This integral goes on forever 

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So it's 5φ/6? Interesting that phi shows up here, although it sort of makes sense because phi likes to show up in Fibonacci-esque problems with sums and recursion.

You got to love how calculus is able to deal with the infinite and make it manageable

Wonderful video! I got the same answer in a different and admittedly slower way. I set y equal to the function then noticed that y = sqrt(x + y). From there, you can isolate y by completing the square and you get y = sqrt(x+ 1/4) + 1/2. From there, it's smooth sailing to integrate.

You're the king of exotic integrals.

Anybody have a proof for why the infinite thing converges at all for x in that range? Technically, before he showed what the infinite thing equals, he first needs to prove it equals a finite number at all. I used my computer to estimate it, and values seem to converge pretty quickly for all positive x that I tried (including x = 1000), but that's not a proof. Edit: I came up with a proof. It's hard to read as a youtube comment but I assume no one will read it anyways. part 0: simple setup 1. Define a function T(n,x) = sqrt( x + T(n-1, x) ), and T(1,x) = sqrt(x) 2. Assume x is a real number & x >= 0 part 1: T(n,x) is increasing as n increases for all n 1. show if T(n,x) > T(n-1, x) then T(n+1,x) = sqrt(x + T(n,x)) > T(n,x) = sqrt(T(n-1,x)). This is true for all x >= 0 2. show T(2,x) = sqrt( x + sqrt(x) ) > T(1,x) = sqrt(x). This is true for all x > 0 Point 1 basically means that if T ever increases as n increases, then T will continue increasing forever as n increases. Point 2 shows that as n increases from n=1 to n=2, T(n,x) increases (unless x = 0, which we will get to later) Points 1 & 2 together show that if x ≠ 0, then T(n,x) is increasing as n increases for all n. part 2: if T(n,x) =< 1/2 + sqrt(x + 1/4) for all (n,x) in the domain 1. Assume there is a value (n,x) such that T(n,x) >= 1/2 + sqrt(x + 1/4) T(n,x) > 1/2 + sqrt(x + 1/4) T(n,x) - 1/2 > sqrt(x + 1/4) (T(n,x) + 1/2)^2 > x + 1/4 (T(n,x))^2 - T(n,x) + 1/4 > x + 1/4 (T(n,x))^2 - T(n,x) > x (T(n,x))^2 > x + T(n,x) T(n,x) > sqrt(x + T(n,x)) From the definition, we know sqrt(x + T(n,x)) = T(n+1,x) T(n,x) > T(n+1,x) Make the substitution 2. This means that if T(n,x) > 1/2 + sqrt(x + 1/4), then T(n+1,x) < T(n,x). This would mean that T decreases as n increases, which is impossible as shown in part 1. Therefore, T(n,x) =< 1/2 + sqrt(x + 1/4) for all (n,x) conclusion: T(n, x) must converge to a number 0 1. part 1 shows that T(n,x) is increasing for all n if x > 0 2. part 2 shows that T(n,x) is never greater than 1/2 + sqrt(x + 1/4) 3. this means that T(n,x) must approach a number less than or equal to 1/2 + sqrt(x + 1/4) as n goes to infinity. if x < 0 What if x = 0? Then T(n,x) = 0 for all n, so T(n,x) "approaches" 0 as n goes to infinity 1. if T(n-1, 0) = 0, then T(n,0) = sqrt( 0 + 0 ) = 0 2. T(1) = sqrt(0) = 0 3. points 1 and 2 together show that if x = 0 then T(n,x) = 0 for all n Real conclusion: limit as n -> infinity of T(n,x) = a finite number if x >= 0.

I think the minus on the plus or minus is valid only when x is 0, which happens to be the lower limit of the integral.

Brooo that was amazing

Of course, this all assumes that the sequence defined by x(n + 1) = sqrt[x + x(n)] converges for appropriate values of x(0) and x.

