Oxford University Mathematician takes Admissions Interview (with @AnotherRoof)

[@AnotherRoof]

Thanks for having me, Tom!

[@ArzerMe]

Give this man a pin

[@proloycodes]

@@ArzerMe and a heart (not that one)

[@lambda653]

Did you guys really struggle on this? These questions were easy, and I'm 19... does that mean I could potentially get into Oxford lmao???

[@JoyoSnooze]

@@lambda653 Pfft. I found these easy and I'm 18. Does that mean I could potentially get into Oxford lmao??? Maybe bullshit college, if not. See you there!

[@rikmaitra]

U had ur revenge it seems 😂

[@cheerfulken]

Really pleased that you showed a question going wrong. Seeing that even a maths fellow can stumble will surely help a lot of students who take these interviews and feel they struggle with the questions. And commendable bravery on your part to put yourself under that pressure!

[@gildedbear5355]

I second this sentiment. It's always good to see that the Knowledgeable people on a subject ALSO struggle sometimes. I'm reminded of what my first public speaking teacher said, "every person who has ever gotten up to speak in public has been nervous. If you aren't, then your speech will be terrible."

[@empathogen75]

My entire math education is watching math videos and I did know it was an exponential function just by the properties of exponents.

[@toppinzr3743]

He doesn't seem to be a math professor, just a teacher there. A math professor would figure this out a lot more quickly. His fumbling was surprising, even if he were a grad student. It's obvious that an exponential function will work.

[@empathogen75]

Not at all@@toppinzr3743 If you're specialized, it's pretty easy to forget stuff you learned years before. Sometimes younger people do better at this kind of stuff than people with tons of experience in the field -- it's pretty well known that recent CS grads do a lot better at coding interviews that professional software developers with 20 years of experience do, because the algorithms and data structure classes are fresher for them.

[@aquilabamigbade3473]

@@toppinzr3743pretty sure he has his maths phd from Cambridge and undergrad in maths from Oxford. Wild idea I know but people make mistakes, even professors make errors

[@RomanNumural9]

16:56, you can take y as fixed but arbitrary for the first question. Then differentiate wrt x (i.e. partial w.r.t x), (f(y) here is just a constant) then you have f'(x+y)=f'(x)f(y), then set x=0 and vary y, and you have f'(y)=f'(0)f(y) which is a common differential equation with solution f(y)=Ce^(f'(0)), which then leaves us with f(y)=1*e^(3x) as a unique solution to the differential equation, no guesswork needed :)

[@nairanjith]

That's nice!

[@RomanNumural9]

@@nairanjith thanks!

[@pelledanasten1615]

Another solution is using a definition of e. We know f(x+y)=f(x)f(y), which can be generalised (by induction, if neccessary) to f(sum(set))=product(f(elements) in set). Essentially, f(a+b+c...+x+y+z)=f(a)f(b)f(c)...f(x)f(y)f(z). We can describe any number x as n*x/n=x/n+x/n...+x/n+x/n. As such, f(x)=f(n*x/n)=f(x/n)^n. Lim as n->infinity of x/n is 0, meaning lim as n->infinity of f(x/n)^n=lim as n->infinity of (1+3x/n)^n from the derivative and value being 3 and 1 respectively, which means f(x)= e^3x from e^ab=lim n->infinity (1+a/n)^bn from the definition of e.

[@irober02]

My maths lecturers were no doubt brilliant at maths but almost all were terrible at teaching. It's fascinating to watch a brilliant mathematician wrestling with a problem - a point of common humanity after all. So often solutions are presented as "one I created earlier". Nice to watch the sweat. Thanks to both of you. PS: And you CAN do it without chalk!

[@rmsgrey]

Yeah, there's a tradition going back to Euclid of presenting the simplest, most elegant version of a proof, including things like "let epsilon equal three delta" that you only do if you already know that factor of three is going to cancel perfectly later on. That's arguably better for proving your result - the more elegant the proof, the less there is to check - but it's also a horribly distorted picture of the work you actually did, and gives no hint at the reason you were even trying to prove whatever it is in the first place. If you're looking to educate, rather than just convince, you want to include the messiness - the initial experimenting with a problem to get a handle on it; the developing intuition and insight; and then the possibly somewhat rocky solution and maybe also a polished version of the solution.

[@bastiaanstapelberg9018]

Wereldwijd speelt dit, wiskundeleraren zijn de slechtste leraren betreft onderwijskunde

[@obiwanpez]

But Euclid also struggled and taught students via Socratic Method, showing students the counter-examples to their logical conclusions. Presumably, he did not feed them precompleted examples, as has been the teaching mainstay since only around the Industrial Revolution.

[@rmsgrey]

@@obiwanpez Maybe when teaching directly, Euclid taught better, but Elements is full of the briefest, most elegant, polished constructions, with no hints of how anyone would think of them...

[@obiwanpez]

@@rmsgrey Because Elements was never meant to be a textbook or a guide. It's the sum total of their society's knowledge. It's condensed because Euclid was trying to compile a step-by-step ex axiomata construction of all of known mathematics. I would NEVER teach out of Bertrand Russell's Principia. Euclid is only slightly more possible because the subject is more reachable.

[@kruksog]

Love @AnotherRoof. This was really fun. I love these kind of math problems where if you have some general math education, you can solve them, but they're not what I'd call "easy." More of these kinds of problems on the channel!

