This integral goes on forever

[@BriTheMathGuy]

🎓Become a Math Master With My Intro To Proofs Course! (FREE ON KZbin) kzbin.info/www/bejne/aZTdmJl-irGNedU

[@robloxguy1268]

I’m a kid

[@trbz_8745]

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.

[@maxthomas-bland4842]

Its not surprising at all really, the quadratic involving f(x) is the form of the solution that gives phi, x^2-x-1, exactly the same, considering the solution of f(x) was 1+sqrt(1+4x) all on 2 subbing in x as 1 gives phi.

[@thatssomethingthathappened9823]

HOW DO YOU GET THE PHI SYMBOL?!

[@Ghost____Rider]

@@thatssomethingthathappened9823 Greek keyboard I assume

[@targetiitbcse1761]

@@Ghost____Rider ερτθπφδσζψωβηκμλξ here are all the symbols

[@loneranger4282]

@@thatssomethingthathappened9823 Just search up "Phi unicode"

[@dissonanceparadiddle]

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

[@BriTheMathGuy]

Right?!

[@naivedyam2675]

Here it was made manageable by algebra not calculus

[@dissonanceparadiddle]

@@naivedyam2675 this is true.

[@spudhead169]

Algebra can exist alone, Calculus cannot without algebra. Calculus is like a clingy girlfriend.

[@hydropage2855]

This is algebra

[@mintsjams8862]

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.

[@MatteoDolcin-ye8xm]

Same, although I'm not that good at integration yet, i was mostly interested in the algebra

[@josephcohen734]

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.

[@theophilearnould9868]

Could you write it on a paper and make a photo and post the link in this comment it really interestes me but in this form it is unreadable

[@theophilearnould9868]

This is exactly what I was looking for in the comments

[@josephcohen734]

@@theophilearnould9868 Sure. Kinda busy rn but I'll get to it in the next few days. U familiar with the idea of a function that takes multiple inputs? Kinda important to the proof but I know it confuses a lot of people.

[@theophilearnould9868]

@@josephcohen734 maybe not familiar but I know them and know how to use them, and if I have any trouble I'll juste take courses on the internet 😁

[@theophilearnould9868]

@@josephcohen734 thank you 😁

[@jensknudsen4222]

You're the king of exotic integrals.

[@BriTheMathGuy]

I hope you enjoy them!

[@jensknudsen4222]

@@BriTheMathGuy I sure do! Thanks for making these videos.

[@aashsyed1277]

@@BriTheMathGuy I do!

[@maalikserebryakov]

Bri is decent but no. Maths505 is the King of Integration. If there was a KZbin integration bee im telling you that guy would destroy everyone. Bri, blackpenredpen, Papa flammy.,.. Even the unhinged Autism of Dr Peyam stands no chance.

[@jtris01]

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)

[@LogosNigrum]

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.

[@uskilla489]

Brooo that was amazing

[@BriTheMathGuy]

Glad you enjoyed it!

[@abelhivilikua8735]

Excellent vídeo dude! Your channel is amazing!!

[@BriTheMathGuy]

Thanks a ton!

[@angelmendez-rivera351]

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.

[@Bennici]

I wonder if you could find an infinitely nested representation of e^(-x²) that could be "solved" by this method.

[@angelmendez-rivera351]

@@Bennici That wouldn't work, because the antiderivative is nonelementary

[@dneary]

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!

[@mickschilder3633]

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

[@lutingchen8958]

bprp had a good video on this: kzbin.info/www/bejne/qHy5dn6upNV1qdU

[@mickschilder3633]

Okay i see, thanks!

[@Happy_Abe]

Great video!

[@BriTheMathGuy]

Glad you liked it!

[@Mehraj_IITKGP]

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 $\frac{5+5\sqrt{5}}{12}\approx1.34836165729158$.

[@justinpark939]

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).

[@mathieuaurousseau100]

At that point both functions are equal though

[@angelmendez-rivera351]

No, he takes away the negative version because, given how the square root radical symbol is defined, he knows for a fact that the function is nonnegative.

[@justinpark939]

@@angelmendez-rivera351 oh, of course! that actually is obvious at first glance

[@jonathandambrosio4628]

That was awesome!

