Infinitely Nested Integral Problem

We solve a fun infinitely nested integral problem. Posing the problem in a more rigorous way leads to an interesting discussion of fixed-point iteration.

00:00 What does it mean?
00:30 A neat solution
02:15 More rigorous approach
04:37 A solution using iteration
06:37 More solutions?

[zephyrred]

I wouldn't say we have an infinity otherwise. Consider t=1, then the series is 0, -1, 0, -1... The series is clearly diverging, but we cannot say the limit is equal to infinity, as 0 is the upper bound for the series.

[DrBarker]

Correction:
Thanks for so many insightful comments about exactly when the iteration tends to infinity/oscillates! Firstly, the limit also exists for t = -ϕ and t = 1/ϕ, because you then get ϕ and -1/ϕ, respectively, as your next term, and the limit also exists for other values which give -ϕ and 1/ϕ as their output, and values which give these as their output, and so on.

But if you start with any other t in (-ϕ, ϕ) which doesn't lead back to one of the values above, you get "stuck" in the interval (-1, ϕ), and might "converge" to an oscillation between -1 and 0, or get stuck in a loop (so the limit still doesn't exist). And finally, the iteration tends to infinity if you start with t>ϕ or t<-ϕ.

[sharpnova]

nice problem.

it's always bothered me how when you simply (naively) go at continued fractions recursively and arrive at what is usually a principal branch, a bunch of other convergent fixed points get overlooked.

(this integral is essentially a continued fraction btw)

so you addressed that nicely (albeit incompletely) with your second method

[РођакНенад]

Fun, elegant and easy to follow, you've got yourself a new subscriber! No fancy animations or editing needed.

[anandarunakumar]

Beautiful. Golden ratio in integral form, is a beautiful realization.

[itays]

Well, it t=0 then the integrals alternate between 0 and -1, so the series doesn't have a limit.
Edit: even more generally, we can find a point that has a cycle of n for every natural number n, we just take n iterations of the function f(x) = x^2 - 1 and set that equal to x, that will give us a polynomial with degree 2n, that generally has a couple of real solutions. (It is possible that all the roots are imaginary, and I assume that we work over the reals, but there are at least two real numbers that have have a 2 cycle, one of them is x = 0, and after dividing the polynomial f(f(x)) - x by x, we get a polynomial with degree 3, which always have at least one real solution)
What I'm saying is, there is a discrete set of point, either finite or countably infinite, that if the initial t was was a part of, the series would alternate and the limit wouldn't even be defined.


Other than that, great video, I was afraid that you won't talk about where the expression converges or not, and just assume it converged

[michaelzumpano]

This is the first Dr. Barker video I’ve watched. It was in my youtube recommendations. That was a great topic and excellent narrative. I subscribed and enabled notifications. Now I want to learn more about fixed points and stability.

[jimenezluis]

Excellent video, very interesting problem! this is the principle of fixed point iterations, x[n+1] = f( x[n] ) whose limit is solution to the equation x = f(x). The convergence depends on initial value x0, such that |f ' (x0)| < 1. Here | 2*x0 | < 1 for a starting value of x0 satisfies this condition, then the convergence is guaranteed

[alicesmith]

A familiar problem with an unusual setup! The video was quite a delight, especially for the discussion on whether it's even well defined. I didn't pick up on any of the other solutions noted in the comments, but thank you for presenting an analysis that goes beyond the simple trick in the opening.

[mackenziekelly]

2:00 This is a completely arbitrary conclusion. There's nothing wrong with having the upper limit smaller than lower limit, therefore both answers are completely valid until you do more assessment.

[angelmendez-rivera]

Essentially, what we are doing is studying a family of sequences, specifically, the family of sequences that satisfies the recursion that a(n) is the Lebesgue integral of 2·x on [1, a(n + 1)]. The Lebesgue integral of 2·x on [1, a(n + 1)] is equal to a(n + 1)^2 – 1, by the fundamental theorem of calculus, so we are studying the family of sequences satisfying the recursion a(n + 1)^2 – 1 = a(n). These family is subdivided into two families: the family of sequences satisfying a(n + 1) = –sqrt[a(n) + 1], and the family of sequences satisfying a(n + 1) = sqrt[a(n) + 1]. This can be writen as a(n + 1) = f[a(n)] and a(n + 1) = g[a(n)], with f : [–1, ♾) —> (–♾, 0], f(x) = –sqrt(x + 1) and g : [–1, ♾) —> [0, ♾), g(x) = sqrt(x + 1). This means that the sequences in the f-family necessarily must have a(0) lie within [–1, 0].

