Understood every step ✅ Makes sense to do it this way ✅ Could have dreamed up the solution ❌ Not today, maybe in my youth. A troll of a question to put on an exam.
I do not follow how the 1/sqrt(k) inequality at about the 6 minute mark implies the a_k inequality. The inequality on the cubic term goes in the wrong direction due to the negative sign.
5:56 Awesome video! Just a minor question. What justifies the last line? Since a_k >= 1/sqrt(k), we can replace a_k with 1/sqrt(k) in the first term. But we cannot just do the same replacement for the second term since we are subtracting.
Well, it’s not a constant power, but (assuming I’m not making an early morning error), if you add an nth power at each step to make it a_n = sin(a_(n-1))^n, then since sinx < x for all x in (0,1) and therefore sin(x)^n < x^n for all positive natural n, it follows that a_n < a_(n-1) ^n < a_2 ^n for all n. But the sequence a_2 ^n is just a geometric series with constant term a_2, so its sum converges, and thus a_n is a positive decreasing series that is less than a series whose sum converges and so a_n also converges. I’m not sure about sinx^c for constant c. Maybe it converges for larger c, or maybe it always diverges? It’s still too early in the morning for me to get my head around, other than that the technique I used above doesn’t quite work here. 🤷♂️
@Khemiri Moez Cool. Do you have a proof? I’m curious how to prove it. P.S. Never mind, I think I got it. Using the ratio test for convergence. Assume a_(n+1) = sin(a_n)^2. Then L = (a_(n+1)/a_n) = sin(a_n)^2 / a_n Since sinx < x for all x in (0,1), and since a_n < 1 implies a_n^2 < a_n, we get L < a_n ^ 2 / a_n < a_n Therefore L < a_n so the limit of L as n grows to infinity is 0, and thus the sum converges. Thanks!
@@Bodyknock take (a_n+1)^-2-(a_n)^-2=1/sin^2(an)-1/an^2 then expand sin since an goes to 0 you'll get a an equivalent to that series then use cesaro's lemma which states sum k=1 to n of u_k/n has the same limit as u_n you'll get that a_n+1 is equivalent to some constant over sqrt(n) then you can get a_n and by the p series test you can prove that the series of (a_n)^p converges iff p>2
This one made my head hurt. You lost me when I realized you weren't talking about Sinex, the nasal spray. It seemed the approach of comparing to the harmonic series wasn't terribly intuitive, predicated as it was on the assumption that the series would diverge. If you first assumed otherwise, you'd spin your wheels for quite a while trying to find a way to tackle the problem. I'll definitely have to watch the video again.
Take the sum of something squared. Something squared is always positive. The sin function cycles, so nothing decreases faster than it's being added as N trends towards infinity => the series diverges.
Besides being good, when you squared the inequality you made both sides positive so the inequality may not work, e.g. -2^2 > -1^1 but -2 < -1. If both sides must be positive then it won't matter.
The terms a(n) which are squared in the series are generated through a ricursive formula a(n+1)=F(a(n)) , a(0)=π/2, where the function F=Sin has a "contractive" behaviour, since its derivate Cos has absolute value strictly minor than 1(only for 0 argument Cos(0)=1, but 0 is the fixed point, i.e. Sin(0)=0 and this would definitively set the sequence to zero, vfr infra). As a consequence we can for instance apply the Caccioppoli-Banach Theorem and so we can rigorously prove that the terms a(n) tend to zero as n tends to infinity. In fact, whenever the CB theorem can be correctly applied, the limit of the sequence a(n), as n tends to infinity, is the so-called "fixed point" a that verifies the equation a=F(a); in our casa F=Sin, and so we obtain the equation a=Sin(a), whose only solution is a=0. The abovementioned proof is omitted in the video,(replaced by an implicit intuitive claim?) Obviously, the zero limit of the terms is only a necessary condition for the convergence of the series. In fact, the actual divergent behaviour of the series is rigorously studied in the video by means of comparison of terms with the well known harmonic series; the approach is cleverly based on induction, and this is coerent with the nature of the ricursively-defined sequence. The convergence behaviour of infinite series which may be built with higher power of a(n) can be easily studied through the generalization of the described method (a claim of comparison with a known series to be inductively proved;Taylor approximatio approximation of Sin...)
