Before 15:20, you had an (1/x-1) term and when you send all up, you replace it for (1-1/x), adding the sign error in 19:55 (changing (-1)^m to (-1)^(m+1), you obtain the same result. I don't know if you did another error that I didn't see but you get by coincidence the same result. (Reply if I'm wrong)
@@tobysmith5857 If x approaches 1 even if from the negstvie side isnt the term in parentheses always 1 plus 1 divided by 1 plus 1 aince 1 raised to any power is 1?
Beautiful solution. After watching more of these problems, I'm more impressed by the people who come up with these problems. Surely constructing these kinds of problems is far more difficult than solving them and requires absolute mastery of the subject to even conceive of such problems
The specific problem pretty much utilises the series integration trick together with the use of partial fractions and the use of the sum of heometric series what really buffles me is how to construct a problem which utilises all these tricks and done many times as well
I think I said "Holy shit!!" about 8 times while watching this video. I said it twice when you converted that one term to an integral. Mind-blowing and amazing.
This best part about this channel is that you explain the motivation for every step. No other math yt channel does this, and it’s *so* helpful for solving problems. Keep up the good work!
No one explains the motivation better than blackpenredpen. I love when he presents the paths that lead nowhere before presenting the correct path to solve the problem.
Really nice problem. The tricky part is replacing that term by an integral to then get a geometric expansion. Once you get that, the rest isn’t too bad, but it does take a leap of faith to pursue that avenue .
Stumbled across this channel in my recommended videos a couple weeks ago. I really like your explanations. Kind of reminds me of the Khan Academy, but a pace that I can follow without getting bored. Oh, and an actual chalkboard. Thank you.
highly impressed for changing ln(1+x^m) to integration. Only that part was what I missed first time and after noticing that, I could solve this before I finish your writting. Thanks^^
You changed (1 /x - 1) to (1- 1/x) at 15:20. That caused the sign error. This involves too many tricks, but it was an easy solution once you know what to apply.
Since the infinite product whose limit is being taken in the original problem actually evaluates to 1 (and not 2/e) when x=1, it would appear that this product is discontinuous at x=1.
Another simple way: set t_n= ln (1+x^n) we have x^n= e^{t_n}-1 and max |t_n-t_{n+1}|-->0 ln (L) =lim sum [ e^(t_n)-1] [t_{n+1}-t_n] Right-hand sum of the integral = int_{ln 2}^{0} [e^t-1] dt = 1-ln 2. Hence, L=e/2.
It was awesome! A massive display of different techniques found throughout an entire calculus course. It seems feasible in the sense that a person with an average knowledge of calculus (I consider myself as being below the average) could entirely solve it after several hours of playing around with it, but to solve it under the Putnam Exam conditions definitely requires an outstanding preparation and mathematical intuition, in my opinion. Btw great video Prof.
The double limit(NOT the iterated limit) as N-> infinity x -> 1- x^{m-1} (1-x) log(1+x^m) + x^N log(1+x^N) can be interpreted as a Riemann sum for integral of ln(1+x) from 0 to 1. Consider the partition of (0,1) induced by this Riemann sum for a finite N and fixed x - effectively, we're dividing (0,1) in the logarithmic scale rather than a linear scale. The largest piece of the partition has size max(x^N, 1-x). As N-> infinity AND x-> 1-, we have a refinement of this partition; in the limit, we hence get the integral of ln(1+x) from 0 to 1 = 2 ln(2) - 1. Since the double limit exists, the iterated limit does as well.
in my opinion, this problem was surprisingly easy. no really unexpected or unusual tricks. (and i think in the putnam, you can just write down the series expansion of ln(1+x) without proving it). also you could've written the alternating sum of 1/(m(m+1)) nicely as a double integral
I would've expected the answer to be 1, since if you plug that into x, you get product of (2/2)^1. But that turned out to be wrong. Just shows that you cannot be so careless with infinity.
Honestly surprised at this question. Using approximation of log(1+x) and known infinite series, like alternating harmonic series, you can avoid the integrals, and this question is guided pretty nicely by telescoping series.
at 17.13 or therabouts we have negative ln 2 + limit x tends to 1- of (1-1/x)*( ). why cant the limit be applied at this stage and calculate (1-1/x) as zero giving the final answer as -Ln 2 or Ln(1/2) or for that matter earlier itself.
Focusing on the 2nd sum, adding 1 makes it so the sum now starts from m=0 and goes to m=inf. Now doing the substitution m->m-1, which changes the bounds into m=1 to m=inf and it changes the summand into (-1)^(m-1)/(m). (-1)^(m-1) is the same as (-1)^(m+1) so you arrive at (-1)^(m+1)/(m), which is the same as the 1st sum, thus the two sums are combined.
Loved the video but had a question. Why can’t you, at 10:14 for eg, just substitute in x = 1 in the equation since it will give a defined number? I guess the better question would be why would this be wrong but the straight substitutions at other points (such as 8:35) be correct?
