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I apologize if there may be any mistake in my solution for A1 and for my English (because I am too weak at that) So, first of all, after seeing a bunch of product rules involved, I took a natural log of fn to create summations. Like lnf(x)=Sum from r=1 to n ln(cosrx). Then, differentiating it created f'(x)/f(x)= Sum [-rsin(rx)/cos(rx)] =>f'(0)=0 Then proceeded for f"(0) which comes out Sum r² Sum r² =n(n+1)(2n+1)/6 A/q let a fn g(n) = n(n+1)(2n+1)/6 - 2023 g(17)0 By Intermediate value theorem there exist a point c bw 17 and 18 such that g(x)=0 =>Minimum value of n=18
This brought back memories... mostly good, I promise! 😄 I participated in Putnam 1988 as a freshman representing Rensselaer Polytechnic Institute, which apparently was one of the "easier" years. I believe there were 7 or 8 of us that took it that year. I don't recall my exact score, but I do recall it was in the high 20s, and that I scored the 2nd highest at RPI that year, behind one of the seniors, which really floored me. It was a very cool experience, and one I wish I'd experienced more than once! (Only once as I left RPI my sophomore year for personal reasons.) Thank you for bringing this very difficult but fascinating competition to a wider audience!
I took this exam and was able to solve both questions A1 and B2. It wasn’t easy though, it took me 2 hours of thinking and writing to get to each of those answers. It’s definitely nothing to sneeze at
I am a competitive programmer And you really is my inspiration 🥹 Because of you I started studying mathematics again and I am becoming more logical and better programmer..!
With the trig problem, you could realise that 1/2f "(0) is the coefficient of x^2 in the maclaurin series. Expanding all the cosines up until second order term, we just need the x^2 term of (1 - 1/2 x^2)(1 - 1/2 (2x)^2)....(1 - 1/2 (nx)^2). To get x^2 terms, we must take the "1" term in n-1 brackets, and the x^2 term of the remaining bracket. This yields - 1/2 (1^2 +2^2 + 3^2 + .... n^2) as required. (*Also, you could just realise the function is even, so it's derivative must be zero at x = 0 in your analysis)
I read lots of articles that praise Indian mathematicians Ramanujan's genius but somehow I can't buy that that guy was a math genius, the adulation seems to arise more as a need to give praise to a poor guy from a poor country. At least not at the level of the Europeans. The first skepticism I have is how he simply produces his formulas from the thin air, without showing the receipts. He doesn't do a step by step derivation of the formulas, it's almost like simple crystal balling, which to me is very puzzling. LOL. Really the proof that would convince me once and for all that Ramanujan really was a math genius would be for him to solve a Putnam exam within the 6 hours and with a score of at least 90%.
I took the Putnam exam twice and surprisingly one of the hardest things about it was not being allowed a calculator. Even if you get a question mostly correct, which is rare, it would take me a lot of the allowed time to get to the final answer as a number. For example, A1 2023, which you did in the video, was one of the more manageable problems I have encountered on the practice papers and the two I took. But it took me longer to find the cutoff of 17/18 than to get to the final stage of the cubic inequality 😅
For those wondering, I got 12 and 18 in 2022 and 2023 respectively. Some people in my university 60+ which was awesome. For context, they were all on IMO teams for their countries before university.
For the first one, it seems you could think of multiplying the Taylor series for cos(kx) together. You would only need to care about the x^2 terms in the product. It’s immediate then.
I beg to differ. Since each cos(k x)=(Exp[k I x]+Exp[-k I x])/2, you take the product for k=1...n and the 2nd derivative at 0, at the same time. The product is then -(n!)^2. With trial and error, knowing that sqrt(2023) ~ 45, n! > 45. That has to be n=5. It's no wonder it's the first exercise, it's easier.
I took the Putnam as a physics major, and at the time had very little idea of how to do formal mathematical proofs. The results were predictable, though to be fair to myself, I think only one student at my (relatively small) university got even a single answer correct. Oddly, I don't remember the test being long (though I know it was indeed six hours), but I definitely remember the free donuts 😂 In retrospect, I'm happy I took the chance to at least attempt it.
