In my opinion, to show the existence of the limit it would be simpler using the fixed point theorem, showing that the mapping x-> 1/4 - x^2 is a contraction for -1/2 < x < 1/2, and therefore also for the values of the sequence (bounded by 0< a_n

@complexquestion3601

Жыл бұрын

That is the correct method. Though technically, the expression in the wikipedia article is ill-defined, since its value depends on the seed you chose to compute the limit (the "a_0" value). Nothing in the expression actually implies that you should use 1/4, and although Michael tries to justify it by making it seem like he's computing the parentheses from the outside in, what he's mathematically doing is computing them from the inside out (since this is how iterated composition of functions work). In practice, this means that his initial 1/4 value is actually "hidden" in the "..." in the middle of the parentheses rather than being the outermost 1/4 in the expression. More formally, the only way to compute the expression is to define it as lim n-> inf f^n(x_0) where f is x -> 1/4 - x^2 (or (1/4 - x)^2). But this requires choosing an x_0 value, which Michael arbitrarily defines as 1/4. In this specific case, the fact that the expression is ill-defined is not that bad, since there is only one stable fixed point, and taking any initial value outside the convergence domain you pointed out will make the sequence diverge (except for the other, unstable, fixed point). My point is that these expressions are well-defined only when the convergence domain of the iterated sequence is the entire domain of the function to iterate, as there is an ambiguity if it's not the case.

Жыл бұрын

Yes, exactly. But you should consider that recursive sequences are taught in the first year of most science oriented university programmes while fixed point theorem is taught at later years of math oriented university programmes.

@MizardXYT

Жыл бұрын

If the initial value is between (−1−2√2)/4 (≈ −0.957) and (3+2√2)/4 (≈ 1.457) the sequence converges to (3−2√2)/4 (≈ 0.04289). If it is outside, the sequence diverges to positive infinity.

@ingiford175

Жыл бұрын

Yah, when i saw the thumbnail, i mentally put in the value of 10, and watched the terms just quickly get larger and knew there must be some bounded area.

At 12:40 you say: "but, since they both satisfy the same recursion, this means they'll converge to the same number." This statement seems wrong, since you can clearly see that the sequence defined by a_0 = 1, a_{n+1} = -a_n has both the odd and even terms converging to different values, despite the fact that they satisfy the same recursion. A slightly more complicated example using absolute values could also have the odd/even terms strictly increasing/decreasing, while still approaching different limits. Perhaps I'm missing something?

@lefty5705

Жыл бұрын

You are correct

@okra_

Жыл бұрын

Yea, any subsequence of a convergent sequence will converge to the same value, but the opposite (that having convergent subsequences means your sequence also converges) is not necessarily true like he implied

@Alex_Deam

Жыл бұрын

If instead of setting a=lim(1/4 -(a_n-1)^2), you do it as a two step and solve a=lim(1/4-(1/4-(a_n-2)^2)^2), you get a quartic with only two real solutions. One is purely negative and discounted as before, and the other is the same one that Michael got. Since this two step version is definitely true for both even and odd terms, then both converge to that same value.

@jimcameron6803

Жыл бұрын

True; but both the odd and even terms are defined in terms of the same 2nd-order difference that gives rise to a quartic with two real and two imaginary solutions: the two you get from the 1st-order recurrence and (1 +/- i.sqrt(2))/2. (If I haven't made some error in my calculations, which is possible.) So there is only one real solution within the appropriate range and both odd and even subsequences do in fact converge to it. Michael is playing it a bit fast and loose here though.

@rkPixie

Жыл бұрын

True; for example, let sequence {a_n} s.t. a_0 = 1 and a_{n+1} = a_n - (a_n^2 + 3)/(2a_n), then you'll get a_{2k} = 1 and a_{2k+1} = -1 for all integer k. note: it's an invalid example of newton's method; x^2+3=0.

12:40 "since they both satisfy the same recursion, it means they converge to the same limit" this is not necessarily true. Take a_0=1 and a_n=-a_(n-1).

@MrMctastics

Жыл бұрын

Yeah that part confused me

@user-jc2lz6jb2e

Жыл бұрын

We already proved the sequence is convergent, so any subsequence of a convergent sequence converges to the same limit. Your example is not convergent.