Because it's in a 0*infinity format which is indeterminate. Plug x=1 and see for yourself: (1/x-1) will give 0 and the sum to infinity will give infinity
First he transforms the logarithm into an integrale with 1/(1+y) end then we know that 1/(1+y) is thé result of the Infinite sum (because |y| < 1 ) I'm Hope it's clear enough If not juste write the sum of x^n (from 0 to infinity) with |x| < 1 then you should get the result
Your maths is correct but you're misunderstanding what he did. He wanted to increase the power of x^n to x^n+1, which means he also needs to cancel out this extra factor, which he did so by dividing by x at the front. So he changed two things at the same time but in effect nothing changed at all.
Here's a Christmas-themed number theory problem from the University of Gothenburg! Santa's elves are sitting on a carpet, consisting of 36 equally-sized squares. However, they're called back to work, and are asked to assemble the 36 squares into a lidless box. (So a bottom and four walls.) What are the possible dimensions of the box, assuming all 36 squares have to be used? I solved it through an exhaustive search, both with a quick Python program and through more rigorous case studies, but I want to see how y'all would approach this problem. Happy holidays!
Why is that we take x to be in the interval (0,1) instead of (-1,1) at 8:40? Don't we shift over the domain of the ln function since we have the +1 term?
At 8:32 you're using assumption that lim_x->x0 lim_y->y0 F(x,y) = lim_y->y0 lim_x->x0 F(x,y). But in general you can't swap order of limits. Here's a simple counterexample lim_x->inf lim_y->inf x/y = 0 while lim_y->inf lim_x->inf x/y = inf.
About a third of the way through the video, I randomly guessed that the answer was 1/e (figured it'd be fairly simple). Can't believe I was only off by a factor of 2.
The clever tricks used here are fairly well known in maths competitions. Since this is just an infinite product for a B5 problem, I imagine the bigger concern will be something like "Oh this can't possibly have a straightforward answer. It's probably gonna rely on some strange inequality obtained from some set of rotations in 7 dimensions or something. I'm not gonna waste my time - I'll just focus on one of the other problems".
Also, several of the participants came up with a clever substitution that leads to a much shorter solution. {Second solution:} (by Greg Price, via Tony Zhang and Anders Kaseorg) Put $t_n(x) = \ln(1 + x^n)$; we can then write $x^n = \exp(t_n(x)) - 1$, and \[ L = \lim_{x \to 1^-} \sum_{n=0}^\infty (t_n(x) - t_{n+1}(x))(1 - \exp(t_n(x))). \] The expression on the right is a Riemann sum approximating the integral $\int_0^{\ln 2} (1-e^t)\,dt$, over the subdivision of $[0,\ln(2))$ given by the $t_n(x)$. As $x \to 1^-$, the maximum difference between consecutive $t_n(x)$ tends to 0, so the Riemann sum tends to the value of the integral. Hence $L = \int_0^{\ln 2}(1 - e^t)\,dt = \ln 2 - 1$, as desired.
I have run the following code in Matlab: L = 1; x = 0.999999; aux = 0; for n = 0:10000000 aux = (1+x^(n+1))/(1+x^n); aux = aux^(x^n); L = L*aux; end It yields: L = 0.7358 which is approximately 2/e. However, if I put more 9s after the comma in x, L will eventually approach 1. After checking the value of x^n I found that I need a bigger value of n for the x^n-type terms to start decreasing from 1 in the last steps of the product, but that takes an eternity to run. Just a curious thing I wanted to share. Since there are two variables and the expression has a discontinuity at x=1 there is no sure way to know the limit just by putting an arbitrary amount of 9s in x and making n abitrarily high.
AT 19:38, the proof is not rigorous, since the exchange of limits are not proven. One can calculate limit as follows ln (L) = lim sum x^n ln [1+x^n (x-1)/(1+x^n)] Using ln(1+t)= t + O(1) t^2, |O(1)| 0 = lim sum x^n/(1+x^n) [x^{n+1}-x^n] = - int_0^1 t/(1+t) dt using definition of Riemann integral with points t_n=x^n, n=0,1,2,... = ln 2 -1 Thus L= 2/e.
I lost track of this about half way through, because I'm not clever enough. But why not just evaluate the product at x=1, in which the whole thing equals 1? (Unless I have missed something). If the whole thing is 1, how does the lim approaching x = 2/e?
Because we only approach 1, and do it from left. Consider the graph of this function: it goes closer and closer to 2/e and at x=1 there is a discontinuity and it equals to 1. The simplest example: function 1/x. If you approach from left (0-) you will get -inf, but if you approach from right (0+) you will get +inf.
Please include this question in one of your videos, Let x,y,z be reals in the interval [4,40] such that: x+y+z=62 xyz=2880 find all such triples (x,y,z)
The equation is symmetric with respect to x, y and z. Start by noticing that xyz factorizes as (2^6)*(3^2)*(5^1). This should give you a hint on how to progress. Good luck :) Edit: Reals or integers?
@@Div1nePiece And by the way we are working on reals. So, symmetry will not work here. And why does one have to consider factoring when we are working on reals.
@@imobusters4049 Oh my bad, I thought you were looking for the solution. I edited and asked if it was reals or integers/naturals. I assumed a typo and integers/naturals at first but that wasn't the case, apologies. Symmetry still applies however.
