How Hard is an Oxford Maths Interview? Feat. Tom Rocks Maths 

Pretty sure that probability problem is on the first problem sheet of the year 1 probability :D

the probablity problem confused me because i thought "Tom writes question, Alex answers". so i thought Tom was trying to help you simplify the answer about D_p by asking about their complements, not asking another question altogether.

You guys are awesome!!! My degree is in Math, as well. While probability was never my strongest area, I thoroughly enjoyed seeing these problems done out. I can definitely see the advantage of the Oxford tutorial system. I went to school, and now teach, in the US where the course work is much heavier on the lecture side.

@@davis.yoshidaYes, but it’s an intersection, not a union. So if X isn’t 1, there’s some other prime q that X is divisible by, so X is not in D(q)’ and so is not on the intersection as a whole.

😮

@@davis.yoshida Other than 1, yes. I’m not entirely clear on what you’re trying to say.

@@davis.yoshida And I’m saying that the number 1 is in the intersection, so it’s not empty. If 1 isn’t in the intersection, what set D(p) do you think it belongs to?

You're awesome mate!! The way you're also supportive is impressive. You're lowkey bossing it

What is bri yapping about

That is what I was thinking too

I'm pretty sure I cover this in part 4 on defining every number!

@@AnotherRoof Ah, I must have missed or forgotten that. Thanks, I'll go take a look!

​@@AnotherRoofeven if that topic is covered already I would love more topics in that series!

Well congratulations to your wife for completing her Dr. Phil! … which is definitely how my brain processed your first sentence upon my initial read.

Well congratulations to your wife for completing her Dr. Phil! … which is definitely how my brain processed your first sentence upon my initial read.

I ended up taking a couple of extra steps getting to the recursive pattern / cycle, and found it in general.

2:37 I paused the video there, lol (I would never solve this in an interview setting, I'm sure it's usually less amicable than this video)...I tried to plug stuff in, x=0, etc, lol...then I tried what I think the guy is trying to do, lol, to get both inputs to be equal, which would give f(a)=b, etc, a specific value...got the Golden ratio number or something, not even sure that's supposed to be that number, etc, [1+-(sqrt(5))]/2 or something, lol, pointless...then tried to get the second argument to be equal to 3, etc, x=2/3 ...so got f(2/3) + 2f(3) =2/3, so had that along with f(3) + 2 f(-1/2)) = 3, if I didn't make any mistakes...then try to see if there is a way to cancel anything out, f(-1/2) would be nice, so plugged it in, etc, got f(-1/2) + 2 f(2/3) = -1/2, so somehow those equations can be used to cancel out f(-1/2), etc, get something specific for f(2/3), and then one can get f(3) using that original equation, but I might have made some mistake...

@@archangecamilien1879and yeah, that was the answer

Ah, ok, lol, I'm sure they get harder after that...

I've only taken O level maths but I was able to solve the problems just fine. Sure, I know some out-of-syllabus math like complex numbers and extremely basic linear algebra, but that didn't help me here at all

For the first problem I suppose you mean to look at 1/(1-x) as an automorphism of H and use the matrix representation M= 0 1 -1 1 And through some argument about eigenvalue or companion matrix of 1+x+x² (or computation) you show that M³ = -Id and corresponds to the identity ?

That's essentially what he did, but he used the number 3 instead of x. It's kinda hard to expect that putting 1/(1-x) into itself twice yields back x if you've never seen it before. The reason I was able to solve this is because I once encountered the anharmonic group when studying projective geometry.

@@Noname-67 Yeah. I realized that after watching his solution. It was a surprise for me as well that it worked back on its own. However, I must admit that I had seen such a problem 25-odd years ago in my good old days.

