Prime Reciprocal Series with
ฝัง
- เผยแพร่เมื่อ 29 ก.ย. 2024
- Steve from blackpenredpen answers a real Oxford University maths admissions interview question set by Oxford Mathematician (and interviewing tutor) Dr Tom Crawford.
The question looks at the divergence of the sum of the reciprocals of the prime numbers, using the Fundamental Theorem of Arithmetic, the divergence of the Harmonic Series (1/n), and the Basel Problem (sum of 1/n^2 equals pi-squared over 6).
This is part 2 of the interview - you can find part 1 on Gabriel's Horn here: • Oxford Maths Admission...
Check out Steve’s brilliant channel blackpenredpen here: / blackpenredpen
Produced by Dr Tom Crawford at the University of Oxford. Tom is an Early-Career Teaching and Outreach Fellow at St Edmund Hall: www.seh.ox.ac....
For more maths content check out Tom's website tomrocksmaths....
You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.
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Get your Tom Rocks Maths merchandise here:
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Watch part 1 of Steve's Oxford interview here: th-cam.com/video/htB_NGmPKVI/w-d-xo.html
Can you do this with Mathologer next?
Steve is just adorable. He's so excited and he is having so much fun with math. I remember watching him since long before the beard, and he's grown up with no loss of that excitement. Thanks for motivating us, both of you:)
man, i've been watching your videos for long time. Never commented tho. You are the live image that the image doesn't matter. so normally people would think you're some kind of person with no studies or anything for your image. That happened to me a lot, but you're frecking brilliant. I really love your stuff, you're a genius on math and magnificient with teaching. I really love your content and enjoy it. Thanks man, you are the boss!
I don't want to offend anyone just wanted to say that because I've been thinking it for too long
Wawwwwwwww this is real maths !!!!! Wwwwwaaaaaaawwwwwwwwwwww no words to say . Just wonderful maybe the best of all times ... Thnx both of you ... You were excellent prof steve and you have shown how smarter you really are . Thanks Dr CRAWFORD .
you're welcome :)
Cor, that's a stinker to solve in interview conditions! I have Maths A Levels (1983) but I would have struggled with this.
You and me both!
@@TomRocksMaths That's kinda reassuring, at least!
Its hard for me although i cracked jee advanced
Hello Tom, please response to my email
Only one word, WOOWW
Haha thanks Marcos!
XD el profe john escribiendo en ingles
@@litiginuyt6272 Es la primera vez que lo veo XD
Lmao
Although I understood nothing, it was freaking awesome to watch
With hard work and lots of practice you'll be able to understand it eventually :)
@@TomRocksMaths What made you fall in love with maths?
What I love about this is that you are two guys that have come together to have a bit of fun with a subject you both love. There is nothing more admirable than when someone does something well and enjoys every second of it. Really inspiring. Good job guys and well done Steve for surviving that 😂
Awesome thanks Graham :)
The last «pretend interview» with Gabriels Horn that has a finite volume but infinite surface area was awesome, but this one was even more interesting. I learned a lot from this video, very funny to see how an actual Oxford Interview would have been. And Steve did excellent!
Don't know how I (CS major) end up here and watching both 2 parts... amazing series !
Glad you enjoyed them :)
There was a mistake at the end in the second use of comparison test when (1+1/p) was replaced with just 1/p, but I guess Tom didn't notice, and it's kinda hard to see how you get back to the 1/n series unless you write out some terms. Notice that the partial product (1+1/2)(1+1/3)(1+1/5) = 1+1/2+1/3+1/5+1/6+1/10+1/15+1/30, so you're getting the reciprocals of all square-free numbers when you continue with more primes. Then including the 1/k^2 part, you're getting back the entire 1/n series using the representation of a general n as a product of a perfect square and a square-free number.
Yes. The penultimate sum should be equal to zero.
Gold comment. I also think that part was a mistake. Your explanation makes sense. Basically we cannot just replace (1+1/p) with (1/p), otherwise the formula would just converge, since the product of 1/k^2 and 1/p is smaller than 1/k^2. Taking every possible combination of product of 1/p and 1/k^2 would for sure contains all 1/n, therefore the final inequality stands, and this should come from the formula with (1+1/p).
