This series of six videos reg. Shor's and prerequesites is soooo awesome. I'm teaching myself Quantum Mechanics and QC and there's lots of material around. These videos belong to the top of my personal list! THANKS A LOT !!!
I tried learning QC like 6 years ago and gave up after two weeks because I had no clue what the books were talking about. God I wish I had this legend teaching me back then.
Same for me I have heard about this quantum computing in 2016 and only managed to learn basics at that time.....now learning slowly by digging deeper with the help of these lectures
22:22 According to your definition, the sine function would not be periodic. I think you should replace the "iff" by a simple "if". Actually the "iff" should be elsewhere, namely: f(x) is periodic with period p "iff" {f(x) = f(y) "if" |x - y| = k p for all k in N"}
I'm confused at around 23:30 when you define the strict definition. Surely when x = pi/3 say, and y = 2pi/3 then f(x) = f(y), x does not equal y, but mod(x-y) does not equal an integer multiplied by the period? Should the implication only be one way?
The moment you move slightly away from pi/3 in either direction, you'll see that the same thing doesn't hold at sin(2pi/3). In other words, sin(pi/3 + eps) =/= sin(2pi/3 + eps), but instead sin(pi/3 + eps) = sin(pi/3 + 2*pi + eps). That's why the implication goes both ways, but in this case the period p is 2pi and not (2pi/3-pi/3 = pi/3) as you have in your comment.
Is the If and only If right? I can see that f(x) = f(y) if |x-y| is a multiple of the period. But I don't see that this is true only if |x-y| is a multiple of the period.
In this case it goes both ways, for every time you start at x and cross a distance of k*p , you end up at the same value of the function given by f(x) irrespective of what x was to begin with.
@@danielraymond3045 Right. But the moment you move slightly away from 0 in either direction, you'll see that the same thing doesn't hold at sin(pi). In other words, sin(0 + eps) =/= sin(pi + eps), but instead sin(0 + eps) = sin(0 + 2*pi + eps).
@@ethiopianqubit I think I'm missing something. f(x) == f(y) ONLY if |x - y| = kp. So, if there was a case where f(x) == f(y) and |x - y| =/= kp, then the double implication breaks, no?
U Shape Wave Thanks for your informative and well produced video. Hello,You and your viewers might find my quantum-like analog interesting and or useful. I have been trying to describe the “U” shape wave that is produced in my amateur science mechanical model in the video linked below. I hear if you over-lap wave together using Fournier Transforms, it may make a “U” shape or square wave. Can this be correct representation Feynman Path Integrals? In the model, “U” shape waves are produced as the loading increases and just before the wave-like function shifts to the next higher energy level. Your viewers might be interested in seeing the load verse deflection graph in white paper found elsewhere on my RUclips channel. Actually replicating it with a sheet of clear folder plastic and tape and seeing it first hand is worth the effort. ruclips.net/video/wrBsqiE0vG4/видео.htmlsi=waT8lY2iX-wJdjO3
hmm handwriting is hard to read for people with disabilities and foreign students. IBM should hire a Educational Technology to not create unnecessary cognitive overload.
![The moment we stopped understanding AI [AlexNet]](/img/1.gif)
I reckon when he first published his paper on this, loads of people were like "are you Shor?"
You are truly a great teacher! I love the way you make complicated stuffs look so easy :-)
This series of six videos reg. Shor's and prerequesites is soooo awesome. I'm teaching myself Quantum Mechanics and QC and there's lots of material around. These videos belong to the top of my personal list! THANKS A LOT !!!
I tried learning QC like 6 years ago and gave up after two weeks because I had no clue what the books were talking about. God I wish I had this legend teaching me back then.
Same for me I have heard about this quantum computing in 2016 and only managed to learn basics at that time.....now learning slowly by digging deeper with the help of these lectures
What are you doing now?
This the best explanation, I have found so far. Thank you!
This is literally invaluable, thank you so much
22:22 According to your definition, the sine function would not be periodic. I think you should replace the "iff" by a simple "if". Actually the "iff" should be elsewhere, namely: f(x) is periodic with period p "iff" {f(x) = f(y) "if" |x - y| = k p for all k in N"}
Awesome! Thank you for the free content!
Thank you for very good explanation!
Thank you. An excellent tutor
If you consider orthogonal matrices the same as Unitary than OK. But in the space of real numbers Orthogonal matrices preserve norms.
I'm confused at around 23:30 when you define the strict definition. Surely when x = pi/3 say, and y = 2pi/3 then f(x) = f(y), x does not equal y, but mod(x-y) does not equal an integer multiplied by the period? Should the implication only be one way?
The moment you move slightly away from pi/3 in either direction, you'll see that the same thing doesn't hold at sin(2pi/3). In other words, sin(pi/3 + eps) =/= sin(2pi/3 + eps), but instead sin(pi/3 + eps) = sin(pi/3 + 2*pi + eps). That's why the implication goes both ways, but in this case the period p is 2pi and not (2pi/3-pi/3 = pi/3) as you have in your comment.
how can we see the animation is there a new link?
Thanks so much for the video, it really helps!
Super easy..so far😀
Thanks a lot. Saved my time
Thank you for the content!
Is the If and only If right? I can see that f(x) = f(y) if |x-y| is a multiple of the period. But I don't see that this is true only if |x-y| is a multiple of the period.
4G telecom uses information spreading and FT >>> send ... receive
23:46 The implication does not go both ways right?
In this case it goes both ways, for every time you start at x and cross a distance of k*p , you end up at the same value of the function given by f(x) irrespective of what x was to begin with.
I think you're right. sin(0) == sin(pi) but that is not an integer "k" multiples of "p" (2*pi) away from the cycle.
@@danielraymond3045 Right. But the moment you move slightly away from 0 in either direction, you'll see that the same thing doesn't hold at sin(pi). In other words, sin(0 + eps) =/= sin(pi + eps), but instead sin(0 + eps) = sin(0 + 2*pi + eps).
@@sagardollin475 You got it Sagar
@@ethiopianqubit I think I'm missing something. f(x) == f(y) ONLY if |x - y| = kp. So, if there was a case where f(x) == f(y) and |x - y| =/= kp, then the double implication breaks, no?
Thank you !!!
35:41 FOR QFT
God bless. You saved my ass.
U Shape Wave
Thanks for your informative and well produced video.
Hello,You and your viewers might find my quantum-like analog interesting and or useful.
I have been trying to describe the “U” shape wave that is produced in my amateur science mechanical model in the video linked below.
I hear if you over-lap wave together using Fournier Transforms, it may make a “U” shape or square wave. Can this be correct representation Feynman Path Integrals?
In the model, “U” shape waves are produced as the loading increases and just before the wave-like function shifts to the next higher energy level.
Your viewers might be interested in seeing the load verse deflection graph in white paper found elsewhere on my RUclips channel.
Actually replicating it with a sheet of clear folder plastic and tape and seeing it first hand is worth the effort.
ruclips.net/video/wrBsqiE0vG4/видео.htmlsi=waT8lY2iX-wJdjO3
hmm handwriting is hard to read for people with disabilities and foreign students. IBM should hire a Educational Technology to not create unnecessary cognitive overload.