@MrMctastics

Жыл бұрын

@@user-jc2lz6jb2e no

@TedHopp

Жыл бұрын

@@user-jc2lz6jb2e That's wrong. We're trying to prove that the sequence converges. At this point، we've only proved that each subsequence converges, not that they converge to the same value.

I always solve such infinitely nested expressions f(f(f(...))) by letting x = f(f(f(...))), then noting f(x) = f(f(f(f(...)))) = x and solving that equation. In this case, f(x) = x = (1/4 - x)^2, which has solutions x = (3/2 +- sqrt(2))/2, numerically 1.4571 and 0.042893. Discard the larger one because it's obviously too large for the iterated function to converge to (or because plugging it into f(x) = x shows that it doesn't satisfy the equation).

@eroraf8637

Жыл бұрын

Exactly what I did before even watching the video. I think I’ve watched too many videos involving the Golden Ratio.

@aaronson4340

Жыл бұрын

"Discard the larger one because it's obviously too large ... or because plugging it into f(x) = x shows that it doesn't satisfy the equation" Really?

@keyboard_toucher

Жыл бұрын

@@aaronson4340 I made a mistake when evaluating f(x) = x for the larger root, so it turns out that isn't a reason to reject it as a solution. The method I stated does not provide a reason to reject either solution. I don't remember why I didn't like the larger root when I did this work. It may have begun with expecting a unique solution and wanting x 1/4, but seeing (1/4 - x)^2 primes you to expect x 0, since if x > 1/4, it would have been more natural to write (x - 1/4)^2.

@erichgustphysicslectures6858

Жыл бұрын

@@keyboard_toucher Look at it as a fixed point iteration. It is an unstable fixed point of f(x) = x because |f'(x)| > 1. The associated sequence will not converge to an unstable fixed point.

@sussybawka9999

Жыл бұрын

@@keyboard_toucher when plugged into the original equation the larger root gives -sqrt(2)

Around 6:00 induction is not needed to show a_n > 0. We already know a_n = 0 So a_(n+1) = 1/4 - (a_n)^2 >= 1/4 - (1/4)^2 = 3/16 >= 0. That along with the starting value a_0 = 1/4 >= 0 completes the proof. It is not necessary to assume a_k >= 0 to prove a+(k+1) >= 0, which is what the induction step would do.

In one step, you decompose the sequence into two monotone bounded subsequences, which converge. Don’t you also need to show they converge to the same thing to conclude the original sequence converges?

@sjswitzer1

7 ай бұрын

The only thing that needs to be shown is convergence. If the two sequences converge, their sum does too. However symmetry considerations show the must be the same.

Set s = (1/4-(1/4-(1/4- ...)^2)^2)^2 Since s is nested infinitely within itself: s = (1/4 - s)^2 = (s - 1/4)^2 = s^2 -1/2s + 1/16) Subtracting s from both sides: 0 = s^2 - 3/2s + 1/16 yielding quadratic roots s = [ 3/2 + - Sqrt(2) ] / 2 Substitutite into the original expression for Sqrt(2) and using the negative option for the pair of roots Sqrt(2)=3/2 - 2s=3/2 - 2( [3/2 - Sqrt(2) ]/2) = Sqrt(2)

@jcsjcs2

Жыл бұрын

That's what I did -- but I had a hard time justifying that only the negative option is valid. I hand-waved that x must be smaller than (1/4)², but that didn't feel quite right. Did you find a better justification?

@skylardeslypere9909

Жыл бұрын

That is skipping the most important step, namely proving that s actually exists. You can't assume that.

@nozack5612

Жыл бұрын

@@skylardeslypere9909 True, not a proof of convergence, but getting the 'correct' answer is a strong indicator of convergence.

@ProfessorDBehrman

Жыл бұрын

I like your solution. I did it slightly differently, setting your s = y^2. Then take square root of both sides to get: y = 1/4 - y^2 , taking the positive root. Then solve by completing the square, again taking the positive root.

@skylardeslypere9909

Жыл бұрын

@@nozack5612 In this case it is, because we all know there is a solution. Take a(n) = (-1)^n. Then we have the recursive relation a(n) = -a(n-1). If we just assume it has a limit then L = -L so L=0. There are probably more examples that aren't as obvious, but I'm just trying to make a point.

hey michael, i'm a physicist who really enjoys your channel, and i'd love a video on special relativity!

