An awesome infinitely nested radical.

[@goodplacetostop2973]

0:16 It’s me :) 13:25 Good Place To Stop

[@MathSolvingChannel]

There is a shortcut to solve it 😉 Here is the link: kzbin.info/www/bejne/rXurcn2tp71rhtk

[@GaryFerrao]

hello Good Place to Stop. good to see you're featuring in Michael's videos. (hey wait a minute…)

[@goodplacetostop2973]

@@MathSolvingChannel Oh really? Post it on your channel then 😀

[@goodplacetostop2973]

@@GaryFerrao 😂

[@MathSolvingChannel]

@@goodplacetostop2973 Sure, this is a general method, and you can use this method to create your own infinite nested sqrt puzzle :) Here is the link: kzbin.info/www/bejne/rXurcn2tp71rhtk

[@a52productions]

It's very neat how using inequalities and applying absolute values makes it seem like you'd be losing information, but it's what makes the solution possible at all

[@aryadebchatterjee5028]

Yeah it somewhat seemed unnatural, works though (Intriguing but not what someone does in the first try) but it just seems like what someone does in hindsight , Ramanujan's solution was very interesting it was based on the algebraic identity: n+k+1=sqrt(1+(n+k)(n+k+2)) now take n(n+2) and keep writing the terms in the above form like n(n+2)=n sqrt(1+(n+1)(n+3)) theb writting (n+1)(n+3) in the above form we will get the infinite nested root now replacing n with just 1 we get it evaluates to 3. The hindsight solution he provided was even more elegant : 3 can be written as sqrt(9) which can be written as sqrt(1+2*4) which can be written as sqrt(1+2 * sqrt(1+3*5)) which can be written as sqrt(1+2 * sqrt(1+3*sqrt(1+4*6))) and so on and when repeated infinitely many times this gives us 3. P.S. Call me a spammer if you will but this solution is really cool :)

[@firefly618]

Maybe the real treasure was the information we lost along the way.

[@woody442]

One might reframe this as concentrating on the important part, and ignoring the unnecessary distraction. In particular, that by designing another nested radical in a backwards fashion, that converges to the desired limit, one can observe, that the difference of our desired limit and 3 converges to zero, without asking explicitly what the desired limit is. Still you are left with a single possibilty for a true answer.

[@zunaidparker]

It's essentially the Squeeze Theorem / Sandwich Theorem for limits, and a very clever use of it!

[@stevenp7991]

thats what i was thinking as well

[@patrickbanzet7488]

Indeed, there is a little mistake at 11:06, last factor is (n-1)/(n+1) and not n/(n+2), but it doesn't change the result. If we keep at the left side a_m,n instead of a_2,n, we get the more general result : lim a_m,n = m+1

[@dneary]

This looks like Ramanujan's famous continued radical sequence which you can "prove" if you already know the answer as: 3 = sqrt(9) = sqrt(1+2sqrt(16)) = sqrt(1+2sqrt(1+3sqrt(25))) etc.

[@dragonlog7842]

f (x)= a+x , f^2 (x)=(a+x)^2=a^2 +2ax +x^2 =a^2+x (2a+x) =a^2+xf (x+a) , so f (x)=a+x=sqrt (a^2 +xsqrt (a^2 +(x+a)sqrt (a^2 +(x+2a)sqrt (a^2 +(x+3a)sqrt (....)))))) , let a=1 and x=2 , 1+2=sqrt (1+2sqrt (1+3sqrt (...))) =3

[@eulertoiler9774]

or k^2 = 1 + k^2 - 1 = 1 + (k-1)*sqrt((k+1)^2) = 1 + (k-1)*sqrt(1 + k*sqrt((k+2)^2)) = 1 + (k-1)*sqrt(1 + k*sqrt(1 + (k+1)*sqrt((k+3)^2)))) ......

[@zunaidparker]

@Creatotron I believe you still need to prove that the infinite radical converges. I don't think you get that fact "for free" by this construction method. Happy to be corrected.

[@peresztrollka4572]

@@zunaidparker I am late to the party, but you are correct. Herschfeld (1935, American Mathematical Monthly) pointed that Ramanujan's proof was incomplete.