[@sqlexp]

I don't see what the big deal about this problem is. For constant y, d f(x+y)/dx = f'(x)f(y) d f(x+y)/d(x+y) = f'(x)f(y) f'(x+y) = f'(x)f(y) f'(y) = 3 f(y) by substituting x = 0

[@mathnerd5647]

The first problem is a type of a question Dr. PK is an expert for. Good video Tom

[@TomRocksMaths]

Don't forget to watch part 1 on Alex's channel where I ask him some of my own admissions interview questions: kzbin.info/www/bejne/rqPJfXerms2tpck

[@AylaTheQueenIdk]

The creativity of problem solving (the thing tested in these interviews) is my favourite part of mathematics. It’s really unfortunate to see how little creativity and problem solving is actually valued in academia, from my experience at least. The problems were great to try go through myself while watching along, in both videos. Thanks for the collaboration, I hope to see more of these in the future maybe

[@hanshans387]

Loved this! really enjoy hearing the thought process for problems you don't know the answer too already, it's really useful to see a problem solved organically rather than be being told the logical steps from someone who already knows the path! For the second question, you can completely ignore the 10s, because all the 10s have a factor of (5x2) in them, and because there's a surplus of factors of two, we can always pair factors of 5 with a 2 so we only need to know how many factors of 5 exist in the elements of 1000! - so count multiples of 5, then 5^2 (because e.g. 25 contributes 2 factors of 5, not just the frist one which is captured by finding all the multiples of5), then the multiples of 5^3 and 5^4. We can ignore multiples of 5^5 because that's greater than 1000 anyway. There are 200 elements divisible by 5, 40 by 5^2, 8 by 5^3 and just one by 5^4 for a total of 200+40+8+1 = 249 factors of 5 = 249 zeroes at the end!

[@GaborRevesz_kittenhuffer]

26:22 ooh i got this one almost instantly! note that 10=2·5. given that every other factor in a factorial is even whilst only every 5th factor is a multiple of 5, there will be a vast abundance of factors of 2 compared to number of factors of 5. so the number of trailing 0s will be limited by (i.e. equal to) the number of factors of 5. thus counting trailing 0s is tantamount to counting factors of 5: ⌊1000/5⌋ + ⌊1000/5²⌋ + ⌊1000/5³⌋ + ⌊1000/5⁴⌋ + ... = 200 + 40 + 8 + 1 + 0 = 249.

[@kaspervestergaard2383]

Nerd.

[@mahiruddin3304]

@@kaspervestergaard2383 or it is just you being stupid

[@professorzoom7800]

i’m so fucking confused

[@methatis3013]

​@@kaspervestergaard2383mate, you just watched 20 minutes of that exact problem 💀

[@kaspervestergaard2383]

Don't understand it tho. Still like to watch people explain things. @@methatis3013

[@toppinzr3743]

His first impulse to differentiate f(x + y) = f(x)f(y) works great! Differentiating wrt y: f'(x + y) = f(x)f'(y) Setting y=0, f'(x) = f(x)f'(0) = 3f(x) f(x) = c*e^(3x) f'(0) = 3, so c=1. So f(x) = e^(3x).

[@JesterFlemming]

No sorry, I don't like your solution. From f'(x)=3f(x) you used your knowledge, that f has to be an exponential function. But the same knowledge told me immediately, we have to deal with an exponential, because a^(x+y)=a^x*a^y and there are no other functions that fullfill that. You just use this knowledge with an extra step. So I wonder, what kind of knowledge is even allowed in these tests and why can I not directly start with an exponential function?

[@PedroCristian]

25:39 I'm glad to see I still remember formulas seen years ago.

​@@JesterFlemming The difference in stating an exponential solution from the beginning and the procedure of the previous comment is that the linear first degree diff. equation has a 1D vector space as solutions, so you have a more compelling argument that the exponential is the ONLY solution. I suppose you can bring every possible mathematical knowledge as far as you use it correctly and precisely (the so-called rigor).

[@TheJmebe]

Although I have no real understanding, (a bad GCSE student) I'm fascinated by the breakdown and analysis of the problems.

[@WatchingTokyo]

That was a lot of fun! I'd love to watch more of such videos with the two of you :)

[@MaxCareyPlus]

I got to f(x)=1/f(-x) and then sketched the graph of a smooth function that satisfies that constraint. Seeing the picture really helped me identify the function!

[@pinkfloyd1979]

Another way to do this would be to differentiate f(x+y) for x and y. You will see that f'(x+y) = f'(x)f(y) and f'(x+y)=f(x)f(y). Then, you have f'(x)/f'(y) = f(x)/f(y). So, intuitively and very quickly, it is possible to think about an exponential function since it is the only form that the function itself, and its differentiation are the same. Since, f'(0)=3, it is straightforward to say that f(x) = e^3x.

[@andrearaimondi882]

5 minutes in and I can tell this video is incredibly useful to Maths students. Amazing professor, great question.

[@slly8611]

Saw you at the maths in action event in london! I loved everything you did and that football game made me giggle

[@TomRocksMaths]

Glad you enjoyed it!

[@conradcox4371]

Good luck to everyone with your interviews in the upcoming month!! We've got this!!!

[@nol2521]

you too my friend, we got this

[@GainesSteve]

The "non-10 5's". Great mathematics! :) Great video, both - thanks

[@Megathescientist]

In considering the initial question, let x and y vary independently. Our goal is to derive the expression D_(x+y)(f(x+y)) = f(y)D_x(f(x)), where f(y) = f(x+y)/f(x). It is crucial to note that f(x) cannot be 0 at any point, as this would imply f(x) being null, and consequently, its derivative D_x(f(x)) would also be 0. Initiate the derivation with D_x(f(x+y)) = D_(x+y)f(x+y)*(1+0) and D_x(f(x+y)) = f(x+y)/f(x)D_x(f(x)) by substituting f(y) with f(x+y)/f(x). Introduce a new variable t = x+y, resulting in D_t(f(t))/f(t) = D_x(f(x))/f(x). This expression must be a constant since t varies independently from x. Set D_x(f(x))/f(x) = c (a constant), yielding f(x) = c'e^x, where c' = e^c. By imposing the appropriate boundary conditions, we can swiftly determine c', thereby concluding the solution to the problem.

