What a great video, I knew how to use the formula but never understood the concept. You made me enjoy math a lot more, I wish all my teachers would explain exactly how the concepts work just like you did. You are amazing!
In the previous approach, we could fit a higher degree polynomial too with extra DOF. How to do it using Lagrange Polynomial Interpolation? Choosing input points randomly and putting constraints to have STRAIGHTish behavior e.g. P5(5)=1.1, P6(6)=1.3 etc?
I'm an ignoramus when it comes to history and I would not dare rank them, by my favorite five are Euler, Archimedes, Hadamard, Fermat, Descartes, Lax, and Strang.
So am I to understand that he 4 cubic polynomials p_1, ..., p_4 that were constructed in this example are linearly independent? Otherwise they could not form a basis for the space of cubic polynomials. I suppose their linear independence follows from constructing each of them in such a way that p_n(x=n) = 1 and p_n(x) = 0 at x=/=n for n=1,2,3,4? Something about having to take a multiple p_n in order to make sure the curve goes through a given y-value y_n=a at x=n, I.e., demanding the term 'a*p_n' to be included as part of the sum that you would get by decomposing the curve p(x) into a linear combination of the 4 p_i functions... This seems likely to imply linear independence but I'm having a hard time explaining to myself the exact details of how the linear independence follows. Maybe by tomorrow evening I'll have the answer, but it's out of reach for my tired mind tonight. Very interesting video!
kniv....ouse - A suggestion, first think why intuitively we know that 3 basis vectors [ 1,0,0 ] , [ 0,1,0 ] , [0,0,1 ] must be lin.indep. But if you then think of a reason it'll unswer also your question. In each vector the single non-zero entry (here a 1 ) sits in different position, in which all others have nothing (only 0-es).And any linear comb.of 0-es can yield only 0. Thus the non -zero entry of one vector cann't possibly be produced by lin.combination of some others,therefore he is is lin.indep. Now returning to polynomials,firstly k p_i ("k" a coefficient ) has the same roots as p_i (looking at the formula is obvious that multiplying p_i by a coefficient doesn't alter its roots,the x-es for which it's value is 0). So now try to obtain any p_i as lin.comb.of the others,for example p_2 as a lin.comb.of p_1,3,4 . Consider that ( o n l y ! ) p_2 has a non-zero value at x=2.The value of all the others , k p_i i=1,3,4 is there (at x=2) 0 by design. The parallel to above mentioned basis vectors is obvious,therefore polinomials p_1,2,3,4 must be lin.indep.
@@sergelifshitz1034 thanks for the reply! I ended up figuring it out the next day, but this may help someone else later. Thinking of polynomials and other fuctions in terms of linear algebra is pretty interesting!
I think that the most difficult part to digest is to have Polynomials that are linear independent but JUST for some particular values of x! Which makes me think... Can two quadratic polynomials be linear independent for any value of x ??? or is this just another "fancy math trick" ?
Hello Professor, nice lectures, thank you so much for them. Question: does the "shape" of the cloud of points has to do with the choice of basis decision? Is there some intrinsic information on the relative position of the points that would help in the decision?
+Omar Taha There is a unique polynomial of the appropriate degree that passes through those points, therefore the two polynomials are the same. Why is the polynomial unique? Because the Vandermonde matrix is invertible. Why is the Vandermonde invertible? Because there is a unique polynomial p(x)=0 that passes through 0 at the prescribed x's.
+Omar Taha Observe the difference. Since they are both at most n-th degree polynomials one may observe that their difference, which is at most an nth degree polynomial, has n+1 roots. It therefore implies their difference is trivial i.e zero.
Bruh , y man. Y we not live in peace. Just like u how hate Indians, I hate u. U and ur white peepee. Y so small , become Indian , become big. Thank you -Allen Saldhana(I'm 19 and I study in vit. Don't @ me)
Go to LEM.MA/LA for videos, exercises, and to ask us questions directly.
Thank you so much for this absolute gem of an explanation of Lagrange polynomial interpolation Professor Grinfeld.
What a great video, I knew how to use the formula but never understood the concept. You made me enjoy math a lot more, I wish all my teachers would explain exactly how the concepts work just like you did. You are amazing!
Best math content on youtube, primarily because the concepts are explained so well. Thank you very much!
Wow, that was by far the best explanation of Lagrange polynomials, thank you!
Awesome! This video also proves how math is beatiful not just lagrange interpolation
Thank you for thinking so!
Absolutely fantastic lecture!!!
Very clear and helpful, thanks for making internet a better thing
Such a simple concept but I just couldnt understand the derivation before this!
Thank you for my favorite kind of comment
This was so wholesome. We must protect him at all cost.
amazing
Lagrange polynomials are sooo cool!
Agreed.