The fixed point of f is given by x = –sqrt(x + 1) = (x + 1) – 1 = sqrt(x + 1)^2 – 1, implying sqrt(x + 1)^2 + sqrt(x + 1) – 1 = 0 = sqrt(x + 1)^2 + sqrt(x + 1) + 1/4 – 5/4 = [sqrt(x + 1) + 1/2]^2 – [sqrt(5)/2]^2 = [sqrt(x + 1) + 1/2 – sqrt(5)/2]·[sqrt(x + 1) + 1/2 + sqrt(5)/2] = [sqrt(x + 1) – 1/φ]·[sqrt(x + 1) + φ] = 0. Therefore, sqrt(x + 1) – 1/φ = 0, which is equivalent to x + 1 = 1/φ^2 = 1 – 1/φ, which simplifies to x = –1/φ. If –1 =< x =< –1/φ, then 0 =< x + 1 =< 1 – 1/φ = 1/φ^2, which implies 0 =< sqrt(x + 1) =< 1/φ, which implies –1/φ =< f(x) =< 0. The criterion for convergence occurs if |f(x) + 1/φ| < |x + 1/φ|, which implies f(x) + 1/φ < –(x + 1/φ) or –f(x) – 1/φ < x + 1/φ, which is equivalent to f(x) + x + 2/φ < 0 or f(x) + x + 2/φ > 0. Notice that f(x) + x + 2/φ = x – sqrt(x + 1) + 2/φ = x + 1 – sqrt(x + 1) + 2/φ – 1 = sqrt(x + 1)^2 – sqrt(x + 1) + 2/φ – 1 = [sqrt(x + 1) – 1/2]^2 – [5/4 – 2/φ] = [sqrt(x + 1) – 1/2]^2 – [5/4 + 1 – sqrt(5)] = {sqrt(x + 1) – 1/2 – sqrt[9/4 – sqrt(5)]}·{sqrt(x + 1) – 1/2 + sqrt[9/4 – sqrt(5)]} < 0 if and only if sqrt(x + 1) < 1/2 + sqrt[9/4 – sqrt(5)] and sqrt(x + 1) > 1/2 – sqrt[9/4 – sqrt(5)], and > 0 if sqrt(x +1) < 1/2 – sqrt[9/4 – sqrt(5)] or sqrt(x + 1) > 1/2 + sqrt[9/4 – sqrt(5)]. This implies 1/4 – sqrt[9/4 – sqrt(5)]/2 + 9/4 – sqrt(5) < x + 1 < 1/4 + sqrt[9/4 – sqrt(5)]/2 + 9/4 – sqrt(5), which is equivalent to 3/2 – sqrt[9/4 – sqrt(5)]/2 < x < 3/2 + sqrt[9/4 – sqrt(5)]/2; or it implies x < 3/2 – sqrt[9/4 – sqrt(5)]/2, or x < 3/2 – sqrt[9/4 – sqrt(5)]/2. This implies that for every –1 =< a(0) =< 0, a(n) converges to –1/φ.

The fixed point of g is given by x = φ, and this is easy to see from φ = sqrt(φ + 1). For –1 =< x =< φ, f(x) >= x, since sqrt(x + 1) >= x implies sqrt(x + 1) >= sqrt(x + 1)^2 – 1, which implies 0 >= sqrt(x + 1)^2 – sqrt(x + 1) – 1 = [sqrt(x + 1) + 1/φ]·[sqrt(x + 1) – φ], which only occurs when sqrt(x + 1) – φ =< 0, which implies x =< φ. Conversely, for φ =< x, we have that x =< f(x). So for every –1 =< a(0), a(n) converges to φ.

[TheLukeLsd]

De two values of the relation who the positive result is the golden ratio. Very nice!

[as-qhqq]

The analysis is more interesting than just the original problem.

[PuerinTheHunter]

That was cool! From what I've seen, many of those infinitely nested problems result in ( 1 +- V5 ) /2

[Xammed]

Im so happy I stumbled across this channel
!

[jimschneider]

@9:20 - I(t) does not converge when t is not equal to (1 +/- sqrt(5))/2, but that does not mean that it always diverges to infinity. The iteration implicit in I(t) will cycle between 0 and -1, if t is equal to any of 0, 1, -1, sqrt(2), or -sqrt(2).

[Filip-pzwu]

Nice video, fascinating subject and nice presentation. Subscribed. Hope to see more fun problems like it.

[olbluelips]

Really illuminating, good video! :) I find infinite nested problems like this really beautiful

[aliberro]

Underrated channel
It's amazing!!

[wuyqrbt]

Hello Dear Dr Barker.
It was new; at least for me. Therefore I like it a lot (I want to watch it again).
Well done Dr. Barker
((And also, Let me tell you something about your presentation; your presentation is getting better and better)).