AN UNBELIEVABLE INTEGRAL!!!

It's amazing how an infinitely nested integral like this is connected to a special constant like (sorry no spoilers 😂). Enjoy the video.

[daddy_myers]

It's crazy just how φ came to be this fairy-esque constant among this community; not gonna lie, I like it.

Great video, audio was top-tier!

[The_Shrike]

I love how you paused after the “well-rooted” pun just to let it sink in haha

[lexinwonderland]

WOW! That was such a cool result indeed!! And the pun at the beginning "well-rooted" got an audible chuckle out of me. Awesome, dude!

[zunaidparker]

One rigorous step that you should have mentioned (can be left as homework for the viewer) is to prove that the nested radicle actually converges.
Actually this gives me an idea for a new category of video: convergence proofs of interesting sequences and limits. A bit of variety from integrals and infinite sums, but with the same level of cleverness and entertainment value.

[MathOrient]

I wanted to say that you just have to find the denominator and then compute the integration, but then I saw you did the same 😄Great job Sir 🙏

[rohitashwaKundu]

It can also be solved by substituting the infinite root/denominator as u and then solving the integral in the u world.
Just you have to be cautious about one thing: that u approaches 1(and not 0) when x approaches 0(the lower limit).
And you'll get the exact same answer then.

[grafrotz]

referring to your handwriting: What is the differernce between the letter zeta and the integral sign?

[rk]

Sweet. Never saw that coming. Also. Yes, clever convergence proof please.

[adityasingh]

“You've got such an amazing work🙂

[RobertGabor]

Root of substitution Zeta on positive x=1 = golden ratio. (1 +/- sqrt( 1-4x ))/2, where x = 1, Could it be generalized as limits of integration od Zeta?

[ІгорСапунов]

How can we see that infinitely nested roots functional sequence converges to (1+1sqrt(1+4x))/2 uniformly? We have to proove this fact, otherwise, we will not be able to figure this integral out

[MrWael]

savage Integral, but the talent makes it domestic. Thank you very much for your effort.

[CM_France]

Ok, great!

When I saw this nested radical, I guessed there were some thing to do whith the golden ratio! Very nice!

But please finish writing the expressions before eliminating or simplifying them, it would be better.

I guess this comes from φ = sqrt(1+sqrt(1+sqrt(1+...)), can we do something similar with φ = 1+1/(1+1/(1+1/...)) ?

[floatingturtle]

Bruh this is crazy. I literally did a problem similar to this where i differentiated a continued fraction function and came to a similar result

Next solve a differential equation involving nested derivatives of increasing order or show that it is unsolvable

[아이고이거참]

The function of the denominator looks continuous at zero. In particular, the value at 0 is probably zero. However, the derived function is non-zero at zero. It's really confusing. haha

[CryToTheAngelz]

I would write the final answer as 1/φ+ln(1/φ)+1/φ

[cryptinum]

Totally agree with the statement ‘Math is a lot easier than English’😵‍💫

[kewalmer]

I solved it in my head, an easy yet a satisfying question

[adamhe]

x=z^2-z, dx=(2z-1)dz. Substitute z and dx into the integral, the integration could be more straight forward.

[angelmendez-rivera]

Technically, the integral as presented in the thumbnail is undefined, because the domain of integration presented is the closed interval [0, 1], but 0 is not in the domain of the integrand function. Instead, what we can do is integrate on the interval [t, 1] for 0 < t =< 1, and let t —> 0 at the end. On that note, we actually need to rigorously define the integrand here. What we need to consider the functions f[n] : (0, ∞) —> R, such that f[n + 1](x) = sqrt(x + f[n](x)) for all natural n, and all x in the domain. Notice that if g[x] : (0, ∞) —> R, g[x](s) = sqrt(x + s), where x is in (0, ∞), then simply f[n + 1](x) = g[x](f[n](x)). This means that for all natural n, f[n](x) = (g[x]^n)(f[0](x)). Here, f[0] has not been specified at all, so the requirement for the integrand to be well defined is that lim (g[x]^n)(f[0](x)) (n —> ∞) = L *for all* choices of f[0] possible.

Notice that g[x](s) = sqrt(x + s) = s = (x + s) – x = sqrt(x + s)^2 – x, which is equivalent to sqrt(x + s)^2 – sqrt(x + s) – x = 0, which is equivalent to sqrt(x + s)^2 – sqrt(x + s) + 1/4 = (sqrt(x + s) – 1/2)^2 = x + 1/4, which is equivalent to sqrt(x + s) – 1/2 = sqrt(x + 1/4), which is equivalent to s = (1/2 + sqrt(x + 1/4))^2 – x = 1/4 + x + 1/4 + sqrt(x + 1/4) – x = 1/2 + sqrt(x + 1/4). This is the fixed point of g[x]. If 0 < s < 1/2 + sqrt(x + 1/4), then g[x](0) = sqrt(x) < g[x](s) < 1/2 + sqrt(x + 1/4). In general, this means (g[x]^n)(0) < (g[x]^n)(s) < 1/2 + sqrt(x + 1/4). If h[x](n) := (g[x]^n)(0), then for all x in (0, ∞), we know that h[x] is a monotonically increasing sequence bounded from above by 1/2 + sqrt(x + 1/4), and therefore, it converges. It turns out that h must therefore converge to the fixed point 1/2 + sqrt(x + 1/4). Now, by the squeeze theorem, it follows that for all 0 < s < 1/2 + sqrt(x + 1/4), lim (g[x]^n)(s) (n —> ∞) = 1/2 + sqrt(x + 1/4). Using an analogous analysis, a very similar proof with monotone convergence applies if 1/2 + sqrt(x + 1/4) < s, and so, for all s in the domain of g[x], lim (g[x]^n)(s) (n —> ∞) = 1/2 + sqrt(x + 1/4). This is key, because this means that for all f[0], lim (g[x]^n)(f[0](x)) (n —> ∞) = 1/2 + sqrt(x + 1/4), as long as f[0](x) > 0, which is required anyway for the recursion presented earlier to be well-defined.

This means that the integrand function is j : [t, 1] —> R such that j(x) = 1/(1/2 + sqrt(x + 1/4)). Here, one can substitute x + 1/4 |—> x, so that [t, 1] |—> [t + 1/4, 5/4], and 1/(1/2 + sqrt(x + 1/4)) |—> 1/(1/2 + sqrt(x)) = 2/(1 + 2·sqrt(x)) = (2·sqrt(x))/(1 + 2·sqrt(x))·1/sqrt(x). This sets up 2·sqrt(x) |—> x, so [t + 1/4, 5/4] |—> [2·sqrt(t + 1/4), sqrt(5)], and (2·sqrt(x))/(1 + 2·sqrt(x))·1/sqrt(x) |—> x/(1 + x). By this point, the integrand has become such that t = 0 is possible, and so taking t —> 0, as we said in the beginning, we now can simplify so that the domain of integration is [1, sqrt(5)]. Notice that x/(1 + x) = ((1 + x) – 1)/(1 + x) = 1 – 1/(1 + x), which antidifferentiates to x – ln(1 + x) + c, so the integral is equal to sqrt(5) – ln(1 + sqrt(5)) – 1 + ln(2) = sqrt(5) – 1 – ln((1 + sqrt(5))/2) = 2/φ – ln(φ)