For the first question my first thought was taking the natural log on both sides for the function then differentiating it once and finding f'(0) then differentiating again to find f''(0).this seems pretty straight forwards and I'm questioning myself if it is working out.could you help out if there is any error in the method
@@user-se2pl5hd5s It's fine to take the log. You are only interested in evaluating the derivative at x = 0, and in a sufficiently small the neighbourhood of x = 0 the given function AND all the functions in the product are positive!
I took this exam in grade 13 when there was a grade 13 in Ontario. I won a gold pin. There was no fanfare just my teacher dropping it off at my desk as if to say "By the way, this is for you!" I wish I still had it because it was real gold.
A1. n=5 . (n!)^2 > 2023. Since each cos(k x)=(Exp[k I x]+Exp[-k I x])/2, you take the product for k=1...n and the 2nd derivative at 0, at the same time. The product is then -(n!)^2. With trial and error, knowing that sqrt(2023) ~ 45, n! > 45. That has to be n=5. It's no wonder it's the first exercise, it's easier.
I took the Putnam three times. The first two times I got a 0. The third time I got a 10 having solved exactly one problem. I'm really happy with that lol.
Congratulations! This reminds me of the most difficult math exams in my local country, Uganda: The Mathematics Contest by the Uganda Mathematics Society. I sat for it the first year and failed to qualify with a mere 15% pass mark, despite consistently scoring 80s and 90s in our syllabus-based curriculum. However, when I retook it, I was more determined with the help of God, and in the end, I managed to qualify, scoring 21%-quite an achievement considering the modal score of the contest was 10%. As I continued reflecting on this accomplishment, I realized I was nowhere near the level of great mathematicians. There are even tougher competitions, like the International Mathematics Olympiad, where each country sends its best six students, and students from China regularly achieve perfect scores as a team. Just when I thought that was extreme, I heard of the Putnam competition, with a median score of 0%. And then, as if that were the limit, boom-ChatGPT 4, capable of solving 60% of the Putnam mathematics questions.
23:11 I know I am sounding foolish, but why don't we just put n=1 and then bob would just chose odd no 1, which would result in him getting k=1 that is odd which he selected.
for the first question, you could multiply and divide the function by sinx, and use the double angle formula for sin repeatedly. this would result in sin(2^n.x)/[(2^n)sinx] and then take the derivative of this twice. would be simpler
Actually that only works if the scalar multiple inside the cosine function is a power of 2. The formula is cosx*cos2x*cos4x*cos8x...cos2^(n-1)x = (sin2^nx)/2^nsinx
I'm an engineering student, but I think that with the knowledge I have, I can't solve any question in this exam 😂😂😂, with the training I have so far. Apart from the fact that in an exam you have to solve this question in a maximum of 5 minutes, your explanations are great, although it is still challenging for me, but I am familiar with the rules of derivation and integration as well as summations and their properties, however I am not a mathematician . I'm really excited to get to your level, or at least evolve with your teachings. ❤👍
if n is odd, we do the same pairing, all numbers except n are in the pairing, and bob have to not to select n and leave it to alice. It is always possibe because alice goes first and n is odd. so Bob choses the goal "odd" and start the game chosing the numbers corresponding to the chosen pairing of alice. if alice in a turn chooses n, bob chose any random number in a the remaining paring, if alice choses the other number in the pairing we r ok, if not, Bob continue to chose the numbers according to the rule, and in the end Alice has to chose the remaining number of the random pairing chose by Bob.
