How Hard is an Oxford Maths Interview? Feat. Tom Rocks Maths
2023 ж. 25 Қар.
109 439 Рет қаралды
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• Oxford University Math...
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After failing to score an interview with the University of Oxford back in the day, I've always wondered what the process is like. Luckily for me, Dr Tom Crawford is a fellow maths KZheadr who was all too happy to ask me some interview questions!
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To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/AnotherRoof/ The first 200 of you will get 20% off Brilliant’s annual premium subscription. Be sure to check out part 2 over on Tom's awesome channel! kzhead.info/sun/mMqRkqV6rpp5p30/bejne.htmlsi=My_3wwraqOSyOK5B
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.
This was a LOT of fun :)
@@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
I love the statement "I'm a pure mathematician, I know a solution exists" When you got to the equation f(3) + 2(-1/2 - 2f(3)) = 2/3 I personally felt like the question was already done and you could move on 😂
OH. MY. GOD. The 2nd exercise on probability using the zeta function is so BEAUTIFUL ! It lends perfectly on the euler product representation of the zeta function almost without noticing it, such a great idea ! Also, bravo @AnotherRoof, you did a really good job in a stressful situation like this PUBLICLY :)
What is bri yapping about
This is the coolest thing I’ve ever seen: the man is solving problems he’s never seen before, in real time.
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...
18:23 me screaming at the video: zeta(s) is independent of n, you can oull it out of the sum!!!! 😅😂
Just watched the first problem. You two really mash well together and it's fun. I hope if you do stuff like this more often, you become more relaxed.
What is remarkable is how interesting the questions are presented such that even with limited math knowledge it can be worked through in a fairly straightforward way. I think that having more advanced math knowledge was actually a disadvantage because without it you are forced to trust your intuition because you don't have as many tools to bring to the table.
The probability exercise was simply beautiful! Something we did in our first year undergrad. as an example sheet question - such beauty
This is awesome! It's also really cool to see a mathematician as bright as you admitting that you didn't even get an interview at Oxford. It sort of humanizes the world of academia a bit just knowing that you don't have to go to an Ivy League school to be successful in the academic world. Also you guys have chemistry lol. You should do more crossovers!
That was a lot of fun! I'd love to watch more of such videos with the two of you :)
The probability questions reminded me a lot of a paper by Solomon W. Golomb, titled "A Class of Probability Distributions on the Integers" -- You might find this paper interesting after having worked on these problems. :)
Tom is great and what a nice test!
Really loved your videos on set theory. Is there any chance you'll do one on the Axiom of Choice, Zorn's Lemma and the Well-Ordering theorem? As I found your explanations really clear.
Man, that getting into your own head and over analysing hit home!
I wanted to follow along but I'd seen both before! Question 1 was asked to me during my Oxford interview and question 2 was asked to my friend during their Oxford interview 😂
I hope you both did well
This is the Collab we've all been waiting for 🎉🎉🎉 I have my notifications turned on, I'm hyped!
7:49 damn it that's such an excellent approach. Normally I wouldn't check if the function is valid for the substituded value.
Amazing video! I had fun trying to solve the problems myself so where can I find more stuff like this?
I think you were confused by the Zeta function question because the notation provided by the question is awkward. First n is established as the summation variable for the zeta function, then its used as a separate variable outside the summation. When I tried solving the problem myself I just replaced the n’s outside of the summation with another letter.
One of my favorite problems in classe préparatoire (MP) among the few problems where I really felt that waow and the "mathematical elegance"
Nice team work, pleasant nocturnal entertainment to watch the solution being developed here. For a real student, things would be harder, of course, due to the pressure (much is at stake) and immaturity (they don't already have a maths degree on their belt).
Mind blown by the last question 🤯
The first one i initially tried to find x=1/(1-x), and then i tried to find x=1/(1-(1/(1-x))), and then i did it nested three times which simpplified to x=x so i knew it would be a three cycle and then i just solved it as a system of equations
I found these pretty straight forward.
I proudly managed the first question ... and the second and the third one ... and half of the fourth one (the zeta part was cool ;-))
by way of comparison, my wife interviewed for 3 colleges when she was completing her D.Phil. Her holy grail was a question that was random, where there was no single/expected 'correct' answer, but it could be asked to a candidate to show their logical progression based upon information presented. She is a geographer, but I think many Oxford Interviews have this format - it is not so much a pub quiz as a test of enthusiasm for pub quizzes!
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.