I just saw this problem posted on the “actually good math problems” facebook page a few days ago i commented and didn’t understand why the answer wasn’t just 1... i assumed that, if we call that big product f(x), that f(.99999) ~1 and hence the limit = 1 i still kinda don’t understand it tbh. i understand that there do exists sequences a_n in (0,1) such that a_1a_2a_3... > 0 [ex a_n = (.5)^(.5^n)]] but this problem really hurts my head. why is that function not continuous at x=1? why is it not the case that limit of a product = product of limits?
lim(n to inf) of x^n is not continuous at 1, anything below becomes 0, at 1 it equals 1, and anything above goes to inf. Since the product contains this term, it is also not continuous at 1
@@angelmendez-rivera351 firstly, thank you for this explanation. i really appreciate people who take the time out of their day to help a random stranger online with something. (1) so if we call f_k(x)= prod_{n=1}^{k} [stuff], the functions f_k converge (pointwise) to, say, f. thus the problem is asking you to evaluate lim_{x->1^-}( lim_{k->infty} f_k(x) ) = lim_{x->1^-} f(x). you're saying that those limits don't commute, which i totally understand from a rigorous standpoint. it's the intuition that hurts my brain. although, now that i remember this, in analysis, we learned about the example f_k(x) = x^k on [0,1]; f_k converges pointwise to f(x) = {0 if x in [0,1), and 1 if x=1}. each f_k has that lim_{x->1^-} f_k(x) = 1. however, lim_{x->1^-} f(x) = 0, which proves that limits don't commute. by the way, if the convergence was uniform, would those limits commute? ^ I think the answer is yes by the "uniform limit theorem"; each f_k in the given problem is continuous and thus, if the convergence was uniform, f would also be continuous (at 1). hence this convergence is not uniform did i get all of that correct? (2) lol I know technically almost all functions (R->R) are discontinuous. my point wasn't that i was surprised that a random function was discontinuous, my point was that i was surprised that this particular one was discontinuous at x=1... feels like it should be but it's not
Hello Micheal, is it necessary to convert the series into limit of partial sum at 2:31? As I reach the same result without using the partial sum. Or I missing so technically here?
One can calculate limit as follows: since a^x = e^(x ln a), ln (L) = lim sum x^n ln [1+x^n (x-1)/(1+x^n)] Using ln(1+t)= t + O(1) t^2, |O(1)| 0 = lim sum x^n/(1+x^n) [x^{n+1}-x^n] = - int_0^1 t/(1+t) dt using right-hand rule for integral with points t_n=x^n, n=0,1,2,... = ln 2 -1 Thus L= 2/e.
Wow. I missed, what happens when you got 2Summ of 1/m and then again gain 1/(m+1). Should you try to change Summ of 1/m to 1/(m+1) instead? Idk. Then it become complicated to understand. But it is sooo cool)) I'd never try to substitute Integral instead of this summs. That's a lot) Thanks a lot))
I had a hard time understanding it too, but figured it out. He adds the + 1 to the series. Which is the same as the 0th term of the second sum, namely (-1)^0 / (0+1) = 1. So the second sum becomes the same sum but from m=0 to inf. The argument of the sum doesn’t change. Then he transforms the sum by setting m->m-1. Therefore it becomes a sum of m=1 to inf. The sum will become (-1)^(m-1)/(m-1+1) which is (-1)^(m+1)/m. Which is the first sum.
I am not satisfied with taking the log of an infinite product, resulting in a infinite sum of logs. I would write the product from n=1 to n=N, take the log, resulting in a sum from n=1 to n=N. Then taking the two limits N----->infinite, X----->1. And then take the exp of that result.
at 8.29 prof mentions that x goes to 1 from below AND adds that x is between 0 and 1. I did notice this when i raised the question. my doubt was - the statement x goes to 0 from below means it can take any value from minus infinity to 1 through 0 from the left. hence the absolute value could be greater than 1 in this domain. x goes to 1 from below is not the same as x is between 0 and 1. i am not a trained mathematician but majored in physics. i am unable to reconcile the dichotomy here. please dont misunderstand me i am not nitpicking. i will learn some thing if i am able to get my doubt cleared. thanks.
the limit of f(x) as x -> 1^- means you are looking at the behaviour of f(x) as x approaches 1, where x < 1. Its the limit on the left hand side (of the number line), and x>1^+ would be looking at x -> 1, x > 1 (from the right hand side of the number line)
Finished in much less steps I guess... Take the log, then develop log(1+x^n) and log(1+x^(n+1)) into Taylor series and combine the two sums. We get sum_n(sum_k((-1)^(k+1)/k * x^n(k+1) * (x^k-1) By Fubini (edit: fubini does not apply here, see comments below), swap the sums. We get a geometric sum on n of (x^(k+1))^n = 1/(1-x^(k+1)) New sum becomes sum_k((-1)^k/k * (1-x^k)/(1-x^(k+1)) Taking limit x to 1- of (1-x^k)/(1-x^(k+1)) = k/(k+1) using l'Hopital Finally I get sum_k((-1)^k/(k+1), k=1..inf) = sum_k((-1)^k/k, k=1..inf)-1 = log(2)-1 Then we get S=2/e
@@angelmendez-rivera351 I believe you can reprove fubinni for lebesgue integral which takes Dirac's measure. Then just use Dirac's measure, concentrated at points from natural numbers. And this is practically a sum
@@angelmendez-rivera351 Regarding Fubini, agreed there was no need to mention it here, though technically it works for infinite sums. I disagree about time. the steps above took me roughly 7min or so. The methods are very similar of course, but here manipulating the sums was a bit different and got me to the result faster. Anyway, no big deal man...