@JosephSullivan777

Жыл бұрын

This comment needs to be higher imo. A topic as difficult to conceptualize as pure mathematics discussed by someone very talented at communicating and thinking. Yes please.

I really enjoy watching the various strategies you employ in your proofs/solutions.

“I was looking through the Wikipedia page of the square root of two…as one does…” Thank god it’s not just me.

"...as one does."🤣 Michael Penn's sense of humour is drier than the Sahara.

This was a really nice problem. I did it slightly differently. Define a function that will be applied to all elements of a set to derive a new set. Like this: f(x) = (1/4 - x)² f : S[n] → S[n+1] where S[0] = (0,1/4) Then claim: S[n+1] ⊊ S[n] ∀ n ≥ 0 (*) Prove it by first breaking the claim down into two points: 1.) S[n+1] covers a smaller range than S[n] 2.) S[n+1] contains no elements that aren't in S[n] Proof of 1.) let h = a-b where 0

12:50 OK, we've shown that the even sub-sequence starts at .25 and decreases, but never goes below 0. And we've shown that the odd sub-sequence starts below a0 and increases, but never above .25. Thus both subsequences converge. However, we haven't shown that they converge to the *same* value. This does *not* follow from both sub-sequences being defined by the same recurrence. As a counterexample: the bifurcation of the logistic map. This is a very similar recurrence that, for certain parameters, can converge to 1 value, converge to any number of alternating sub-sequences, or become chaotic. Also, this sequence method is not specified by the original problem. We can get to the same quadratic solution *without* the sequences at all, so the other solution remains valid (which produces -√2 instead of √2).

This was very nice, thanks Prof Penn!

Special relativity video suggestion: It would be really cool to see a video on the relationship between special relativity, hyperbolic geometry, and the Lorentz group from the perspective of algebra. I’ve studied the Lorentz group in physics but I’ve never seen much on the algebra behind it.

@edwardlulofs444

Жыл бұрын

I'm a big fan of hyperbolic geometry. There is a unified field theory (causal fermion system) that appears to indicate that the universe has hyperbolic geometric shape. There is one KZread video that seems like a very good visualization of this shape. And all of this has left me with the idea that if our 4 dimensional (hyperbolic) universe was embedded into a higher dimension, maybe every point in our 4 D space would be adjacent to every other point in our 4 D universe if you could go through an infinitesimal traversal distance through the 5D space. It's just an idea, and so I'm probably wrong. So if a mathematician says that I am wrong, I'll concede.

@SimonClarkstone

Жыл бұрын

I like how this means the rule for combinig tanh x and tanh y to find tanh (x+y) is the same as velocity addition formula from Special Relativity.

@AsdfAsdf-uo1rj

Жыл бұрын

Coming from Complex Analysis and linear algebra, the Lorentz group is isomorphic to the group of Mobius transformations of the Riemann Sphere, and the various preservations of physical quantities under them are equivalent to the things that Mobius transformations preserve under their actions. There's a bijection between Minkowski Spacetime and 2x2 Hermitian Matrices, and so it's not hard to equivalently define things in one in terms of the other.

Each of the nested series is identical when all the way to infinity is computed, so let's assume they are equal to S. Then, S = (1/4 - S)^2 => 16 S^2 - 24 S + 1 = 0 (that's a quadratic). Therefore, S = (12 ± √(144 - 16 = 128))/16 = (3 ± 2√2)/4 = (3/2 ±√2)/2 => 2S - 3/2 = ±√2 (Hence, proved)

@TheEternalVortex42

Жыл бұрын

You have to prove it converges first. Otherwise this kind of thing can go wrong, such as the famous example 4 = x^x^x^x^…

"as one does" lol

Good video. I think I finally got my head around it.

I was having a hard time understanding how two subsequences could have different behaviors(one increases and one decreases) but the more I thought about it, I eventually realized that (-1)^n/n is a simple example of the same concept.

