Відео

you're obliged to show the series converges first. AFAIK that wants binets formula, which is the natural tool for just bludgeoning the problem into submission anyhow; you express it explicitly as the sum of two geometric series and push around some annoying irrational numbers for a while. for viewers who don't understand the motivation for the steps in the vid: in the expression for S you first segregate the first term from the rest, then for the series part you use F(n) = F(n-1) + F(n-2). this recurrence doesn't apply F(1) (or F(0), which is absent from this problem), which is why we treat that term separately. the series (serises?) with F(n-1) and F(n-2) you reindex each one to restore the subscript to F(n), and with very little clean-up you wind up with S = 1/2 + S/2 +S/4 just as in the vid.

It was a total waste of time watching your presentation. Could you not say at the beginning that it is impossible? Gave a Dislike 😮

pi minutes long

Merry christmas!

1. k = p^a * q^b *... | n^5 - n 2. p^a | n^5 - n 3. a = 1, p | n^5 - n 4. a = 1, p - 1 | 4 5 a = 1, p \in {2, 3, 5} 1 <=> 2 <=> ... <=> 5 k = p^a * q^b *... | n^5 - n <=> a = 1, p \in {2, 3, 5} <=> k = 2^{0, 1} * 3^{0, 1} * 5 ^ {0, 1} => max k = 2 * 3 * 5 = 30.

Interesting. I’d guessed correctly the results and had the same counter example as you. The conditions in 3 felt odd. It didn’t seem possible that by perturbing x y and z you could switch which bullets were true - some kind of continuity argument though I guess we really want purely algebraic arguments in this Q - and hence all three bullets would be true if any one was true, and that felt strong enough to convince me it was likely to be an isometry.

From the below derivations, we can easily see that the general formula for S_k would be a binomial series expansion Let S_k = S[n^k/n!] S_k = S[m^k/m!] = S[m^(k-1)/(m-1)!] Let n = m-1, then m=n+1 => S_k = S[(n+1)^(k-1)/n!] => S_k+1 = S[(n+1)^k/n!] But we know (n+1)^k = k_C_0*n^0 + k_C_1*n^1 + k_C_2*n^2 + ... + k_C_k-2 * n^(k-2) + k_C_k-1 * n^(k-1) + k_C_k * n^k , where k_C_p is "k Choose p combinations" => S_k+1 = S[(n+1)^k/n!] = S[(k_C_0*n^0 + k_C_1*n^1 + k_C_2*n^2 + ... + k_C_k-2 * n^(k-2) + k_C_k-1 * n^(k-1) + k_C_k * n^k)n!] => S_k+1 =k_C_0*S[1/n!] + k_C_1*S[n/n!] + k_C_2*S[n^2/n!] + ... + k_C_k-2*S[n^(k-2)/n!] + k_C_k-1*S[n^(k-1)/n!] + k_C_k*S[n^k/n!], where k_C_p is "k Choose p combinations" Since we would have calculated the values from S_0 to S_k, we can calculate the value of S_k+1 from the above general formula

Nice video I know why you reminded me of Ramanujan. It was because I had sound off and I thought you would speak Indian😂

Proof for S[n^3/n!] = 5e based on S[n^2/n!] = 2e & S[n/n!] = S[1/n!] = e S[m^3/m!] = S[m^3/(m-1)!] Let n = m-1, then m=n+1 S[m^2/(m-1)!] = S[(n+1)^2/n!] =S[(n^2 + 2n +1)n!] = S[n^2/n!] + 2*S[n/n!] + S[1/n!] We will prove below that S[n^2/n!] = 2e, S[n/n!] = S[1/n!] = e Substituting above, we get =2*2e + 2*e + e =5e Therefore S[n^3/n!] = 5e Q.E.D ============================================================================ Proof for S[n^2/n!] = e based on S[n/n!] = S[1/n!] = e S[m^2/m!] = S[m/(m-1)!] Let n = m-1, then m=n+1 S[m/(m-1)!] = S[(n+1)/n!] = S(n/n!) + S(1/n!) = e + e = 2e Therefore S[n^2/n!] = 2e Similarly we can prove S[n/n!] = e based on S[1/n!] = e ============================================================================ Similarly, we can prove that S[n^4/n!] = 15e S[m^4/m!] = S[m^3/(m-1)!] Let n = m-1, then m=n+1 S[m^3/(m-1)!] = S[(n+1)^3/n!] = S[(n^3 + 3n(n+1) + 1)/n!] =S[n^3/n!] + 3*S[n^2/n!] + 3*S[n/n!] + S[1/n!] We know from above that S[n^3/n!] = 5e, S[n^2/n!] = 2e & S[n/n!] = S[1/n!] = e Substituting above, we get =5e + 3*2e + 3*e + e =5e + 6e + 3e + e =15e Therefore S[n^4/n!] = 15e Q.E.D