Summations actually are deceiving. Their answers are so... so gorgeous. Wonderful that adding and subtracting a bunch of things gives you an elegant solution
Let f(x) = x - x^3/6 We have f'(x) = 1 - x^2/2, which is greater than zero for all x in the range [0; 1]. So if a_k >= 1/sqrt(k) then f(a_k) >= f(1/sqrt(k)) since both a_k and 1/sqrt(k) are in the range [0; 1] so he is right.
I came up with a different proof. Since a_n goes to 0 we can make a_n small enough so that a_n > a_(n+1) = sin(a_n) >= a_n - a_n^3/6 Now think about how many terms it will take for a_n to half. For a conservative approximation suppose that a_(m+1) is always a_n^3/6 less than a_m. Then it will take at least [a_n/2]/[an^3/6] = 3/an^2 terms to get to an/2. In this time we always are adding at least (an/2)^2 to the sum. So before halving a_n we need to add at least 3/an^2 *an^2/4 = 3/4 to the sum. Since we can half infinitely many times, we add 3/4 at least infinitely many times, so the sum diverges
I understood everything up until he said that because k=1/3, 1/8 that it implies that the induction hypothesis is true for all natural numbers. I don’t understand how two fractions prove that it works for all natural numbers
you kind of messed up your notation when doing the algebra part. you used implication arrows, but obviously you would want to use equivalences, since you started with the claim and ended with a true statement.
1:25 Here the remainder is in the Lagrange form. See Wikepedia page for Taylor's theorem: en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder
Oh this seems fun. Since sin x is less than x for positive x we know the sequence is always positive and decreasing. Limit of ratio might get weird though since (sin x)/x approaches 1. Should be a fun video.
5:28 "If we can show [a sub k minus a sub k cubed over three factorial was greater than one over square root of k plus one], then that would [imply our previous step, which is what we want to show]". Me: (almost catching up) 5:54 "Now we want to show this: [one over root k minus one over root k cubed over three factorial is greater than one over square root of k plus one]. It's a bit of a mouthful and a bit clunky, but if we prove this, that would [imply our previous step which would [imply the previous step, which is what we want to show]]." Me: uh… um… is this the domino effect of induction?
The lines of reasoning for these problems are always so unintuitive. You'd have to suspect that the series diverges to even consider comparing to the harmonic series. And choosing the truncation of the Taylor series seems kinda arbitrary until you see it work out later. It amazes me that people can somehow tackle these problems under time constraints, even if this is one of the hardest math exams out there.
Let's Consider the ratio test: If the following inequality holds then the given sum is absolutely convergent limit k->inf [(sin( ...(k+1 times)...(1))/sin(..(k times)..(1))^2] < 1 Base case: (sin(sin(1))/sin(1))^2 ~ 0.785 < 1 OK. Induction step: Consider, bwoc, there exists some n such that (sin( ...(n+1 times)...(1))/sin(..(n times)..(1))^2 >= 1, then (sin( ...(n+1 times)...(1)) >= sin(..(n times)..(1) or (sin( ...(n+1 times)...(1)) = 1 (contrad.). C2: Since sine is an odd function, -sin(..(n times)..(1) = sin(..(n times)..(-1), and applying arcsines as in C1, we have sin(1) = 1, therefore (sin( ...(n+1 times)...(1))/sin(..(n times)..(1))^2 < 1 So each term in the sequence is strictly less than 1, but it is not clear that the limit is so. This analysis assumes such a limit exists and is strictly less than 1. Your analysis would suggest that this is not the case. I suppose it is plausible that the limit is 1, in such case the ratio test would be inconclusive. One could show this using an epsilon-N type limit proof.