Me thinking I'm smart for taking pre-calc: Also me not knowing a god damn thing in this video:

f x = √(x + √(x + …)) -- we can rewrite it as the follwing: f x = fix g where g y = √(x + y) -- `fix g` is the fixpoint of g -- (also notice, that `x` is a known value at this point) fix g = g (fix g) fix g = √(x + fix g) -- now, it's pretty important to decide, what are the domain and range of the square root. -- if it's a function and we want to stick to reals, then both better not be negative g : ⟨-x,∞) -> ⟨0,∞) -- this also narrows down the value of `fix g` fix g : ⟨-x,∞) ∩ ⟨0,∞) fix g : ⟨max 0 (-x),∞) (fix g)² = (√(x + fix g))² (fix g)² = |x + fix g| (fix g)² = x + fix g -- cannot be negative (fix g)² - fix g - x = 0 fix g ∈ {(1-√(1+4x))/2,(1+√(1+4x))/2} -- or let's the `±` give us a set fix g ∈ (1±√(1+4x))/2 fix g ∈ ½±√(¼+x) -- this also means, that if we want to stay in real numbers, we need to constraint `x` x : ⟨-¼,∞) -- now it gets pretty tricky, because we can get various values of `x` and for each there might be zero, one or two different fixpoints satisfying all the conditions -- let's try to inspect some cases individually case x of -¼ -> fix g ∈ ½±√(¼-¼) fix g ∈ ½±0 fix g ∈ {½} ½ ∈ ⟨max 0 ¼,∞) ½ ∈ ⟨¼,∞) fix g = ½ x : (-¼,0) -> fix g ∈ ½±√(¼+x) -- again, we can consider two cases case fix g of ½-√(¼+x) -> ½-√(¼+x) ∈ ⟨max 0 (-x),∞) ½-√(¼+x) ≥ max 0 (-x) ½-√(¼+x) ≥ -x ¼+x - √(¼+x) + ¼ ≥ 0 4(¼+x) - 4√(¼+x) + 1 ≥ 0 (2√(¼+x) - 1)² ≥ 0 -- and this is always true, thus fix g = ½-√(¼+x) -- is consistent ½+√(¼+x) -> ½+√(¼+x) ∈ ⟨max 0 (-x),∞) ½+√(¼+x) ≥ max 0 (-x) ½+√(¼+x) ≥ -x ¼+x + √(¼+x) + ¼ ≥ 0 4(¼+x) + 4√(¼+x) + 1 ≥ 0 (2√(¼+x) + 1)² ≥ 0 -- and this is also true, thus fix g = ½+√(¼+x) -- is also consistent 0 -> fix g ∈ ½±√(¼+0) fix g ∈ ½±½ fix g ∈ {0,1} case fix g of 0 -> 0 ∈ ⟨max 0 (-0),∞) 0 ≥ max 0 (-0) 0 ≥ 0 -- thus fix g = 0 -- is consistent 1 -> 1 ∈ ⟨max 0 (-1),∞) 0 ≥ max 0 (-1) 0 ≥ 0 -- thus fix g = 0 -- is also consistent x : (0,∞) -> fix g ∈ ½±√(¼+x) case fix g of ½-√(¼+x) -> ½-√(¼+x) ∈ ⟨max 0 (-x),∞) ½-√(¼+x) ≥ max 0 (-x) ½-√(¼+x) ≥ 0 ½ ≥ √(¼+x) √¼ ≥ √(¼+x) -- `¼+x` is assumed here to not be negative ¼ ≥ ¼+x 0 ≥ x -- contradiction! -- and thus this branch is impossible ½+√(¼+x) -> ½+√(¼+x) ∈ ⟨max 0 (-x),∞) ½+√(¼+x) ≥ max 0 (-x) ½+√(¼+x) ≥ 0 -- but because both ½ ≥ 0 -- and √(¼+x) ≥ 0 -- thus ½+√(¼+x) ≥ 0 -- is always true in this branch, and thus fix g = ½+√(¼+x) -- is consistent -- and this whole tree simplifies to case x of x : ⟨-¼,0⟩ -> fix g ∈ ½±√(¼+x) x : (0,∞) -> fix g = ½+√(¼+x) {- But this is problematic in the following way: Since for some values of `x`, `g` has several fixpoints, there are uncountably many different functions that would fit the original expression for `f`. There are two 'simple' values ``` f : ⟨-¼,0⟩ -> ⟨0,½⟩ f x = ½-√(¼+x) ``` and ``` f : ⟨-¼,∞) -> ⟨½,∞⟩ f x = ½+√(¼+x) ``` but in principle, we could combine these two by picking points from one or the other, or even skipping some of them. -} {- But since the original goal was to calculate the integral of `f` from 0 to 1, there are only two possiilities for this interval. It could just be ``` f x = ½+√(¼+x) ``` or ``` f 0 = 0 f x = ½+√(¼+x) ``` but this has no influence on the intagral, so we can choose either. Of course, in the `⟨-¼,0)` region the function is still pretty undefined, but there's really nothing more we can do about it. So let's finally get the solution: -} I f (0,1) = I (λ x -> ½+√(¼+x)) (0,1) = I (λ x -> ½(1+√(1+4x))) (0,1) = ½ * I (λ x -> 1+√(1+4x)) (0,1) = ½ * I ((λ x -> 1+√x) ∘ (λ x -> 1+4x)) (0,1) = ½ * I ((λ x -> 1+√x) ∘ (λ x -> 1+4x) * 4*¼) (0,1) -- (ugly) shorthand: f*a = λ x -> a * f x = ½*¼ * I ((λ u -> 1+√u) ∘ (λ x -> 1+4x) * 4) (0,1) -- at the core of the 'u substitution' lies regular function composition = ⅛ * I (λ u -> 1+√u) (1+4*0,1+4*1) -- with indefinite integrals, we could just move the composition outside of the integral = ⅛ * I (λ u -> 1+√u) (1,5) = ⅛ * I (λ u -> 1+u^½) (1,5) = ⅛ * ((λ u -> u + ⅔*u^1.5) 5 - (λ u -> u + ⅔*u^1.5) 1) = ⅛ * (5 + ⅔*5^1.5 - 1 - ⅔*1^1.5) = ⅛ * (4 + (10√5 - 2)/3) = ½ + (5√5 - 1)/12 = (5 + 5√5)/12 = 5/12 * (1 + √5)