@DeanCalhoun

Жыл бұрын

precisely. the even terms are greater than the limit and decreasing to it, and the odd terms are less than the limit and increasing to it

@jeffreylantz546

Жыл бұрын

What about a sequence 1/2, -3/4, 7/8, -15/16, 31/32, etc? It can be written as a_1 = 1/2, a_n = -a_{n-1} - (-2)^(-n)

@GreenMeansGOF

Жыл бұрын

@@jeffreylantz546 Equivalently, (-1)^n(1-1/2^n). However, this example does not work since it oscillates and does not converge.

@jeffreylantz546

Жыл бұрын

@@GreenMeansGOF my point exactly. It is a sequence defined recursively that is bounded, the even terms are monotone decreasing, the odd terms are monotone increasing, so both subsequence converge. This is the exact same as in the video until this point, except we can see these subsequences converge to different values, which Prof. Penn doesn't account for in his video.

@GreenMeansGOF

Жыл бұрын

@@jeffreylantz546 Oh my! He claimed that since both sequences used the same recursion, then the limits matched. However, you’ve proven him wrong. Someone please fix the mistake!

12:40 the statement is not true. For example, consider : 1, -1, 1, -1, ... (or a_n = (-1)^{n+1}) Both odd terms and even terms are bounded and monotone, so both of them will converge. They satisfy the same recursion a_n = - a_{n -1}, but they are not converges to the same limit.

@fullfungo

Жыл бұрын

Yeah, the statement on its own is false. But for this video it can be easily fixed. We can say that a_(2k) converges (to B) and a_(2k+1) converges (to C). Then, by the same logic as in the video, we conclude that B= (-1+r2)/2 and C = (-1+r2)/2. Therefore, B=C. So the limit exists.

@xxxx015

Жыл бұрын

@@fullfungo sir you dropt this 👑 I just noticed that you replied to most of the confused people in the comment section and i am just happy to see smart people like you contributing to the world

@fullfungo

Жыл бұрын

@@xxxx015 Thanks. I love helping in math video comments whenever I understand the topic.

Although it is presented here as self-evident, is there a formal theorem that if you can partition an infinite sequence into several infinite sequences, and each of these partitions converges to the same limit, then the original sequence converges (and does so to that limit)? Does it have a name?

@skylardeslypere9909

Жыл бұрын

Let L be the limit of a(2n) and the limit of a(2n+1). We know they converge to the same limit. Let ε>0. Because we know a(2n) and a(2n+1) converge to L, we can find some N₁, N₂ such that |a(2n)-L| N₁ and |a(2n+1)-L| N₂. Now let N = max{N₁, N₂} and let n>N. If n is even, we know |a(n)-L| N₁. If n is odd, we know |a(n)-L| N₂. Thus, we can conclude that for all n>N, we have |a(n)-L| This means precisely that L is also the limit of a(n).

That is GREAT!! You are genius!

Do we have an error by 12;48? The idea that "since they obey the same recursion they converge to the same number" appears to be evident only if the series converges in the first time, and that is what we were trying to proof, so it seems we have a circular reasoning error. I will say it's not clear that it's monotone (unless it converges) . Am I missing something? A very beautiful proof.

15:57

@Milan_Openfeint

Жыл бұрын

In the good old times, it would be 15:47

Doing a bit of math before watching using my knowledge of dynamical systems : (3-2√2)/4 x=(1/4-x)² = (x-1/4)² x-1/4=(x-1/4)²-1/4 X=x-1/4 X=X²-1/4 So X is the stable fixed point of Qc(x)=x²+c with c=-1/4. It's between -3/4 and 1/4, so every bounded orbit converges to its negative fixed point. That means the solution is X=(1-sqrt(1-4(-1/4)))/2=(1-√2)/2 x=X+1/4=(3-2√2)/4 I also did the long way of solving directly for x and checking the derivatives of each, but then it hit me. Edit : Wow, different answer, but it's correct too, all from that initial recursion definition. I included the squaring step at the end, he did it at the beginning. (-1+√2)²/4 = (1-√2)²/4 = (3-2√2)/4

It looks like the point of this identity is not just that it approximates the square root of two but that it approaches the limit very quickly

My first idea would be to show that the sequence converges, perhaps by the contraction argument of Massimiliano Martinelli, and then to show that the expression is positive and algebraically indistinguishable from root 2, namely that its square is 2.