[@aniruddhnls]

Yeah I started thinking along these lines but I have no idea how to differentiate this. Do I take y as a variable? But this is a univariate function, so both x and y must be the same “variable”. But then they are just dummies to indicate what happens when you plug in 2 numbers. So then how do I differentiate the two sides? I am completely confused about this. Basically if I do d/dx on both sides, what do I do then? Product rule? Not product rule coz it’s the same function in the product in RHS…????

[@DumbMuscle]

24:13 I got into Oxford (physics, rather than maths) pretty much with this kind of "interview hijacking" - the professor finished off a mechanics problem with a quick "and just as an illustration, here's the kind of thing we'd do with it in first year", to which I said something like "that doesn't sound too bad..." and proceeded to get guided through the more advanced bit as I figured it out with some hints from the professor. Which was essentially adding some rotational mechanics into what had previously been a problem only about lateral movement - so a lot of "well if I just assume that this variable works kinda the same as the non-rotating equivalent, does that work?"

[@olivetree7430]

I have interviews coming for physics and I don't know what to do 😭

[@noohp7305]

@@olivetree7430 look at past interview questions to get in the mindset. here's an example: a full bottle of water (you can assume it is a cylinder) is attached by a string to a fixed pivot, such that it acts as a pendulum. i open the lid of the bottle of water so that water exits at a fixed rate, and displace the bottle so it starts swinging. describe the period of the pendulum over time.

[@peterhall6656]

The first problem can be abstractly solved in your head once you know that the exponential function establishes an isomorphism from the additive group of reals to the mulitplicative group. Cauchy did theorems on functional equation stuff like this in the early 1800s. Students steeped in functional equation stuff salivate when they see f(x+y) = f(x) f(y) ! But Tom is a bare hands guy as we know and he got there in the end! There are many, many roads one can take.

[@egwenealvereiscool7726]

That's how I did it. I started with e^(ax) since e^x is nice with derivatives and realized that if a =3 then f'(0) = 3 and that led me to e^(3x)

[@lachydauth4248]

I am so proud of how quickly I got the first question and beating Tom to the solution. I started with setting y = 0 like Tom. But I quickly remember that exponentials turn addition into multiplication like shown in the question. I remembered this fact from a matt parker video I watched on the gearbox calculator (maybe watching a video explaning a key concept in the problem literally only days before doing the problem is cheating, but I still got the answer quicker). I also got the second question quicker but I had litterally done it in an olymiad a couple months prior.

[@ramskiramski]

11:55: Similarly, you would get f(xn) = f(x)^n; especially with x=1, you find f(n) = f(1)^n for a natural number n. Using f(-1) = f(1)^(-1), you find f(z) = f(1)^z for integers z and f(r) = f(1)^r for a rational r by multiplying the rational with its denominator. Rewriting f(1)^r = exp(r ln f(1) ) and taking a rational sequence approaching a real x shows f(x) = f(1)^x for a real x, since the exponential function is continuous. The value of f(1) can be found from the constant derivative log(f)’(x) = f(x)^(-1) f’(x) = log(f(1)) and the derivative of f(1) = f(1-x) f(x) at x=1, which is 0 = -3 f(1) + f’(1), yielding log(f(1)) = log(f)’(1) = f(1)^(-1) f’(1) = 3.

[@parameshwarchamarthi1920]

Notice that f(x+y) = f(x)f(y); this property is unique to the exponential functions-therefore, the solution would be in form of f(x) = a^x. Now, consider that f'(0) = 3; therefore (ln a)a^0 = 3; ln a = 3; a = e^3- substitute back in f(x) = a^x = e^3x

[@perakojot6524]

1000! is a famous Math Olympics question and has been explained so many times on different channels.

[@larsvaneekeren]

In the first question you can diffenerentiate f(x+y) =f(x)f(y) to x and get by the chain rule f'(x+y) = f'(x)f(y). Fill in x = 0 and get f'(y) = f'(0)f(y) = 3f(y). This is a differential equation which is very easy to solve.

[@chloeli880]

For the question with 1000!, one can use p-adic values taking v of 5. The floor of 1000/5 + 1000/5^2 + 1000/5^3 + 1000/5^4 =249. Hence, there are 249 zeros.

[@herghamoo3242]

Ooh, the second problem sort of "triggered" me... In high school, I participated in a national maths contest, and one of the questions was how many zeroes 100! ends in. I had figured out the "easy way", as they called it in the video, but for some reason I had forgotten the extra factor of 5 in 25 and 75. And I only missed one correct answer to get to the finals... It's the only question from the contest I still remember because it still annoys me somewhat

[@siddharth8334]

that's so sad

[@baerrach]

This is why stand up maths doco on the old legends of mathematicians where he talks about they had journals of paperwork on the problem before its even reduced to the rule is so helpful. Students learn the formula but haven't had the hard labour deriving the formula, and they get despondent they dont know maths. The problem is they haven't put enough hours of practice in and have been tricked into thinking they know some stuff because they can remember some formulas.

[@DavidBooercheese]

For the first problem, just set y constant and differentiate both sides wrt x and set y=0 to get f’(x)=3f(x), meaning f(x)=Ae^(3x). f’(x)=3 means A=1, meaning f(x)=e^(3x)

[@nenadrajcic737]

Answer on the first question is f(x) = a^(3*x/lna), a>0 for a = e, f(x) = e^3x.