Bob can win if n=3. Choose "odd". If Alice picks 1, you win. If Alice picks 2, pick 1 and you win. If Alice picks 3, pick 2 and you win. I haven't figured out a solution for higher odd numbers. For even, I found the same strategy as Ellie.
ellie todays question was very interesting, I enjoyed every part of it. but I am a little bit upset because u didn't put the link of the paper in the description other than that I love ur videos and I enjoy it and I expect more and more videos from u
@@EllieSleightholm respect and appreciate every videos because they made me more thoughtful and I recommend if u start a series at which u work at least 3 questions like tutorial, I know it is hard especially being scientist is continuously getting through hard works and I really I appreciate that because it is also my dream, (I am 18 years old kid in university trying my best to do well math) so thank you
I solved A1 a moment ago by expressing f[n](x) = f[n-1](x) * cos (nx). Then differentiate twice. Put x = 0. f[n](0) = 1 for all n. Note that the difference of second derivatives is "- n^2". Prove by mathematical induction that second derivative of f[n](x) at 0 is "- sum of squares". Here I had to look for the sum because I didn't remember. Check a couple of values. 20 is too much by far. 18 is too much by a little bit. Calculate 17 and you have the answer. Is it expected of those people to remember the equation for a sum of squares of integers? Edit: I came up with the winning strategy for Bob if n is even and I fail to see the math problem here. It is a trial and error path to the solution.
Great video, somewhat boring question on my part. What graphics tablet are you using there? I'm looking at remotely teaching some maths to my daughter-in-law over the summer, and looking at how clear your writing came out there it looks to be exactly what I should be getting. Many thanks!
I took log and solved it, it makes differentiating each term seperately easier and did'nt require to check any pattern. In addition since we need to evalute at 0 a lot of terms just cancel out.@@michaelperry886
wow, did not know A1 is from putnam. it came in one of my tests and I didn't think much of it. I was able to solve it there possibly because I have solved similar questions (taking log and diff.) mom, I solved a putnam question in under 5 minutes!!!!!! edit: I just saw the video and the video didn't use the log. So I will just brief what I did. taking ln on both sides ln(fn(x)) = sum of ln(cos(nx)) of diff, f'(x)/f(x) = -[sum of n*tan(nx)] f(0)=1 (clearly) f'(0)=0 (tan0=0) f'(x) = -f(x) * [sum of n*tan(nx)] again diff, but we can ignore the diff of the sigma as it will be multiplied by f'(0) which is zero, all the tan(nx) will become sec^2(nx) and at x=0, all become 1, we finally get |f''(0)| = sum of (n^2)
Bob's winning strategy for when n is odd: If n is odd in the Alice and Bob problem, we could see the list of numbers in terms of pairs plus the element of {1}. This would be in the following manner: {1}+{2k, 2k - 1} = {1} + {2,3} {4,5}, ..., {2k, 2k-1}. Now, if Alice selects in the first turn a 2k - 1 not equal to 1, then Bob should select the 2k of the same pair, and if after Alice selects a 2k of another pair, Bob should select the 2k - 1 number of the same pair. On the other hand, if Alice selects 1 in the first turn, in the second turn Bob should select a 2k - 1 number of a pair. And if in the third turn, Alice selects a 2k number in a pair different from Bob's, Bob should pick the 2k - 1 number of the pair Alice has selected. And every time Alice does not select a number from the pair that Bob selected in the second turn, Bob should select the other number from the pair that Alice picks. If Alice completes a pair (like, to say, selects the 2k number of the pair Bob selected in the second turn), Bob should always choose an odd number from the remaining pairs. PD: If my answer is wrong, please let me know why :D
@@talwaar007 simple. Dump the religion 😀. Unless there is mathematical proof for the religion. So far none that I am aware. Although there is high mathematical probability for a God or high entity using Bayesian forecasting, so that needs belief. However, no religious book has mathematical evidence or any high probability.
@@user-se2pl5hd5s but we just care about the solution at x=0 which gives positive value for cos, we just don't define log at negative values of cos(x). You do this in pre-calculus as well you restrict your domain so that you don't get undefined values.
@user-se2pl5hd5s But we evaluate the function at x=0. Therefore the cosine terms are never negative. It is generally incorrect to say that logarithms give us an expression for the needed double derivative, but here they do.