@tiripoulain

Жыл бұрын

Nevermind. Once you know it converges it’s fairly easy to see that the expression evaluates to root 2.

I, too, partake in such avocations in my free time.

The two converging subsequences may converge to different limits since it was not proven that the original sequence converges.

Move terms around to get ((3/2)-x)/2=The iterated square Let u=((3/2)-x)/2 Take the square root of both sides sqrt(u)= (1/4) - iterated square But u=iterated square So we get sqrt(u)=(1/4)-u Square both sides u=(1/16)-(1/2)u+u^2 Multiply both sides by 16 and rearrange 0=16u^2-24u+1 This is a simple quadratic in u, solutions are 3/4 plus or minus 2sqrt(2)/4 But u must be non-negative, so u=(3+2sqrt(2))/4 Undo our substitution ((3/2)-x)/2=(3+2sqrt2)/4 -x=sqrt(2) x=-sqrt(2) I've done this twice and gotten -sqrt(2) both times. Must have missed a sign somewhere? st have missed a sign somewhere?

From the initial formula we can see that if a value exists, it has to satisfy (1/4-x)^2=x, which gives solution x=3/4 +/- sqrt(2) /2. This gives 3/2 - 3/2 +/- sqrt(2) = +/- sqrt(2) for the value of the original expression. Near the end of the video I think there was a mistake, because (-1+sqrt(2)^2 is 5-2sqrt(2) and not 3-2sqrt(2).

@ernestzurek5884

Жыл бұрын

its 1^2 - 2sqrt(2) + sqrt(2)^2 so 3-2sqrt(2)

@koenth2359

Жыл бұрын

@@ernestzurek5884 you're right, my bad, sorry.

I didn't understand in this one how to carefully show that if you have a bounded, sequence and you can split it into two subsequences based on even and odd indexes of the terms, which each are monotone and bounded and hence converge, that the limits of the subsequences match the limit of the overall sequence

this was a fun one

I can't really find a good way of thinking about this one. Gonna have to play with it some more.

Does this converge faster than other numerical algorithm to compute sqrt(2)?

Let x = the infinite expression squareroot both sides then x^1/2 = 1/4 - x, set y = x^1/2 then y^2 + y - 1/4 = 0, y = (-1 +/- sqr(2)/2 = x^1/2 square both sides, x = ((-1 +/- sqr(2))^2)/4 = (1 +/- 2sqr(2) + 2)/4 = (3 +/- 2sqr(2))/4.

a simpler proof: let x be the expression inside the square then x=1/4-x² so x=-1/2 ± 1/sqrt(2) then 3/2-2x² gives sqrt(2)

Very nice 🤘🏻🤘🏻👏🏻👏🏻

This is a nice review of some real analysis.

Is it correct to assume that this could be generalized that for any n >=2: sqrt(n) = (n + 1)/2 - 2 * ((1 - n) / 4 - ((1 - n) / 4 - ....)^2...)^2? Just naively replacing -1/4 with (1-n)/4 seems to work out...

Fixed point theorem: Suppose x = g(x) and for x in [a,b], g is defined and continuous and g(x) in [a,b]. Then g has at least 1 fixed point in the interval. In this case, assume x in [0, 1/4] and g(x) = 1/4 - x^2, so g exists, is continuous, and g in [0, 3/16] is within [0, 1/4], so at least 1 solution in [0, 1/4]. Further, if on [a,b], |g'(x)|

Definitely do a special relativity video

"...and so on, and so 1/4th..."

A video on special relativity would be really nice!

I solved the equation by first finding that nested square thingy, (since its iterated) I set ((1/4) - x)^2 = x, and then solved for x, (which ended up being a quadratic) and got 3/4 (+-) sqrt(2)/2, and then just did the rest of the equation out there. Everything cancels out and you get sqrt(2)

Yes on spec-rel!

suppose a_n converges to x then x^2 + x - 1/4 =0 so x = [√2 - 1]/2 => 2x + 1 = √2 => 2(1/4 - x^2) + 1 = √2 => √2 = 3/2 - 2x^2

I'm curious to how something like that was postulated in the first place to be then proven. Someone would have to take √2 apart and into that kinda sequence. That's the process I'd be very interested in learning.

Pls do an episode on special relativity with an emphasis on time phenomena.

