French Math Olympiad | 1999 Q2

[@martiniquevodka5574]

Hello this exercice is not from french math olympiad but from a contest called "Concours général". It's a contest where the participants are in their last year of highschool. The difference between the olympiad and this contest is that the exercices are based on the french highschool maths curriculum

[@cauchy2012]

Est-ce que après le bac vous faites ce concours

[@martiniquevodka5574]

@@cauchy2012 Non c'est seulement pour les élèves en Terminale. En France il y a pas vraiment de concours postbac de type olympiade comme Putnam au USA

[@cauchy2012]

C'est ça aussi dans notre pays comme, c'est pour ce la quand j'ai vu votre commentaire je l'ai trouvé bizarre '' un concours de terminal''

[@robertgerbicz]

Complicated part, easier when you need: (m+4)^(m+1)>=2*(m+3)^(m+1) for m>=5 then just divide the inequality by (m+3)^(m+1): (1+1/(m+3))^(m+1)>=2 rewrite this as: (1+1/(m+3))^(m+3) / (1+1/(m+3))^2 >=2 But here in the left side both term is a monotone increasing function [the first is a known fact, the 2nd is trivial]. Hence (1+1/(m+3))^(m+1)>=(1+1/8)^6 but that is bigger than 2. Done.

[@beautifulworld6163]

How you know that function is monotone increasing

[@CM63_France]

Hi, For fun: 2:41 : 4 is not equal to 3, 3:15 : a nice pythagorian triple, 6 "let's go ahead and", including 2 "let's go ahead and do that" 1 "so I'll just go ahead and, 1 "I'll may be go ahead and, 1 "may be we'll go ahead and" 2 "so let's may be go ahead and", 2 "great", 1 "what we want to do", 1 "so may be the first thing that we want to do", 1 "so the first thing I would do", 1 "and now what I can do".

[@sunnycatt201]

1 "that's a good place to stop"

[@pingpongfulldh2308]

Hey Michael! Why don't you do a complete lecture series on Fourier Analysis & PDE please. Would be awesome. Thanks and keep it up.

[@yamahantx7005]

Fourier Analysis is a VAST field and is typically a graduate course onto itself.

[@athysw.e.9562]

Im, French, not at all in rockclimbing, so I haven't climbed up this cliff of course but I used to go in vacation around Gap and heard of this place for sure.

[@CM63_France]

A good place to climb!

[@goodplacetostop2973]

19:20 C’est un bon endroit pour s’arrêter

[@InstigationMex95]

Merci

[@InstigationMex95]

Vôtre nom !

[@shivansh668]

I guess , you have written in French

[@CM63_France]

Very good, with the accents s'il vous plaît!

[@keedt]

@@CM63_France *plaît ;-)

[@kinenveu1]

Ceuse, beautiful mountain, often good thermals to soar!

[@Pradowpradow]

Well I'm french, I just learnt about the existence of Céüse

[@hellosquirrel7271]

Funny xD

[@CM63_France]

so do I (both things)

[@tonyhaddad1394]

Mst michael penn i have a question : can we find the critical points of f(x) at f(x)'=0 and from this we can now if the fucntion is increasing or decreasing ???

[@rikhalder5708]

Please make a video on sum of binomial coefficients

[@skrill500]

(1+1)^n = nC0+nC1+nC2+...+nCn = (2)^n

[@tgx3529]

For the proof (m+4)^(m+1)>=2*(m+3)^(m+1) we can proof that (m+4)^(m+1)/(m+3)^(m+1)>2.It's the same as (1+1/(m+3))^(m+1).For m=5 is the value of this function >2, the limit of this function is e.

[@premanand4043]

yes, a much more natural approach

[@Charliethephysicist]

This reasoning is insufficient. How do you know your last function in m does not dip below 2 for some n>5?

[@mrmathcambodia2451]

I like Math, and I try to learn more about Math also. This video so good for me .

[@ghislaindebusbecq8864]

Anybody having holidays near Gap is a person of good taste.