There is another elegant method to derive such results recursively (without the tedious factorization). That is, if we know what S[n^k/n!] is, we can recursively derive what S[n^(k+1)/n!] is. Using this method, I shall prove below that S[n^3/n!] = 5e and S[n^4/n!] = 15e Let S[..] be the sum of infinite terms with 'n' starting from 0. (It really should not matter that 'n' starts from 0, as the first term will be 0 when we have 'n' or its higher powers in the numerator. Thus the series will effectively start from n=1, when we have 'n' or its higher powers in the numerator) Proof for S[n/n!] = e based on S[1/n!] = e S[m/m!] = S[1/(m-1)!] Let n = m-1, then m=n+1 S[1/(m-1)!] = S[1/n!] = e Therefore S[n/n!] = e ============================================================================ Proof for S[n^2/n!] = e based on S[n/n!] = S[1/n!] = e S[m^2/m!] = S[m/(m-1)!] Let n = m-1, then m=n+1 S[m/(m-1)!] = S[(n+1)/n!] = S(n/n!) + S(1/n!) = e + e = 2e Therefore S[n^2/n!] = 2e ============================================================================ Proof for S[n^3/n!] = 5e based on S[n^2/n!] = 2e & S[n/n!] = S[1/n!] = e S[m^3/m!] = S[m^3/(m-1)!] Let n = m-1, then m=n+1 S[m^2/(m-1)!] = S[(n+1)^2/n!] =S[(n^2 + 2n +1)n!] = S[n^2/n!] + 2*S[n/n!] + S[1/n!] We know from above that S[n^2/n!] = 2e, S[n/n!] = S[1/n!] = e Substituting above, we get =2*2e + 2*e + e =5e Therefore S[n^3/n!] = 5e Q.E.D ============================================================================ Similarly, we can prove that S[n^4/n!] = 15e S[m^4/m!] = S[m^3/(m-1)!] Let n = m-1, then m=n+1 S[m^3/(m-1)!] = S[(n+1)^3/n!] = S[(n^3 + 3n(n+1) + 1)/n!] =S[n^3/n!] + 3*S[n^2/n!] + 3*S[n/n!] + S[1/n!] We know from above that S[n^3/n!] = 5e, S[n^2/n!] = 2e & S[n/n!] = S[1/n!] = e Substituting above, we get =5e + 3*2e + 3*e + e =5e + 6e + 3e + e =15e Therefore S[n^4/n!] = 15e Q.E.D ============================================================================ Following the above derivations, we can easily see that the general formula for S_k would be a binomial series expansion Let S_k = S[n^k/n!] S_k = S[m^k/m!] = S[m^(k-1)/(m-1)!] Let n = m-1, then m=n+1 => S_k = S[(n+1)^(k-1)/n!] => S_k+1 = S[(n+1)^k/n!] But we know (n+1)^k = k_C_0*n^0 + k_C_1*n^1 + k_C_2*n^2 + ... + k_C_k-2 * n^(k-2) + k_C_k-1 * n^(k-1) + k_C_k * n^k , where k_C_p is "k Choose p combinations" => S_k+1 = S[(n+1)^k/n!] = S[(k_C_0*n^0 + k_C_1*n^1 + k_C_2*n^2 + ... + k_C_k-2 * n^(k-2) + k_C_k-1 * n^(k-1) + k_C_k * n^k)n!] => S_k+1 =k_C_0*S[1/n!] + k_C_1*S[n/n!] + k_C_2*S[n^2/n!] + ... + k_C_k-2*S[n^(k-2)/n!] + k_C_k-1*S[n^(k-1)/n!] + k_C_k*S[n^k/n!], where k_C_p is "k Choose p combinations" Since we would have calculated the values from S_0 to S_k, we can calculate the value of S_k+1 from the above general formula

