Solving Mathematics in X+Y (A Brilliant Young Mind)
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- Опубликовано: 4 окт 2024
- Looking at the mathematics in yet another mathematics film! Let me know what film you want to see next 👀
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For those of you that are new here, hi there 🌞 my name is Ellie and I'm a Part III Mathematics Graduate from the University of Cambridge and current Astrodynamics Software Engineer! This channel is where I nerd out about maths, physics, space and coding so if that sounds like something you're interested in, click the subscribe button to follow along ☺️
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I love that movie. It's a favorite. The boy actor was also in Hugo.
Make a video on how can you design aircrafts ..! [Also physics behind them]
always love your videos
I think if you allowed the third rule the sequence would never terminate. I'm guessing here but it looks like an infinite loop.
I'm thinking that if you're allowed to overflow, then you can no longer assure decreasing behaviour; then if you start with an odd number of cards, you'll always end up with a remainder and hence overflow (if no overflow is allowed, you'd stall since there may be no card to the right to flip/digit to set).
@@beancount811that's what the ultimate result
Dear Ma'am
Ma'am please bring a video about "how you make a career in space engineering " "What exam you give for this ?" ..
From my family , no one experience about this ,if you guide me it would be big help for me.🙂
This is queen shit 🙌🏻👑
Hey Ellie! What app do you use when writing out these equations, on an iPad I presume? It looks so nice and hopefully is free 😢
I second this. I crave this app!
I'm agreeing with those that think the loop would be infinite if we allow the 3rd sequence, hence I think this is why it was a good idea to reject it for the sake of the criteria of the question( should terminate is not equal to infinite). For the Odd part : if we use your system of Binary then we're still looking at 0 as our final answer.( Terminates )
plzz post some videos on application of coding in astrodynamics
So to start with the odd number of cards question: something to notice is that whether you do move 1 or 2, the number of 1s in the binary number will have the same parity as at the beggining ( we either decrease it by 2 by doing move 1, or not decreasing it at all by doing move 2), so, in the case we have an odd number of cards, we will always end up with an odd number of 1s , and since 0 is an even number, the process will never terminate.
Now regarding the move 3... if we are allowing that to happen, then it doesn't matter how many cards we have and the parity of that number won't matter, the process will always terminate! The problem is quite tricky now and I'm not actually sure that my answer is correct... One first thing to note is that no move can be done on the same card 2 times in a row. At the same time, the parity doesn't matter because by doing move 3, the number of 1s will decrease by 1. So the binary number that we have will always decrease after a move like it does in the video, so the process will eventually end since we don't have the problem we had in the odd number of cards before: i believe it's obvious that, no matter what is the sequence of moves, the number of 1s will either decrease or at least will remain constant until a point where it will have to decrease (we can't do move 2 an infinite number of times), so we'll have 2 cases:
1. We'll get to the point where we only have 2 ones, and in the worst case scenario, we'll always do move 2 until we have both the 1s on the last places on the right side (we actually can guarantee they will get there beacause when we do move 2, the 1 will move right by one place), at which point, considering we can do move 3, the process will terminate.
2. We'll get to the point where we have only 1 one, at which point , the only move left to do is 2, which will move the digit 1 right until it reaches the far right spot, where the only possible move to do will be move 3, and the proccess will terminate.
If move 3 is valid, then 1st case is true, but if move 3 is not allowed, then 1st case will never occur
Thanks PROF UR IQ is above 200+
KISSI KAA IQ ka ptaa lgaane ka tumhara kabiliyat ko dekh ke mere ko lagta tumhar IQ 0. Hai😂😂😂
introduction to calculas next?
Why u dont reply any of the comments????????/👀👀👀💔💔
Cause of creeps and weirdo
Did you graduate yet?
Permuattaions and combinations and probabbility were my enemy in my JEE preparation.....I always secured 0 or negative marks in these chapters....After 1 year today also feeling like that...
Try starting by revisiting basic sets, logic and counting techniques first. It may also be helpful not to look for direct solutions to most JEE type problems first, as these can often be more labour intensive or not very easy to deliver under exam conditions. There's normally a trick that's required to make life easier.
@@beancount811 Bro I have Cracked Jee Advanced exam.....I am right now in IIT bombay.....Here in B-Tech Course there is no use of P&C and Probablity....I was Just Telling my experience with these chapters...........and Are u An INDIAN ??/
@@KaustubhTiwari-n7v No, but I like seeing people tackle JEE problems on RUclips. It's a tough test. What area of engineering are you going to specialise in?
@@beancount811 I am in mechanical engineering branch
ELLIE CHANNEL IS NOTHING WITHOUT JEE CONTENT . NO ONE KNOWS IT UNTIL YOU PUT JEE CONTENT
Those cards are massive!🪄🎩With overflow rather than chop, shouldn't it loop forever?
X0011XX0011XX001-> 1 cannot return as 0 if X equals 1, like the game show in America, the price is right, enough life rafts on the Titanic, but enough whistle blowers to hike inflation, American graffiti