[@Terence3184]

For n >=4 MOD n , LHS-RHS no matter n=even or odd , the residue will not be zero

[@corentinchabanol9256]

I knew you was a climber ! The arms put me on the idea ;) greetings from France ! If you come back to ceuse I will maybe see u ;)

[@leif1075]

Am I the only one who found the replacement of m from n at 9:00 confusing..why not just make thenineqaulity by replacing n with 5 on one side and with 5 on the other..much simpler and clearer...

[@SlipperyTeeth]

15:17 How exactly would you go about showing that f'(6) = ln(9/8)-1/12 > 0? Did they allow calculators?

[@SlipperyTeeth]

By playing around with it, I see that it follows from 3^19 > 2^30. These numbers are within a persons grasp, but certainly not a simple calculation that someone is likely to do unless they're already pretty sure that they're right.

[@SlipperyTeeth]

Actually, it's not necessary, since you show that the derivative itself is strictly decreasing (the second derivative is negative) and has a limit at infinity of 0, then it must be positive. I actually glossed over the thing that does matter, which was showing that f(6) > ln(2), which amounts to showing (9/8)^6 > 2 or equivalently that 3^12 > 2^19. Orders of magnitude easier.

[@E1Luch]

​@@SlipperyTeeth i think you can prove that first hard inequality by raising second easier one to 19/12: 3^19 > 2^(19*19/12) = 2^(361/12) > 2^30

[@SlipperyTeeth]

@@E1Luch Ooooh. It's neat that they're related. And by such a close margin too.

[@E1Luch]

@@SlipperyTeeth its kinda lucky, 12 is the smallest power of 3 for which this trick worked

[@turdferguson3400]

It's so fun doing these my own way and seeing your way was completely different but we arrived at the same answer. It is the most fun part of math for me! For me I used a parity argument to show that n must be either 2 or 3 mod 4, and I used an upper bound condition to show that (n+3)^n

[@elgourmetdotcom]

Why was the second derivative necessary?

[@HagenvonEitzen]

Divide (m+4)^(m+1) > 2 (m+3)^(m+1) by (m+3)^(m+1) to get new target (1+1/(m+3))^(m+1)> 2. Naive application of Bernoulli's inequality would be too weak as it only gives (1+1/(m+3))^(m+1) > 1+(m+1)/(m+3)=2-2/(m+3). However, with the Bernoulli inequality generalized to non-integer exponents >=1, we have (1+1/(m+3))^((m+1)/6) > 1+(m+1)/(6m+18) = 7/6 - 1/(3m+9) >= 7/6 - 1/24 = 9/8 for all m >=5. Then verify (9/8)^6 > 2 to complete

[@Charliethephysicist]

Nice.

[@neilgerace355]

What was special about the case n = 4?

[@pierrot31511]

Enfin !

[@JalebJay]

Merry Christmas!

[@tgx3529]

Easy proof that x*ln(x+3)-x*ln(x+2)>=ln2 x*ln(1+1/(x+2)) we can use that ln(1+y)>y/(1+y) for y>0 [it's from Lagrange], then is it true for x>=7. For x=5 and x=6 "handly"

[@smoked8471]

When mathematicians draw a wiggly line inside of a square, what does that mean??

[@girianshiido]

What was to be shown.

[@smoked8471]

@@girianshiido I don't get it how does a square show what was to be shown??

[@neilgerace355]

@@smoked8471 I think it's a tradition.

[@ahzong3544]

Draw a box at the corner of a proof indicates that everything else is a result of thinking outside the box. Just kidding, don't quote me on that. However I personally like to think about this.

[@mooncowtube]

A square, sometimes filled and sometimes hollow, is often used as a "here ends the proof" marker. This is particularly useful in textbooks where the maths then continues. Notice that it's a bit more general than "QED", which only really makes sense if the final statement of the proof matches the thing to be proven, which in complicated proofs isn't always the case. However, Michael has made the square, filled in with a chalk wiggly line, a kind of trademark of his presentational style.

[@leif1075]

Why does he replace m with m plus 1 why not just work with m??

[@ericaurissergues6306]

Il y a beaucoup plus simple : faire passer (n+2)^n à droite et écrire ( n+3)^n-(n+2)^n sous la forme d'une somme de a^k x b^(n-k) La comparaison de la somme de gauche avec celle de droite donne immédiatement le résultat.

