a Putnam-exam-level system of equations
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- เผยแพร่เมื่อ 1 ส.ค. 2024
- This question is B3 from the 1969 Putnam exam. Very suitable for calculus 2 students.
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blackpenredpen
Legend has it that the bunny can solve the Reimann Hypothesis in four different ways.
blackpenredpen video in 2024: How to solve the Riemann Hypothesis (two ways), Calc 3 Practice
是吗？Are you sure?
Twin primes will be found at multiples of 15.
Odd numbers will find twin primes ending in 1,3 and 7,9
Even numbers will find twin primes 9,1.
Euclid PNT Q=P1,P2,P3...... P24
First three primes are 2,3,5=30.
So multiplying any prime after will only find spots at intervals 30.
This is only partial proof of the riemann hypothesis. I have the full proof written down.
But no friends or family want to talk about it... because to them math is pointless. And went I went to a college the professor wasn't there and when they asked what it was about. I told them it was essentially a thesis... but after they found out I had no college. They said it was probably nothing before even looking.
I don't have the skills to take it even higher because I cannot afford college. And I have been working on this for two years alone. But if I go to college or not won't take away the meaning of this proof.
And after comprehending the full scope of this. I am going to post my
I just want to be a productive member of society and have a voice. This is why many mathematicians die before they are recognized.
I would really like a response before I post it. But... I will probably post it within the next 24hours.
@@noahzuniga I guess I am the quantum bunny
I am the last mimzy. (A movie) I have the other 4 proofs too
Hahaha!
2:54 "sometimes he likes to be on the bottom, sometimes he likes to be on the top"
I guess you could say the variable is pretty versatile then huh?
Definitely a switch
Disappointed it's not 69 😂
Can also say that he's the evil version of i.
Algebruh moment at the end there
5:55
I'm sensing a double factorial here, not yet sure, i'm not yet done watching
You were right!
Wow you must have like 200 IQ to have seen that coming
@@blackpenredpen yes, it seemed to be so, i noticed that they are just like a factorial, but instead of decreasing by 1, they are decreasing by 2. Hehehe
Are you 140 IQ
You should check limit of a_(2n+1)/a_(2n+2), in case the limits are not same.
I thought about that too
::confused face:: "square root... pi/2???"
::long pause::
::cut to real solution:: "its obviously plus or minus the square root of 2/pi!"
thank you for that. i really needed that laugh. xD
this truly made my day. :D
Yay! You finally uploaded!
Sorry, took too long to decide “which mic” to use. : )
i love these types of problems thanks to you to make me able to see fun in mathematics!
Hi bro!! Love your maths videos, they keep me motivated!!!
Did it via a bunch of factorials and stirling's approximation, was very satisfying cancelling everything out to get to the right answer!
Just found out about your channel! I love math
This video, is why the pie notation for multiplication was invented.
Nice microphone, i wanna use one like that in my math videos too..
This is a great help for students on lock down but still needing to learn...
our blckpenredpen will be doing goldback conjecture also in 2 methods till 2025
Jajajaja no me esperaba que pusieras tu pequeño error de cálculo, pero quedó bien. ¡¡Buen video!!
Genial!!, como siempre.
Me encantan tus videos, a pesar de no hablar tu mismo idioma entiendo la mayor parte del video. Muchas gracias y exitos.
Psd: me gustaría que añadas subtítulos al español :)
Printed the numbers up to a1000 and the solution is almost certainly correct. Good job! Although the convergence of an/an+1 to 1 is not mononically decreasing. Around a93 it starts oscillating a little. Could be some strange floating point error.
I recognized the Wallis product only at 9:17 . Shame on me !
Very interesting video. I didn't think that pi woud come here !
pi always like to come to math parties
It's hard to remember some formula :D
@@blackpenredpen pi always come to parties that he shouldn't go :D
@@blackpenredpen i always like to come to math parties
Hi! I've been watching your videos for weeks and I am really fascinated. I love math and I really want to study it at advanced level and be good at it, but I don't know how to begin. Have you got any advices(maybe some videos or sites or books)?
I hope you read this message
Thank you, you're amazing
Hi. Where do you buy your microphone covers from? They look so cool!
That's a pokeball, with asH inside it😂
I want one too... 😊
A big fan sir
Great ✌️✌️, I was suspecting π right from the start!!!