I read lots of articles that praise Indian mathematicians Ramanujan's genius but somehow I can't buy that that guy was a math genius, the adulation seems to arise more as a need to give praise to a poor guy from a poor country. At least not at the level of the Europeans. The first skepticism I have is how he simply produces his formulas from the thin air, without showing the receipts. He doesn't do a step by step derivation of the formulas, it's almost like simple crystal balling, which to me is very puzzling. LOL. Really the proof that would convince me once and for all that Ramanujan really was a math genius would be for him to solve a Putnam exam within the 6 hours and with a score of at least 90%.
Education should provide the necessary tools to solve these problems. There should be complete books available providing the skills for this level of performance. We must reach this level of education if we want to stay relevant against Russia and China.
Not the hardest mathematics Exam, the hardest ones are French exams ENS and Polytechnique, take a look at them, you will see what is a mathematics exam.
The math is practically trivial and is not that hard to understand, instead the hard part of these problems is the problem solving and I cannot overstate how useful problem solving skills are
STEM? Love stem. Einstein was wrong, do you know how physics was lead down a dead end in the 80's...? Relativity/ Astrophysics will never explain reality. I think you'll find Einstein divides by zero... and gets infinity. (Err, considered incorrect)
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Lol! I don't even understand how levers work.
I apologize if there may be any mistake in my solution for A1 and for my English (because I am too weak at that)
So, first of all, after seeing a bunch of product rules involved, I took a natural log of fn to create summations.
Like lnf(x)=Sum from r=1 to n ln(cosrx).
Then, differentiating it created
f'(x)/f(x)= Sum [-rsin(rx)/cos(rx)]
=>f'(0)=0
Then proceeded for f"(0) which comes out Sum r²
Sum r² =n(n+1)(2n+1)/6
A/q let a fn g(n) = n(n+1)(2n+1)/6 - 2023
g(17)0
By Intermediate value theorem there exist a point c bw 17 and 18 such that g(x)=0
=>Minimum value of n=18
This brought back memories... mostly good, I promise! 😄 I participated in Putnam 1988 as a freshman representing Rensselaer Polytechnic Institute, which apparently was one of the "easier" years. I believe there were 7 or 8 of us that took it that year. I don't recall my exact score, but I do recall it was in the high 20s, and that I scored the 2nd highest at RPI that year, behind one of the seniors, which really floored me. It was a very cool experience, and one I wish I'd experienced more than once! (Only once as I left RPI my sophomore year for personal reasons.) Thank you for bringing this very difficult but fascinating competition to a wider audience!
I took this exam and was able to solve both questions A1 and B2. It wasn’t easy though, it took me 2 hours of thinking and writing to get to each of those answers. It’s definitely nothing to sneeze at
a1 was easy enough. i'm a high school senior and i solved it within 20 minutes.
@@KartikBalla are you flexing ?
@@KartikBallawomp womp
Congratulations for both
@@KartikBalla happy birthday
I am a competitive programmer And you really is my inspiration 🥹
Because of you I started studying mathematics again and I am becoming more logical and better programmer..!
Hello Bro, Are You from tier 1 or 2 college?? Are you graduate??
He's from IIT Delhi ,mnc department..@@rajrajnish3136
@@rajrajnish3136 I’m graduated
With the trig problem, you could realise that 1/2f "(0) is the coefficient of x^2 in the maclaurin series. Expanding all the cosines up until second order term, we just need the x^2 term of (1 - 1/2 x^2)(1 - 1/2 (2x)^2)....(1 - 1/2 (nx)^2). To get x^2 terms, we must take the "1" term in n-1 brackets, and the x^2 term of the remaining bracket. This yields - 1/2 (1^2 +2^2 + 3^2 + .... n^2) as required. (*Also, you could just realise the function is even, so it's derivative must be zero at x = 0 in your analysis)
"its"
exactly what I had in mind, its simpler than product rule
I read lots of articles that praise Indian mathematicians Ramanujan's genius but somehow I can't buy that that guy was a math genius, the adulation seems to arise more as a need to give praise to a poor guy from a poor country. At least not at the level of the Europeans. The first skepticism I have is how he simply produces his formulas from the thin air, without showing the receipts. He doesn't do a step by step derivation of the formulas, it's almost like simple crystal balling, which to me is very puzzling. LOL. Really the proof that would convince me once and for all that Ramanujan really was a math genius would be for him to solve a Putnam exam within the 6 hours and with a score of at least 90%.