The William Lowell Putnam Mathematical Competition was founded in 1927 by Elizabeth Putnam in memory of her late husband. en.m.wikipedia.org/wiki/William_Lowell_Putnam_Mathematical_Competition
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Understood every step ✅
Makes sense to do it this way ✅
Could have dreamed up the solution ❌ Not today, maybe in my youth.
A troll of a question to put on an exam.
That's usually how Putnam questions go.
I like how he "researched" For a while through yt polls, and now is constantly bringing these great videos
Doing my best! Thanks for watching!
@@BriTheMathGuy Really appreciate you making these videos 😊
I do not follow how the 1/sqrt(k) inequality at about the 6 minute mark implies the a_k inequality. The inequality on the cubic term goes in the wrong direction due to the negative sign.
It should have been shown in the video that x-x^3/3! is increasing in (0,1).
@@tavishu jep, he should have mentioned that indeed
Same remark
a(k) is in the (0,1] interval, he demostrared its true before because the remainder theorem of taylor series at that interval.
great video!! ive always struggled with induction, i like how much you hammered the steps in, i really started to get it
Great to hear!
5:56 Awesome video! Just a minor question. What justifies the last line? Since a_k >= 1/sqrt(k), we can replace a_k with 1/sqrt(k) in the first term. But we cannot just do the same replacement for the second term since we are subtracting.
More precisely, a_k >= 1/sqrt(k), BUT
- (a_k)^3 / 3! = 1/sqrt(k) - (1/sqrt(k))^3 / 3!
is true or not.
@David Schmitz That resolves it! Thanks!
had the same question thank you
I wonder if there's a power we could take a_n to to make the series converge
4?
Well, it’s not a constant power, but (assuming I’m not making an early morning error), if you add an nth power at each step to make it a_n = sin(a_(n-1))^n, then since sinx < x for all x in (0,1) and therefore sin(x)^n < x^n for all positive natural n, it follows that a_n < a_(n-1) ^n < a_2 ^n for all n. But the sequence a_2 ^n is just a geometric series with constant term a_2, so its sum converges, and thus a_n is a positive decreasing series that is less than a series whose sum converges and so a_n also converges.
I’m not sure about sinx^c for constant c. Maybe it converges for larger c, or maybe it always diverges? It’s still too early in the morning for me to get my head around, other than that the technique I used above doesn’t quite work here. 🤷♂️
@@Bodyknock litterally any power strictly bigger than 2
@Khemiri Moez Cool. Do you have a proof? I’m curious how to prove it.
P.S. Never mind, I think I got it. Using the ratio test for convergence.
Assume a_(n+1) = sin(a_n)^2. Then
L = (a_(n+1)/a_n) = sin(a_n)^2 / a_n
Since sinx < x for all x in (0,1), and since a_n < 1 implies a_n^2 < a_n, we get
L < a_n ^ 2 / a_n < a_n
Therefore L < a_n so the limit of L as n grows to infinity is 0, and thus the sum converges.
Thanks!
@@Bodyknock take (a_n+1)^-2-(a_n)^-2=1/sin^2(an)-1/an^2
then expand sin since an goes to 0 you'll get a an equivalent to that series then use cesaro's lemma which states sum k=1 to n of u_k/n has the same limit as u_n you'll get that a_n+1 is equivalent to some constant over sqrt(n) then you can get a_n
and by the p series test you can prove that the series of (a_n)^p converges iff p>2
I feel like home when I watch these videos
What a nice comment! Thank you for watching and have a great day!
This one made my head hurt. You lost me when I realized you weren't talking about Sinex, the nasal spray.
It seemed the approach of comparing to the harmonic series wasn't terribly intuitive, predicated as it was on the assumption that the series would diverge. If you first assumed otherwise, you'd spin your wheels for quite a while trying to find a way to tackle the problem. I'll definitely have to watch the video again.
Keep going mate, this channel is destined to grow.
Hope so!
