When i see the material represented in a such cool way, i feel a need to write a comment. Thank you for the time and the work you have done.
@aviralsood81412 жыл бұрын
These videos are incredible! Great work.
@DrWillWood2 жыл бұрын
Thanks a lot!
@Mothuzad2 жыл бұрын
Nice video! When I thought through this at the beginning, I imagined that the convexity would rely on a definition of a mean for the given set. However, when I saw you give the linear interpolation definition, I thought a little more and realized that my idea would allow fractals which aren't convex, such as the set of all numbers n/(2^k) where n and k are non-negative integers. It would include every whole number and yet it would miss infinitely many rationals (such as 1/3), despite clearly including the midpoint of any two elements.
@hadidwindoro2 жыл бұрын
This is very helpful! thank you!
@BRYDN_NATHAN2 жыл бұрын
Thank you. 0:05:01 👍
@CricMania00015 күн бұрын
this is awesome
@minari32744 ай бұрын
Thank youㅠㅠㅠㅠ
@bernardofitzpatrick54032 жыл бұрын
Hope you will be posting more vids🤙🏽 . subscribed
@DrWillWood2 жыл бұрын
Thanks a lot! definitely will be :-)
@amritawasthi70302 жыл бұрын
I want to ask u ....why are u soooooooooo good. You've become my fav now. More health to you 🙇🏻🙇🏻🙇🏻. A great thanks from India.
@DrWillWood2 жыл бұрын
thanks a lot! :-)
@llllllllllllllll777 Жыл бұрын
Thank you a lot❤
@biswanathpaul7813 Жыл бұрын
@Dr. Will Wood, at time 3:40 the example you have shown will also satisfy for Rhombus(with respect to one norm prospect)
@abhinavkumarverma10172 жыл бұрын
I think we need to ensure A is not zero else the C will become a polynomial of degree 1 or less. Maybe you have to say set of polynomials of degree 2 or less is convex.
@arindambasu4616 Жыл бұрын
Nice video. Surprised why view is so less.
@zorm_2 жыл бұрын
the set of polynoms of degree d=2 is not convex the set of polynoms of degree d
@2false6373 жыл бұрын
This series is superb. Keep it up!
@DrWillWood3 жыл бұрын
Thanks so much! I will do!
@8685mike9 ай бұрын
This is great! Is there an example of a concave set? What set of functions won't work the way polynomials of degree 2 did?
@matematicaspanish83012 жыл бұрын
What if you have two quadratic polynomials with main coeficients 1 and -1? there will be a polyinomial middway that loses one degree. Shouldn't we consider the set of polynomials of degree less or equal than 2 as the convex one?
@DrWillWood2 жыл бұрын
Hi, yes sorry I wasn't clear there! generally I consider "the set of nth order polynomials" to be those including those with zero as the coefficient of x^n. I'm not sure whether this is a common definition but it works well in this kind of situation!