I just got my first 10/10 on a Putnam problem and have been beaming all day (I found out this morning), so glad you’ve made a video about it!
This is very STEP-like-esque + interview. Nice video 👍🏽
I took the Putnam exam twice and surprisingly one of the hardest things about it was not being allowed a calculator. Even if you get a question mostly correct, which is rare, it would take me a lot of the allowed time to get to the final answer as a number. For example, A1 2023, which you did in the video, was one of the more manageable problems I have encountered on the practice papers and the two I took. But it took me longer to find the cutoff of 17/18 than to get to the final stage of the cubic inequality 😅
For those wondering, I got 12 and 18 in 2022 and 2023 respectively. Some people in my university 60+ which was awesome. For context, they were all on IMO teams for their countries before university.
For the first one, it seems you could think of multiplying the Taylor series for cos(kx) together. You would only need to care about the x^2 terms in the product. It’s immediate then.
That’s right! Immediate
I beg to differ. Since each cos(k x)=(Exp[k I x]+Exp[-k I x])/2, you take the product for k=1...n and the 2nd derivative at 0, at the same time. The product is then -(n!)^2. With trial and error, knowing that sqrt(2023) ~ 45, n! > 45. That has to be n=5. It's no wonder it's the first exercise, it's easier.
I took the Putnam as a physics major, and at the time had very little idea of how to do formal mathematical proofs. The results were predictable, though to be fair to myself, I think only one student at my (relatively small) university got even a single answer correct. Oddly, I don't remember the test being long (though I know it was indeed six hours), but I definitely remember the free donuts 😂 In retrospect, I'm happy I took the chance to at least attempt it.
For the first question my first thought was taking the natural log on both sides for the function then differentiating it once and finding f'(0) then differentiating again to find f''(0).this seems pretty straight forwards and I'm questioning myself if it is working out.could you help out if there is any error in the method
Yeah I thought the same
Taking log won’t be correct since cos function can take 0 and negative values
@@user-se2pl5hd5sbrother then take the mod first and then take log and then differentiate
Yeah i did it the same way and it came out neat and quick my answer was also 18
@@user-se2pl5hd5s It's fine to take the log. You are only interested in evaluating the derivative at x = 0, and in a sufficiently small the neighbourhood of x = 0 the given function AND all the functions in the product are positive!
For Q1 an alternative way, first we can ln both sides converts to log series then differentiate first gets f' then similarly f"
I took this exam in grade 13 when there was a grade 13 in Ontario. I won a gold pin. There was no fanfare just my teacher dropping it off at my desk as if to say "By the way, this is for you!" I wish I still had it because it was real gold.
A1. n=5 . (n!)^2 > 2023. Since each cos(k x)=(Exp[k I x]+Exp[-k I x])/2, you take the product for k=1...n and the 2nd derivative at 0, at the same time. The product is then -(n!)^2. With trial and error, knowing that sqrt(2023) ~ 45, n! > 45. That has to be n=5. It's no wonder it's the first exercise, it's easier.
I took the Putnam three times. The first two times I got a 0. The third time I got a 10 having solved exactly one problem. I'm really happy with that lol.
Congratulations! This reminds me of the most difficult math exams in my local country, Uganda: The Mathematics Contest by the Uganda Mathematics Society. I sat for it the first year and failed to qualify with a mere 15% pass mark, despite consistently scoring 80s and 90s in our syllabus-based curriculum. However, when I retook it, I was more determined with the help of God, and in the end, I managed to qualify, scoring 21%-quite an achievement considering the modal score of the contest was 10%. As I continued reflecting on this accomplishment, I realized I was nowhere near the level of great mathematicians. There are even tougher competitions, like the International Mathematics Olympiad, where each country sends its best six students, and students from China regularly achieve perfect scores as a team. Just when I thought that was extreme, I heard of the Putnam competition, with a median score of 0%. And then, as if that were the limit, boom-ChatGPT 4, capable of solving 60% of the Putnam mathematics questions.
