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.
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!
@@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?
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. :-)
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?
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
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!
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.
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.
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.
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?
@@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.
![โรงเรียนดัดสันดาน นักเรียนเถื่อน..!! [โรงเรียนไดเม่]](http://i.ytimg.com/vi/vRKPeXn4Fl4/mqdefault.jpg)
These videos are very helpful and they deserve more recognition from the TH-cam algorithm
It honestly makes me so happy to hear that people find these videos useful! Thank you!
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.
I would honestly love to see you do more on approximation theory. You are the best in this field. Your videos are awesome.
you deserve wayyyyyy more attention these videos are insanely good
Honestly, I tried to understand from various places, but you have done an amazing job explaining this. I hope you upload more content
Thanks a lot! glad it was helpful. Of course, plenty more vids to come :-)
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
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!
@@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?
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. :-)
I discover this chaîne today and it’s an amazing one! I’m wonder why I never see it before!
Very good job! The norm of ||y' - y||, where your channel is y' and y is the vector of TH-cam numerical analysis videos should be minimum! :)
Thanks a lot! :-D
Wow, you are an awesome teacher. Please make more videos. Thanks
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?
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
@@DrWillWood wow. Amazing!
I took this class with bamberg in 2016 first time I'm actually understanding this lol
Great explanation, thank you!
Would be nice to see some hilbert spaces, fourier transform
Ah, so that's the L2 norm! Thank you!
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!
I love your content. Thank you for the wonderful visuals.
Thanks a lot!
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.
Excellent spot, you're right! Thanks a lot for pointing that out i'll pin a correction to the top :-)
Maybe you meant something else with the equation, since it makes sense that y hat is found in a plane. I was unsure myself : )
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.
Maybe it should be a(1,2,3)+b(1,1,1) where a and b are any real numbers?
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?
You did an amazing job!
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.
This is easy to see: a metric (distance) does not need to be defined by a norm necessarily.
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.
Nice
Well done
Who are u man? U gooood ❤
Thank you sir ❤️😊
Why is it that the set of all possible Y hat form a 2-D vector space rather than a 3-D vector space?
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?
@@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.