@@angelmendez-rivera351Not quite so. 19min (26-9) intro and explaining does not prove your argument. Plus you compare apples and oranges. When someone presents a solution about a problem he already knows, he does not waste time trying to find it or thinking about it , versus someone else who sees the problem first time, so the 7min period includes trial and what not.
@@angelmendez-rivera351 Obviously you are stuck to your opinion and I am to mine, so let's not waste our time, this debate is becoming pointless and especially here, there is nothing to prove.
@@angelmendez-rivera351 Regarding Fubini and absolute convergence, you are correct actually, it would not apply to this case anyway. Here you start to make sense. I gladly concede this one.
If you can commute the limits, the problem would be trivial, as it would be just evaluating the product of exact ones. And you always can commute limits, come on, the counter examples are so rare to come by and have to be specifically constructed, this one clearly does not look like one where you cannot commute limits.
And come on, you do exactly that at 19:02!!! Why couldn't you use the uniform convergence from the beginning, and not do the whole shuffling of the sums in the first place?
@@angelmendez-rivera351 One shouldn't be throwing recommendations about what others should study, as you never know the credentials of a random person over the internet.
@@angelmendez-rivera351 *This is just false.* *I have no time to explain to you why this is the case, you are better off just taking a real analysis course and reading a textbook, because I have no idea how much of a summarized explanation from me you would understand anyway.* Because apparently, telling basically "you are wrong because I am right. I have no time or wish to explain why" is a way to go in maths.
Before 15:20, you had an (1/x-1) term and when you send all up, you replace it for (1-1/x), adding the sign error in 19:55 (changing (-1)^m to (-1)^(m+1), you obtain the same result.
I don't know if you did another error that I didn't see but you get by coincidence the same result.
(Reply if I'm wrong)
This is correct!
The other mistake was where he changed (-1)^m into (-1)^(m+1)
Double negative = positive
Ok
@@tobysmith5857 If x approaches 1 even if from the negstvie side isnt the term in parentheses always 1 plus 1 divided by 1 plus 1 aince 1 raised to any power is 1?
Beautiful solution. After watching more of these problems, I'm more impressed by the people who come up with these problems. Surely constructing these kinds of problems is far more difficult than solving them and requires absolute mastery of the subject to even conceive of such problems
The specific problem pretty much utilises the series integration trick together with the use of partial fractions and the use of the sum of heometric series what really buffles me is how to construct a problem which utilises all these tricks and done many times as well
boy, I love the sound of the chalk knocking on the board. thanks for good quality audio!
I was thinking that I was only I
My solution: Take the log. That was a far as I got.
and that's a good place to stop.
@@rahulmistry5019 So I did!
I think I said "Holy shit!!" about 8 times while watching this video. I said it twice when you converted that one term to an integral. Mind-blowing and amazing.
This best part about this channel is that you explain the motivation for every step. No other math yt channel does this, and it’s *so* helpful for solving problems. Keep up the good work!
No one explains the motivation better than blackpenredpen. I love when he presents the paths that lead nowhere before presenting the correct path to solve the problem.
Every math channel i have ever watched explains the motivation for each step
Really nice problem. The tricky part is replacing that term by an integral to then get a geometric expansion. Once you get that, the rest isn’t too bad, but it does take a leap of faith to pursue that avenue .
@@angelmendez-rivera351 Sure, one mans inspiration is another’s routine.
@@hfix307 that's a great quote
25:50 Loading 96.7%...
Thanks!
Stumbled across this channel in my recommended videos a couple weeks ago. I really like your explanations. Kind of reminds me of the Khan Academy, but a pace that I can follow without getting bored. Oh, and an actual chalkboard. Thank you.
highly impressed for changing ln(1+x^m) to integration. Only that part was what I missed first time and after noticing that, I could solve this before I finish your writting. Thanks^^
You changed (1 /x - 1) to (1- 1/x) at 15:20. That caused the sign error. This involves too many tricks, but it was an easy solution once you know what to apply.
"And that's a good place to stop."
Well, yes, you solved the problem. It's usually okay to stop there. Lol
Sheer endurance and foresight - it's amazing how people can solve this under pressure
And in less than 30 min per question
Since the infinite product whose limit is being taken in the original problem actually evaluates to 1 (and not 2/e) when x=1, it would appear that this product is discontinuous at x=1.