Really good explanation of the concept. Thanks a lot
Wow👏thanks a million🤭that's perfectly clear
Great video !!!!
Thanks!!
By the way which program do you use to plot your grapha / sovle you matricies? I looks like a Mac screen. Thank you
Perfect explanation!
Great lecture, thank you very much.
Thanks, great video!
I found your content very helpful.
Thank you very much and God bless. :)
Thanks a lot!
In the previous approach, we could fit a higher degree polynomial too with extra DOF. How to do it using Lagrange Polynomial Interpolation? Choosing input points randomly and putting constraints to have STRAIGHTish behavior e.g. P5(5)=1.1, P6(6)=1.3 etc?
Thanks ! Pretty well explained ! :)
Thank you so much, amazing explanation!
THANK YOU SO MUCH I FINALLY UNDERSTAND
Thank you, that's my favorite kind of comment!
Great explanation
Thanks a ton!
thank you!
anyone know what that software is?
Scientific Workplace
@@MathTheBeautiful thank you!
thanks professor, you saved my butt
Who are your top 5 mathematicians of all time?
I'm an ignoramus when it comes to history and I would not dare rank them, by my favorite five are Euler, Archimedes, Hadamard, Fermat, Descartes, Lax, and Strang.
So am I to understand that he 4 cubic polynomials p_1, ..., p_4 that were constructed in this example are linearly independent? Otherwise they could not form a basis for the space of cubic polynomials. I suppose their linear independence follows from constructing each of them in such a way that p_n(x=n) = 1 and p_n(x) = 0 at x=/=n for n=1,2,3,4? Something about having to take a multiple p_n in order to make sure the curve goes through a given y-value y_n=a at x=n, I.e., demanding the term 'a*p_n' to be included as part of the sum that you would get by decomposing the curve p(x) into a linear combination of the 4 p_i functions...
This seems likely to imply linear independence but I'm having a hard time explaining to myself the exact details of how the linear independence follows. Maybe by tomorrow evening I'll have the answer, but it's out of reach for my tired mind tonight. Very interesting video!
kniv....ouse - A suggestion, first think why intuitively we know that 3 basis vectors [ 1,0,0 ] , [ 0,1,0 ] , [0,0,1 ] must be lin.indep. But if you then think of a reason it'll unswer also your question. In each vector the single non-zero entry (here a 1 ) sits in different position, in which all others have nothing (only 0-es).And any linear comb.of 0-es can yield only 0. Thus the non -zero entry of one vector cann't possibly be produced by lin.combination of some others,therefore he is is lin.indep.
Now returning to polynomials,firstly k p_i ("k" a coefficient ) has the same roots as p_i (looking at the formula is obvious that multiplying p_i by a coefficient doesn't alter its roots,the x-es for which it's value is 0).
So now try to obtain any p_i as lin.comb.of the others,for example p_2 as a lin.comb.of p_1,3,4 . Consider that ( o n l y ! ) p_2 has a non-zero value at x=2.The value of all the others , k p_i i=1,3,4 is there (at x=2) 0 by design.
The parallel to above mentioned basis vectors is obvious,therefore polinomials p_1,2,3,4 must be lin.indep.
@@sergelifshitz1034 thanks for the reply! I ended up figuring it out the next day, but this may help someone else later. Thinking of polynomials and other fuctions in terms of linear algebra is pretty interesting!
I think that the most difficult part to digest is to have Polynomials that are linear independent but JUST for some particular values of x! Which makes me think... Can two quadratic polynomials be linear independent for any value of x ??? or is this just another "fancy math trick" ?
thank you very much !!!! :)
Hello Professor, nice lectures, thank you so much for them. Question: does the "shape" of the cloud of points has to do with the choice of basis decision? Is there some intrinsic information on the relative position of the points that would help in the decision?
Yes, I would say so. Or some other knowledge about the problem you are dealing with.
Thanks god....that I found you.....
wow
I want the theoretical solution
+Omar Taha Give us an example of what you mean by a theoretical solution.
How prove Pn(x) of newton equal Pn(x) of Lagrange
+Omar Taha There is a unique polynomial of the appropriate degree that passes through those points, therefore the two polynomials are the same. Why is the polynomial unique? Because the Vandermonde matrix is invertible. Why is the Vandermonde invertible? Because there is a unique polynomial p(x)=0 that passes through 0 at the prescribed x's.
+Omar Taha Observe the difference. Since they are both at most n-th degree polynomials one may observe that their difference, which is at most an nth degree polynomial, has n+1 roots. It therefore implies their difference is trivial i.e zero.
not an indian badly explaining engineering math. thank you
Bruh , y man. Y we not live in peace. Just like u how hate Indians, I hate u. U and ur white peepee. Y so small , become Indian , become big. Thank you
-Allen Saldhana(I'm 19 and I study in vit. Don't @ me)