The first equation, in the solution to the conundrum is reimansum n goes to i =1, i . 1+2+3...n = n(n+1)/2
23:11 I know I am sounding foolish, but why don't we just put n=1 and then bob would just chose odd no 1, which would result in him getting k=1 that is odd which he selected.
for the first question, you could multiply and divide the function by sinx, and use the double angle formula for sin repeatedly. this would result in sin(2^n.x)/[(2^n)sinx] and then take the derivative of this twice. would be simpler
Actually that only works if the scalar multiple inside the cosine function is a power of 2. The formula is cosx*cos2x*cos4x*cos8x...cos2^(n-1)x = (sin2^nx)/2^nsinx
Please do a video on the entrance exam to ENS Paris (the school that has the most Fields medalists among its alumni)
I'm an engineering student, but I think that with the knowledge I have, I can't solve any question in this exam 😂😂😂, with the training I have so far. Apart from the fact that in an exam you have to solve this question in a maximum of 5 minutes, your explanations are great, although it is still challenging for me, but I am familiar with the rules of derivation and integration as well as summations and their properties, however I am not a mathematician . I'm really excited to get to your level, or at least evolve with your teachings. ❤👍
Thanks my favorite most prestigious UNIVERSITY MATH graduated PROFESSOR👍
Please also the second part thanks very much)
to differentiate in the first question
we can take log on both sides it simplifies things easily..
Did you have any advice for a person in 1st year of math to deal with all this abstract concept ?
The video starts here 3:23
if n is odd, we do the same pairing, all numbers except n are in the pairing, and bob have to not to select n and leave it to alice. It is always possibe because alice goes first and n is odd.
so Bob choses the goal "odd" and start the game chosing the numbers corresponding to the chosen pairing of alice. if alice in a turn chooses n, bob chose any random number in a the remaining paring, if alice choses the other number in the pairing we r ok, if not, Bob continue to chose the numbers according to the rule, and in the end Alice has to chose the remaining number of the random pairing chose by Bob.
I think for the last question, Bob always loses when n is odd except for the case when n=1. Bob has a winning strategy when n is even or n=1.
Bob can win if n=3. Choose "odd". If Alice picks 1, you win. If Alice picks 2, pick 1 and you win. If Alice picks 3, pick 2 and you win.
I haven't figured out a solution for higher odd numbers. For even, I found the same strategy as Ellie.
really enjoying your videos
25:02 in the last case its 1 and not 0
How does it compare to the Math Tripos exam?
I took the one for 2023. I maybe got upwards of 10 points at the absolute most.
very good explanation
ellie todays question was very interesting, I enjoyed every part of it. but I am a little bit upset because u didn't put the link of the paper in the description
other than that I love ur videos and I enjoy it and I expect more and more videos from u
Thank you! Will post the link in the description box :)
@@EllieSleightholm respect and appreciate every videos because they made me more thoughtful and I recommend if u start a series at which u work at least 3 questions like tutorial, I know it is hard especially being scientist is continuously getting through hard works and I really I appreciate that because it is also my dream, (I am 18 years old kid in university trying my best to do well math) so thank you
I solved A1 a moment ago by expressing f[n](x) = f[n-1](x) * cos (nx). Then differentiate twice. Put x = 0. f[n](0) = 1 for all n. Note that the difference of second derivatives is "- n^2". Prove by mathematical induction that second derivative of f[n](x) at 0 is "- sum of squares". Here I had to look for the sum because I didn't remember. Check a couple of values. 20 is too much by far. 18 is too much by a little bit. Calculate 17 and you have the answer. Is it expected of those people to remember the equation for a sum of squares of integers?
Edit: I came up with the winning strategy for Bob if n is even and I fail to see the math problem here. It is a trial and error path to the solution.
Congratulations to you. Keep it up.🇬🇾
The highest math I've taken was Calculus 2 for my BSEET degree, and this exam looks extremely rough.