[@lazbn90]

There is no need for differentiability assumption, continuity at x=0 is sufficient. With this, you can prove successively: 1) f(0) = 1, f(-1) = 1/f(1). 2) f(m) = f(1)^m and f(1/n) = f(1)^(1/n) for m, n integers, and from this conclude that f(m/n) = f(1)^(m/n). 3) Finally, prove that f is continuous everywhere, and prove f(x) = f(1)^x for all real numbers by approaching x by a sequence of rational numbers.

[@anjapeggy727]

Thanks that was so loveable human!!!!!!!!

[@jffrysith4365]

I always consider numbers by prime factorisation, so it took me like 2 seconds to solve the second problem, and a few minutes to prove my intuition formally. Though I probably wouldn't have solved the first question in a reasonable timeframe. Great questions all around though, really fun to watch these awesome mathematicians take on an interview.

[@pinkfloyd1979]

For the second question, my approach would be to write the factorial term such as (1000)(1000-1)(1000-2)......(1000-998)(1000-999). Then, this would give you (1000^999) - 1000.sigmasum (x) (x=1 to x=999). It would be easy then to find the sum after the minus sign. with the formula n(n+1)/2.

[@toppinzr3743]

Multiplying out that expression would have a lot more terms than *that*. And the first term would be 1000^1000, not 1000^999.

[@mathcanbeeasy]

To avoid partial differentiation, you can replace y with a c. Then, f(x+c)=f(x)*f(c). Now we differentiate with respect of the variable, which is x, and we get f'(x+c)=f'(x)*f(c). Now if we make x=0 we get f'(c)=3*f(c). But this is available for all real c, so f'(x)=3*f(x) for all real x. And from here is easy to find f(x)=e^(3x).

[@rmsgrey]

With the advantage of sitting at home watching rather than being on camera, I pretty much immediately clocked the first question - it probably helps that I've recently been reminding first years about the rules for exponents. Converting multiplication problems into addition is where logs came from in the first place, so a function that does the reverse is going to be an inverse logarithm, or an exponentiation. So I immediately knew it's going to be f(x)=e^(ax), and it only takes a little more thought to get to e^(3x) and then to check that it's a solution. Proving uniqueness, I did need to watch the proof in the video. For the second question, I recognised the general question (number of 0s at the end of n!) so, aside from forgetting the 4th 5 in 625, I got to the answer almost immediately. On another note, when it comes to notation, even pure mathematicians work at roughly three tiers of rigour: - formal proof is the stereotypical version where every i must be dotted, every t crossed, and the whole thing immune to any nitpick. Though even there, you're allowed to invoke standard results rather than going back to axioms at every step. - conversation/collaboration is the level this sort of interview works at. You need to show enough details, and be rigorous enough that you both agree about what's being done, and that your steps are probably valid (some things can be handwaved in the moment to come back to later if you get somewhere interesting along your current path) - personal/exploratory work can use whatever notation you want to whatever extent you feel like. You don't even need to be able to come back and understand it a week from now - you just need to get down the details you don't have room for in your head so you can keep going, and then you can come back and fill in enough detail to reconstruct a full argument later.

[@Bethos1247-Arne]

I once wrote a program which computed 100.000!. Took some 4 hours on an AMD K6-2 @400 MHz. I implemented string-endcoded numbers addition and uses Russian multiplication to multiply with implementing only additions.

[@joseph_donovan]

Made my ear lobes tingle with delight! Helped me work out the numer of zeros in 10,000 (2,499). Thank you! 10,000/5 - 2,000 10,000/25 - 400 10,000/125 - 80 10,000/625 - 16 10,000/3125 - 3 ============= 2,499

[@federicofallucca1952]

I would like to add a comment on the first beautiful exercise. We don’t need actually to assume f being differentiable everywhere, but it is sufficient to require just that is differentiable at 0. Indeed, using the same approach as the solution, you can prove that the limit of (f(x+h)-f(x))/h exists, and its exactly f(x)f’(0). At this point you obtain that the function is defferentiable everywhere and you can go on to prove that the only function that is solution of the problem is e^3x

[@SoumilSahu]

The way I solved the first one was setting y to be a constant, say k. Then you get f(x+k) = f(x)f(k) . Differentiate this on both sides and you get f'(x+k) = f'(x) f(k) now set x = 0 => f'(k) = f'(0) f(k) => f'(k) = 3 f(k). This is true for all real k, and therefore is a general equation, ie; f'(x) = 3 f(x) This can now simply be integrated to give you the result e^(3x) (combined with the boundary condition that f(0) = 1 )

[@drewmiddleton9715]

I haven't done any serious math since I graduated college over 5 years ago and this was such a pleasure

[@thrilhousesf]

For the first question, after f(0) = 1, you get f(x) = f(0+x/n+x/n + ... + x/n) = f(x/n)^n, immediately giving that f is exponential on rationals, and hence everywhere by continuity. Differentiabilty isn't even needed except for giving the value at f'(0)

[@DrR0BERT]

OMG, thank you for saving me from writing my OCD driven detailed proof (that everyone would say tl;dr). Yours gives a great and succinct answer. And you actually use differentiability to show continuity. Awesome.

[@decare696]

The first problem has a really elegant solution actually: You can use induction to show that the function must satisfy f(n x) = f(x)^n for any natural n. Take that together with f(-x) = f(x)^-1 and it holds for any integer n. Plug in x = y/n to see that it holds for reciprocals of integers as well, so f(q x) = f(x)^q for any rational q. Finally invoke continuity to see that f(r x) = f(x)^r for any real r, and in particular, for x = 1, f(r) = f(1)^r. The derivative then easily lets us determine that f(1) = e^3. The thing I'm not sure about tho is whether someone who is just applying to a university has the necessary prerequisites to give this answer...

