I reckon when he first published his paper on this, loads of people were like "are you Shor?"
@Jamblox-nm5er3 ай бұрын
bro ongl i was expecting all formal comments this caught me off guard 🤣
@tomgraupner1713 жыл бұрын
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 !!!
3 жыл бұрын
You are truly a great teacher! I love the way you make complicated stuffs look so easy :-)
@saikrishnasunkam43444 жыл бұрын
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.
@satishchandramedi97293 жыл бұрын
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
@sigma72085 ай бұрын
What are you doing now?
@pranshisaxena60632 жыл бұрын
This the best explanation, I have found so far. Thank you!
@sakuranooka3 жыл бұрын
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"}
@_oubliette3 жыл бұрын
This is literally invaluable, thank you so much
@fleurette2104 жыл бұрын
Awesome! Thank you for the free content!
@xShadzHD4 жыл бұрын
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?
@ethiopianqubit4 жыл бұрын
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.
@infinity-and-regards4 жыл бұрын
23:46 The implication does not go both ways right?
@sagardollin4754 жыл бұрын
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.
@danielraymond30454 жыл бұрын
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.
@ethiopianqubit4 жыл бұрын
@@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).
@ethiopianqubit4 жыл бұрын
@@sagardollin475 You got it Sagar
@danielraymond30454 жыл бұрын
@@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?
@user-jq5bz4wp2e Жыл бұрын
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.
@mavihs268 ай бұрын
how can we see the animation is there a new link?
@ferashamdan42524 жыл бұрын
Thank you. An excellent tutor
@nastyavicodin6229 Жыл бұрын
Thank you for very good explanation!
@anthonyvenable30482 жыл бұрын
Thanks so much for the video, it really helps!
@jxchtajxbt533 жыл бұрын
If you consider orthogonal matrices the same as Unitary than OK. But in the space of real numbers Orthogonal matrices preserve norms.
@yevonnaelandrew95534 жыл бұрын
Thank you for the content!
@vardhan2548 ай бұрын
35:41 FOR QFT
@senkum803 жыл бұрын
Thanks a lot. Saved my time
@pghislain3 жыл бұрын
4G telecom uses information spreading and FT >>> send ... receive
@aaroningram94654 жыл бұрын
Super easy..so far😀
@saadhassan8813Ай бұрын
Nice Lecture
@pragneshprajapati7743 жыл бұрын
Thank you !!!
@BigDaddyGuagg3 жыл бұрын
God bless. You saved my ass.
@SampleroftheMultiverse11 ай бұрын
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 KZbin channel. Actually replicating it with a sheet of clear folder plastic and tape and seeing it first hand is worth the effort. kzbin.info/www/bejne/raOlpKSfepWpfZYsi=waT8lY2iX-wJdjO3
@alextran89062 жыл бұрын
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.