Came here to say the same. You already have the harmonic series in the line with the (1+1/p) term, as the product over all p creates a sum where every prime can either be “turned on or off”, giving you exactly what you need to be multiplied by every possible k^2 in order to achieve every natural number exactly once.
i'm a bachelor undergraduate student in mathematics from Brazil and its so amazing!
I still think it's kind of harsh expecting a student to answer this. It might just end up killing their love of math.
this wouldn't be used today - it's from 50 years ago
Didn't understand a word of it - loved every minute of it! Gobsmacking and its clear you both enjoyed the fun too !
YES I was waiting for the second part
It is cool how powerful the taylor-series is. You can make e^sum(1/p) and then with taylor you can split it all up
Seeing BPRP "struggling" gives me a small hope that I may not be as dumb as I think I am...but then I haven't been in uni for a year now cause of covid so all my math knowledge these days are from youtube
Arent you doing university online now?
@@donovanb8555 unfortunately, no. I'm from Myanmar and if you look up recent news from Myanmar you'd find out the huge reason why...
@@atraps7882 oh, ok I understand. How is it there with all this going on?
@@donovanb8555 thank you for asking 🙂 Well in short, we are all just trying to stay safe, avoid any trouble and doing our best to stay positive.
I hope everything will go fine with you and your city. Stay safe!
A video with zer0 dislikes. 😀👍
Love the video, but you're abusing that k aren't you? The i_1, i_2, ... numbers should stop at an i_j, where j
For 9:30, I thought that this could prove the rest of the answer.
As N->inf, P->inf because there are infinitely many prime numbers. Thus, sigma (1/P) {p
The first sum states that sigma 1/n goes to infinity, not 1/p.
Just because the primes are an infinite subset of the naturals, doesn’t mean that sigma 1/p must also tend to infinity. For example consider the sequence (1/(2^n)) (so 1, 1/2, 1/4....) . This is an infinite subsequence of (1/n) but, as n ranges from 0 to infinity it approaches 2.
I didn't understand anything - but that's oki :)
as long as you had fun :)
How did Steve pass the interview if he did less than 30% of the second problem unaided? Just kidding. Anyways, one way to pass the real interviews to these elite institutions if you happen to be "unfortunately rich, connected and adorable" is to tutor with the same people who will design the interview questions. Meritocracy! -) After you were admitted to the said institution you'd feel very filled up with guilt so you'd have to become "progressive". Another way is to be "Chinese" (not necessarily literally Chinese) and train 7 days a week 18 hours a day for 10 years ... Which way you prefer ? Hmmm, ... so hard to chose -)
As a year 11, is it bad that I don’t understand much of the maths involved.(Watch the video because the maths is really cool to watch)
Hi, I’m in year 11 as well and understand how you feel. Just keep in mind that so much new and important maths is taught at A level, so we will understand it in not too long. If you are doing further maths GCSE, and/or are interested in maths enough to watch further videos (which you seem to be as you are here), some things will be more familiar to you than others so although this looks like gibberish now, you will understand it well in the future. Hope this helps :)
Edit: Just think of it like this - when you were in primary school, you probably didn’t even know what algebra was. When you were first introduced to factorisation, you probably found it hard. Yet now it is easy. Most high school level maths will be like that. When I first looked at matrices for further maths GCSE, I had know idea how to do them. But now I have been taught so I do understand. So of course most people below A level won’t understand this as we haven’t been taught these concepts or rules yet.
Thanks. I really want to do maths so I was getting kinda scared that I really couldn't follow much of it. It's great to know there are other year 11s that are like me :)
I’m in year 13 and takes maths and further maths - that’s perfectly fine if you didn’t understand this stuff. There’s lots of concepts you’ll learn at A-level. Once you start having half of your lessons just be maths your mathematical knowledge just explodes. That being said, I also found this proof quite tricky.
I found this proof to be a bit more accessible:
th-cam.com/video/u5EzjE5fXjU/w-d-xo.html
No one in Y11 should have any idea what is going on - so don't worry! Awesome you are watching though :)
isn't there a problem about the last step? not all n have a square as a diviser, for many numbers they could be just a product of primes, or even big prime numbers?