[@agentdarkboote]

To me the first identity given just immediately screamed exponential identities which highschool students are very familiar with. So if you use as an ansatz an exponential it's just a matter of figuring out the base which taken with the derivative at zero just requires another familiar identity. I was very happy to have beaten Tom to the answer... Although I wasn't under the same pressure as him!

[@ZuhairKhouri2003]

Such a fun video to watch, math is so cool

[@thibautstefani9879]

Thank you for the video. For the first question, we could fix y and define function g_y as g_y(x) = f(x+y) = f(x)f(y). Then g’_y(x)= f’(x+y) = f’(x)f(y). I think this is valid because y is fixed hence f(y) is constant. Then with x=0, we have f’(y) = 3 f(y) which holds for every y in R since y was fixed without any conditions. Solving the differential equation, f(x) = e^(3x). I don’t know if the method is completely rigorous or if I get the correct answer by coincidence?

[@albertwood8836]

this is rigorous and the best way to solve the question as stated.

[@lucanina8221]

I did sort of the same but I put y=0 and found f'(x)=f(x)*f'(0) for every x. Is it correct anyway?

[@thibautstefani9879]

I think you first have to differentiate, then set y=0, because in the other way, you get f’(x) = f’(x)f(0) since f(0) is a fixed value. But otherwise, it is correct

[@lucanina8221]

@@thibautstefani9879 you are right in the other way around it does not work thanks

[@Jonathan_Corwin]

Oh I wish I had this video 30 years ago when I had my interview!

[@MrJdcirbo]

The mathematical way to solve the second problem is fascinating, but there's a quick python program you can use to get the number of ending zeroes for any number. It's a TOTAL cop out (I know), but it's a quick way to check your math, and it shows an interesting pattern. Here's the program: import math number = 1000 factorialStr = str(math.factorial(number)) for i in range(1, len(factorialStr)): # start at last digit and count backwards if int(factorialStr[-i]) != 0: # current i will be at one digit past the last zero. Subtract 1 from answer print(f"{i - 1} number of ending 0's") break

[@rafazieba9982]

I figured out that the answer is "f(x) = e^3x" within a couple of minutes of looking at the question but: 1) you need to guess that it is an exponent function 2) proving that it is the only solution is way beyond the high school graduate I ask a question similar to the number of zeros at the end of 1000! in IT interviews. I usually start with what is the last digit of n!, then how many zeros at the end of n!

[@alexanderkoutouzis7302]

You can solve the same problem assuming just the continuity of your function, by just accepting the familiar exponential property given. Of course, a trivial solution is the identically zero function (which is dismissed by the assumption of a non-zero derivative at a point, is the differentiable case). So, for the sake of avoiding trivialities we are going to assume that f is non-zero everywhere, equivilantly at a point, say 0. A direct consequence is that f is positive. Also, for every integer n, we can easily check that f(n) = (f(0))^n Furthermore, for every rational n/m it is easily seen that f(n/m)=(f(0))^(n/m) Because the rationals are dense in R and f is continuous, we get that for every real x f(x)=(f(0))^x Or f(x)=e^(x*ln(f(0))) This is, of course, the same result that we see in the video. It is interesting to restate the problem as follows; the only homomorphisms between the additive group of reals and the multiplicative group of non-zero reals are the exponentials.

[@dimitriskliros]

yet another solution: differentiate f(x+y)=f(x)f(y) wrt y, set y=0, use f’(0)=3 and solve the resulting diff eqn: f’(x)=3f(x) subject to f(0)=1 (which follows from f(0)=f(0)f(0)).

[@toppinzr3743]

One does need to observe that f(0) can't be 0.

[@dimitriskliros]

@@toppinzr3743 indeed, thank you

[@beautyofmath6821]

Heres an math olympiad approach to problem 1: We first note that f is continuous and positive over R as f(2x)=f(x)^2>=0. Then taking ln, ln(f(x+y))=ln(f(x))+ln(f(y)) Let g(x)=ln(f(x)), we have: g(x+y)=g(x)+g(y) Now note that g satisfies cauchy’s functional rquation and is indeed continuous over R. Hence G is linear and the result follows.

[@toppinzr3743]

You'd have to rule out f(x) = 0 for some x first.

[@AllisterMyth]

Hi professor thanks for replying on twitter. I am now better at problem solving thanks to your guidance. Thank you for creating such interesting videos as always. 😊

[@charlesarthurs]

He is not, so far as I am aware, a professor. If am correct about this he really should stop claiming that he is. Also, there is something almost infantile about mature mathematicians going back and taking A-levels, entrance exams, etc. It's quite creepy, actually.

[@AllisterMyth]

@@charlesarthurs Mate he is an outreach professor. If he creates content at graduate level, it will not exactly be ideal for high school students would it ?

[@Sir-Bane]

@@charlesarthurs He literally did 3 years teaching IN Uni as a prof and is now a out reach prof... 😅

[@charlesarthurs]

He is not a professor. Look, he is NOT a Professor of Mathematics at one of the world's leading universities and I wish to Christ he's stop pretending that he is. He disgusts me.

[@Sir-Bane]

@@charlesarthurs *facepalm*

[@roudyelhaddad1841]

For the first question, let f(x+y)=f(x)f(y). Differentiate with respect to y to get f'(x+y)=f(x)f'(y). Evaluate at y=0: f'(x)=3f(x). Solve to get f(x)=e^3x.

[@jacobcable7729]

You can get the answer directly by differentiating using the product rule wrt y: f'(x+y) = f'(y)f(x) and you know f'(0)=3, so set y=0 and you get the differential equation f'(x) = 3f(x), with f(0)=1 you can separate variables and integrate to get log( f(x) ) = 3x

[@reu.mathematicsacademy8566]

Dr Tom is a very cool guy sacrificing his time to go through rigorous and challenging entry exam questions to Oxford which many wouldn't want to be challenged. The good thing is that he's a smart Dr. Keep up!