Sir you are awsome
Thanks : )
I have found by myself, but it took me quite a long time. finding explicit formulas for a(2n+2) and a(2n+3) with factorials, then using Stirling Formula and many calculus manipulations. I didn't remember Wallis Product, clearly it would help me find faster the answer
I need a rabbit like that for when I get stuck explaining to my students.
hahaha that part at the end was funny "底下 底下 底下"
Oh i said “hold on hold on”
等一下 not 底下
Yea lol
Series! Cool haha I like to cover topic related to the patterns, also like that lil thing you are holding~ great
Thanks. It’s a pokéball. : )
Where is the error in the following ?:
a_n*a_n+1 = n
--> a_n = n/a_n+1
multiply with a_n and use a_n/a_n+1 --> 1 for n --> infty
--> a_n**2 = n*a_n/a_n+1 --> n
--> a_n --> sqrt(n) (*)
a_1 = (n-2)!! / (n-1)!! *a_n --> sqrt(2/pi) * sqrt(n) --> infty with (*) and square root of wallis product.
Long time no see, sir!
a1 is being a chaotic vers and I'm here for it 😂
AMAZING!!!!! That was delightful, how can you deny math’s platonic qualities when you glance at stuff like this?
phenomenal 🙀🙀
13:50 that is such a math teacher moment :D
at first i was confused it looks to me like at the end of the wallis product it would be
(2n-1)/(2n-2) * (2n-1)/(2n) but then what about the other (2n-2)?
well before the (2n-1) term, there is (2n-3), so the other (2n-1) goes to that.
🤣“等一下”
0:33 and don't worry ,if I get stuck 小兔兔 will help us out
Amazing..
Please somebody explain me at 7:58 why did he put the square on (2n-2) in the denominator.
Between 2, 4, 6, ... 2n there is a (2n - 2).
Edit: ok, that was a bit terse.
We are multiplying (2 x 4 x 6 x ... x (2n - 2)) with (2 x 4 x 6 x ... x 2n). Now, 2n is bigger than 2n - 2. You match up all the terms to 2n - 2, so they get squared. There is still one term 2n left, so multiply that in too.
AWESOME
Nice to see Nijntje came to the rescue
@blackpenredoen
Could you pl find the limit of x tending to infinity
Root(x^2+2x)-x
6:06 may i ask what is he calling that term? can't quite hear it well.
"The limit as n goes to infinity of a-whatever over a-sub whatever (the same whatever) and then plus one"
曹老師的頻道越來越多朋友了🤣
13:55 (發現怪怪）
13:57 等一下等一下！
應該要找小兔兔來幫忙才對
哈哈哈 是啊
@@blackpenredpen WHY??????????????????
stupid question, why is the (2n-2) squared in the equation outlined in red? Shouldn't it just be to the first power times (2n) ?
We are multiplying all the evens up to (2n - 2) with 2 x 4 x 6 x .... x (2n - 6) x (2n - 4) x (2n - 2) x 2n, so the (2n - 2) is squared, too.
Nice year, btw.
I solved this myself and got the same answer! Despite the fact I started to forget all this math (and now trying to remember it as a way to massage my brain, while others solve sudokus and crosswords).
My guess : alternatively divide and mulitply, wallis product style,
so a_1 = a_1 * a_2 / (a_2 * a_3) * (a_3 * a_4) ... =1 /2 * 3 / 4 ...
= lim_{n -> inf} (2n+1)!! / (2n)!! or = lim_{n -> inf} (2n-1)!! / (2n)!!
But the limit is infinite for the left one and zero for the right one 🤔
I also thought about this method of solving it, though I couldn't find a way to include the condition imposed by the limit in the proof!
If you square both sides in your last expression you should get:
(a_1)² = 1²×3²×5²×•••/2²×4²×6²×••• = 2/π
and therefore a_1 = ±√(2/π)
But, again, how does the limit come into play in this proof? must be wrong :c
Varad Mahashabde You are probably not carrying out the product correctly. You are supposed to ultimately just have a ratio of double factorials multiplied by some extra normalizing factor.
Also, I am not quite sure that your claim that lim (2n + 1)!!/(2n)!! (n -> ♾) is true.
EDIT: The claim is true, but irrelevant, since a(1)^2 = lim (2m + 1)!!(2m - 1)!!/(2m)!!^2 (m -> ♾), it is not the case that a(1) = lim (2m - 1)!!/(2m)!! (m -> ♾).
Valerio Bertoncello The idea is that lim a(n)/a(n + 1) (n -> ♾) = 1 implies that 1/a(1)^2 = lim (2m)!!^2/[(2m - 1)!!(2m + 1)!!] (m -> ♾) .
gotta solve 'em all
But what if you had started with lim (a_(2n-1)/a_(2n)) instead?
1969. Nice.
the fact that lim a_(n)/a_(n+1) = 1 is just a consequence of the definition, right ? It's easy to see that a_(2n)/a_(2n+2) and a_(2n+1)/a_(2n+3) both converges to 1 by using the definition of a_(n).
