Lagrange Polynomials

Lagrange Polynomials for function approximation including simple examples.

Chapters
0:00 Intro
0:08 Lagrange Polynomials
0:51 Visualizing L2
1:00 Numeric Example
1:11 Example Visualized
1:27 Why Lagrange Works
1:47 Lagrange Accuracy
2:12 Error
2:59 Error Visualized
3:20 Error Bounds
4:08 Notes
4:25 Thanks For Watching

Further Viewing:

#LagrangePolynomials #NumericalAnalysis

[alexandertaffe]

I finally figured out what I was doing wrong after I found this video. Thank you!!

[times]

Excellent, concise explanation. Thanks very much!

[speedyspud]

you could not have uploaded this at a better time!

[pitito]

Excelent source. Best work i have ever seen. Keep up the good work!

[ProtoG]

Great video! Helped me study for my exam.

[Mulkek]

Thanks, and it's so easy & simple!

[Trackman]

yeah, really nice explanation. Thanks!

[jettgreen]

Thanks for the video, it'll help me with my maths paper for school.

[AJ-etvf]

Awesome concise vid about Lagrange polynomials. I love relearning numerical methods. What are the differences between piecewise lagrange polynomials vs spline interpolating polynomials? Which is more accurate? I feel that conceptually, piecewise lagrange polynomials are the same or almost the same as spline polynomials since splines also approximate the function piecewise using lower order polynomials.

[fuck-you]

Good video.

One thing you should note is that the denominator can never have 0 as a factor (Because each factor in the denominator polynomial does not have the jth node as a root), and the denominator doesn't have a free variable x so it can be treated as a constant that is never 0. Otherwise the Lagrange basis polynomials would not be holomorphic, and also wouldn't be polynomials!

When the jth node for the jth polynomial is inserted, it equals 1 because the denominator polynomial (the constant term) will equal the numerator since the free variable is the jth node, but it is 0 for the free variable for all other nodes, which has similar behavior to the Kronecker delta, and has the sifting property useful in approximating the interpolating polynomial. That "sifting" property is the key here.

We know that need to calculate this only at the k+1 nodes because the Fundamental Theorem of Algebra tells us that each polynomial of degree k is represented uniquely by k+1 roots, which can be worked further off of to prove uniqueness for k+1 non-zero solutions. So k+1 Lagrange basis polynomials in Lagrange are enough to interpolate a degree k polynomial.

Error happens when the original polynomial is greater than degree k, or is not a polynomial. If the original polynomial is at most degree k, and all node/value pairs (or points) are distinct, then there is no error.

Understanding things this way is the gateway to understanding convolution and other interpolating basis functions, such as the basis trig polynomials and the Fourier transform.

I guess I'm also posting this comment to also help myself understand that I am learning this correctly too.

[amrragheb]

In the error evaluation (at 2:55) why do we divide by 3! not 2! ?

[johrahussain]

why is it 0.4 apart, im still confusedm there are 5 points, so shouldnt it be (0.1)^(5)

[wbuchmueller]

2:23 "if we do some fancy math"
thats not a very good explanation of what is shown there