@@Noname-67 I think you can sort of get a suspicion of what is going on by looking closely at the operation x -> 1/(1-x). This is a composite of two order 2 operations on the real line (minus 0 and 1). One is just taking the reciprocal x->1/x and one is the reflection about 1/2: x -> 1-x. It is not guaranteed that the composite of two order 2 operations will have finite order (i.e it will eventually return to the identity if you apply it enough number of times), but it is a bit suspicious and worth playing around with. And then when you play with it a bit, you realize pretty quickly that it has order 3. Anyone who has taken some group theory would have some inkling of where this problem is going, but for a high school student, it is likely a reasonable challenge.

yep the function that associates x to 1/1-x has order 3 meaning f(f(f(x))=x or f^3(x)=x

But I think you can't substitute the input. It's not a real valued f(x)? Can someone please explain this to me?

I'm trying! It's important to me that I'm proud of every video I upload, and I hold myself to a pretty high standard. I do everything myself so it's a lengthy process! Thanks for watching though and look forward to more videos soon 🙂

It is a bit a shame that he didn't mention this in the video, because constructing such question is just as interesting (and sometimes even more) than solving it. Here there is a Mobius action x->1/(1-x) in the background which as you noted is the important part of the question (and the fact that when composed with itself 3 times is the identity). It means that there is a 2x2 matrix hiding in the background, which has an eigenvalue which is a root of unity of order 3. If you wanted to do something similar with order n, then you need a 2x2 matrix with eigenvalue which is a root of unity of order n, which is not really possible in general when you want all the coefficients to be integers. In a sense, a 2x2 matrix is too "small" to have an n-th root of unity eigenvalue if n is large.

I think that's a legitimate concern but I think people like Tom are fully aware of the fact that the interview is a daunting and uncomfortable process and wouldn't mark a student down for being awkward or because of poor social skills!

@@AnotherRoof Yes, of course he's not marking for social skills, but feeling self conscious, shy or pressured in that situation is bound to inhibit your ability to think clearly about the maths. Plus if you're not used to articulating your thoughts then 90% of your brain power is doing that instead of the maths. I can think of a several influential but introverted mathematicians throughout history who could struggle in a situation like this.

Yes, it is an empty event

@@eofirdavid thanks for confirming, I thought I was going crazy. I don't know why Tom didn't point it out. The reason it really caught my eye is that I was studying some ring theory recently, and the intersection of two principal ideals is a principal ideal generated by the lcm, so for finitely many it's fine, but the intersection of infinitely many, such that there are infinitely many coprime elements, is just the zero ideal.

Dude as a NS you're chatting shit, stem isn't that woke

Thanks for watching! I remember seeing different solutions to Q1 in the comments so maybe have a scan through, and if we didn't say it outright the probability mass function is defined for any s so all the working holds regardless of the choice of s. Tom also says we don't have to worry about convergence -- it's given that they converge. Hope that helps!

I mean id be terrified too tbh ☠️

They don't confirm anything of the sort. Mathematics is one of the most testing subjects at university level, and especially at elite universities like Oxford, where it is a very different discipline to what students will be familiar with at school. The purpose of the interview is to give the applicant the opportunity to show their problem solving approach, without prior advantage being assumed. Overall, about ⅔ of Oxbridge undergraduates are from state schools, and at postgraduate level there is a great diversity of international students, including students from China and other developing nations in Asia. Tom Crawford hardly conforms to the clichéd stereotypes of either Oxbridge don, or mathematician; but he was himself, I believe, a state school pupil. In any case he does a significant amount of outreach work to encourage state school students to apply to Oxbridge, as well as the public maths education he undertakes though his presence on RU-vid and the various programmes he has done for the BBC and other media. Oxford and Cambridge continue to rank among the best universities in the world, whether measured by academic research output or graduate earnings, so there is stiff competition for places; but they have greatly diversified the socioeconomic & ethnic profile of their student intake, without becoming radically more expensive than other British universities. So unless you've got some evidence to offer, you probably need to find a different place to park your grievance-sodden polemic.

Actually, you are correct. And it makes sense to say it is equal to the poduct of probabilities. We are just multiplying infinitely many values less than 1, this product should be less than any product made of finitely many probabilities, no matter how many you include. Thus it should be equal 0. So everything is alright.

