Correction: L³ = L + 6 implies L³ - L - 6 = 0. THe rest is correct in the sense that L³ - L - 6 = (L - 2)(L² + 2L + 3) so 2 is an actual root.

I really liked the rigor of using the monotone sequence theorem to justify how the limit of the sequence was calculated. So many times these kinds of details get skipped over in math proofs leaving an uneasy feeling of not quite knowing why certain moves are justified beyond handwaving.

@noahtaul

Жыл бұрын

Lmao your complaints describe the second example perfectly

@charlesbrowne9590

Жыл бұрын

These kinds of details are never left out of proofs. Never.

@Rory626

Жыл бұрын

If a proof skips over these details, then it is not a sufficient proof

@leif1075

Жыл бұрын

Where in God's name does the 8 :07 equation come from..there isNO basis for thisnguess..whybsont everyone else asking this..surely everyone else is just a flummoxed by it..kts the kind of thing thst makes math stupid and annoying or infuriating and I think should stop altogether..

@charlesbrowne9590

Жыл бұрын

@@angelmendez-rivera351 The comment at the top to which I was replying stated “these kinds of details get skipped over in math proofs”. No they don’t, or they are not proof. I’m not sure how it is possible to misinterpret my comment, but apparently this happened twice.

Lovely second problem. The Penn fact is just an AMAZING addition to the videos. Keep it up, prof. Penn

11:00 Thank you for clearly explaining the motivation for making the guess. I think that intuition benefits from learning the motivation behind different guesses.

Re 2nd nested root. The lesson here is: problems that are not solvable in a given context become solvable in a more general context. "Generalise to solve": how true that often is in Mathematics

The 2nd problem can be done rigorously. The sequence of functions that you constructed are, around x=1, monotonous, increasing, bounded above and analytic. Therefore it converges to an analytic function around x=1. Therefore it has a Taylor series expansion. Square the Taylor series and compare the coefficients with the right hand side, you get the linear function as the only solution.

@apopet

Жыл бұрын

I don't get it... Can you please explain the Taylor series part?

@TheEternalVortex42

Жыл бұрын

@@apopet Analytic functions mean they can be defined by a power series

@apopet

Жыл бұрын

@@TheEternalVortex42 Thanks, but I didn't mean explain what a Taylor series is. I didn't get how the series expansion will be used.

@fayard..

Жыл бұрын

The second problem must be solved much more rigorously than it has been. In particular, nothing proves that the limit F(x) is a polynomial function. First consider a sequence F_n(x) = sqrt(f(x) + ...+ sqrt(f(x + n - 1))), with f(x) equal to x^2 + x - 1. Then, prove by induction that the sequence is increasing, and bounded above by x + 1. Finally, let F(x) be the limit. Prove by induction that for every positive integer k, F(x) >= a(k), where a(k) is the following sequence : a(1) = 1/2 and a(k+1) = sqrt (a(k)). Not so simple as that, if we want to stay at a reasonable mathematical level. Sorry for my broken English, I'm French. Hi.

The "Penn fact box" is a nice idea, I tried to click on it to see if there was a link to a sketch of a proof or to some other video.

For the first sequence, wouldn't replacing the "last" 6 with an 8 just collapse the whole sequence like a house of cards to the number 2, proving this is its upper bound?

I’m taking my first rigorous proofs based class next semester and one of things in the syllabus completely foreign to me was working with sequences, i.e finding their limit, proving they converge, etc. so it is very helpful to see videos like this to get a feel for some of the tools used. These problems were also just quite fun and novel to me as well.

Nice, clean, logical & interesting problems. Thanks.

6:10 Hey Michael, Is it not L^3-L-6=0 instead of L^3-L+6=0

@Predaking4ever

Жыл бұрын

Yes, but his work after that line was correct.

@Horinius

Жыл бұрын

Another usual "glitch" from Michael.....

Awesome video. Thank you

Michael: constructs a very nice real analysis proof that the first sequence converges, then computes the limit in a nice rigorous fashion. Also Michael: Well let’s just let this infinite radical be a function, probably well-defined, who knows. Let’s get a polynomial equation, I dunno, probably a linear function. Hey! This linear function satisfies the recurrence, we have no other choice but to declare this the answer!