Nice.
Shouldn't there be a (2n-1) before 2n at 8:21?
sir I am your great fan n regular viewer.Where are you from? whats ur job ,qualifications i wanna know plzzz
How can I send you questions ???
Twitter is the best place since u can just tweet me a pic.
13:56 「等一下等一下等一下」wwwww
13:56 Chinese? 等一下!
So in the end you asked 小兔兔 for help right?
Lol yea
You should confirm that also lim_{n->∞} [a_{2n-1} / a_{2n}] = 1 with those values of a_1. Otherwise there would be no solution for a_1.
We are NOT done: you have only demonstrated that if there is an answer, then it must be $\pm \sqrt{2/\pi}$. The problem is that you only computed the limit of a subseries, namely the one obtained by taking every other term of $a_n/a_{n+1}$. You can easily finish the proof by showing the same result for the subseries of the remaining elements. (Or an elegant finish: it is clear from the definition of the $a_i$ that $a_n/a_{n+2}=n/(n+1)$ tends to 1, so if the above half of the quotient series tends to 1, then the other half does too.)
András Pongrácz This is a moot point. The limit of the ratio of the subsequence is the same as the limit of the ratio of the sequence itself. This is because if lim a(n)/a(n + 1) = 1, then n = 2m, and as n -> ♾, m -> ♾. Since this transformation is bijective, which is trivial, the implication holds in the other direction.
@@angelmendez-rivera351 The limit of a subsequence is the same as the limit of the sequence... provided that the sequence has a limit. It is not a moot point, but a theoretical mistake (which is of course easy to fix, as I wrote). Imagine that every other term is 1, and every term wit odd index $n$ is $n$. If you only focus on the even indices, you might think that the limit is 1.
András Pongrácz Okay, but we already knew the sequence has a limit at all, it is literally part of the premise of the problem. So, it is indeed a moot point. There is nothing theoretically incorrect that he did in this video. He didn't use induction to prove the form of the terms a(2k) and a(2k - 1), but to be fair, proving it is so trivial, it's not as if anyone watching the video is confused about why he didn't do it.
13:45 Me on every single fucking test... u.u
I was expecting a limit of a[n]. You can show an equivalence a[n] ~ √n , that is limit a[n]/√n = 1
Great
+-sqrt(pi/2) could be also an answer
Did you sing the song at the end bprp?
It says on the bottom-right "outro song by Zach & Jonah (check out MotorMusic)" very clearly, so...
Shouldn't the pi over 2 be squared on that last part?
Jose de Arimatea No, only a(1) needed to be squared.
Wouldn't the higher degree of n in the denominator cause the limit to go to 0?
Avraham Radin The denominator does not have a higher degree of growth with respect to n. Just to be precise, a(1)^2 = lim (2n - 1)!!(2n + 1)!!/(2n)!! (n -> ♾), which is the reciprocal of the Wallis product. (2n - 1)!!(2n + 1)!! and (2n)!!^2 have the same growth. This can be proven by using double factorial identities to convert into factorials and exponentials, and then compare the numerator and denominator by using the Stirling asymptotic approximation.
Can you do a video on washers on disks? Can you please do the same problem in both washers and discs? I have a problem with setting it up. Can you please do more than just 1 problem? Please explain in your video why 1 method would be easier than the other method. Please, explain the setup. Also, can you please do the same problem if it was rotated about the x-axis and then about the y-axis. In some problems, I must solve for x and in other problems and in other problems I must solve for y. Can you please explain when to solve for x or y? I have a problem with setting them up. You don't need to solve the problems. Just the setup is where I am running into a problem. I can integrate the problem(s) myself. The integration is the easy part. The setup is the HARD PART. Thank You, Sir.
哈哈哈，，，我听到了“等一下等一下”
Wow ese producto está curioso
Pi always being on strange places
DJ VALENTE DO CHP It isn't as strange as it seems. There are a bunch of double factorials in this product, and the double factorials come from the Gamma function being evaluated at half integers, which is equivalent to integrating a cylindrically symmetric bell curve. The cylindrical symmetry is where π becomes relevant.
@blackpenredpen today I uploaded a Putnam 1992 functional equation problem try it
It shows a pattern=n!!/n-1!!
Don't forget the square :D
Not n!!/(n - 1)!!, but (2n)!!/(2n - 1)!!, and you must square the ratio.
Love the pokeball mic.
Thanks.
BrownWater nice mr robot pfp
@@Diaryofaninja Thanks :)
Thought he captured the bunny in the pokeball... By the way, does the bunny have a name?
@@idontwantanamethx th-cam.com/video/zbZsdllCFQw/w-d-xo.html
At 0:05
Do you ever feel like you are inadequate?