Great video, somewhat boring question on my part. What graphics tablet are you using there? I'm looking at remotely teaching some maths to my daughter-in-law over the summer, and looking at how clear your writing came out there it looks to be exactly what I should be getting. Many thanks!
19:16 shouldn't it be 2 -m1m2^2
I took this exam, the exam was so hard that it took me 30 mins for me to complete all the answers...
Please explain me about hodge conjectures
I'm just enjoying the ride with my high school math :)
In the first question cant we just take log and then differentiate?
Why would taking the log make things easier if you have to differentiate the original f(x) regardless?
I took log and solved it, it makes differentiating each term seperately easier and did'nt require to check any pattern. In addition since we need to evalute at 0 a lot of terms just cancel out.@@michaelperry886
wow, did not know A1 is from putnam. it came in one of my tests and I didn't think much of it. I was able to solve it there possibly because I have solved similar questions (taking log and diff.)
mom, I solved a putnam question in under 5 minutes!!!!!!
edit: I just saw the video and the video didn't use the log. So I will just brief what I did.
taking ln on both sides
ln(fn(x)) = sum of ln(cos(nx))
of diff,
f'(x)/f(x) = -[sum of n*tan(nx)]
f(0)=1 (clearly)
f'(0)=0 (tan0=0)
f'(x) = -f(x) * [sum of n*tan(nx)]
again diff, but we can ignore the diff of the sigma as it will be multiplied by f'(0) which is zero,
all the tan(nx) will become sec^2(nx) and at x=0, all become 1, we finally get
|f''(0)| = sum of (n^2)
Perhaps you could try the Hong Kong ALE(HKAlevel) exams, much harder than uk alevel haha
Can you try the suken test?
You can be a great teacher 👏🏼☺️
would the solution to A6 be for all values of N ?
Why didn't you calculate 2*du/dx *dv/dx to find f''(x)?
Bob's winning strategy for when n is odd:
If n is odd in the Alice and Bob problem, we could see the list of numbers in terms of pairs plus the element of {1}. This would be in the following manner: {1}+{2k, 2k - 1} = {1} + {2,3} {4,5}, ..., {2k, 2k-1}.
Now, if Alice selects in the first turn a 2k - 1 not equal to 1, then Bob should select the 2k of the same pair, and if after Alice selects a 2k of another pair, Bob should select the 2k - 1 number of the same pair.
On the other hand, if Alice selects 1 in the first turn, in the second turn Bob should select a 2k - 1 number of a pair. And if in the third turn, Alice selects a 2k number in a pair different from Bob's, Bob should pick the 2k - 1 number of the pair Alice has selected. And every time Alice does not select a number from the pair that Bob selected in the second turn, Bob should select the other number from the pair that Alice picks. If Alice completes a pair (like, to say, selects the 2k number of the pair Bob selected in the second turn), Bob should always choose an odd number from the remaining pairs.
PD: If my answer is wrong, please let me know why :D
Should be 2k+1 everywhere instead of 2k-1, but for the rest, this solution should work!
I attempted the first question only. My score
10/120
Can you please try BUET admission test questions
You are really good at maths .keep it up sister ❤
Ellli you are example of sincere girl good girl
what device and app do you use for handwriting ?
Looks like Goodnotes on IPad
What do you use to write?
Chalk
Subbed girl!
Should have taken MSC financial mathematics. Can make millions in quant trading applying stochastic calculus
I was thinking about it once upon a time but religious impediments stopped me.
@@talwaar007 simple. Dump the religion 😀. Unless there is mathematical proof for the religion. So far none that I am aware. Although there is high mathematical probability for a God or high entity using Bayesian forecasting, so that needs belief. However, no religious book has mathematical evidence or any high probability.
Didn't understand a damn thing but it's cool
I was able to answer 3 questions correctly this year as a freshman😄
Try to take the hardest japanese or chinesese math exam. I think it would be a nice experience.
Amazing 😊
Where can i find it that test, just for curiosity?
I adore also British English 😊
Suggest me some Math questions book.