[@charlesarthurs]

He's not "sacrificing" his time. What would he do otherwise? Why don't you ask him?

[@Catalinawolff]

Best thing i saw today (Even Though i had maths at University) you are just awesome ! and the other guy (from another roof) is very funny, i want to See more of you 2 Talking about maths !❤ greetings from Karlsruhe Germany

[@AnotherRoof]

Thanks for the kind words 🙂

[@DaniFadli]

Great example of an interview for a strong candidate 💪🏼thank you for this great video

[@AndrewTaylorPhD]

This video and the whole "if the limit exists" sequence just makes me want to know when @anotherroof's MAFS series will do Mean Girls.

[@serban1979]

The approach could be more simple: what operation has the property of multiplying the terms to be the sum of entry data: power --> a^x*a^y= a^(x+y). a^0 = 1. So fist step is done -> f(x) is a power function. f'(x) = (a^x)'= a^x*ln(a) --> for f'(0) = 3 ==> ln(a) should be natural that means that a=e. We know that (e^nx)' = ne^nx ==> n=3 in order that f'(x) = 3. It is a more heuristic approach, but in the end the result is the same :P.

[@nairanjith]

Hi guys! I think there is a subtle issue with the L'Hospital rule approach at 21:10. The LH rule says that the limit (as x -->0) of the quotient N(x)/D(x) equals that of N'(x)/D'(x) when N(0)=0 and D(0) =0 *if the functions in the second expression and the second limit exist*. In our case, N(x) = f(x) - 1 and D(x) = x. We are given f is differentiable, so N'(x) = f'(x) and D'(x) = 1 so the functions are well-defined. However, we cannot assume a priori that lim (x -->0) f'(x) exists. Moreover, we then set this limit to f'(0). Thus we are implicitly assuming that the derivative f' is *continuous*, which is not part of the problem statement. Mark's approach doesn't make this extra assumption.

[@Ntt903]

If f is differentiable, then by definiton, f' is continuous.

[@nairanjith]

@@Ntt903 If only that were true, sweet summer child ..

[@harrybarrow6222]

The first line of the problem statement gives a heck of a big clue: Multiplying two things on the right corresponds to adding them on the left. That screams EXPONENTIAL! Now we just need to determine the details…

[@daxhausentraveller8488]

I really enjoyed this vlog

[@mohad12211]

question 1 is what Michael Spivak's Calculus book (which is basically an intro to real analysis) uses to motivate and actually derive the notion of an exponential function on the reals. I recognized that immediately from the f(x+y) = f(x)f(y).

[@YyNRCyY]

good job

[@highviewbarbell]

More proof that using more sources than your school suggests is almost always beneficial

[@Sidnv]

Interestingly, you can also solve this problem by using a little bit of group theory and continuity. The differentiability of the function is actually an overly strong requirement, you would get the same answer if you just assume the function is continuous, except that you would need to give the value of f(1) instead of f'(0).

[@SangaPommike]

Yeah, I immediately knew it was an exponential function from the same. And within a minute figured out the complete answer. I'm actually kind of surprised it took so long in the video. But then again I don't really know Tom's exact background so maybe he doesn't have the experience to immediately jump at that.

[@johnshortt3006]

we know without using l'hopital's rule that the limit is 3. it's written on the line above the line written in red at 23:11

[@Bethos1247-Arne]

2*5 = 10 giving a 0, so this should be considered. edit: But I am glad of all the extra thoughts. It helps to understand the problem in new ways. Hearing a math professor thinking out loud when he is not instantly finding the answer, teaches me SO MUCH MORE than math vidoes loosing to time explaining the right answer.

[@theorigin8537]

Weird how my brain just strangely knew 1000! has 249 zeros. I didn’t even do a process consciously.

[@timpani112]

There is a really simple way of deducing the limit of [f(h)-1]/h as h tends to zero in the first question. We can argue as follows: 1. As f'(x)=f(x)*lim([f(h)-1]/h) for every x, and since f'(x) is well-defined for every x, lim([f(h)-1]/h) must exist as a real number. 2. lim([f(h)-1]/h) does not depend on x. 3. We know that f'(0)=3 and that f(0)=1. 4. Hence, lim([f(h)-1]/h)=f(0)*lim([f(h)-1]/h)=f'(0)=3. After this, we know that f'(x)=f(x)*lim([f(h)-1]/h)=3*f(x) for all x, which together with the boundary condition f(0)=1 determines f uniquely.

[@martinscheunemann]

But why is lim f‘(h) = f‘(0) , because we don’t know whether f‘(x) is continuous or not ???

[@timpani112]

​@@martinscheunemann I don't see how that matters to the above argument. We never claim that f'(0) is equal to lim f'(h), but rather, from the given information we deduce that f'(x)=f(x)*lim([f(h)-1]/h) for every x. Since it was given that f'(x) exists for every x, and we know what f'(0) and f(0) are, we can compute lim([f(h)-1]/h) using this. At no point in the argument do we invoke continuity of f'.

[@vdeave]

I feel like the first question is testing if you recognise that addition inside the function being equivalent to multiplication outside the function is a sign of an exponential. If someone has never seen it put like that before, it's very difficult even if they're good, but if you have seen it, it takes a minute or two.

[@toppinzr3743]

The point is to prove that *only* an exponential could satisfy the conditions, and which one.

[@vadim64841]

20:20 - just recognize that (f(h) -1)/h is the same (f(h) - f(0))/h since f(0) = 1. This is how you’d compute f’(0) which is 3. So, you got f’(x) = 3f(x) and the rest is trivial.