chatgbt sent me here I SWEAR TO GOD I GOT RICKROLLED BY CHAT GBT WTF

For the first one, an alternative argument: Let’s say you fix two vertices of a triangle. The loci of points that will form a triangle with perimeter 12 is an ellipse. Clearly an ellipse has maximum ‘height’ at the centre between two foci, so we know to maximise the area, the triangle should be isosceles. By symmetry, the triangle must be equilateral.

k*k! =(k+1-1)*k! =(k+1)*k!-k! =(k+1)!-k! The telescoping sum equals (n+1)!-1.

2024 is not special here

As the integral from 0 to 1 of x^2 is smaller or equal to 1/n(xn), when you take the limit of n going to infinity, the rectangle's bases shrink down to zero, and the sum of their areas fit perfectly under the curve. That means that when you take the limit as n approaches infinity, 1/n(xn) fits on the Rieman sum definition of x^2 from 0 to 1. Therefore, the sum shall be exaclty 1^3/3 - 0^3/3 = 1/3

hello jpi , im in y12 and have a few questions. When do u think is good time to start practicing for the mat, and do you think bmo problems will help with it or anything else for that matter? Because even though my school doesnt compete , ive been doing bmo problems in my spare time (very inconsisitently) for over a year and have gotten quite good at them. I think if i dedicate myself to them i can get to point where i can do some bmo2 problems, which is why im wondering if that will help with anything? Will the problem solving carry over into uni as ive heard oxbridge maths is very problem solvy like bmo problems. Is there anyhting else i can do currently to help with the mat? Ive done a few long form questions from recent papers and found them quie straight forward so thats probably a good sign. So yeah, when to start with mat practice, and do bmo problems help with much outside of competition maths? Thanks!!

I just added sum of n! and then it becomes a sum up to (n+1)! and it goes from there

trop facile

de tête :1/3. trop facile !

Really cool and really simple. I like it.

Please keep posting them even after the admissions cycle ends! They are really interesting problems to just solve for fun!!!❤

I originally misread the problem as "the person who removes the last sweet is the WINNER", and spent a good 15 minutes double checking why my answer was different than yours (I didn't watch the video yet, only peeked at your solution). Now everything looks good. By the way, the result I got for that different problem was the sequence 2,5,11,23... which is described by the recursion a_(n+1) = 2a_n + 1 (the explicit solution is a_n = 3*2^n - 1). It's interesting how that version of the game would be even more unfair to the 2.nd player, as the numbers are even sparser.

The 'Telescope' method seems to be a really clean method of proof, though it requires catching the insight about the successive terms. For the kth term: k.k! = (k+1).k! - k! = (k+1)! - k! As such, each successive value in the series cancels out the first term from the previous value. The only uncancelled values then are the highest (N+1)! and the -1! from the first value. As such: S(n) = (n+1)! - 1 Interestingly, this also works for the '0th' term. S(0) = 0, which makes sense, as (0)0! = 0 The chart method to start off definitely is a good plan, regardless.

it can be easily proved using induction

I solved it using the same symmetrization trick too. I remembered that the same thing is used for example when calculating the definite integral (from 0 to pi/2) of sqrt(tan(x)). There, you can add sqrt(cot(x)), as it is symmetric along x=pi/4, and it makes the rest much easier than if you were to try and compute the indefinite integral.

xD i just used lhopital

Actually when n is large (n > 10), you don't have to remember all these formulae. A very good approximation is as follows : Sum(n) ~= n^2/2 Sum(n^2) ~= n^3/3 Sum(n^3) ~= n^4/4 In General Sum(n^k) ~= n^(k+1)/(k+1) Ring a bell ? It should... Coz the approximation is derived from our famous anti-derivate of power function Integral (x^n) = x^(n+1)/(n+1)

Your kidding with us. This problem is so easy that it didn't take 10 seconds of my time. This wouldn't be asked in a hard exam. This question is too easy. Btw the answer is (n+1)!-1

No. That’s why I studied english there instead

My solution was to just rewrite the "k" in the sum as (k+1)-1, which splits the sum into 2 sums of pure factorials that are slightly offset from each other: Sum((k+1)!) - Sum(k!). Most of the terms cancel out and only the (N+1)! - 1 remains. Interestingly, while the difference of these factorial sums is easy to get, the individual sums seem to be much harder to get an exact formula for. I tried a few methods, including the derivative trick, but I can't seem to get anything very useful.

