotsingh2251
    @navjotsingh2251 2 роки тому +1

    I would honestly love to see you do more on approximation theory. You are the best in this field. Your videos are awesome.

  • @ALEX-us8fx
    @ALEX-us8fx 3 роки тому +5

    Very good job! The norm of ||y' - y||, where your channel is y' and y is the vector of UA-cam numerical analysis videos should be minimum! :)

  • @olympunk4212
    @olympunk4212 3 роки тому +3

    Honestly, I tried to understand from various places, but you have done an amazing job explaining this. I hope you upload more content

    • @DrWillWood
      @DrWillWood  3 роки тому

      Thanks a lot! glad it was helpful. Of course, plenty more vids to come :-)

    • @olympunk4212
      @olympunk4212 3 роки тому

      I had a doubt, around 5:40, are we saying that the sey consisting of (1,2,3) and (1,1,1) forms a basis to that 2D vector space ? If yes, then how

    • @DrWillWood
      @DrWillWood  3 роки тому +1

      Yeah, although the vectors (1,2,3) and (1,1,1) are 3-D vectors, we only have two of them. So taking any linear combination of a x (1,2,3) + b x (1,1,1) for arbitrary constants a and b will form a 2-D plane. The plane will exist inside a 3-D vector space (as shown graphically in green at 5:40) but the plane itself is 2-D. Hope that helps!

    • @olympunk4212
      @olympunk4212 3 роки тому

      @@DrWillWood yup I got it. Thanks a lot. Also not really sure how the best fit line works but I think you are taking any two points and forming the line equation right? So shouldn’t the best fit line be y = x? Like why are we getting for x = 3
      y hat = 3.1 and not 3?

    • @DrWillWood
      @DrWillWood  3 роки тому

      I guess I didn't specify what the line of best fit is in this case only tried to look at the problem of finding a line of best through the eyes of an NLS. In this case, if you have N data points it's equivalent to finding the distance from the data Y to the subset of all the approximations you could make of it (the approximations are constrained to a subset of this space eg in this case it was constrained to a straight line). I don't think I was very clear at the motivation looking back so let me have another go! so lets say you have (x,y) data (1,0), (2,3), (3,2). a straight line could never go through these points so Yhat could never equal Y. (Y in this case is [0,3,,2] the vector of y values). This is equivalent to the picture of Y being outside the Yhat plane since its outside the scope of data points which can be perfectly fit by a straight line! I should say too we don't need to worry about the x values because they're the same in the Y and Yhat cases. If we had data points (1,2),(2,4),(3,6). then this could be fit by a straight line y=2x, or in the form of the vector equation at 5:30, Yhat = 2 [1,2,3], and Yhat = Y. :-)

  • @patmichel4724
    @patmichel4724 5 місяців тому

    I discover this chaîne today and it’s an amazing one! I’m wonder why I never see it before!

  • @dariosilva85
    @dariosilva85 Рік тому +1

    Wow, you are an awesome teacher. Please make more videos. Thanks

  • @robmarks6800
    @robmarks6800 3 роки тому +5

    Would be nice to see some hilbert spaces, fourier transform

  • @robin2080
    @robin2080 Рік тому +1

    I took this class with bamberg in 2016 first time I'm actually understanding this lol

  • @hey.guitarbjorn
    @hey.guitarbjorn Рік тому

    Great explanation, thank you!

  • @meccamiles7816
    @meccamiles7816 3 роки тому

    I love your content. Thank you for the wonderful visuals.

  • @anthonybernstein1626
    @anthonybernstein1626 Рік тому

    Ah, so that's the L2 norm! Thank you!

  • @guilhemescudero9114
    @guilhemescudero9114 2 роки тому

    You did an amazing job!

  • @VolumetricTerrain-hz7ci
    @VolumetricTerrain-hz7ci 4 місяці тому

    There are unknown way to visualize subspace, or vector spaces.
    You can stretching the width of the x axis, for example, in the right line of a 3d stereo image, and also get depth, as shown below.
    L R
    |____| |______|
    TIP: To get the 3d depth, close one eye and focus on either left or right line, and then open it.
    This because the z axis uses x to get depth. Which means that you can get double depth to the image.... 4d depth??? :O
    p.s
    You're good teacher!

  • @vaishnav4035
    @vaishnav4035 3 роки тому

    Thank you sir ❤️😊

  • @PS-dw5qo
    @PS-dw5qo 3 роки тому +2

    How does the equation at 5:40 span a plane? Is this not a line for a given a and b? Thanks for the video.

    • @DrWillWood
      @DrWillWood  3 роки тому +2

      Excellent spot, you're right! Thanks a lot for pointing that out i'll pin a correction to the top :-)

    • @PS-dw5qo
      @PS-dw5qo 3 роки тому

      Maybe you meant something else with the equation, since it makes sense that y hat is found in a plane. I was unsure myself : )

    • @PS-dw5qo
      @PS-dw5qo 3 роки тому

      That y hat is rotated in the direction of the vector (1,1,1) also makes sense, since this is, presumably, the first column vector in the independent variable X and its column space spans the plane in which y hat is found. I’d be grateful for a clarification about the statement regarding the vector equation spanning a plane at 5:22.

    • @PS-dw5qo
      @PS-dw5qo 3 роки тому

      Maybe it should be a(1,2,3)+b(1,1,1) where a and b are any real numbers?

    • @PS-dw5qo
      @PS-dw5qo 3 роки тому

      One more small detail, but at 7:37 it says VxV=||x-y||. Isn’t the left hand side a set whereas the right hand side a real number?

  • @محمدابراهيمعمارارحومه

    Who are u man? U gooood ❤

  • @abhinavkumarverma1017
    @abhinavkumarverma1017 2 роки тому

    Hello. Thank you for all the effort you are putting. Can you provide an example of metric spaces that is not normed linear space. This would clarify the difference. Thanks in advance.

    • @Evan490BC
      @Evan490BC 2 роки тому

      This is easy to see: a metric (distance) does not need to be defined by a norm necessarily.

    • @-minushyphen1two379
      @-minushyphen1two379 Рік тому

      This example was the one shown in the video, but I’ll repeat it here:
      Consider the unit disc in the plane. It is a metric space since every two points in it have a defined distance from each other, but it is not a normed linear space since you can scale a vector and it will go out of the disc.

  • @austinbristow5716
    @austinbristow5716 2 роки тому

    Why is it that the set of all possible Y hat form a 2-D vector space rather than a 3-D vector space?

    • @austinbristow5716
      @austinbristow5716 2 роки тому

      Is it because of the constraint of Y hat being composed of the linear regression of the set of Y? Does this lower the dimension from 3-D to 2-D? If so, why? Do constraints like these, in general, reduce the dimension of a vector space?

    • @anthonybernstein1626
      @anthonybernstein1626 Рік тому

      ​@@austinbristow5716A line is fully determined by just 2 values: its slope and y-intercept, so it doesn't have enough "degrees of freedom" to form a 3D vector space.

  • @thecarlostheory
    @thecarlostheory 2 роки тому

    Hello. First of all I want to say that ur 2 first video since I´ve seen them, are amazing. I´m with joy to see the rest. One particular thing I liked a lot it´s the animations. How do you do them?

    • @DrWillWood
      @DrWillWood  2 роки тому +1

      Thanks a lot! All the animations are made in Apple Keynote which has lots of functionality for manipulating and animating shapes like arrows, curves, squares etc

    • @thecarlostheory
      @thecarlostheory 2 роки тому

      @@DrWillWood wow. Amazing!

  • @syedfaizan5841
    @syedfaizan5841 2 роки тому

    Well done