[@BriTheMathGuy]

Glad you liked it!

[@mskiptr]

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)

[@smrpkrl]

great effort

[@anshumanagrawal346]

@@smrpkrl I mean in the video he does some pretty stupid shit

[@jadepinto4321]

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

[@BriTheMathGuy]

Great to hear!

[@BriTheMathGuy]

Another Incredible Integral! kzbin.info/www/bejne/onnMZmaHmteYfqM

[@chispun2]

Well yes, but actually no. If you plug in both functions + - , you get an even better solution, you get the result is equal to one :)

[@chispun2]

Nevermind, I am wrong, the - part of the function is introduced by squaring

[@sktbizmathprogphy433]

Nice video bro.

[@BriTheMathGuy]

Thanks and thanks for watching!

[@Calculusgoat]

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

[@romajimamulo]

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

[@birdieman_1180]

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

[@SlimThrull]

(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.

[@aryagerami9022]

The answer can br written as (-5/6)phi where phi is the golden ratio. Wouldn't be surprised if it was related but I strongly doubt it.

[@colbyforfun8028]

The value of the integrand at x=1 is the golden ratio. I'm not sure how that relates to the sum once you integrate it though.

[@nidhiagrawal3354]

@@aryagerami9022 Why the NEGATIVE ⅚. I think it should be POSITIVE ⅚.

[@vezulykamarari]

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

[@logan9093]

You'll get there! We all were there at one point

[@bowenjudd1028]

Which grade are ya in? The way you said your comments suggests you’re in a gifted class.

[@vezulykamarari]

@@bowenjudd1028 I’m finishing 11th while taking ap classes but I promise I am not gifted lol

[@bowenjudd1028]

@@vezulykamarari, OK, that makes sense. Calc is fun though.

[@kshitijsharma2200]

Yeah man the recursion part is confusing. The first time I heard about this I was like "you can do that???" This started to make more sense for me when I started finding limits of recursive sequences in calc II. Just hold on.

[@manucitomx]

Easy on the eye and great with the maths!

[@BriTheMathGuy]

Thank you! Cheers!

[@MohamedSHbib]

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

[@Miju001]

If you're referring to the things used for writing, I think it's just a glass pane and regular markers (also the video gets flipped in editing to make it so the writing isn't backwards)

[@MohamedSHbib]

@@Miju001 that makes sense. Thank you Marta!

[@Miju001]

@@MohamedSHbib No problem! c:

[@bot24032]

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.

[@samyaspapa]

That was my logic too. I tried solving it before watching the video and basically followed the same steps, but I chose the negative square root because at x=0, f(x) should be zero.

[@samyaspapa]

On the other hand, the function looks like it should be monotonically increasing. If you try plotting this in Excel (or some computer program), you'll find that the function takes the route of the positive root. I think this is a case where the infinite expression has a discontinuity at x=0. But since this is only a single point, it doesn't contribute to the value of the integral and the evaluation in the video is correct. (But as you see from my other comment, I made the same mistake as you. It was only after watching the video did I look further into the how the function actually acts.)

[@lutingchen8958]

bprp had a good video on this: kzbin.info/www/bejne/qHy5dn6upNV1qdU

[@bot24032]

@@lutingchen8958 have seen that video, it proves that function is not continuous in 0

[@goyaldev09]

Where did u buy that orange glass marker?

[@thatssomethingthathappened9823]

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).

[@thatssomethingthathappened9823]

If x = 1, y = phi

[@lyrimetacurl0]

Most abrupt ending.

[@ankitbasera8470]

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

[@Firigion]

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

[@BriTheMathGuy]

Very Cool!

[@kasuha]

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?

[@kasuha]

Okay I did some estimaties and it really looks like the original function limits to 1 when x gets close to zero. Weird. And the integral is not affected by the function being defined or not at zero.

[@reeeeeplease1178]

@@kasuha maybe consider what happens to values like 0.01 when square-rooted (since it's close to 0) 0.01 = 1/100 => sqrt(0.01) = 1/10 = 0.1 so sqrt(0.01) is way bigger than 0.01 itself and iterating the sqrt function will eventually bring all numbers (except for 0 itself) to 1 16 -> 4 -> 2 -> 1.414... -> .... -> 1 (in the limit) and as demonstrated earlier, the same thing happens for all 0 < x < 1 because if x

[@minhanhnguyenngoc972]

Amazing!! Love your videos a lottt@

[@BriTheMathGuy]

Thank you so much!!