[@saranshkhare2658]

Can you please discuss more questions like that 👍🏻

[@kqual4614]

first problem was trivial by cauchy's functional equation and f'(0) = 3 implying f is continuous in an open interval around 0 and hence by induction over Q it is exponential

[@diskgrinder]

I remember my maths problem set in an Oxford university interview (vaguely, it was the 80s). It seemed to be about a World War One biplane catching up with bullets it fired. I spent half the time drawing a picture of sopwith camel with machine guns mounted behind the propeller, and went down a rabbit hole of timing bullets to fire through the cadence of the propeller without shooting it off. I got accepted at imperial college. The other story is how that didn’t happen either

[@jamescamacho3403]

If you've done competition math, these were super well-known (and easy) problems.

[@przemysawkwiatkowski2674]

Great video. Watching his step after step was an adventure! 😁 On the other hand I'm really really surprised it took him so long. First question is not trivial, but come on... he is math professor! I Immediately noticed "it's something exponential". A function that changes addition into multiplication is nothing unusual and noticing that it is e^3x should be rather easy for math professor! The only not-so-easy part is proving that it's the only function satisfying given conditions. (But it wasn't required.) And second guestion was... easy. Anyone interested in math should immediately notice that prime factorization is the way.

[@toppinzr3743]

Yes it would have been required to show that it's the only function. Giving an example of a function that will work doesn't answer the question.

[@przemysawkwiatkowski2674]

@@toppinzr3743Ah, right. The question was "determine function...". Anyway - I would expect him to find the working example sooner. 😊

[@toppinzr3743]

@@przemysawkwiatkowski2674 Also, his "proof" using l'Hospital's rule is no proof ... It isn't given that f'(x) is continuous.

[@olayinkaanifowose5099]

happy I guessed the solution to the first problem, although my first guess was 3* e^x instead e^3x, rusty maths.

[@GaborRevesz_kittenhuffer]

9:32 i'm thinking try taking the derivative of f from scratch using f(x+δ) = f(x)f(δ) in the difference quotient... yep: you'll get f(x)(f(0+δ)-f(0))/δ, which, taking δ→0, yields f’(x) = 3·f(x). So f(x)=exp(3x). ... well whaddya know! i got the "book solution" 😊

[@murk1e]

In the first one, I wouldn’t have gone through all the intermediate results, but the original function screamed logarithmic to me, and I guessed early and saw it worked. Physicist approach, rather than mathematician going for full proof!

[@julianbruns7459]

You meant exponential, right? Because logarithmic would be f(x) +f(y)=f(x•y).

[@thezerothandtheinfinite]

​@@julianbruns7459he just said it reminded him of logarithms, that thought process could very easily lead to exponentials

[@Chr15T]

@@julianbruns7459 Define g(x)=log f(x) and the functional equation reads g(x+y)=g(x)+g(y), so g(x) is a linear function of x and thus MUST be expressible by g(x)=ax. Rest is straightforward. Using this, you can also prove the uniqueness of the result f(x)=e^(3x). Of course, for taking the log you have to prove that f(x) > 0 everywhere - which is straightforward because f(x+x) = f(x)^2 >= 0, and f(x) can also nowhere be zero since that would easily lead to f(x)=0 everywhere, in contradiction with f'(0)=3.

[@Sidnv]

The first problem is actually also a pretty classic elementary group theory problem (with a pinch of analysis in the form of continuity thrown in) and actually the problem only requires that f be continuous (although if it is not assumed to be differentiable, you need to give the value of f(1), not f'(0)). The property f(x + y) = f(x)f(y) is the same as requiring f to be a group homomorphism from the additive real numbers to the multiplicative positive real numbers. The fact that all values must be positive is easy to see because f(x) = f(2 x/2) = f(x/2)^2. Here you can use a nice trick, any continuous function on the real numbers is determined by the value on the rational numbers. So you just need to figure out what f(x) is for x = m/n with m, n integers. From here, you use the idea of determining a group homomorphism on the rational numbers from the value at 1. You don't even need any of the machinery of group theory, it is just where the idea comes from. f(m) = f(m . 1) = f(1)^m. And f(m) = f(n m/n) = f(m/n)^n. So f(m/n) = f(1)^m/n. So, if you set a = f(1), then f(x) = a^x for all values of x. From here, you can use the value of f(1) to determine a, or if you know f is differentiable, you can differentiate to get f'(0) = ln(a), and hence, f(x) = e^(f'(0)x).

[@adavirus69]

This feels more like olympiad questions

[@sSpaceWagon]

I mean, the first one, the f(x+y)=f(x)f(y) describes the entire family of exponential functions, like that is one of the fundamental rules? b^(a+c) = b^a*b^c

[@manaiakilani3837]

I think the easiest way is to define at first a function g / g(x) = ln(f(x)), then we get this equation : g(x+y) = g(x) + g(y), which is a well known Cauchy functional equation. So we immediatly get g(x) = a.x (a is a real to define) => f(x) = exp(ax) We notice f'(x) = a.exp(ax) => f'(0) = a = 3 Finally the solution is f(x) = exp(3x)

[@mitancheyflorian6570]

I think you would have to prove that g(x) = a.x and not only giving the result

[@victorrangel5274]

Guys tell if am I wrong, but I did the First one in a different way that is showed in the vídeo. If We defferenciate the Function in terms of x, we get fprime(x+y) = fprime(x)×f(y), do x =0 then we get a seperavable differential equation in terms of y, fprime(y) = 3f(y), and the solution is e^y, than you say that y=x and you get the function.

[@JonathanBrandes-l5o]

The problem is actually easy: the only basic functions satisfying the property this are log and e-function. The function is differentiable, thus the function has to be an e-function. It’s actually e^3x.