Great idea for a video! Both questions were really nice, especially the 2nd one. However, isn't the question posed at 26:30 a side track? The intersection of the sets D_p over all primes p is the null set, because there is a prime q bigger than any given positive integer n. n is therefore not divisible by q, and is therefore not in D_q. Thus, that intersection is the null set and has probability zero.
That is what I was thinking too
I haven't seen x ↦ 1/(1-x) used as a cube root of the identity map before. Very neat.
That was flippin' brilliant. The final question was amazing. Thanks.
that was amazinggggg
Not related to this video, but are you planning on continuing your "Mathematics from the ground up" series? I'm really looking forward to more of that - in particular, division. It has occurred to me that the definition/functional description (not sure which is more appropriate) of division that I know is just bad, at least taken by itself: Multiply by the reciprocal of the divisor and then reduce. It runs into a lot of problems. For example, reduction is just dividing the numerator and denominator by a common factor. Also, there's no way to get decimals by that definition. On top of that, since fractions are, themselves, statements of division, you end up with tautologies and/or infinite recursions( ie: 1 / 2 = 1/2), depending on how you want to look at it. So, I would love to know how to properly define division to not run into those problems, or if there are other processes going on that fixes them.
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!
I can imagine you going over question 2 on your own on the train back to Manchester.
That second problem, was maybe one of the best problems I have ever seen.
Well in the first functions question you could just do that more easily just by first giving the input in the function of 1/(1-x) . Now you have two equations but three variables so for now for the third equation input (x-1)/x. so now you have three equations and three variables that are f(x), f(1/(1-x)) and f((x-1)/x) so solving them you will get f(x) put x=3 and you get the answer
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?
This video is excellent; why ? …. Because it shows that ANYONE confronted with a problem they have never seen before, is in the same boat as everyone else. This also means that even a learned professor can be bamboozled by radical advances in thinking. I’m referring more to Physics than Mathematics when I say this
Beautiful proof that is kept simple (no heavy math required). Thank you! 😉
great vid
I am heartened that @AnotherRoof took the same paths that I did.
I ended up taking a couple of extra steps getting to the recursive pattern / cycle, and found it in general.
Godspeed
That second question is so juicy.
Is this exam for students who want to study at oxford?
is this for undergrads/high schoolers? seems pretty reasonable tbh
I know Tom says one doesn't need any maths beyond the normal A-level maths (bad advice) but this sort of problems are easier with more maths knowledge which most Oxford applicants secretly do know. So how is that selecting purely on mathematical ability. If some students have done group theory and some linear algebra then question one is easy. If the student has seen the zeta function in its different formulations question 2 is easier for them.
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
I'm slightly confused, if anyone can clarify I would be very grateful, in the prob question, the event defined by the intersection of all D_p's, that will be the empty event won't it. There is no positive integer that is divisible by every integer. So that probability should just be zero. The next part is alright.
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.
Increase your frequency of uploading. Your videos are great!
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 🙂
That was a dope problem and conclusion!
Got the first one in under 30 seconds ;)
These videos confirm that if you are already in the select group you are allowed to hesitate, be creative and enjoy the process. In the real life, the commoner, be it the Chinese student in a failing "socialist" country or a native of the failing Western meritocracy, would be expected to be on the right track to solution in a maximum of 10-30 seconds. That is achieved by the commoner by years of expensive tutoring, the role of the tutor is to feed the candidate with all types of puzzles the candidate might encounter. Once inside the select group one might indulge himself in "enjoying experimentation". Both would have failed a real admission exam based on their performance in these videos.
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 KZhead 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.
Riemann Hypothesis is my problem
Brutal
damn i oughta apply to oxford lol
For the first one, finding f(x) was possible. f(x) +2f(1/(1-x)) = x is given. Plugging in x = (1/(1-x)) yields f(1/(1-x)) + 2f(1/(1-1/(1-x))) = 1/(1-x) simplifies to f(1/(1-x)) + 2f((x-1)/x) = 1/(1-x) ; Finally putting x = (x-1)/x, we get f((x-1)/x) + 2f(x) = (x-1)/x. Thus, we have 3 equations that can be solved to yield, f(x) = 4/9(1- 1/x) + (1/9)(x^2-x+2)/(x-1). Indeed f(3) = 20/27. Edit: The Reimann zeta function problem was pretty trivial too. I was wondering where it was headed until he mentioned the event of not being divisible by any prime, and the Euler formula popped out from there. I am unsure as to how that mass function comes about, though. It is rather hard putting yourself in front of the camera and doing math on the fly. I recommend against it. I have professionally tested exam papers for their accuracy and ease/difficulty of solving. It is much easier if you are doing this off-camera than on-camera since every mistake of yours becomes visible to viewers. Kudos to you for your courage!