[@ostdog9385]

You can actually start with any number and continue the same pattern. Its just at any finite step the last radical will be a complicated number rather the nice ones you get using 3

[@laszloliptak611]

This problem also appeared as problem A6 in the 1966 Putnam Competition. It was originally proposed by Ramanujan.

[@Reliquancy]

This is the one Ramanujan sent into an Indian journal.

[@aryadebchatterjee5028]

Here is Ramanujan's take on this: n(n+2)=n*sqrt(1+(n+1)(n+3)) [he is a genius he could see through this in the blink of an eye] now he iterated this fact and wrote (n+1)(n+3) = (n+1) sqrt(1+(n+2)(n+4)) and replaced it in the first equation now at the end after as many iterations we like but the root still evaluates to n(n+2) which in our case is just 1 and we get 3 of course! It was so amazing it is there in ramanujan's second notebook P.S.: Ramanujan's birthday is very near only 2 days left, it is celebrated as National Mathematics Day here in India I am Very excited

[@Alex_Deam]

When you take it all the way to n at 11:06 why is it n/(n+2) and a_n,n? The right term in the inequality is m/(m+2) and a_m+1,n, so if you want a_n,n, shouldn't the fractions finish at (n-1)/(n+1)?

[@JPK314]

Thought the same thing. It ends up mattering very little because it just leads to 6/(n(n+1))|sqrt(1+n)-n-1| which still obviously goes to 0

[@bosorot]

You are right but it does not matter at the end. Both limit will goto zero.

[@krisbrandenberger544]

Yes.

[@AnshumanMondal-u4e]

Being right doesn't harm, does it?

[@robertapsimon3171]

This is a really nice derivation! Very elegant!

[@mtbassini]

A technical suggestion: try to record on mono, or blend the channels from your mic. Every time you swivel your head, the sound comes from a different earphone. It gets distracting and makes me think if my earbud is broken.

[@MichaelPennMath]

I forgot to "click" the mono button when prepping the file for my editor... oops!

[@Craznar]

@@MichaelPennMath Oh good, it wasn't my ear infection this time :)

[@aryadebchatterjee5028]

@@MichaelPennMath Here is Ramanujan's take on this: n(n+2)=n*sqrt(1+(n+1)(n+3)) [he is a genius he could see through this in the blink of an eye] now he iterated this fact and wrote (n+1)(n+3) = (n+1) sqrt(1+(n+2)(n+4)) and replace it in the first equation now at the end after as many iterations we like but the root still evaluates to n(n+2), and n in our case is just 1 and we get 3 of course! It was so amazing it is there in ramanujan's second notebook P.S.: Ramanujan's birthday is very near only 2 days left, it is celebrated as National Mathematics Day here in India I am Very excited

[@leif1075]

​​@@MichaelPennMathWhat youbdo at 5:30 PLEASE DON'T YPU AGREE IT'S NOT JUST RIDICULOUS BUT IT'S CPNTRIVED and no one wouldmever think of it right so can you please solve without it?? It's just very frustrating. I don't see Ramanujan or me or anyone else thinking of that wver. Hope you can respond

[@stevenp7991]

This is spectacular. THANKS for this

[@Drk950]

Excelent video; i had no clue about how to treat this problem!

[@teambellavsteamalice]

I solved it as follows: To have a finite answer you need a nice mechanism that resolves the roots into a finite part that relates to the integer in the step (I'll use the number n). You can immediately see this part f(n) = √ (1+n*f(n+1)) must be larger than √n and most probably is in the order of magnitude of n+c. After all, in the limit n>>1 and n is roughly equal n+1. If the 1 is negligible and f~f+1, f is similar to √(n*f). This works if f is roughly equal to n (plus a small constant). If f ~ n^k then f~ n^(1+k)/2. This is only finite for k=1. Not ignoring the 1, with f(n+1) = n+c+1 and f (n) = n+c we get f(n) = √(1 + n*(n+c+1) ) or f(n-1) = √(1 + (n-1)*(n+c) If familiar with (a-b)(a+b)=(a^2-b^2) it's easy to see c = 1 and f(n) = n + 1. This resolves the formula and f(2) = 3.

[@howmathematicianscreatemat9226]

The cool thing about international mathematical Olympiad is that it equalizes all mathematicians to their core problem solving capacity. A professor can know extremely much and still have little advantage over a student and vice versa if he doesn’t think out of the box. Theoretically it would be fair if everyone joined the mathematical Olympiad: students and professors at the same time. Note that only a minority of even math professors were successful in such competitions. The reason why it’s fair if professors join is that higher mathematical theorems aren’t allowed to be used unless it can be proven that the result was found accidentally within the examination of one’s options for action to solve the problem.