@adamnevraumont4027

Жыл бұрын

The math speaks for itself

@noahtaul

Жыл бұрын

@@adamnevraumont4027 but the logic connecting the math together does not.

@dnuma5852

Жыл бұрын

i mean he did say he wasnt gonna be as rigorous with that one he left it as an exercise to the viewer

@noahtaul

Жыл бұрын

@@dnuma5852 If he said that, it doesn’t make any sense, because unlike the first one, the second infinite radical is not the limit of a sequence with a simple recurrence. Instead, it’s the limit of one chain in some doubly-indexed sequence a_{m,n}, with a_{m,m}=sqrt(m^2+3m+1) and a_{m,n}=sqrt(m^2+3m+1+a_{m+1,n}), namely the limit of a_{1,n} as n goes to infinity. This is a totally different type of problem, and can’t be handled with the exact ideas as the first one. It would be useful and instructive for Michael to have covered this one rigorously too.

@MyOneFiftiethOfADollar

Жыл бұрын

Noahtaul(know it all): you are among many who appear to be quite knowledgeable/proficient mathematically who have “empty math channel” Noahtaul: So why don’t you produce your own math videos and publish them on your channel?

Hope you keep the Penn Facts as a regular part of the channel, they're very helpful!

I really liked the Penn Fact box containing a formal statement of the Monotone Convergence Theorem. This kind of thing should be very helpful to someone like me who doesn't use math too often but enjoys your videos.

Thank you, i've managed to figure out how to solve a problem that i was stuck on for a while! 😀

How amazing and how beautiful the answers are

Thanks for writing the actual sequence. One thing I really hate about "..."-expressions is that the actual sequence usually is not obvious (to me), in particular the starting value (which is important for the limit in the general case). I guess convention is "..." starts at zero in sums and at one in products? There are other places it gets put though, for instance roots, what then?

The second series of numbers is also the convergents to the continuous fraction of sqrt(2)

The first example is a typical case of an iteration of the form f[f[f[ .....].If this iteration converges , then the limit is a solution of x = f[x]. In our case f[x]= (x+6)^(1/3). This is easy to see if one looks a the graphs of x ,and f[x] , where one can see a sort of spider web which arises if one looks at the points f[f[x] ,f[f[f[x]]] etc. around the point of intersection of the two graphs.The cubic equation is not hard to solve since one can easily guess a solution.

The method used for the second radical was very clever

For the second one, if you pick f(x)=x^2-x-1=x^2-(x+1) you get sqrt(x^2-(x+1)+sqrt((x+1)^2-(x+2)+sqrt((x+2)^2-(x+3)...)))=F(x) and the value we're looking for is F(3) but in this case, F(x)=x

@henriquepinto

Жыл бұрын

or you just notice Fˆ2(x) - F(x+1) = xˆ2 + 3x + 1 = (x+2)ˆ2 - (x+3), i.e., F(x) = x+2.

Here's how I did the second one: I noticed that the numbers can be written as 2^2+1, 3^2+2, 4^2+3, ..., so I write a1=(the original formula), a2=(formula starting from 11), a3=(formula starting from 19),... and observe that a2=a1^2-2^2-1, a3=a2^2-3^2-2, ... so that a_n ^2 - a_{n+1} ^2 = (n+1)^2 + n = (n+2)^2 - (n+3) and just guessed a_n = n+2... so a_1 = 3. Of course it's not rigorous, but it works.

For the top one I figured if a(n+1) for infinity is cubert(6 + an), I could substitute it as such: x = cuberoot(6+x) where the obvious answer is 2

It is interesting to note that if the square root is used in the radical above instead of the cube root, it gives the same result as the radical below

Very nice

F(x) is always rational. that blows my mind. no matter how deep we step into this sequence, we can start there and the tail will be rational.

For the second problem, if you started the x^2 + 3x + 1 function with f(0) = 1, then the solution would be 2, just like the first problem.

It's the Christmas Three!

fun! thank you!! And why don't you give some other similar questions?

Penn fact is the best

🙂 I would found it helpful to prove the functional form of f using the fact that its increment form sequence 6, 8, 10, 12, ...