Another simple way: set t_n= ln (1+x^n) we have x^n= e^{t_n}-1 and max |t_n-t_{n+1}|-->0
ln (L) =lim sum [ e^(t_n)-1] [t_{n+1}-t_n] Right-hand sum of the integral
= int_{ln 2}^{0} [e^t-1] dt
= 1-ln 2.
Hence, L=e/2.
It was awesome! A massive display of different techniques found throughout an entire calculus course. It seems feasible in the sense that a person with an average knowledge of calculus (I consider myself as being below the average) could entirely solve it after several hours of playing around with it, but to solve it under the Putnam Exam conditions definitely requires an outstanding preparation and mathematical intuition, in my opinion. Btw great video Prof.
The double limit(NOT the iterated limit) as N-> infinity x -> 1- x^{m-1} (1-x) log(1+x^m) + x^N log(1+x^N) can be interpreted as a Riemann sum for integral of ln(1+x) from 0 to 1. Consider the partition of (0,1) induced by this Riemann sum for a finite N and fixed x - effectively, we're dividing (0,1) in the logarithmic scale rather than a linear scale. The largest piece of the partition has size max(x^N, 1-x). As N-> infinity AND x-> 1-, we have a refinement of this partition; in the limit, we hence get the integral of ln(1+x) from 0 to 1 = 2 ln(2) - 1. Since the double limit exists, the iterated limit does as well.
Instead of reverse integral one would have simply expanded log(1+x) and got same results.
That was also my first thought, because it converges between 0 and 1
Kudos to prof Penn. A huge shoutout to the unknown person who devised this sinister problem.
Hey Michael! PLZ GIVE LESS ADS!! I am getting one literally every 3 minutes and this can be a possible source of future audience negator
i find it interesting that you read (lim x->1-) as "x approaching 1 from *below*" instead of "x approaching 1 from *the left*"
@ゴゴ Joji Joestar ゴゴ do magnitudes go vertically? i tend to visualize magnitudes horizontally as well
in my opinion, this problem was surprisingly easy. no really unexpected or unusual tricks. (and i think in the putnam, you can just write down the series expansion of ln(1+x) without proving it). also you could've written the alternating sum of 1/(m(m+1)) nicely as a double integral
I would've expected the answer to be 1, since if you plug that into x, you get product of (2/2)^1. But that turned out to be wrong.
Just shows that you cannot be so careless with infinity.
Honestly surprised at this question. Using approximation of log(1+x) and known infinite series, like alternating harmonic series, you can avoid the integrals, and this question is guided pretty nicely by telescoping series.
Wouldn't log's taylor's expansion and Geometric progression obviate the need for those 2 integrations respectively ?
Instead of doing the whole integral thing, couldn't you just have directly used the Taylor series expansion of ln(1+y) to arrive at that double sum?
19:55 (-1)^m became (-1)^(m+1) when you moved it up the board
@@angelmendez-rivera351 Ah yes, that's him fixing the sign mistake. Good point.
I hope you'll get your 100.000th subscriber still in 2020
omg this is really hard, thank u mr penn
this man is doing great...thanks Michael for educating us on this type of problems.👍😊
at 17.13 or therabouts we have negative ln 2 + limit x tends to 1- of (1-1/x)*( ). why cant the limit be applied at this stage and calculate (1-1/x) as zero giving the final answer as -Ln 2 or Ln(1/2) or for that matter earlier itself.
I like Math, I try to study about Math also.
Around minute 23:00 you add a 1 and subtract a 1. Got it. But then you combine the two sums and I don't quite see how you got that. Any ideas? Anyone?
Focusing on the 2nd sum, adding 1 makes it so the sum now starts from m=0 and goes to m=inf. Now doing the substitution m->m-1, which changes the bounds into m=1 to m=inf and it changes the summand into (-1)^(m-1)/(m). (-1)^(m-1) is the same as (-1)^(m+1) so you arrive at (-1)^(m+1)/(m), which is the same as the 1st sum, thus the two sums are combined.
Loved the video but had a question.
Why can’t you, at 10:14 for eg, just substitute in x = 1 in the equation since it will give a defined number?
I guess the better question would be why would this be wrong but the straight substitutions at other points (such as 8:35) be correct?
Because it's in a 0*infinity format which is indeterminate.
Plug x=1 and see for yourself: (1/x-1) will give 0 and the sum to infinity will give infinity
@@alekisighl7599 Ohh, I understand. Sorry for the late reply and thank you
Genius. Thank you!
ooo rewriting 1 as ln(e), i completely forgot about that trick
Michael, the cubic factorization requires correction
can someone explain what he did at 10:57 when he introduced y and turned it into an integral??
ln(1+a)=integral of 1/(1+x) from 0 to a (a>0)
First he transforms the logarithm into an integrale with 1/(1+y) end then we know that 1/(1+y) is thé result of the Infinite sum (because |y| < 1 )
I'm Hope it's clear enough
If not juste write the sum of x^n (from 0 to infinity) with |x| < 1 then you should get the result
If you mean introducing y, he just changed it to something literally equal.