Can you analyze a mass on a spring in simple harmonic motion but taking special relativity into account?

@koenth2359

Жыл бұрын

Since it involves accelerations, general relativity would be needed.

@landsgevaer

Жыл бұрын

@@koenth2359 Not per se. SRT can do accelerations in principle by transitioning to different coordinate systems continuously (e.g. the twin paradox). en.m.wikipedia.org/wiki/Acceleration_(special_relativity) But the *mass* part is the tricky bit. You would indeed get gravity waves. Maybe you mean the same idea though.

Special Relativity would be a good topic.

This one gets much more interesting: explore every real a_0

Thank you Prof. Penn! That was something I never have seen before. I love how irrationality appears from rational numbers.

Please cover special relativity

I let x = the entire iterated square, and noted that x = (1/4 - x)^2. From there a bit of algebra and the quadratic formula got me to x = (3+-2sqrt(2))/4, and the rest fell out naturally.

Wow. I feel like this math problem was a little more unusual than normal. But still I found it interesting and a learning experience.

While I like the math behind this, what must also hold, however is that irrational numbers are numbers that cannot be expressed in the form a/b where a and b are integers. However, the nested 1/4's would be a rational expression. This demonstrates that limits are not the same as equivalencies. This goes to the age-old argument that 0.9999... repeating = 1, which is erroneous. The limit of 0.9999... repeating = 1, which is different than it being equal to 1.

Do a video on special relativity

At 6:24 you say "ak+1 is bigger than or equal to 1/4", but I think you meant to say, "ak+1 is bigger than or equal to 0"

This looks like something Ramanujan came up with

The choice of a_0 = 1/4 is arbitrary. If we instead choose a_0 = (-1 + sqrt(2))/2, then the proof of convergence is a lot easier :)

I was trying to solve this problem on my own from IMO 2022 problem 5: Find all triples of (a,b,p) of positive integers with 'p' prime & a.p=b!+p I tried to solve this with 2 cases (Case 1 a is even, Case 2 a is odd). With Case 1 I am able to find 1 set of solution which is (2,2,2) In the Case 2, a being odd ap is odd which mans b!+p is odd; which can be further divided into 2 sub cases: Sub-Case(a): as b!+p is odd Let b! odd and p be even which gives no solution Sub-Case(b): I cant solve with b! being even and p being odd. I need some help with this or suggestions if I can solve this with this approach.

@TheEternalVortex42

Жыл бұрын

If b If b >= p then you always get a solution by taking a = b!/p + 1

x=(1/4 - x)^2 = x^2 - 1/2x + 1/16 x^2 - 3/2 x + 1/16 = 0 (x-3/4)^2= 1/2 x= 3/4 ± √(1/2) 3/2 - 2x = 3/2 - 3/2 ± 2√(1/2) = ±√(4/2) = ± √2

@-ZH

Жыл бұрын

Let’s see, how to justify where it’s + or -

@-ZH

Жыл бұрын

Depends whether it’s alternation positive negative, or staying positive

@-ZH

Жыл бұрын

√x = 1/4 - x 1/4 - x > 0 x x = 3/4 - √(1/2)

@-ZH

Жыл бұрын

I’m assuming √x ≠ -(1/4 - x) = -1/4 + x Since that changes the nature of the infinite sum, and that doesn’t work, right?

topic request: Heisenberg uncertainty principle

12:49 that both sequences converge is not sufficient to prove that both sequences converge to the same value. Even finding a v=1/4-v^2 doesn't prove that. A simple example: use the same series with a_0=0, it's just alternating between 0 and 1. Odd and even terms each are constant, so obviously both series have a limit, and both are part of the same series, and there's the same v. One could prove that |a_(n+1)-a_n|=(a_n+a_(n-1))*|a_n-a_(n-1)| goes to 0 as n->oo since (a_n+a_(n-1))

One content, two languages. What I have now written may have a perfect mirror in another language.!!!!!!!!!!! Hack in the brain

nice

I would love to see you do a video on relativity. This would be the first time, maybe the only time, to see the topic presented by a non-physicist. I am not a physicist, but I bet many regular viewers are and they should like it too.

Formally, if you call the parentesis by x, then you have x=(1/4-x)^2. then x=3/4+-sqrt(2)/2.