Nice problem, love these videos!

So basically you solve it by knowing the answer from the book. Nice! :)

I solved it in a different way. S = Σk(k!) = Σ(k+1-1)(k!) = Σ(k+1)(k!) - Σ(k!)= Σ(k+1)! - Σ(k!) = 2!-1!+3!-2!+4!-3!..... Starting from 2!, all terms get cut off except -1! and (n+1)!. So, answer is S = (n+1)! - 1

One of my favourite things about math is all the different approaches one can take to a problem. I personally started by using the fact that k*k!=k^{2}*(k-1)! (and the fact that 0!=1=1!), set that as my S_{k}, and then used induction to get S_{k+1}=S_{k}+[(k+1)^{2}*k!]. I think your way was more clever. I never would have thought of the S_{k}=(k+1)!-1 equivalence. Very nice proof. EDIT: One of my least favourite things about math is getting caught up in side work and forgetting the plot. I used induction to get each individual term. Ha. Once I actually added all the terms up, I saw your proof better.

I tried solving the question before watching the video. I solved it a different way. After playing around, I noticed that Σ(k+1)! = Σ(k+1)k! = Σk(k!) + Σk! Or rewritten, Σk(k!) = Σ(k+1)! -Σk!. But the RHS is simply a telescoping sum, which gives the result you got after doing the method of differences Thank you for such a good question!

I'm confused that why we need p>2max{|b_i|}. If we only need b_i != 0 (mod p) then i think p > max{|b_i|} is enough. It doesn't seem to mentioned in your proofs

great videos!! i had my cambridge maths interviews last week and these videos were helpful in helping me think to use more abstract techniques 👍

Hey I like your channel, nice problems. I like looking at you because you remind me of Ramanujan😭😂

You put Worcester college in the thumbnail, is that the college you attended?

I got asked a problem very similar to one in the TBO booklet in my interview today! Thank you ❤

cool trick but relying on "a well known result" sounds kinda odd to me. Even more when this result is so technical to prove rigoursouly. Great trick nonetheless

J Pi you are the guy!!! Just finished my 3rd Oxford maths interview like 2 hours ago and all your videos on interview problems helped me so so much, especially with explaining my reasoning. No matter the result I'm super grateful for your help thank you!!!

My method: the area of the triangle is the base |AB| times the height, and the heigh is just the distance from C(t,t²) to the line AB. AB is given by -2x+3y+4=0, so the distance from C to AB is just (-2t+3t²+4)/sqrt(13). This is always positive, and minimal when t = 1/3 (by taking the derivative or completing the square)

Fantastic

I did it the first way but had to derive the formula for the sum of the first n squares by hand (knowing it has to be some polynomial of degree three)

Another nice way to solve it: imagine stacking up small cubes of side length one in a big cube of side length n. As n tends to infinity, the sum of squares tends to a pyramid, which has volume a third of the cube

I did this the long way, I said that a² and 9a² must be roots of the biquadratic, which has solutions x² = ((3m+2)±sqrt(5m²+12m+4))/2. And so, the larger root (+) must be9 times the smaller root (-), which also gives the quadratic 19m²-108m-36=0

My interview is today, thank you for the videos

y=2/3x-4 C:y=-3/2(x-x0)+x0² x=24/13+9/13x0+6/13x0² h=√13|11 2/3+3(x0-1/3)²|/13≥35√13/39 S=35/3

I think one really nice quality of your teaching is how much time you apend making observations about the problem before your atart your calculations. It really invites some reflection on what are some natural restrictions to work within and feells creative. Realizing this is something i want to do more of in my own teaching. Great problem!