Insted of this, just take the limits inside then simpify. Will get the ans in 4 or 5 steps
21:30 is there no way to make this into a telescoping series?
you need use the dominated convergence theorem to interchange int and series
in minute 19:50. Why can you say, that x^m is one. (even vor m = 1) but (x-1) is not zero?
Great channel. Greetings from Italy
Wouldn’t (1/x)*x^n be x^n-1? Not x^n+1? 3:39
Your maths is correct but you're misunderstanding what he did. He wanted to increase the power of x^n to x^n+1, which means he also needs to cancel out this extra factor, which he did so by dividing by x at the front. So he changed two things at the same time but in effect nothing changed at all.
Wait so what's the correct answer because you did make a sign mistake at the end
2/e is the correct answer, I believe he fixed his error before the end.
@@yoto60 ok
Here's a Christmas-themed number theory problem from the University of Gothenburg!
Santa's elves are sitting on a carpet, consisting of 36 equally-sized squares. However, they're called back to work, and are asked to assemble the 36 squares into a lidless box. (So a bottom and four walls.) What are the possible dimensions of the box, assuming all 36 squares have to be used?
I solved it through an exhaustive search, both with a quick Python program and through more rigorous case studies, but I want to see how y'all would approach this problem. Happy holidays!
Why doesn't (1/x - 1) collapse the limit down to 0? Great video btw
Why is that we take x to be in the interval (0,1) instead of (-1,1) at 8:40? Don't we shift over the domain of the ln function since we have the +1 term?
Does the domain of the ln function even have anything to do with the action?
Huge proof,...BUT FANTASTIC !!!!
Sir in the last step instead of integral after partial sum, we can just write partial sum and apply log(1+x) expansion.
Beautiful question! I really thought it was somewhat over when we got to the numerical serie, but not quite.
At 8:32 you're using assumption that lim_x->x0 lim_y->y0 F(x,y) = lim_y->y0 lim_x->x0 F(x,y). But in general you can't swap order of limits. Here's a simple counterexample lim_x->inf lim_y->inf x/y = 0 while lim_y->inf lim_x->inf x/y = inf.
About a third of the way through the video, I randomly guessed that the answer was 1/e (figured it'd be fairly simple). Can't believe I was only off by a factor of 2.
I refuse to believe people can fully solve this with time pressure.. there were far too many "clever trick" bs
Well, it’s putnam. Most contestants actually solve none fully
I mean... the are geniuses, so...
The clever tricks used here are fairly well known in maths competitions. Since this is just an infinite product for a B5 problem, I imagine the bigger concern will be something like "Oh this can't possibly have a straightforward answer. It's probably gonna rely on some strange inequality obtained from some set of rotations in 7 dimensions or something. I'm not gonna waste my time - I'll just focus on one of the other problems".
You would be wrong. kskedlaya.org/putnam-archive/putnam2004stats.html
Also, several of the participants came up with a clever substitution that leads to a much shorter solution.
{Second solution:}
(by Greg Price, via Tony Zhang and Anders Kaseorg)
Put $t_n(x) = \ln(1 + x^n)$;
we can then write $x^n = \exp(t_n(x)) - 1$, and
\[
L = \lim_{x \to 1^-} \sum_{n=0}^\infty (t_n(x) - t_{n+1}(x))(1 - \exp(t_n(x))).
\]
The expression on the right is a Riemann sum approximating the
integral $\int_0^{\ln 2} (1-e^t)\,dt$, over the subdivision of $[0,\ln(2))$
given by the $t_n(x)$. As $x \to 1^-$, the maximum difference between
consecutive $t_n(x)$ tends to 0, so the Riemann sum tends to the value
of the integral. Hence $L = \int_0^{\ln 2}(1 - e^t)\,dt = \ln 2 - 1$, as
desired.
have you tried to differentiate log(L) so then you have geometric series, etc, or it wont work?
I have run the following code in Matlab:
L = 1;
x = 0.999999;
aux = 0;
for n = 0:10000000
aux = (1+x^(n+1))/(1+x^n);
aux = aux^(x^n);
L = L*aux;
end
It yields: L = 0.7358 which is approximately 2/e.
However, if I put more 9s after the comma in x, L will eventually approach 1. After checking the value of x^n I found that I need a bigger value of n for the x^n-type terms to start decreasing from 1 in the last steps of the product, but that takes an eternity to run.
Just a curious thing I wanted to share. Since there are two variables and the expression has a discontinuity at x=1 there is no sure way to know the limit just by putting an arbitrary amount of 9s in x and making n abitrarily high.
AT 19:38, the proof is not rigorous, since the exchange of limits are not proven.
One can calculate limit as follows
ln (L) = lim sum x^n ln [1+x^n (x-1)/(1+x^n)] Using ln(1+t)= t + O(1) t^2, |O(1)| 0
= lim sum x^n/(1+x^n) [x^{n+1}-x^n]
= - int_0^1 t/(1+t) dt using definition of Riemann integral with points t_n=x^n, n=0,1,2,...
= ln 2 -1
Thus L= 2/e.
I lost track of this about half way through, because I'm not clever enough. But why not just evaluate the product at x=1, in which the whole thing equals 1? (Unless I have missed something). If the whole thing is 1, how does the lim approaching x = 2/e?