@Pestrutsi

Жыл бұрын

Yeah this, the only question I had after seeing this video was where the hell did he fart out the explicit form for f from, but the sequence of numbers in the radical being a quadratic one makes it clearer. The calculations themselves feel like they are rightfully omitted as trivial for the purposes of this video but the motivation behind finding f and the general logic of getting there would've been a good addition.

@txikitofandango

Жыл бұрын

I would've liked this too. I know it's related to the fact that adding successive odd numbers gets you the squares. There's a video by Mathologer about these kinds of sequences and how to derive a polynomial from them, but it escapes me

@txikitofandango

Жыл бұрын

It has to do with discrete derivatives

The second one is very cool~ But I have a question (10:53) about how to explain that F is a polynomial. If F is not a polynomial then you can't suppose it as F(x) = ax + b.

@mrynoplanetashka8988

Жыл бұрын

There isn't any reason behind F being polynomial. It's just a smart assumption.

@reeeeeplease1178

Жыл бұрын

It is more like: "Wouldn't it be nice if F were a linear function? Let's check if that could work." If it doesn't work, then try another approach/guess. If it does work, then there you are. You found a solution.

Could you provide a counterexample of a nested sequence that has a solution to the equation we always solve, but does NOT converge.

@cmilkau

Жыл бұрын

You can cook that up pretty easily. Choose a function f with exactly one fixed point and a derivative outside the range [-1,1]. The fixed point is a solution to the recursive equation but the sequence a, f(a), f(f(a)), f(f(f(a))), ... will drift away from the fixed point unless you start exactly at the fixed point (then it is stationary). For concreteness, say f(x) = 12x - 1. We have f(1/11) = 1/11. However the sequence 0, f(0), f(f(0)), ... = 0, -1, -13, -157 diverges. So the expression "-1 + 12(-1 + 12(-1 + 12(-1 + ...)))" has no value.

@TheBeesuke

Жыл бұрын

@@cmilkau Thanks, but when I worte "nested" i meant sequences of the form a(n+1)=(ka(n)+m)^(1/p) where p>=2.

Neat fact about the second sequence: If you pick a value of x to terminate it at, i.e. if you take some n to calculate √(f(1)+√(f(2)+√...+√f(n))) but then you replace the final f(n) with (n+2)², your sequence lands at exactly 3. Idea of proof: You'd have f(n-1)+√(n+2)² in the second last root, so that's (n-1)²+3(n-1)+1+(n+2) = n²+2n+1 = (n+1)², so that gives f(n-2)+√(n+1)² in the third last root which reduces to just n², the next one (n-1)², and so on all the way back to f(1)+√4² = f(1)+4 = 5+4 = 9, so the topmost root becomes √9 = 3.

@DavidSavinainen

Жыл бұрын

This gives another, more "backwards" way to prove the second sequence: Start with √9, clearly 3. Extract the number 4: √9 = √(5+4) = √(5+√16) Extract the number 5: √(5+√16) = √(5+√(11+5)) = √(5+√(11+√25)) Extract the number 6: √(5+√(11+√25)) = √(5+√(11+√(19+6))) = √(5+√(11+√(19+√(36))) Keep extracting the integers one by one, and you end up at the sequence above. Since every intermediate step is equal to 3, so is also the limit. (Of course, this is not mathematically rigorous as I have written it here.)

@TheEternalVortex42

Жыл бұрын

@@DavidSavinainen Seems rigorous enough, a sequence of 3, 3, 3, 3, ... obviously has the limit 3 ;)

My biggest problem to solve the second question is to note that 5, 11, 19, ... Come from f(x) = x² + 3x - 1

@doctorb9264

Жыл бұрын

Can find parabola given the 3 pairs : (1,5) , (2,11) , (3,19)

how did you arrive at the. x^2+3x+1 as f(x)/

First one is easy, basically say x = cbrt(6+cbrt(6+...), then notice that x = cbrt(6+x) since it's nested with itself, then just cube both sides and then arrenge the equation like: x^3-x = 6... plugging in small values you can notice that x is equal to 2. Am I wrong? I am being really honest, I'm not that experient in maths... is there any error or detail that I should consider before assume that "nested with itself statement"?