Here is an alternate way: Substitute $t=\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}$ So $t^{2}=x+t$ This means $2tdt=dx+dt$ Whence, $dx=(2t-1)dt$ Notice in the relation $t^{2}=x+t$ when $x=0$, $t=0$ and when $x=1$, $t=\frac{1+\sqrt{5}}{2}$ Substituting these values in the integral: $\int_{0}^{\frac{1+\sqrt{5}}{2}}{t(2t-1)dt}$ Which evaluates to $\frac{5+5\sqrt{5}}{12}\approx1.34836165729158$.

I was on a similar track, but made a mistake somewhere. I used y instead of f(x), and used implicit differentiation from x = y^2-y to dx = (2y-1)dy - then the integral become int(0 .. phi) y(2y-1)dy - I see now that there are two solutions to the equation y^2-y = 0 and I took y=0 when I should have taken y=1, since lim_(x to 0+) y = 1 - but it's a tricky limit to evaluate!

Excellent vídeo dude! Your channel is amazing!!

this is great! makes it easier to apply myself in my calc classes

y = sqrt(x + y) y² = x + y y² - y - x = 0 ½[1 ± sqrt(1 + 4x)] y = ½[1 + sqrt(1 + 4x)], as y cannot be negative. int_{x = 0}^{x = 1} ½ + ½sqrt(1 + 4x) dx = [ ½x + (1/12)(1 + 4x)^(3/2) ] {x = 0}^{x = 1} = (½ - 0) + (1/12)(5^(3/2) - 1) = 5/12 + (1/12)(5^(3/2)) = 5/12 + 5/12(5^½) = (5/12)(1 + 5^½)

I have a good way to do this (i assume x is a certain values so lets say its 4) sqrt of 4 is 2 and sqrt is 2 is 1.41 so it could go to 1

I don't love that choice though, because x=0 should have F(x)=0, but that's probably a point of discontinuity EDIT: It is a point discontinuity, and can be ignored in integration

That's 5/6 times the golden ratio =D

2:59 I am assuming he is thinking to take away the negative version of the function because he knows that the limits of integration do not fit the conditions for (1-sqrt(1+4x))/2 being always a positive value (it is not negative at x=0, where f(0)=0).

Can someone explain the following to me: in the starting notation of f(x), should we not note that f(0)=sqrt(0+sqrt(0+...)), which should then always equal 0 right? But in our representation we get f(0)=1. I must be missing something but that does not sound right to me

I learned this problem in my 11th grade and I think myself a genius after I know the solution to a Bri The math guy vedio😂😂

(1+Sqt(5))/2 is the golden ratio. The answer we get here has 1+Sqr(5) in it. It's not quite the golden ratio, but it is close. I'm curious if there's a reason that it would pop up in here or if it's just a coincidence.

Great video!

I was on my calculator once and I found that sqrt(x*(sqrt(x*(sqrt(x*sqrt(x*(sqrt(x....... seems to tend towards x, which seems quite odd to me.

But what if x is only nested a finite number of times O.o

Most abrupt ending.