[@Lirim_K]

The enigma of our time: When will it ever be a BAD place to stop? Still waiting for this day...

[@premanand4043]

That’s absolutely brilliant

[@CraigNull]

Can we see more problems where the path to the solution is discovered via a plausible sequence of observations. The solution here is slick because it's reverse-engineered from knowing the answer, after all the unseen rough explorations and subsequent whittling down of how to get there efficiently. Like we started once we're 60% of the way there. Without that pre-knowledge I'm left a bit at a loss as to what I'm actually watching. Why are we resorting to absolute values now, how do we know an inequality this loose suffices from here on, etc.

[@zilvarro5766]

Most stuff in maths (university level or higher) is like that though. In the finished proof, you never get to see the hours upon hours of failure before stumbling across something that works. You just get the shiny polished final version of the proof where everything appears out of nowhere but somehow miraculously falls into place to work out at the end. What you get from doing and studying many proofs is some intuition of which things/techniques are worth trying and when some approach turns into a waste of time.

[@xxgn]

With the following inequalities: 0 ≤ x ≤ y n ≥ 0 Then: xn ≤ yn If you're missing n ≥ 0, you can cheat by just taking the absolute value of n. I'll note that he did take the absolute value of both sides, but simplified (without writing it out) by relying on a ≥ 0 implies |ab| = a*|b| . To be fair to, he did explain why he used absolute values, but he introduced the absolute value a couple minutes before he leveraged it to simplify the inequality. In a more organically derived proof, you'd realize you needed the absolute value once you tried to take the inequality, but would write in the absolute value step earlier. In practice, this means that the mental order of deriving a proof will be a bit different than the written order, causing some steps to be justified by future steps.

[@howmathematicianscreatemat9226]

@@zilvarro5766 your reasoning is powerful because it basically reveals the core reason why 80% of mathematics university students stop after they started. It’s like you say: the presentation is too magical and few students believe they can be such wizards without the professors carefully showing them the process rather the end product. Thanks for pointing this out, bro

[@DeJay7]

You don't know how much I would want to remove the absolute value at the very end sqrt(1+n) - n - 1 is obviously negative (or 0 for n = 0), because 1 + n is greater than 1, so its sqrt is smaller than n + 1 or sqrt(1+n) < n + 1 => sqrt(1+n) - n - 1 < 0 So |sqrt(1+n) - n - 1| = 1 + n - sqrt(1+n) I know this is entirely pointless and that it doesn't help, but why not lol

[@henrymarkson3758]

It seems that Michael Penn doesn't say 'ok, great' any more.

[@eulefranz944]

:( that was so much fun

[@ogasdiaz]

Similar to Ramanujan's yet simplier imo. Note that. n + 1 = sqrt(n*n + 2*n + 1) n + 1 = sqrt(1 + n(n + 2)) Let. f(n) = n + 1 f(n) = sqrt(1 + n * f(n + 1)) f(n) = sqrt(1 + n * sqrt(1 + (n + 1) * f(n + 2))) ... Then. f(2) = 2 + 1 = sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(1 + 5... 3 = sqrt(1 + 2*sqrt(1 + 3*sqrt(1 + 4*sqrt(1 + 5... *To complete this proof one should prove that the series of radicals does not diverge.

[@curtiswfranks]

I would say that the more reasonable place to start the sequence is Sqrt(1+1 a_(2, ∞)), which converges to 2. How weird!

[@李宥緯-e7u]

Is there any insight of minus (m+1)^2 from both sides? It's so clever but where does it come from??

[@Zarunias]

At 11:00 you went 1 step too far (the fraction should end at (n-1)/(n+1)). Luckily this doesn't change the rest of the proof.

[@forgor7180]

can i ask how it does not change the proof? Since he used the fact the the lim as n-> inf of n/n^2 goes to zero, but in this case it is n^2/n (actually more than n^2 but still), so his method wouldn't work?