A very nice nested radical will be if you have only 1s inside.

The second differences in the second sequence are identically 2 [ (f(x+2) - f(x+1)) - (f(x+1) - f(x)) = 2], and you ended up with F(x) = x+2. I wonder if that's a coincidence. What sequences produce other formulas for F, such as x+3?

The second one's definition is a bit ambiguous. For me it looked like the second element is the first one plus 6, the third one is the second plus 8, and so on, so a_n = a_n-1 + 2 * (n + 1), which gives the same numbers for the first few n but then diverges from the f(x) in the video.

@wulli_

Жыл бұрын

while the second problem is defined ambiguously, the quadratic in the video is actually the explicit form of the sequence you described.

@bjornfeuerbacher5514

Жыл бұрын

"but then diverges from the f(x) in the video" No, it doesn't, it gives the same values. Apparently you made a calculation error somewhere.

How would one construct a rigorous proof around 11:35 ? That F should be linear. Physicists would forge happily ahead with this, but I wonder if a true proof is hard.

i really like the question but may you explain how did we get the f(x) = x^2 + 3x + 1

@stephenyip5827

Жыл бұрын

same question want to know as well

@TheStickManPainter

Жыл бұрын

@@randomguyontheinternet_69 Yes, and you can see that it's quadratic by noticing that the size of the gaps between the numbers increase linearly! (6, 8, 10, 12, ...) It works a little bit like derivations, since the rate of change is of first order, the function itself is of second

@cmilkau

Жыл бұрын

kzread.info/dash/bejne/pZeFl5Oqj5TUmNo.html

@hoangnguyennguyen6445

Жыл бұрын

@@cmilkau thanks bro !!

It seems kind of out of nowhere, the appearance of the formula in the second one. If this was a comp. problem, I suppose the hardest part would be to figure it out, after which it would be quite straightforward.

@Chalisque

Жыл бұрын

If you look at the sequence 5, 11, 19, 29, 41, ... and look at successive differences, you get 6, 8, 10, ... and if you look at successive differences again, you get 2, 2, 2, 2, ... all 2's. You then work back from this. 6,8,10,... = 4+2n, so if we let the sequence 5, 11, ... be a{n} with a{1} = 5, then a{2} = a{1} + 4 + 2*1, a{3} = a{1} + 4 + 2*1 + 4 + 2*2. So a{n} = a{1} + 4(n-1) + 2(1+...+(n-1)) and since 1+...+(n-1) = (1/2)n(n-1), and a{1} = 5, we get a{n} = 5 + 4n -4 + n(n-1) = 5 + 4n -4 + n^2 -n = n^2 + 3n +1

@HershO.

Жыл бұрын

@@Chalisque ahh thanks for that! I was thinking there had to have been a way to derive it naturally, this is a neat useful trick.

@bjornfeuerbacher5514

Жыл бұрын

@@HershO. Or you notice that the first number is 2² + 1, the second number is 3² + 2, the third number is 4² + 3 and so on. Hence the nth number is (n+1)² + n = n² + 3n + 1. If you don't notice that, yet another approach is to realize that since the differences between the numbers increase linearly, the numbers themselves have to increase quadratically. So you try a_n = a n² + b n + c and determine the coefficients a, b, c by inserting e. g. the numbers n = 1, n = 2 and n = 3.

"The Q*bert of 6"

For the first one, what if we had gotten a cubic that had more than one real root? Would that mean we have multiple values for the limit?

@HershO.

Жыл бұрын

Then I suppose we would be able to find upper and lower bounds to the expression to rule the other values.

@carstenmeyer7786

Жыл бұрын

Great question! The main problem here is -- the notation with "infinite radicals" is just not precise enough to answer the question to begin with. You need a recursive definition for that. If you have multiple possible limits, the one you get (if any) may depend on two things - how you defined the recursion representing the radical -- infinitely many possibilities! - which initial value you chose -- again infinitely many possibilities!