[@yurenchu]

- " écrire ( n+3)^n-(n+2)^n sous la forme d'une somme de a^k x b^(n-k) " You mean: write (n+3)^n - (n+2)^n in the form of a sum of a^(k-1) * b^(n-k) , from k=1 to k=n (with a = n+3 and b = n+2 ). Note that in the sum, the exponent of a is (k-1), not k ; so the highest exponent in the sum will be (n-1), not n .

[@Rahul-ky8mm]

Suggest me a best book consists various countries mathematical Olympiad questions

[@emanuellandeholm5657]

f(6) > log 2 is actually a surprisingly bold claim. log 2 is .69.... and f(6) is .70...

[@jesusalej1]

Great job!.

[@Charliethephysicist]

Here is another proof for f(x):= x\ln((x+3)/(x+2)) is increasing and concave. f(x) = x \int_0^1 1/(x+2+t)dt = \int_0^1 x/(x+2+t)dt. The integrand in the last integral is increasing and concave in x for x>0, thus so is f(x).

[@2kreskimatmy]

got lost in logarithmic inequality

[@wesleydeng71]

To prove (m+4)^(m+1)/(m+3)^(m+1)>2, let x=m+3 and you have LHS = (1+1/x)^(x-2) = (1+1/x)^x / (1+1/x)^2. Easy to see (1+1/x)^x is increasing and (1+1/x)^2 is decreasing. So LHS is increasing .

[@Charliethephysicist]

Here is my way to see (1+1/x)^x is increasing. x ln(1+1/x) = \int_0^1 1/(1+t/x)dt. The integrand increases with x, thus so does the integral. Can you show me your way?

[@kshitijsharma3170]

Loved it

[@tonystarklive9018]

How about we know that xln(x+3)-xln(x+2) equal to ln((x+3)^x/(x+2)^x that mean x+3>2^1/x(x+2)>x+2 which is true?

[@Charliethephysicist]

You should put parenthesis around 1/x. Otherwise it is really confusing. The real question is how you would prove your last inequality.

[@theevilmathematician]

I used a different method that didn't involve any calculus. Notice if n=1, 4=3, so n≠1. If n=2 and n=3, the equation holds. Assume that if n≥4, (n+3)^n= k^n. If n is even, then 2=2p, p>0, which implies that (2p+3)^2p= 3^2p, which means 2p+3= 3, 2p=0. However, p>0, and thus if n≥4 and is even it cannot hold. If n is odd and n≥4, then n=2p+1, where p>0. Then, (2p+4)^2p+1= 3^2p+1, which implies that 2p+4=3, 2p=-1, but ∀p>0, it just cannot hold. Therefore, n={2, 3}, assuming n is a positive integer.

[@adityamohan7366]

At 18:50 Is the function always increasing or that it always remains a non-negative function? If it is increasing function the double derivative would be positive no?

[@virajagr]

Function is increasing, but the speed by which it is increasing, is decreasing

[@prizzoshamza9697]

Can you make a video about the problem 6 in imo 1988

[@jimskea224]

Shortcute (n.): a cute shortcut.

[@djvalentedochp]

awesome

[@algorfrile739]

Another way of solving it is using the fact that for all integers a and n, a^n = a mod n. Using this property, the equation reduces to n+3 = 3 + 4 + ... + n + n+1 + n+2 mod n which is equivalent to 3 = 0 + 1 + 2 + ... + n-1 = (n(n-1))/2 mod n If n is odd, then 2 divides n-1 and we obtain 3 = n*integer = 0 mod n which implies that n divides 3, ie n = 1 or n = 3 If n is even, say n = 2k, we have 3 = k(2k-1) = -k = k mod 2k which can be rewritten as k = 3 + 2kp for some integer p, yielding k(1-2p) = 3 which implies that k is either 1 or 3 This narrows down all the possible solutions to only 1,2,3 and 6, and a quick check shows that only 2 and 3 are solutions.

[@TechToppers]

Mind giving a proof?

[@ivanmitrovic1964]

That just isn't correct, a^n=a (mod n) is not true for all a and n

[@TechToppers]

@@ivanmitrovic1964 Yeah, that property he has used is not written correctly!