[@davidplanet3919]

I knew the answer to the first one straight away but didn’t know how to derive it without just citing the definition. All the pieces Tom discovered helped but he needed to see the big picture. This sort of blindness happens in interviews and tests though.

[@comedyfriendsenglish]

Wait. Why can we not simply look at the function g_y: R -> R, x->f(x+y) for some y in R. This is differentiable as a composition of differentiable functions and we also find g_y(x) = f(x)f(y). Then the derivative of g_y'(x) = f'(x)f(y) which implies that g_y'(0) = f'(0)f(y). Where g_y'(x) = f'(x+y) via chain rule. And thus f'(y) = 3 f(y). Which is only solved by the exponential function and use uniqueness of solution of differential equations. Then the solution is clear (after also getting the initial condition from f(0) = f(0)^2 i.e. f(0) = 1 or 0) so the solutions are e^3x or the 0 function. Shouldn't this work?

[@ftlPhysicsGuy]

So, maybe it’s just not being in the hot seat, but the first equation [f(x+y) = f(x)f(y)] screamed exponential. I mean, turning addition into multiplication…. But since that works for any e^(kx), you need a condition to determine k. That’s where f’(0) = 3. Comes in, and the rest is a piece of cake. Is that not an acceptable approach? Was the requirement to derive the answer without assuming anything?

[@keesutherat1952]

For the 1000! question: Multiples of 5: 1000 / 5 = 200 Multiples of 25: 1000 / 25 = 40 Multiples of 125: 1000 / 125 = 8 Multiples of 625: 1000 / 625 = 1 249

[@popamihai3196]

for the first question, you can find out that it s an exponential without using the derivative. you get that f(n) = (f(1))^n for any positive integer n by induction, that you can also extend for any integer. similarly f(1/n) = f(1)^(1/n) (because f(1)=f(1/n+1/n+…+1/n) = f(1/n)^n so f(p/q) = f(1/q+1/q+…+1/q) = f(1/q)^p = f(1)^p/q for all integers p and q meaning f(t) = f(1)^t for all rationals t any real number r can be written as the limit of a sequence of rational numbers (for example An=floor(n*r)/n) thus, f(An) = f(1)^An, as every term of An is rational. so by continuity of f, f(r)=lim f(An) = lim f(1)^An = f(1)^r, meaning f(x) = f(1)^x for any real number x. by renaming f(1) to some arbitrary base a, f(x) = a^x for any x. This is really interesting as it shows that even if you’re not given that f is differentiable, you are able to prove that it’s an exponential as long as it’s continuous. note: continuity might not even be the weakest requirement for this problem, as having primitives over R might be enough (idk if this is right and even if it is, idk the proof for this variant)

[@thephysicistcuber175]

Continuity at a point, boundedness from above or below by a strictly positive number and monotonicity in an interval all separately imply the same conclusion (excluding the trivial case f=0).

[@davidliu3463]

Basic math operations, but requires such a nuanced understanding of math overall

[@velpz7426]

Great video! Could you check out the HSC extension 2 maths paper? Its past papers are freely available online and it would be wonderful to see you attempt and complete these questions. Thanks!

[@swarnam9592]

these questions are easy if u do practice them . as these are standard questions and are the easiest questions in any math related olympiad and competitive exams (iykyk)

[@martinshoosterman]

once you had f'(x) = f(x) × lim(...) given the limit didjt depend on x, you didnt need to compute the limit at all. At that point you could have just said f' = L×f therefore f is somesort of exponential function, and then you could argue which exponential function it has to be, based on the properties given previously.

[@astronelson]

How I approached the first one was to notice that f(x+y) = f(x)f(y) means that f(x)=f(1+1+... (x times)) = f(1)f(1)... (x times) = f(1)^x. Differentiating this with respect to x gives f'(x) = f(1)^x ln(f(1)). Evaluated at x = 0 this is f'(0) = ln(f(1)) = 3, so f(1) = e^3, and so f(x) = f(1)^x = e^3x.

[@saksham7296]

I have seen this question in my entrance exam (IIT JEE) , but skipped it cus it was taking too much time but turned out it was being solved with a simple concept of limits , it was relieving to see a teacher getting troubled by this type of question too

[@Deoxys_da2]

Can we just differentiate the first question with respect to y and sub y=0 And get f'(x) =3f(x) ?

[@GreenMeansGOF]

I have a question. I have been able to approach these problems partly because I am already a PhD student and therefore already learned a lot of this stuff but also because I watch a lot of math youtube. Seeing these interviews, is it the case that the people more qualified to study math at Oxford are those who have already studied university level mathematics or mathematics beyond high school? It seems like the average high school student would not be prepared for this.

[@aubertducharmont]

An ordinary high student definetly is not prepared. But a student whose goal is to study math should be.

[@regiondeltas]

Remember, we don't go to University after high school - we have an inbetween step. So yes, somebody applying for Oxford Uni would have had 2 years of advanced maths, far in excess of our secondary school syllabus. And of course those hoping to go into Oxford will be the top of even that cohort.

[@Nepenth]

We already know that f(x) != 0 for all x. Differentiate f(x+y)=f(x)*f(y) w.r.t. x and y yielding: f'(x+y)=f'(x)*f(y)=f(x)*f'(y). By rearranging terms we get f'(x)/f(x)=f'(y)/f(y) for all x and y, so f'(x)/f(x)=c=3, since f'(0)=3 and f(0)=1, so we arrive at the ode f'(x)=3*f(x).

[@lforc25]

I immediately saw that it was an exponential function and set f(x) to a^x, but I wouldn't even know where to begin to prove that without guess work