Lecture 1 Part 2: Derivatives as Linear Operators

MIT 18.S096 Matrix Calculus For Machine Learning And Beyond, IAP 2023
Instructors: Alan Edelman, Steven G. Johnson

Description: The key to generalizing derivatives is to realize that they are linear operators, relating small changes in a functions input to small changes in the output. 

License: Creative Commons BY-NC-SA

[ABCNULL]

Really excited to learn more from these lectures!!

[fabioborges]

Nice lecture, but the notation is somewhat confusing to me. 'f(x)' is not a function, it's the value of the function 'f' evaluated at 'x'. 'f' is a function. If 'L' is a linear operator on the space of R-to-R functions, then 'Lf' is another function, specifically, the function the 'L' maps 'f' to. So by doing '(Lf)(x) = f(x^2)' you mean your're saying that 'L' takes a function 'f' and outputs another function which maps x to f(x^2). So if you define the sum function as (f+g)(x) = f(x) + g(x), then (f+g)(x^2) = f(x^2) + g(x^2) = (Lf)(x) + (Lg)(x) = (Lf + Lg)(x). Therefore (L(f+g))(x) = (Lf + Lg)(x), from which you conclude L is linear. In other words, the functions L(f+g) and (Lf + Lg) are identical.

[harshavardhanlakhinana]

lecture 2 I am going to complete for today
Edit: completed the lecture really loved the perspective behind linear operator and understanding why this is important.

[sfratini]

Nice lecture but as with the first part, the choreography has issues, e.g., he keeps writing below where you can see on the screen. Concerning linear transformation (for those not familiar), he meant to show L(av+bu)=a L(v) + b L(u) for scalars a and b, and vectors v and u.

[AmanVerma-uqsh]

Hello sir I am from India . Sir your teaching skill is good

[ramankumar-kksl]

After 30 Minutes of Lecture i noticed echo in audio recording.

[rw]

Steven's technology (both audio quality and video cutoff) really dropped the ball in this lecture. Otherwise it's a good lecture content-wise.