[@TM-ht8jv]

@@forgor7180 I don't really understand how you get to n²/n, in this case it doesn't matter if it's 6/(n(n+1)) or 6/(n+1)(n+2) because it is still from degree 2 while sqrt(1+n)-n-1 is still from degree 1, so the limit as n->inf most definitely still goes to 0

[@forgor7180]

​@@TM-ht8jv dum dum

[@forgor7180]

@@TM-ht8jv oh i see what you are saying now. i did a dum dum

[@hassanalihusseini1717]

Very good video. Even I could not follow all the steps involving the inequalities.

[@peterdecupis8296]

very cool "exercise"! Although I have been deeply studying the theory of infinite nested sequences for years (in particular about convergence behaviour, seee.g. Herschfeld theorem) this was a "new entry" in my horizon! Honestly I quickly guessed the necessity of considering a "double-sided" sequence a_m,n but I would have continued in an insanely intricate evaluation of its closed form with respect to a_n,n... There is only a mistake in the inequality factor, i.e. it must be replaced (n-1)/(n+1) instead of n/(n+2); anyhow, this makes no difference when taking the limit of n to infinity. A straightforward generalization is the evaluation of the limits n to infinity of a_m,n for any value of the "starting" parameter m; we simply obtain m+1 for any finite value of m. This is immediately related to the identity 1+ (m-1)(m+1)=m^2, as one can observe by "inserting" such solution at some point of the infinitely nested radical sequence

[@matthewbusche4547]

This is a total nit, but at 0:35 the first few terms of the sequence look funny. One property of the pattern is that each term of the sequence has one more radical, but that pattern is broken for the first term. I think the sequence should instead be: 1 1*sqrt(1+2) 1*sqrt(1+2*sqrt(1+3)) 1*sqrt(1+2*sqrt(1+3*sqrt(1+4))) I find your videos mesmerizing. I may watch this again to see if I can better understand the motivation for each step. Seems like the heart of it was to replace a bunch of complicated stuff that is known to be positive with a simpler upper bound that goes to zero.

[@kajamix]

When he turns his back the label of the t-shir is out. Also his neck was biten.

[@AndreasHontzia]

I understand the solution. I may have an idea how to get to such a solution. But I am always stomped, how early some "random" choice of solution path is taken. I know, this is just a solution in a short form. The hard part is to "find" the solution. Is there any programmatic approach to this?

[@replicaacliper]

if there was, everyone would be an international olympiad medalist.

[@aryadebchatterjee5028]

@@replicaacliper Here is Ramanujan's take on this: n(n+2)=n*sqrt(1+(n+1)(n+3)) [he is a genius he could see through this in the blink of an eye] now he iterated this fact and wrote (n+1)(n+3) = (n+1) sqrt(1+(n+2)(n+4)) and replaced it in the first equation now at the end after as many iterations we like but the root still evaluates to n(n+2) which in our case is just 1 and we get 3 of course! It was so amazing it is there in ramanujan's second notebook P.S.: Ramanujan's birthday is very near only 2 days left, it is celebrated as National Mathematics Day here in India I am Very excited

[@dragonlog7842]

@@aryadebchatterjee5028 f (x)= a+x , f^2 (x)=(a+x)^2=a^2 +2ax +x^2 =a^2+x (2a+x) =a^2+xf (x+a) , so f (x)=a+x=sqrt (a^2 +xsqrt (a^2 +(x+a)sqrt (a^2 +(x+2a)sqrt (a^2 +(x+3a)sqrt (....)))))) , let a=1 and x=2 , 1+2=sqrt (1+2sqrt (1+3sqrt (...))) =3

[@Anishkrsinghiit-jee]

@@replicaacliper absolutely right 😂😂😂😂😂😂😂😂😂😂😂😂😂😂🤣🤣🤣🤣🤣🤣😅😅😅😅😅😅😅😅😂😂😂😂😂😂😂

[@stephenbeck7222]

The programmatic approach probably takes an hour or more to explain for each problem like this. The real answer is to do a thousand of these and at some point you might start finding them intuitive.

[@nicepajuju3900]

Wow that was clever!

[@literallylegendary]

That's rad.

[@tejarex]

If one can guess that A_m, the n-infinity limit of a_mn, is m+1, others have shown that this satisfies the recursion. If one calculates a_mn for n in range(m+1, m+10), say, the last result is, for m in range(1, 7), m.99+. It is not hard to guess m+1 as the limit and try it. It is an easy program in, for instance, Python. The latter is a great tool for exploratory math.