I saw the sine, and it opened up my eyes, and it’s also good to have your W.T.S. about you. Thanks for another wonderful video!
Thanks so much! Have a great day!
Love your videos 🤗🤗
Thanks so much 😊
Take the sum of something squared. Something squared is always positive. The sin function cycles, so nothing decreases faster than it's being added as N trends towards infinity => the series diverges.
🧐
Besides being good, when you squared the inequality you made both sides positive so the inequality may not work, e.g. -2^2 > -1^1 but -2 < -1.
If both sides must be positive then it won't matter.
The terms a(n) which are squared in the series are generated through a ricursive formula a(n+1)=F(a(n)) , a(0)=π/2, where the function F=Sin has a "contractive" behaviour, since its derivate Cos has absolute value strictly minor than 1(only for 0 argument Cos(0)=1, but 0 is the fixed point, i.e. Sin(0)=0 and this would definitively set the sequence to zero, vfr infra).
As a consequence we can for instance apply the Caccioppoli-Banach Theorem and so we can rigorously prove that the terms a(n) tend to zero as n tends to infinity.
In fact, whenever the CB theorem can be correctly applied, the limit of the sequence a(n), as n tends to infinity, is the so-called "fixed point" a that verifies the equation a=F(a); in our casa F=Sin, and so we obtain the equation a=Sin(a), whose only solution is a=0.
The abovementioned proof is omitted in the video,(replaced by an implicit intuitive claim?)
Obviously, the zero limit of the terms is only a necessary condition for the convergence of the series. In fact, the actual divergent behaviour of the series is rigorously studied in the video by means of comparison of terms with the well known harmonic series; the approach is cleverly based on induction, and this is coerent with the nature of the ricursively-defined sequence.
The convergence behaviour of infinite series which may be built with higher power of a(n) can be easily studied through the generalization of the described method (a claim of comparison with a known series to be inductively proved;Taylor approximatio approximation of Sin...)
Great video as always
Appreciate that!
Summations actually are deceiving. Their answers are so... so gorgeous. Wonderful that adding and subtracting a bunch of things gives you an elegant solution
Glad to hear you enjoyed it! Thanks for watching!
Loved this one. If you mix up the strategy a little, you can prove that the series sum for (a_n/log(n))^2 does actually converge.
I think there's a mistake with the minus sign in the inequality in 5:57.
I also have the same question. Kindly see my comment.
Let f(x) = x - x^3/6
We have f'(x) = 1 - x^2/2, which is greater than zero for all x in the range [0; 1]. So if a_k >= 1/sqrt(k) then f(a_k) >= f(1/sqrt(k)) since both a_k and 1/sqrt(k) are in the range [0; 1] so he is right.
@@vanannguyen5365 Yes. Just to clarify further, since f'(x) > 0 when x is in [0, 1], f is increasing in that interval.
Was able to follow, but there is no way I would be able to solve this alone.
This guy casually pulls out gamma functions out of his pocket but he has to explain a simple method of proving stuff.
Same here. Do u really to need to explain how math induction means when you are talking about a Putnam problem?
I came up with a different proof. Since a_n goes to 0 we can make a_n small enough so that
a_n > a_(n+1) = sin(a_n) >= a_n - a_n^3/6
Now think about how many terms it will take for a_n to half. For a conservative approximation suppose that a_(m+1) is always a_n^3/6 less than a_m. Then it will take at least [a_n/2]/[an^3/6] = 3/an^2 terms to get to an/2. In this time we always are adding at least (an/2)^2 to the sum. So before halving a_n we need to add at least 3/an^2 *an^2/4 = 3/4 to the sum. Since we can half infinitely many times, we add 3/4 at least infinitely many times, so the sum diverges
I understood everything up until he said that because k=1/3, 1/8 that it implies that the induction hypothesis is true for all natural numbers. I don’t understand how two fractions prove that it works for all natural numbers
you kind of messed up your notation when doing the algebra part. you used implication arrows, but obviously you would want to use equivalences, since you started with the claim and ended with a true statement.