Because we only approach 1, and do it from left. Consider the graph of this function: it goes closer and closer to 2/e and at x=1 there is a discontinuity and it equals to 1. The simplest example: function 1/x. If you approach from left (0-) you will get -inf, but if you approach from right (0+) you will get +inf.
Please include this question in one of your videos,
Let x,y,z be reals in the interval [4,40] such that:
x+y+z=62
xyz=2880
find all such triples (x,y,z)
The equation is symmetric with respect to x, y and z.
Start by noticing that xyz factorizes as (2^6)*(3^2)*(5^1). This should give you a hint on how to progress. Good luck :)
Edit: Reals or integers?
@@Div1nePiece I know the answer, I want Micheal Penn sir to include it in one of his vids
@@Div1nePiece And by the way we are working on reals. So, symmetry will not work here. And why does one have to consider factoring when we are working on reals.
@@imobusters4049 Oh my bad, I thought you were looking for the solution.
I edited and asked if it was reals or integers/naturals. I assumed a typo and integers/naturals at first but that wasn't the case, apologies. Symmetry still applies however.
@@imobusters4049
Bruh... Will you give me a hint??
I just saw this problem posted on the “actually good math problems” facebook page a few days ago
i commented and didn’t understand why the answer wasn’t just 1... i assumed that, if we call that big product f(x), that f(.99999) ~1 and hence the limit = 1
i still kinda don’t understand it tbh. i understand that there do exists sequences a_n in (0,1) such that a_1a_2a_3... > 0 [ex a_n = (.5)^(.5^n)]]
but this problem really hurts my head. why is that function not continuous at x=1? why is it not the case that limit of a product = product of limits?
lim(n to inf) of x^n is not continuous at 1, anything below becomes 0, at 1 it equals 1, and anything above goes to inf. Since the product contains this term, it is also not continuous at 1
@@jsimone33 i don’t think “anything below becomes 0” is correct
@@angelmendez-rivera351 firstly, thank you for this explanation. i really appreciate people who take the time out of their day to help a random stranger online with something.
(1) so if we call f_k(x)= prod_{n=1}^{k} [stuff], the functions f_k converge (pointwise) to, say, f. thus the problem is asking you to evaluate lim_{x->1^-}( lim_{k->infty} f_k(x) ) = lim_{x->1^-} f(x). you're saying that those limits don't commute, which i totally understand from a rigorous standpoint. it's the intuition that hurts my brain.
although, now that i remember this, in analysis, we learned about the example f_k(x) = x^k on [0,1]; f_k converges pointwise to f(x) = {0 if x in [0,1), and 1 if x=1}. each f_k has that lim_{x->1^-} f_k(x) = 1. however, lim_{x->1^-} f(x) = 0, which proves that limits don't commute. by the way, if the convergence was uniform, would those limits commute?
^ I think the answer is yes by the "uniform limit theorem"; each f_k in the given problem is continuous and thus, if the convergence was uniform, f would also be continuous (at 1). hence this convergence is not uniform
did i get all of that correct?
(2) lol I know technically almost all functions (R->R) are discontinuous. my point wasn't that i was surprised that a random function was discontinuous, my point was that i was surprised that this particular one was discontinuous at x=1... feels like it should be but it's not
Hello Micheal, is it necessary to convert the series into limit of partial sum at 2:31? As I reach the same result without using the partial sum. Or I missing so technically here?
Thank you so so much
Did not understand a lot of the 'massive' steps you did, but still watched all off it
رائع.الطريقة مفيدة وفيها الكثير من المنعرجات.
Why is the lim of 1-^inf equal to 0? isnt it indeterminated?
Very nice problem. Thanks.
yeah yeah pretty trivial... I mean the part where he says "here we're going to look at a problem"
Nobody:
The students taking the Putnam exam: *Ah yeah, I'm going to use this trivial trick to easily evaluate the limit*
One can calculate limit as follows: since a^x = e^(x ln a),
ln (L) = lim sum x^n ln [1+x^n (x-1)/(1+x^n)] Using ln(1+t)= t + O(1) t^2, |O(1)| 0
= lim sum x^n/(1+x^n) [x^{n+1}-x^n]
= - int_0^1 t/(1+t) dt using right-hand rule for integral with points t_n=x^n, n=0,1,2,...
= ln 2 -1
Thus L= 2/e.
Wow. I missed, what happens when you got 2Summ of 1/m and then again gain 1/(m+1). Should you try to change Summ of 1/m to 1/(m+1) instead? Idk.
Then it become complicated to understand. But it is sooo cool)) I'd never try to substitute Integral instead of this summs. That's a lot) Thanks a lot))
I had a hard time understanding it too, but figured it out. He adds the + 1 to the series. Which is the same as the 0th term of the second sum, namely (-1)^0 / (0+1) = 1. So the second sum becomes the same sum but from m=0 to inf. The argument of the sum doesn’t change.
Then he transforms the sum by setting m->m-1. Therefore it becomes a sum of m=1 to inf. The sum will become (-1)^(m-1)/(m-1+1) which is (-1)^(m+1)/m. Which is the first sum.