[@MathsOnlineVideos]

This is a difficult problem, but much better than some problems I've come across.

[@txikitofandango]

I wonder how many contestants figured out to subtract (m+1)² from both sides. And how many of them just subtracted 1 from both sides to get a difference of squares, and then later realized (m+1)² would get the solution more directly

[@replicaacliper]

hey michael, nice problem! Just a little comment about the production - in the future, could you make the audio mono? On a good pair of headphones your voice panning back and forth as you move around can get a little distracting. Thanks! Edit: Noticed you already replied to a similar comment. Thanks for responding to feedback!

[@OunegNebty]

Nice Dune TEE SHIRT

[@Evil_AsH]

As a more general proof, we have that the limit when n goes to infinity of a(x,n) = x+1 Have a nice day!

[@tonywells6990]

Correct.

[@avronyje]

Amazing

[@marcowen1506]

A very interesting little problem, thanks for this video. P.S. the author's name is pronounced Voyt-yeck Yarn-yick.

[@amirb715]

i think the numerator must end in (n-1) not in 'n' but the final result is the same

[@ronanb3]

At 13:00, at the end, how to you have a less than (

[@lorenzoaste9173]

What is equal to the second line is not |a_(2,n) -3|, what is equal is the second part of the inequality. (for example: 2 < 3 = 2+1)

[@Anishkrsinghiit-jee]

Sir, your video is awesome 👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍

[@ibrahimkachal6759]

Let a,b and c real positives numbers satifying: abc= a²b² + b²c² +c²a² Find the greatest value of real number m where: m= a² +b² +c² can you do this please

[@lavneetjanagal]

m=1/3

[@mericet39]

I think it was in The Simpsons when I first heard the term 'radical' used in mathematics. This is despite being educated to a high level. I guess that the word 'radical' to mean a root is used mainly in the USA, not so much elsewhere. Is this true?

[@jadegrace1312]

I believe so. I think the term is usually "surd" outside of North America.

[@holyshit922]

Seems like Czech version of slavic name Wojciech In Polish this name would be Wojciech

[@manucitomx]

Thank you, professor.

[@redrosin99]

I am missing the mathematical reasoning to attain the solution

[@arpsri9675]

Look at "1+ 2" >> A1 "1+3" >> A2 =4 (can this be with A1? , yes its value should be around 2, but less than 3 when it stay with A1 "1+4" >> A3 =5 "1+5" >> A4. =6 It is Square root ( A1 * (the rest)) Can A2 come out of its own Square root and be with A1? Answer: " yes" , A2 will be at least "2" Can A3 come out of its own Square root ? Yes , will be at least "2" not more than 3 I come to conclusion that A1 value is 5 < A < 9 ....why less than 9? ..because the incremental of A is by "1" , it is less than the incremental of ^1/2. So limit is "3"

[@nin10dorox]

Wow, I've seen this problem many times before, but I've never seen a correct proof before. How do people think of this stuff??

[@pow3rofevil]

Estas a full con Ramanujan me parece genial 🤘🏻🤘🏻🤘🏻

[@rain2001]

what is the red thing on your neck?

[@bilalabbad7954]

10:52

[@michaelslack8900]

I'm not really sure that this solution is particularly well motivated...? Like, it seems to hinge too much on the difference of squares factorisation just happening to be the right one. Or maybe I'm just missing something

[@a_llama]

He's a fan of recursive relations

[@adam-does]

I see what you mean. It seems like the sort of thing you would do if you knew or suspected in advance that a_(2,n) converged to 3. That’s why you’re looking for the |a_(m,n) - (m+1)| term from a DOS. You could have some numerical basis for that, looking at the trend for initial terms of the sequence. But it better would be if you considered the limit as a general function of m, and found (at least for the case m=2) it had to be m+1

[@theamazingworldofgusball1852]

That guy Vojtěch Jarník is mathematician from Czech republic. Not sure why is he included in problem from 2003, since he died in 1970. It could be someone else though.

[@SanityAardvark]

Math competitions are often named after dead mathematicians.

[@vladdu]

Maybe it's a competition dedicated to him?

[@goodplacetostop2973]

The competition is held at the University of Ostrava every year, to "keep the Mathematics alive and aims to increase the interest in Mathematics". It's very likely they chose this name to pay tribute to him and his work.