@alexmore3865

Жыл бұрын

But is this correct? It is different from the answer from the video right?

@HeyHeyder

Жыл бұрын

@@alexmore3865 It is only formal...the video is better...

@alexmore3865

Жыл бұрын

@@HeyHeyder I just tried to solve it this way and thought it should be correct))) So I kinda don't know why the answer doesn't match.

@1conk225

Жыл бұрын

@@alexmore3865 I think everything should match. i.imgur.com/aIDEm7b.jpg

@nathanisbored

Жыл бұрын

@@alexmore3865 it's the same as in the video, if you plug that x value back into the given expression. We aren't solving for just x, we're solving for the whole thing

Endless enigmatic book in all languages. You can write a book with mirrors in all languages of the world. You can speak two languages at once, you just need to find the perfect reflection, same content, different translation. Infinite Mirrors. Pi 3.14 XBooks. Hybrid language!!!!!

I can't help it but I think assigning such infinitely nested sequence an initial value is incorrect. It's infinite, there's no innermost element. Any value that matches it is a correct solution. So the approach should rather be, A0 is set of solutions for x=f(x), A1 is set of solutions for x=f(f(x)), and so on. Any value in limit of these sets is valid value for the nested expression.

I didn't understand why even terms are decreasing and odd terms are increasing, since the formula is the same and there's nothing about evenness in the formula. So.. why does this happen?

@ere4t4t4rrrrr4

Жыл бұрын

Also, you claim that they both converge to the same number, because they are defined by the same recursion - but I think that to prove that you would need to prove that this recursion has only one fixed point. This is stronger than what you're proving when proving the sequence converges (because the sequence begins at 1/4 and not some other value), so you can't use this fact in the proof of convergence.

@fullfungo

Жыл бұрын

The formula is the same, but it doesn’t mean the sequence is monotone. For example, a_0 = 1 a_(n+1) = -0.9 * a_n Here we have only one formula, but even elements decrease and odd elements increase.

@tejarex

Жыл бұрын

Given that the initial value is in [0, 1/4], the iteration values alternate because the derivative of (1/4 - x^x) , -2x, is negative on this interval. Draw the graph of 1/4 - x^ on this interval and start with any initial value in the interval and try the iteration and you will see why.

I'd really like to understand the real mathematics behind the frequently presented result that 1 -1 +1 -1 + ... = 1/2. I realize this is using expected values with simplified notation, but don't understand how that can be used to conclude 1 + 2 + 3 ... = -1/12.

@notanotherraptor

Жыл бұрын

Mathologer has a great video on this : kzread.info/dash/bejne/nJd_tKmpfcy8hNo.html

@TinySpongey

Жыл бұрын

There's a Mathologer video on this topic which gives an idea of the reasoning behind it.

@nathanisbored

Жыл бұрын

That result uses a different definition of infinite summations than the standard one. Typically infinite sums are defined to equal the limit of their partial sums, but you can invent other definitions too. The one for 1-1+1-... uses something called Cesaro summation which is a somewhat natural extension of the limit idea. You just go one step further and take the limit of the averages or you can do the limit of the averages of averages or something like that, etc. I believe 1+2+3+...= -1/12 uses an entirely different definition of summation called Ramanujan summation, which I think can take different forms depending on some parameter. All of these definitions have motivation behind them, but only the original limit definition plays nicely with regular finite arithmetic when it exists.

@carstenmeyer7786

Жыл бұрын

The new "limit" you want to define is essentially a mapping from a set of sequences to the real numbers. Most likely, you want that mapping to be continuous and linear: If two sequences are "very similar", you expect the new "limit" to return similar values as well. Most likely you also want the new "limit" to follow standard limit rules, a.k.a. linearity. _Functional Analysis_ deals with such mappings, though that can easily lead down many rabbit holes, even for simple cases like the set of bounded sequences^^ The two Mathologer videos do a great job giving hints at what happens behind the scenes in the special case you are interested in.

@synaestheziac

Жыл бұрын

Check out this excellent lecture course by Carl Bender: kzread.info/head/PL43B1963F261E6E47. He starts talking about “summing” divergent series in the 3rd or 4th lecture

Always an error in the last line! Here: Wrong factor 1/2.