Nice video bro.

= ( 5 /6 ) * phi

Or if you wanted to be extra cliche, you could multiply the original integral by 6/5 so that you could force in a “golden answer” cliche. Glad you didn’t.

That was awesome!

The lower limit is x=0 ,when x=0 the u=0! not 1 ! .Then the intergal result should be (7+5*srq5)/12

But for x=0 the function should be equal to 0, or (1-√(1+4x))/2. So it's not continuos from 0 to 1.

Small correction - putting the ² with the function itself, as in f²(x), means f(f(x)) The only exception is in trigonometry, because you never really have to take a trig function of something twice, and in the rare case where it's somehow useful, you can write out the function twice. That's also why f^(-1) means the inverse of x, because you can use basic exponent rules this way f(x) = f^1 (x) f^(-1) (f^1 (x)) = f^0 (x) The exponent tells you how many times you apply the function. If you apply it 0 times, nothing happens. f^0 (x) = x

I was able to get the answer before watching the video an it feels great

This is wrong because it is not continuous around f(0) since u can tell from the second function that the lim as x tends towards 0 is 1 but when x=0 f(0) is obviously 0. So there is no solution

I haven’t even learnt calculus yet

If you take the derivative of the equation does it give the equation under integral

The 1 term was NOT divided by 4 and its limits should NOT be changed from 0 to 1.

But the key thing before you evaluate this integral is to verify that it is convergent when x is on this interval

This integral, it doesn't end Yes it goes on and on my friend Some people started writing it Not knowing what it was And they'll just keep on writing it forever just because...

holy shit 5/6 of the golden ratio

1 + √5 is less surprising when you realize that this is two times phi, or 2(√(1+√(1+√(1+√…))))

1+sqrt(5). Phee phi phoe phum, I smell the blood of a golden ratio. I wonder if the integral from [0,1] of an infinitely nested square root function has any geometric relation to (5/6)*phi?

Answer also equals 5/6 of Phi

Super sir

Easy on the eye and great with the maths!

One thing that baffles me is that your simplified f(x) is defined on x >= -1/4 but the original function doesn't feel like it should allow negative values. Also, f(0) = 1 while the original function definitely feels like it should be zero at x=0 ... unless we agree that sum of infinite zeros is one. Or choose the negative option you rejected. Was the simplification really valid?

Is he writing backwards or is the whole video flipped?

I look at the square root of (x + ...) part and I set the value to y. I immediately go into my graphing calculator and graph y = sqrt (x + y).

Why (0.5)! = sqrt(pi)/2 ?

Solved this before your solution

1:08 Wouldn't it be more appropriate to write *{f(x)}²*

1:06 isnt it sqrt(a+b) =/= sqrt(a)+sqrt(b)???

Where did u buy that orange glass marker?

Grea videos! Can I ask you what application you are using for your demonstration?

haha, i took time to calculate it myself and it was the same result!

I don't thing neglecting -ve value is not straightened. More rigor needed.

Nice video but finnish false because (5sqr5+7 )/12

5/6 * phi 😉

For those who didn't see it, the answer is also equal to 5Φ/6 Edit: Sorry, maybe everyone saw it.

I don't get it. The original function seems to be clearly zero when you plug in x = 0. However, your result gives (1/2)(1+sqrt(1+0)) = 1. What went wrong??

does it happen for all recursive function to be a way to write it in a non recursive way?

This stinks to high heaven!

I don't get it. The quadratic formula gives you the roots, not a general solution. Why are you plugging the root back in as a function of x? It is cool that you can write backwards, though.

I just realized that he writes everything in the opposite orientation, so that we see everything.

Amazing!! Love your videos a lottt@

Is there a more rigorous reason to choose the positive solution from the quadratic formula?

solving before watching the video: (spoilers, click read more) (sqrt125 + 5)/12, which is equal to 5/6 * Phi

I just realised he must write this backwards so we can see this from the front wow!

do you actually write all this stuff backwards? or how do you make it correct writing it from behind?

This is quite easy tho

Another Incredible Integral! ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lHi53QereHA.html

Why can we apply the quadratic formula here? I think that because x is not constant, but rather depends on f(x), the quqdratic formula shouldn't apply, right?

this was easy tho

I’m a kid

Love it I am indian 🇮🇳🇮🇳

Everything about the way you move and speak looks mechanical. Just be yourself bro, maybe you've convinced yourself otherwise but that doesn't improve "the experience" for the viewer at all