There’s gotta be more solutions than that because a1 = a2 = 1 also generates the same set of answers...
You may be able to construct a series with a_n * a_(n+1) =n with a_2k and a_(2k+1) like the ones in the video with (a_1)^2 =1, but then the lim (a_n/a_(n+1)) = 2/pi and not 1. So I don't think there are different answers. I'm actually sure there are none
Wow 😲😲
A bit sad the derivation isn't done from the Stirling formula as it's what is expected to be known from Putnam candidates.
Still a nice video :)
Grawuka I think the Wallis product is better known than the Stirling asymptotic approximation formula. In any case, the Stirling approximation can be derived from the Wallis product, and the Wallis product can be derived from the Stirling approximation, hence it really does not matter.
Have you ever heard anything about maryam mirzakhani?
It seems obvious but can anyone tell me why
if the lim (a(n)/a(n+1))=1,
then the lim(a(2n)/a(2n+1))=1.
Otherwise I may have been able to do this
Let u=2n. Then as n goes to infinity, u also goes to infinity. The limit with the 2n and 2n+1 becomes a(u)/a(u+1), which is equivalent to the given limit in the question.
@@stephenbeck7222 that's incorrect because a(n) isn't a function
@@stephenbeck7222 that's incorrect, what if a(n+1)=n and a(n) =1, we have here a(n+1)*a(n)= n but a(n)/a(n+1)=>0
2:54 - 2:58 mood
i should have known that he will come out with a crazy weird formula in the end. but then i clicked in... and saw a weird formula...
evil... formula....
13：56 等一下，等一下 lol
He always starts with
"Hello let's do some math for fun"
It's interesting that a1 = 0.5!
13:52. Hug Diya na!! lol
13:40 LOL! you took a shortcut skipping a step ending with 13:50
Is it pokeball or mic ?
Hahahah
So all the jokes about the Pokéball are now true
I think there is pikachu in that ball
Is every thing equal to pii or what? crazy as pii followed us since primary school
Pi pops up in the oddest places. But as 3Blue1Brown says, wherever you find pi, a circle is involved somehow. It may be hidden, but it's there.
CNVideos Yes. In this case, the sequence {a(n)} is a ratio of double factorials, which can all be expressed in terms of ratios of products of fractional factorials, which are the result of the n-volume of the bell curve, which has polar properties. In those polar properties is where the circle is hidden.
you should prove first the relation that you wrote betwen a2n+1 and n
等一下，差點搞砸了
LMAO
Lolll
For the sake of rigor, you need to construct a concrete sequence to prove that sqrtpi/2 can actually be a1
What did you mean by “prove that sqrt(pi/2) can actually be 1”?
What do you mean? The infinite product is the limit of a concrete well-known sequence.
13:56 等一下XD 突然冒出中文我嚇到
hahaha
That's Wallis product, not Walli's product
Can someone tell me why this formula always generates primes for even values of x?
x^4-x^2(x-1/x)^2
Try 12.
(x - 1/x)^2 = x^2 + 1/x^2 - 2 ==> x^4 - x^2(x - 1/x)^2 = x^4 - x^2(x^2 + 1/x^2 - 2) = x^4 - x^4 - 1 + 2x^2 = 2x^2 - 1. The expression you presented is equal to the much simpler 2x^2 - 1 for all values of x. Anyhow, for x = 0, 2x^2 - 1 = -1, which is not prime.
If x = 2y, then 2x^2 - 1 = 2(2y)^2 - 1 = 8y^2 - 1. For y = 6, 8y^2 - 1 = 287, which is not prime. In fact, there are infinitely many values of y for which 8y^2 - 1 is not prime. This is a famous theorem of number theory: there does not exist any finite polynomial with integer coefficients that generates.
等一下XD 老師是哪裡人啊
我來自台灣喔
我是高中數學老師，關注你的頻道一段時間了，本來以為你是ABC，原來是台灣人啊~老師應該是在大學任教吧
Just about to comment "Hey you missed to square Wallis Product result due to the squares of the limit formula..." but NO NO!! DON'T DO IT!!!! :)
I was too, but later realised. Lol
Didn't you just proof that for even/odd? If an is odd and an+1 is even the answer would be root(pi/2) right? correct me if I'm wrong.
Goatgod No. lim a(n)/a(n + 1) (n -> ♾) = lim a(2n)/a(2n + 1) (n -> ♾). This is the property the question is reliant on.
Plz do a collab on a topic of math with muPrimeMath. plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
We will! Maybe after a few weeks tho since he has AP exams and I have other things to work on as well.
@@blackpenredpen o' Thanks for reply.