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.
Second question is awesome!
This should be how all college admissions work.
Interviewer: "- What is the value of f of three?" "- Thank you for asking! I've always wondered the same thing myself. Now let me write down the bleeding obvious, which probably will be as far as I can get, while I think about why the f-word is used twice here without being censored by KZhead."
26:32 Doesn't this probability of intersection just equal to 0? There's no number which is divisible by ALL the primes is there?
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??
@@KZhead_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.
I don't like the first question. If you go the right way and look for cycles you are done within a minute. If you don't you're stuck. There is no reason to pick the cycles approach. I did it but it feels a bit random. In the second and third question there are so many things that are not said. The "s" is not fixed so it is a set of probabilities. Are they all the same? You can take a factor in front of the sum only if it is a constant in every element of the sum. You can remove those sums only if the sequence converges.
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!
Have to interview just to be a student there?
I was wondering, what's special about the function given in the functional equation and how to create such functional equations. Seems like, g(x) = 1/1-x , is a function which on composed 3 times gives identity. Hence one gets 3 distinct equations. g(x) itself is composition of h(x) - 1-x and k(x) = 1/x each roots of identity function. Maybe these two can be composed to get other nth root fucntions (I might be terribly wrong).
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.
still waiting on a new investigation, please set out a prize this time like investigation 3
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
14:26 I did first on paper. It is intriguing. Past 14:26 is too advanced 4 me
Really pleased I studied engineering not mathematics! So no regrets there 😉
Why do you look so terrified 😅
I mean id be terrified too tbh ☠️
f(x)=1 → f(3)=1
15:31
while not quite Cambridge, Oxford still boast of such MATH titans as Wallis and Hardy. a math interview at Oxford, would be a challenging proposition. otoh, interviewing for this newfangled newage-y math"S" should be quite easy. spout a few platitudes about socio-scientific relativismus, prove your devotion to political correctness you're good to go
Dude as a NS you're chatting shit, stem isn't that woke
When ever he said 'F of', I thought how rude!
is Oxford running out of applications????
I am somewhat proud in solving the first problem ad hoc without ever seeing it before and no supporting tools. Damn I could have aced this shit. However I am already an Engineer so bit late to switch to mathematics.
Man, 2nd and 3rd are one of the most beautiful problem I have encountered so far
For the first probability problem, I think to be a bit more careful, it would have been good to show the two n’s are not the same. Then state that because zeta diverges and P is not a function of s, we can treat 1/zeta as a constant and factor it out. Then what’s left is just zeta. Zeta/zeta = 1.
That probability question is a slightly easier version of question 6 of sheet 1 in first year probability, although I imagine they have changed the questions since 2018.
where can I find the question?
@@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
holy wow, that second question
He may have gotten into Oxford, but you have better control of your arms on camera so I'd call it a draw.
I think I would feel incredibly uncomfortable in an interview like this. Not only do I have to verbalise my thought process (not everyone thinks like that) but I also have someone looking over my shoulder as I think making me feel self conscious. I wonder how many good candidates have been rejected because they don't have the social skills for this kind of exam or are simple not used to it. Up until this point most people's experience of maths exams is working on their own, in their own way and at their own speed.
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.
tom seems to prefer a xi function.
The first one I was thinking geometric series
😂😂😂😂😂😂
A MATH interview would be pretty hard. England, after all, is the home of modern Math giants, like Boole, Hardy, Clifford, Turing. OTOH, interviewing for this gimmicky “maths”, should be pretty easy
Didn’t get an interview? That’s Oxford’s loss. So refreshing to see someone who knows what they’re talking about and not just an expert at juggling rules and symbols in math like an LLM
Didn't know Michael Fabricant was a mathematician....
I wonder- how did they even come up with the first problem? The crux seems to be that the function g(x) = 1/(1-x) has the (to me crazy surprising) property that g∘g∘g = id (mod a removable discontinuity at 1) - which is how you end up looping back on the 3rd iteration. Does the fact that g itself is a composition of two involutions (x^-1 and 1-x) contribute to this property? Thus i guess there's actually nothing special about the mumber 3 here; you should be able to "plug-n-chug" for the value of f(x) for just about any x in exactly the same way... 🤔...indeed, this allows you formulate f(x) explicitly, so here it is: f(x) = (x - 2·g(x) + 4·g∘g(x)) / 9, where g(u) = 1/(1-u). Neat! Also - my gosh! - I'm just blown away by that ζ function reformulation! No words...
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...