Thank you Professor Brian Hawthorn aka Brithemathguy!
You're very welcome!
Well the problem did fool me; I would have thought that sin(a_n)^2 was decreasing fast enough to converge, because sigma(1/x^2) converges.
1:25 Here the remainder is in the Lagrange form. See Wikepedia page for Taylor's theorem:
en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder
Oh this seems fun. Since sin x is less than x for positive x we know the sequence is always positive and decreasing. Limit of ratio might get weird though since (sin x)/x approaches 1. Should be a fun video.
Hope you enjoyed it! Thanks for watching!
5:28 "If we can show [a sub k minus a sub k cubed over three factorial was greater than one over square root of k plus one], then that would [imply our previous step, which is what we want to show]".
Me: (almost catching up)
5:54 "Now we want to show this: [one over root k minus one over root k cubed over three factorial is greater than one over square root of k plus one]. It's a bit of a mouthful and a bit clunky, but if we prove this, that would [imply our previous step which would [imply the previous step, which is what we want to show]]."
Me: uh… um… is this the domino effect of induction?
Papa Flammy is shaking.
At 5:04, what's W.T.S?
Want to show. :)
@@BriTheMathGuy okay... thank you!
Jokes on you mate, all problems fool me.
😅
The lines of reasoning for these problems are always so unintuitive. You'd have to suspect that the series diverges to even consider comparing to the harmonic series. And choosing the truncation of the Taylor series seems kinda arbitrary until you see it work out later. It amazes me that people can somehow tackle these problems under time constraints, even if this is one of the hardest math exams out there.
Wow!
🤯
Nice video :-)
Glad you enjoyed it!
Damn I fell for it too
It's a tricky one!
Those A3 and B3 Putnam questions are like mathematical terrorists.
They're tough!
Let's Consider the ratio test:
If the following inequality holds then the given sum is absolutely convergent
limit k->inf [(sin( ...(k+1 times)...(1))/sin(..(k times)..(1))^2] < 1
Base case: (sin(sin(1))/sin(1))^2 ~ 0.785 < 1 OK.
Induction step: Consider, bwoc, there exists some n such that (sin( ...(n+1 times)...(1))/sin(..(n times)..(1))^2 >= 1, then (sin( ...(n+1 times)...(1)) >= sin(..(n times)..(1) or (sin( ...(n+1 times)...(1)) = 1 (contrad.).
C2: Since sine is an odd function, -sin(..(n times)..(1) = sin(..(n times)..(-1), and applying arcsines as in C1, we have sin(1) = 1, therefore (sin( ...(n+1 times)...(1))/sin(..(n times)..(1))^2 < 1
So each term in the sequence is strictly less than 1, but it is not clear that the limit is so.
This analysis assumes such a limit exists and is strictly less than 1. Your analysis would suggest that this is not the case. I suppose it is plausible that the limit is 1, in such case the ratio test would be inconclusive. One could show this using an epsilon-N type limit proof.
Clearly the limit is greater than 1 since a_1 alone is 1
hmmm These problems are very interesting although it does seem mean to put this on a test
Lol, well it was a Putnam contest problem, so it’s supposed to be tricky. 🙂
I wouldn't do this to my students!
0:36 we see that sin x = x
Gotta zoom in close 🧐
@@BriTheMathGuy it holds for small values of x :D
It sgoudent be by the limit test no conclusion
😄👍
😄
Who is Putnam?
Did you know ??
You can search on Browser too
Put your Vietnam over there
The William Lowell Putnam Mathematical Competition was founded in 1927 by Elizabeth Putnam in memory of her late husband.
en.m.wikipedia.org/wiki/William_Lowell_Putnam_Mathematical_Competition
@@pardeepgarg2640 Don’t be a dick. If you don’t want to answer the question just ignore it.
He is the one from whom this contest is named. **insert genius black guy meme**
first view !
Thanks for watching!
@@BriTheMathGuy welcome