@reeeeeplease1178

Жыл бұрын

As others pointed out, as long as the sequence is well defined, you only have a single solution/limit point (at most). If we get multiple sol's from our cubic, we have to check each of them (rule out negative or complex sol's, rule out sol's bigger than 3 because we have seen that the limit is bounded by 3 etc.)

@zunaidparker

Жыл бұрын

It would have a single "true" convergent limit (if you plug in any initial guess for x1 and iterate the formula, it would converge to this number) and the other solutions would be "valid" in the sense that if you plugged in EXACTLY that value for x1 it would work, but even a small deviation of your first guess from that value would either diverge under iteration or would converge to the "true" solution. This type of recurrence equation falls under the Fixed Point Theorem of which there's a lot of interesting convergence results to learn about. And they are connected with fractals and other interesting phenomena.

@cmilkau

Жыл бұрын

Yes and no. Then the limit would depend on where you start the sequence. So yes, there would be multiple values for the limit of the "..." expression (it would be ambiguous). But no, there can only be one limit of the (explicit) sequence. That's why I don't like "..." expressions without reference to the exact sequence.

Well, engineer me and not a mathematician. Let the value of the first sequence be x. Then x^3-6=x, and then x^3-x-6=0, and x=2 by inspection. But what do I know.

but whoever invented the second exercise really thinks it's so easy to understand that those numbers come from that second-degree polynomial?

from the start, we can deduce: S = (6 + S)^(1/3) thus (S - 2) * (S^2 + 2S + 3) = 0 -- > S = 2 (in R)

@dlevi67

Жыл бұрын

You can only deduce _that_ *if* S exists... which is what Michael does first.

6 minutes to get to L^3 - L + 6 = 0... Should take 6 seconds

As written on the board the definition of the first expression does not make any sense. The dots should be on the left side.

Hey could someone help me to solve one problem? At the school there are 100 boys and 100 girls. Every girl knows at least one boy, every boy knows at least one girl. Once every girl said "Out of all boys i know no less than 2/3 of them studies straightly with F", and every boy said "Out of all girls i know no less than 50% of them studies straightly with C". And we know that all of them told the truth, and also we know that at the school only 10 boys who study straightly with F. What is the minimum possible number of girls that study straightly with C?

X=2 in just a minute mentally

@FaranAiki

Жыл бұрын

I did it in 30 secs while eating. Not even lying.

La prima è facile. 2....t^3-6=t

I solved both of these mentally..I probably got lucky :)

2 doesn not seem to be a root of the equation L3-L+6=0; we should read L3-L-6=0

We only have proven that: 1. *if* F is a first-degree polynomial, than F(x) = x + 2 2. F(x) = x + 2 is *a* solution of F²(x) - F(x+1) = x² + 3x + 1 However we have *not* proven, or even answered the question: 3. Is F(x) = x + 2 the *only* solution to F²(x) - F(x+1) = x² + 3x + 1?

@michaelempeigne3519

Жыл бұрын

did you find the equation n*n + 3*n + 1 from trial and error ??

How do we know the second radical converges？

@demenion3521

Жыл бұрын

you can probably easily find bounds for the sequence considering that under the radicals we have a polynomial function while iterating the square root is essentially an exponential process (sqrt(sqrt(sqrt(x)))=x^(1/2³)).

@AlinaKlein953

Жыл бұрын

Herschfeld's convergence theorem

@Bodyknock

Жыл бұрын

@@AlinaKlein953 It does follow from that but you do have to do a slight bit extra since the sequence 5, 11, 19, ... isn't bounded. (i.e. you have to prove (f(n)¹/²)ⁿ is bounded. Which it is but proving it takes a couple of steps.)

@cedriclorand1634

Жыл бұрын

How do you even compute partial radicals? Given f(n) should be nested first and then from n down to 1? Totally unclear, I am confused...

@Bodyknock

Жыл бұрын

@@cedriclorand1634 You do them from inside out. The inner most radical first, then the one that includes it, then the one that includes that, etc.

I wish everyone would take the time to properly prove convergence for these "..."-expressions. A lot of nonsense would disappear from youtube.

is it just a coincidence that a vid about radicals of 6 has 666 likes?

@NotoriousSRG

Жыл бұрын

Yes. Yes it is