[@tomasstride9590]

@@ivanmitrovic1964 I think it is true for prime n. So may be the result could be changed to show there is a largest prime for which the result is true. The question might then be what does that say when n is non prime.

[@ivanmitrovic1964]

@@tomasstride9590 Yeah, for prime n it is true by Fermat's little theorem

[@abdelilahsalimchatar4653]

This is still too hard for me, I don’t even know what the backwards 3 means, like the n + 2 3 k^n k = 3

[@smoked8471]

When mathematicians draw a wiggly line inside of a square, what does that mean??(like in his channel's logo)

[@virajagr]

That's sigma, used for summation. You substitute k from 3 to n+2 and add those numbers

[@franciscomorilla9559]

@@smoked8471 I guess you mean the end of proof symbol. When you end an exercise or a proof you can write something to make clear that the proof is done, everything after that is not part of it: usually most people draw a square to note this, although other symbols are some diagonal lines like ////, or writing QED which comes from latin and means end of proof.

[@yurenchu]

SOLUTION: n=2 and n=3 CALCULATION: Let L(n) = (n+3)^n R(n) = ∑ k^n , from k=3 to k=(n+2) Solve L(n) = R(n) for integer n: n : 0 1 2 3 4 5 L(n) : 1 4 25 216 2401 32768 R(n) : 0 3 25 216 2258 28975 We see there are solutions for n=2 and n=3 . We can prove by induction that L(n) > R(n) for any n≥4 [*]. Furthermore, for n ≤ 0, RHS = 0, LHS ≠ 0, so there's no solution for negative values of n. So n=2 and n=3 are the only solutions. - - - - - - [*] PROOF by induction that L(n) > R(n) for any n≥4 : (without fiddling with any logarithms!) L(4) = 2401 > 2258 = R(4), so the statement is true for n=4 . Now show that if L(n) > R(n) (and n≥4), then also L(n+1) > R(n+1) . (By the way: obviously, L(n) > 0 and R(n) > 0 for any positive integer n.) L(n+1) = (n+4)^(n+1) = [(n+3) + 1]^(n+1) ... Since (n+1) ≥ 5, the binomial expansion will have at least six terms, and each term is positive. Therefore the total is greater than the sum of the first three terms of the binomial expansion ... > (n+3)^(n+1) + (n+1 choose 1)*(n+3)^(n) + (n+1 choose 2)*(n+3)^(n-1) = (n+3)*L(n) + (n+1)*L(n) + [(n+1)*n/2]*L(n)/(n+3) = (n+3)*L(n) + (n+1)*L(n) + [((n+1)*n)/(2*(n+3))]*L(n) ... for n > 3, we know [((n+1)*n)/(2*(n+3))] > 1 ... ... (because the graphs of f(x) = (x+1)x/2 and g(x) = x+3 show that for n>3, (n+1)n/2 > n+3 > 0 ) ... > (n+3)*L(n) + (n+1)*L(n) + [1]*L(n) R(n+1) = ∑ k^(n+1) , from k=3 to k=(n+1)+2 = { ∑ k^(n+1) , from k=3 to k=n+2 } + (n+3)^(n+1) < (n+2)*R(n) + (n+3)^(n+1) = (n+2)*R(n) + (n+3)*L(n) = (n+2)*[R(n)-L(n)] + (n+2)*L(n) + (n+3)*L(n) ... induction hypothesis: R(n) < L(n) ==> [R(n) - L(n)] < 0 ... < (n+2)*[0] + (n+2)*L(n) + (n+3)*L(n) = L(n) + (n+1)*L(n) + (n+3)*L(n) < L(n+1) Q.E.D.

[@cauchy2012]

3

[@gianpietropalacios8734]

:v 🤗👋

[@soumitrapharikal5503]

First

[@TwilightBrawl59]

Apparently not, Good Place To Stop posted 18 mins ago and you posted 17 mins ago.

[@beautifulworld6163]

Dont use calculus, it is shit

[@3mon3y94]

Hey Michael! Why don't you do a complete lecture series on Fourier Analysis & PDE please. Would be awesome. Thanks and keep it up.