I am sure U can solve also Mensa IQ test problems to score 200+...
i have no idea how i ended up here looking for german recipes but its really interesting 😅
i can solve each faster tet i have atrophied from lack of practice help
Can u solve b6 problem
You could use logarithms for problem 1 as an alternative to the product rule.
No cos can take negative values for which log isn’t defined
@@user-se2pl5hd5s but we just care about the solution at x=0 which gives positive value for cos, we just don't define log at negative values of cos(x). You do this in pre-calculus as well you restrict your domain so that you don't get undefined values.
@user-se2pl5hd5s But we evaluate the function at x=0. Therefore the cosine terms are never negative. It is generally incorrect to say that logarithms give us an expression for the needed double derivative, but here they do.
So, i quit after they showed double differential by two dash. I did not get that. Thanks anyway.
A 6 hour exam?!! Why doesn’t this surprise me from Americans.
I read lots of articles that praise Indian mathematicians Ramanujan's genius but somehow I can't buy that that guy was a math genius, the adulation seems to arise more as a need to give praise to a poor guy from a poor country. At least not at the level of the Europeans. The first skepticism I have is how he simply produces his formulas from the thin air, without showing the receipts. He doesn't do a step by step derivation of the formulas, it's almost like simple crystal balling, which to me is very puzzling. LOL. Really the proof that would convince me once and for all that Ramanujan really was a math genius would be for him to solve a Putnam exam within the 6 hours and with a score of at least 90%.
Extremely difficult exam.
I thought the USAMO was the hardest American math exam.
indians try not to bring up jee advanced challenge: impossible
craziest thing is that jee advanced is way easier compared to this
please make a comaprision video of this exam with indian JEE Advanced
I think this exam wins in terms of difficulty.
Tho jee advanced wins in terms of competition
@@epikherolol8189Have you ever sen jee advanced exam😂.
You need to solve every question within 1:30-2:00 minutes
@@pronabchakrobortty9163jee is nothing before olympiads which dwarfs compared to putnum
आपको टीचर होना चाहिए , आप अछेसे से समझती है , एच लगा वीडियो ,कीप अपलोडिंग ❤
Education should provide the necessary tools to solve these problems. There should be complete books available providing the skills for this level of performance. We must reach this level of education if we want to stay relevant against Russia and China.
Greetings from México 👋
Not the hardest mathematics Exam, the hardest ones are French exams ENS and Polytechnique, take a look at them, you will see what is a mathematics exam.
Literally says “American” in the tittle though
Please also make a video on JEE Advanced exam, it is also freakishly hard
She already made a video on JEE ADVANCED
First one was very easy.
I'm not good at math, I found X tho. Its splattered all over the page!
It’s not hardest math exam…. Anyway, thanks for uploading this video
This is so scary! I gotta run and change my underwear......................
Lo ponen difícil, pero más difícil es la vida.
Mem iam from Bangladesh help some mathe ,i study class,12 higher mathe my problem
Putnam
Nice mathematics and way of solving
From India
Name one reason why anyone would need to know all that ridiculously hard long math, absolute waist of time.
The math is practically trivial and is not that hard to understand, instead the hard part of these problems is the problem solving and I cannot overstate how useful problem solving skills are
It's a freshman's caculus homework level test if you enter a Russia or Chinese university which is above the top 5% among math major circle
Lmaaaooo a freshman in Russia and China can spend their whole life and not solve one problem on this test
You were refering the an Angles one who enter their university by bymasquerading @@raghuvenkatesan6792
look One's on jee advanced
Please try jee advanced question (India)
It's too easy compared to this
Jee adv is given by class 12 students@@raghuvenkatesan6792
Marry me
I have finished phd in mathematics from yale university
😂😂😂😂
🤣😂
Dude 😂
I am honest, I can solve none of them.
STEM? Love stem. Einstein was wrong, do you know how physics was lead down a dead end in the 80's...?
Relativity/ Astrophysics will never explain reality.
I think you'll find Einstein divides by zero... and gets infinity. (Err, considered incorrect)
❤
Maths.
A2 is the easiest i think
A3 question was copied from jee adv. exam
one of the top brutal exam in india
which jee advanced exam