I have just realized that I made a mistake earlier. The factors being ≤ 1 is not enough for the product of finitely many of them to bound the total product and force it to equal 0. The partial products could get smaller without becoming less than some positive value (which would be the value of total product if this happens). In this case we can show that they cover arbitrarily small positive values. We want to find T = Prod( 1/p^s | p ∈ ℙ ) where ℙ is the set of primes and s > 1 *(This is absolutely not a standard notation for products, I am just trying to work with what I have here)* . For all p ∈ ℙ, p ≥ 2 thus 1/p ≤ 1/2 , and because s > 0, 1/p^s ≤ 1/2^s (it would be reversed if s < 0). Furthermore, s > 1 which means for a ≤ 1, a^s ≤ a so 1/2^s ≤ 1/2 resulting in 1/p^s ≤ 1/2. Now let J be a finite subset of ℙ. We get Product( 1/p^s | p ∈ J ) ≤ 1/2^|J|. As J gets bigger, 1/2^|J| becomes smaller and all natural powers of 1/2 will be covered and consequently, Prod( 1/p^s | p ∈ J ) will cover values less than those powers. No matter what positive r you choose, there is always a Prod( 1/p^s | p ∈ J ) < r (This is easier to see if we find some n such that 1/2^n < r then take |J| = n), so T can't possibly be any positive value since T ≤ Prod( 1/p^s | p ∈ J ), it can only equal 0. Note: this argument doesn't work if negative factors get involved.

Hi??

@@RU-vid_username_not_found Sorry for not responding, I just thought that I couldn't give a good contribution to your answer, just a lazy thought: I agree with your argument, but wouldn't it be sufficient to say that since 1/p^s for s > 1 approaches 0 as p approaches infinity, then the whole product approaches 0? Like in the end we are just gonna be multiplying out a bunch of zeros (really rough simplification)?

@@lukandrate9866 Yeah, 😀 that works as well! but I wanted to write everything that way to make motivation for how one should define infinite products and what properties it should have in order to consider it genuinely a *'product'* . Once we agree on what the definition should be (it should also reduce to the same definition in the finite case to be consistent) we can analyse it even further and make results about it. For example as you said, if we can form a sequence out of the factors that converges to 0, then the infinite product will be 0. *Be careful* , there is no order of factors in infinite product, so I don't talk about the limit of partial products of some sequence (analog of the limit of partial sums of a sequence). Those 2 are not the same thing. One can prove that rearranging the terms of a special class of sequences will change the limit of partial products (analog of Riemann rearrangement theorrm for series) And thus, the notion of infinite product doesn't make sense in this case. And in the case of existence of infinite product, one can prove that any sequence you can form (must include all factors) will have a limit of partial products (there is some order) equal to the infinite product.

a great thing for inspiration is ten hundrets last supper arg

@@yours-truely-sir I've really been enjoying your comments lately -- thanks for participating! I want to do more work on this, but my videos take so long to make that usually I don't really have time to set anything like this up. I hope to come back to it in the future though!

@@AnotherRoof thank you very much for your answer I really enjoyed your tresurehund and your videos till now. thank you

@sitterswapper Looks like my memory is failing me, it was actually sheet 2. All the lecture notes and problem sheets for oxford maths are freely available, here is the sheet: courses.maths.ox.ac.uk/pluginfile.php/74404/mod_resource/content/0/Sheet2-2018.pdf

12:19 i tutor students up to calc, and i think i can safely say that being a trained mathematician has nothing to do with a tendency to overcomplicate things... one of my greatest joys in my job is how often i get to respond with "yes" to people incredulously asking "it's that easy?!"

It is not too hard to build such a function. You should read about mobius functions which appear naturally in many places in and outside of mathematics. They also have this nice property that if you know where 3 points go, then you know everything, which relates to your observation that 3 was not important

@@eofirdavid ah thanks i forgot about those. it's been a while...

I hope you both did well