I am not satisfied with taking the log of an infinite product, resulting in a infinite sum of logs. I would write the product from n=1 to n=N, take the log, resulting in a sum from n=1 to n=N. Then taking the two limits N----->infinite, X----->1. And then take the exp of that result.
Why does the ln commutes with the limit?
Because ln is a continuous function, meaning that ln(lim x) = lim ln(x) for all x>0
I wonder if Ramanujan would have solved this one in like 12 seconds, or not.
the geometric series for 1/(1+y) can be used only if asolute value of y is less than 1. where have we defined this in the solution.
He defined y=x^n. Since 0
at 8.29 prof mentions that x goes to 1 from below AND adds that x is between 0 and 1. I did notice this when i raised the question. my doubt was - the statement x goes to 0 from below means it can take any value from minus infinity to 1 through 0 from the left. hence the absolute value could be greater than 1 in this domain. x goes to 1 from below is not the same as x is between 0 and 1. i am not a trained mathematician but majored in physics. i am unable to reconcile the dichotomy here. please dont misunderstand me i am not nitpicking. i will learn some thing if i am able to get my doubt cleared. thanks.
@@angelmendez-rivera351 this is a beutiful reply and precise. thank you so much.
Impressive!
Sir you teach extraordinarily love from india sir
oppa putnam style
I can definitely imagine someone forgetting that pesky ln2 as they progress through the solution
I’m not good at math and English ,so I want to to know that ,what is “1^-“ ...??
the limit of f(x) as x -> 1^- means you are looking at the behaviour of f(x) as x approaches 1, where x < 1.
Its the limit on the left hand side (of the number line), and x>1^+ would be looking at x -> 1, x > 1 (from the right hand side of the number line)
@@Mercur1c I see!!! Thank you very much!!!
Putnam is much better exam than imo.
what do u mean by "better" ? both are tough and prestigious exams
@@prithujsarkar2010 true
Wow!!!!!!
Lol, initially I misread 1- as -1 and so the limit became a cute infinite multiplication of 0/2
Omg a problem I can ask my calc 2 students to complete :p
LOL
Make it worth half the points on the final!
If my Calc 2 teacher puts this in my finals this week I will demand his resignation
evil! hahaha
*NOICE*
tricky
nice nice nice
My solution:
Watch this video
That was an easy B5.
Finished in much less steps I guess...
Take the log, then develop log(1+x^n) and log(1+x^(n+1)) into Taylor series and combine the two sums.
We get sum_n(sum_k((-1)^(k+1)/k * x^n(k+1) * (x^k-1)
By Fubini (edit: fubini does not apply here, see comments below), swap the sums. We get a geometric sum on n of (x^(k+1))^n = 1/(1-x^(k+1))
New sum becomes sum_k((-1)^k/k * (1-x^k)/(1-x^(k+1))
Taking limit x to 1- of (1-x^k)/(1-x^(k+1)) = k/(k+1) using l'Hopital
Finally I get sum_k((-1)^k/(k+1), k=1..inf) = sum_k((-1)^k/k, k=1..inf)-1 = log(2)-1
Then we get S=2/e
@@angelmendez-rivera351 I believe you can reprove fubinni for lebesgue integral which takes Dirac's measure. Then just use Dirac's measure, concentrated at points from natural numbers. And this is practically a sum
@@angelmendez-rivera351 Regarding Fubini, agreed there was no need to mention it here, though technically it works for infinite sums.
I disagree about time. the steps above took me roughly 7min or so. The methods are very similar of course, but here manipulating the sums was a bit different and got me to the result faster.
Anyway, no big deal man...
@@angelmendez-rivera351Not quite so. 19min (26-9) intro and explaining does not prove your argument. Plus you compare apples and oranges. When someone presents a solution about a problem he already knows, he does not waste time trying to find it or thinking about it , versus someone else who sees the problem first time, so the 7min period includes trial and what not.
@@angelmendez-rivera351 Obviously you are stuck to your opinion and I am to mine, so let's not waste our time, this debate is becoming pointless and especially here, there is nothing to prove.
@@angelmendez-rivera351 Regarding Fubini and absolute convergence, you are correct actually, it would not apply to this case anyway. Here you start to make sense. I gladly concede this one.
If you can commute the limits, the problem would be trivial, as it would be just evaluating the product of exact ones. And you always can commute limits, come on, the counter examples are so rare to come by and have to be specifically constructed, this one clearly does not look like one where you cannot commute limits.
And come on, you do exactly that at 19:02!!! Why couldn't you use the uniform convergence from the beginning, and not do the whole shuffling of the sums in the first place?
@@angelmendez-rivera351 One shouldn't be throwing recommendations about what others should study, as you never know the credentials of a random person over the internet.
@@angelmendez-rivera351 *This is just false.* *I have no time to explain to you why this is the case, you are better off just taking a real analysis course and reading a textbook, because I have no idea how much of a summarized explanation from me you would understand anyway.*
Because apparently, telling basically "you are wrong because I am right. I have no time or wish to explain why" is a way to go in maths.