[@kaskilelr3]

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

[@sujalsalgarkar360]

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😂😂

[@4lines633]

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

[@Deeer69420]

I haven’t even learnt calculus yet

[@blidge8282]

= ( 5 /6 ) * phi

[@fengshengqin6993]

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

[@yesminister9511]

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

[@shazullahyusufzai5704]

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

[@synaestheziac]

Answer also equals 5/6 of Phi

[@fahrenheit2101]

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

[@p_square]

It's +ve because our original integrand was sqrt(x *+* sqrt(x *+* sqrt(...))) so it can't be -ve

[@fahrenheit2101]

@@p_square Yes but when was the other solution created? What does it correspond to?

[@geordan6740]

@@fahrenheit2101 The second solution happens because we squared both sides

[@fahrenheit2101]

@@geordan6740 Oh yeah, I should've noticed that. Thanks.

[@albertodiaz364]

holy shit 5/6 of the golden ratio

[@BriTheMathGuy]

Awesome right!

[@calcubite9298]

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?

[@Kapomafioso]

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??

[@samyaspapa]

If you try plotting this in Excel (or some computer program), you'll find that the function takes the route of the positive root. I think this is a case where the infinite expression has a discontinuity at x=0. But since this is only a single point, it doesn't contribute to the value of the integral and the evaluation in the video is correct.

[@lutingchen8958]

bprp had a good video on this: kzbin.info/www/bejne/qHy5dn6upNV1qdU

[@byronwatkins2565]

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

[@nass8899]

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

[@mr.mirror1213]

Legends say he writes backward

[@taemyr]

Write the normal way. Mirror the video.

[@StepanKorney]

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

[@mskiptr]

Or he could be flipping the entire video afterwards

[@stapler942]

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...

[@einsteiniumfusion6658]

Solved this before your solution

[@the_mazer_maker1969]

Is he writing backwards or is the whole video flipped?

[@sandeepkumariaf603]

Super sir

[@BriTheMathGuy]

Thanks!

[@danielbenton5817]

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

[@BriTheMathGuy]

I hope you have a wonderful day!

[@cosmicvoidtree]

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.

[@Seb135-e1i]

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

[@factsheet4930]

5/6 * phi 😉

[@epsilonxyzt]

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

[@2-_]

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

[@NerdWithLaptop]

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

[@NecmettinMarmara-Fizikci]

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

[@subhoghosal7]

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

[@emilpysenisoncrack420]

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

[@alphalunamare]

This stinks to high heaven!

[@yamahantx7005]

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.

[@alphadragonn3685]

The roots are what make a function equal to zero to he rearranged into a quadratic that equals zero because then solving for the root also solves the equation. The function he got is what satisfied his equation therefore it was also the root of the equation since the equation was equal to 0

[@Fillipor]

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

[@gonzalomorislara8858]

Don't you think that mirroring the image in post would be much easier?

[@Fillipor]

@@gonzalomorislara8858 exactly

[@BriTheMathGuy]

I do flip the video during editing :)

[@arcioko2142]

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

[@bodhisatwakundu2226]

This is quite easy tho

[@0xN1nja]

this was easy tho

[@akashkushwaha4559]

Love it I am indian 🇮🇳🇮🇳

[@BriTheMathGuy]

Glad you enjoyed it!

[@enejidjsi5939]

Nationality doesn't matter? Why are you telling us this?

[@akashkushwaha4559]

@@enejidjsi5939 it's my identity any problem

[@enejidjsi5939]

@@akashkushwaha4559 no but it's very weird you have to insert your "identity" into every unrelated topic

[@akashkushwaha4559]

@@enejidjsi5939 what's this behaviour bro not fair it's democracy

[@janekmuric]

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?

[@theophilearnould9868]

I would also like to know !

[@robloxguy1268]

I’m a kid

[@id3655]

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

[@justintroyka8855]

I disagree. I think his lecture style is quite natural and engaging.