(for exemple 23 is just 23*1 )
1 is also a square number
You can always split a number into a square and ‘square free’ part (en.m.wikipedia.org/wiki/Square-free_integer)
It doesn’t matter how big the square free part is, square free numbers have some useful properties.
@@rish5827 what i asked was, how can we get every n from N while mulitplying that number of primes by k^2, that 1/k^2 multiplied by those primes, when k is bigger to 1, isn't going to give us prime numbers , so how can it generate all n in N?
And if that's the case, we still can't tell if that sum diverges or not
Great video just watched for the first time! Would love to see a video on how the prime reciprocals can be transformed into the prime zeta function. Similar to how the harmonic series is a specific instance of the Riemann zeta function. Here, numberphile? Have been looking at zeros of the prime zeta function for possibly 20 yrs so can help in that aspect! Refernces are the wikipedia page and Fröberg, Carl-Erik. "On the prime zeta function." BIT Numerical Mathematics 8 (1968): 187-202.
Thanks for this video! I'm not sure how you go from p1^(n1-1) x p2^(n2-1) x p3^(n3-1) x ... x pk^(nk-1) = k^2 for any n in N, where k is an integer. This is bothering me now. Can you please clarify ? Thanks !
Get super excited about THE PATTERN TO THE PRIMES!!!! Revealed in a little-known video about "Teaching Math for Social Justice" - th-cam.com/video/rVCoT-Z0mBA/w-d-xo.html
I like help me please
I'm a little confused as to how the the weird sum related to the sum of the reciprocal primes. I understand the proof for why the weird sum diverges but how does that prove that the sum of reciprocal primes also diverges?
Really tough. If not seen before is unlikely to figure out at the moment
I did warn you all it was the hardest one I could find!
I got lost from 10:49 with the proof. Nonetheless, I came here from part 1. What an amazing interview! You have a new subscriber and I will watch your PDE video now since PDEs are my favorite :D
Welcome aboard!
I wanna apply to Oxford just to do the interview lol
no youtube you don't understand... I failed all my math courses.
Such a fun video :)
Glad you liked it!!
I mean u guys are playing a game called maths jeopardy
Haven't heard about optic fiber internet?
13:05 this part was the one that didnt make ANY sense, where did he get pi from? Some special property of e^series?
Pi means multiplication here.
As sigma means addition.
Its just a notation
The last step doesn't make sense at all
Is this question asked to students who are coming to get admitted for graduation ??
This is a very old question (from around 50 years ago) but it was used for undergraduate admissions. For a more realistic question that would be asked today check out part 1 here: th-cam.com/video/htB_NGmPKVI/w-d-xo.html
They wanna get more Eulers (perhaps to stack the faculty board). Lol =D
OK never mind my comment on the last one this is way beyond what I can do at home haha
Still enjoyable none the less
Oh
Actually, the terms can be compressed into the "considered" expression upon observation and then split into the individual expressions to check for divergence and convergence. There are also many other ways to prove this question's validity. I wonder if anyone used MI method to solve this... By the way, nice Basal Problem hidden there.
Do one with Bri the Math Guy! I'm sure he'd be down
Tom please keep a live QNA session we wanna ask questions from you .
Also please inform us about the timings in the community section .
we are waiting ................ 😊
I'm afraid I have to go and teach my students! I'll try to do a live Q&A when I post my videos on a Wednesday (and sometimes Thursday) each week.
@@TomRocksMaths OK see you there 😊👍
Superb
It's crazy how someone thought of that consideration in order to make this proof
agreed. very clever stuff.
Is this a interview question for undergrad?
Yes, but a very old and very difficult one! The Gabriel's Horn question is much more representative of what we ask today.
At 11:25 you can't get rid of the 1 in the product. If you do, the proof doesn't work out.
What is with this 30 frames per minute blackpenredpen garbage camera? Are they recording from a laptop or something?
Yes we had to record virtually using Steve's webcam/internet connection
One day when I get admission to Oxford I will get to meet you 😌
fingers crossed!
That might be the coolest tattoo ive ever seen!