42

Can't you just solve it with the equation (1/4-x)^2 = x and eliminate one value? You get from this the same value for the infinite term and its much easier

@fullfungo

Жыл бұрын

You get the same value, but it does not prove that such a value exists. For example, take a_0 = 2 a_(n+1) = (a_n)² If we solve A = A² we get A=0 or 1. However, we did not prove that A exists, so we cannot conclude that the limit is equal to 0 or 1. In fact the limit does not exists, even though a solution to the related equation does.

@YotYot

Жыл бұрын

@@fullfungo Ohhh thanks for your explanation!

Why make it easier, if it can be more complicated.

At a limit point of the infinite thing x = (1/4-x)^2, plug into quadratic formula, solved. Why make. Extra work for yours e of?

I would love to see a crossover with physics! (yes, some of us watch the ads ;) lol)

Square root of 2 you say? Check out the wikipedia page, yes, as one does....😂😂😂

🤯 uhhh, yes - that's exactly how I would do it. Not bad, Dude🤥

What I find so interesting about this is that, at least in this representation, sqrt(2) "inherits" its irrationality from the (1/4 - (1/4 - ...)^2)^2 expression, which appears rational at first glance.

Can you please not write (1\4)² as 1²/4?

2:17 into the video and I'm getting triggered by the fact that you write the square of 1/4 without any brackets, but you write 1^2/4 instead, which is equal to 1/4.

💕💕💕💕💖🎉

There are many ways to define recursive relations for the "..."-term in the thumbnail. Here's another: *x_{n + 1} = ( 1/4 - x_n )^2 =: f(x_n), x_0 := 0 // D := [-1/8; 5/8] complete* We notice *f : D -> D* is a continuous contraction on *D* and differentiable in its interior: *x ∈ D => f(x) ∈ [0; 9/64] ⊆ D* *x ∈ int(D) => | d/dx f(x) | = 2 * |x - 1/4| With these properties *f* satisfies _Banach's Fixpoint Theorem,_ thus the sequence *x_n* converges to a unique limit *x ∈ D.* We can calculate that limit by setting *x = f(x) => x^2 - 3/2 * x + 1/16 = 0 => x_{1/2} = 3/4 ∓ √(2) / 2* The positive case lies outside of *D* and cannot be the limit, leaving *x = 3/4 - √(2) / 2* -------------------- *Rem.:* Notice you could have chosen any initial value *x_0 ∈ D* and still have gotten the same limit. A quick sketch of the graphs of *f(x)* and *x* indicates we could have even chosen any *|x_0 - 1/4| < 1/2 + √(2) / 2*

I need general relativity.... I'm from sri lanka 🙋❤️

@titan1235813

Жыл бұрын

I don't know if Prof. Penn would be able to answer your question, as he is a mathematician, not a physicist. Maybe you should look for that in a physics channel.

RHS is infinitely long. If writing RHS takes infinitely long then you’re never finished so it’s not a valid expression.

@kosterix123

Жыл бұрын

Nice to see university level math. It is what you learn in Analysis A. Induction is taught in computer science 101.

Very, intersting.but.nonpractising!

Proof that right side of equation equals both sqrt(2) AND negative sqrt(2). Let x=the series (1/4-(1/4-...)^2)^2 Therefore, x=(1/4-x)^2 Expand x=1/16-1/2*x+x^2 Make it into quadratic x^2 - 3/2*x + 1/16=0 Get the roots x=(3 + - 2sqrt(2))/4 Plug it back into the original equation 3/2 - 2 * ((3 + - 2sqrt(2))/4) = 3/2-3/2 + - sqrt(2) = + - sqrt(2). In other words the equation equals both sqrt(2) AND negative sqrt(2).

@MrMctastics

Жыл бұрын

He addresses this with the recursive sequence

@forcelifeforce

Жыл бұрын

The right side is not an equation. It is an expression.

@MrMctastics

Жыл бұрын

@@forcelifeforce nobody cares

@borisjerkunica4442

Жыл бұрын

​@@MrMctastics Yeah, I saw that, but look at it from the other side. Since there are clearly at least two "solutions" to the sequence (and maybe more since there is an infinite exponent), there is an error in the proof that the sequence is