[@jesusnthedaisychain]

Why do we have Nobel prizes every year when Nobel has been dead for a long time?

[@TechnoRaabe]

now do this as an infinite fraction

[@Aleksandrus12]

Shouldn't first term be sqrt(1+1) not just sqrt(1)? If it's (n-1)*sqrt(1+n) so for n=1 it's sqrt(1+1)

[@samrachkem2801]

Could you try to do the reverse one? U_{n+1}=√(1+U_{n})

[@dragonlog7842]

f (x)= a+x , f^2 (x)=(a+x)^2=a^2 +2ax +x^2 =a^2+x (2a+x) =a^2+xf (x+a) , so f (x)=a+x=sqrt (a^2 +xsqrt (a^2 +(x+a)sqrt (a^2 +(x+2a)sqrt (a^2 +(x+3a)sqrt (....)))))) , let a=1 and x=2 , 1+2=sqrt (1+2sqrt (1+3sqrt (...))) =3

[@BoringExtrovert]

Hey Michael, is your neck ok?

[@wesleydeng71]

I don't see the need to 'generalize' the sequence. You can simply start with m=2.

[@clangille4351]

You get the fun corollary that a(m,n) converges to m+1. That's about it though. It would simplify a lot of the calculations to just let m=2.

[@ЛюблюТебя-т1у]

So nice

[@ConstantinKubrakov]

Wow

[@martinschulte3613]

Despite it doesn't make any difference I think a_1=1, not a_1=√1.

[@Alex-kp3hd]

why does he has blood on his neck

[@MichaelPennMath]

I think I cut myself shaving and didn't notice.... not a big deal, but it does look sketchy!

[@361Jonel]

love the T-shit, also great video

[@roberttelarket4934]

Vojteck is pronounced Voytek. Jarnik is Yarnik.

[@khalidmohsen3008]

Answer it with f

[@er8061]

I love Ramanujan's root !

[@chris12dec]

You put two 15-20 second adverts that you can't skip at the start, then three more pop-up adverts during the video. Don't you think that's a bit much?

[@NeinStein]

I only got the first two, and the second one was skippable

[@leif1075]

No pne is going to do what he did at 5:30 come on..please solve without it..this isn't intelligent or fair as far as I can tell.

[@charleyhoward4594]

how did the limit = 3 ????????

[@charleyhoward4594]

i see it now ...

[@ycart_tech6726]

Yes, good place to stop, indeed... My sincerest apologies if I belittled your culture. It was unintentional...

[@Pizhdak]

Привет всем, кто смотр т из ЛШ!!!

[@यात्रा-ङ4छ]

Sir, what happened to your neck .

[@tubebrocoli]

Honestly... it kinda sucks that there are frequent mistakes in these video, and while that's not the worst thing in the world there is no correction anywhere unless if you look in the YT comments.

[@Anishkrsinghiit-jee]

Sir, I am from India and 😅😅😅 seeing your lecture from approximately 1 year...... And i found your lecture..... 🤔🤔🤔🤔🤔is good for basic but you are not making video in structured wise ....

[@Anishkrsinghiit-jee]

Sir, in India teacher are doing great effort on you tube video but you only solving one or two question in a lecture..., I think you should make 😅😅 video of long time in a lecture 😁😁😁😁😁😁😁😁😁😁😁😁😁😁😁😆😆😆😆😆😆😆😆

[@jacoboribilik3253]

The guy has a life outside making math-related videos for free. Rather than reproach him, we should show a little bit more gratitude.

[@Anishkrsinghiit-jee]

@@jacoboribilik3253 are you olympiad enthusiasm?????

[@srikanthtupurani6316]

Totally non trivial problem for 9 year old kids.

[@dragonlog7842]

f (x)= a+x , f^2 (x)=(a+x)^2=a^2 +2ax +x^2 =a^2+x (2a+x) =a^2+xf (x+a) , so f (x)=a+x=sqrt (a^2 +xsqrt (a^2 +(x+a)sqrt (a^2 +(x+2a)sqrt (a^2 +(x+3a)sqrt (....)))))) , let a=1 and x=2 , 1+2=sqrt (1+2sqrt (1+3sqrt (...))) =3

[@jacoboribilik3253]

@@dragonlog7842 nice