A Nice Infinite Radical

[@angelmendez-rivera351]

The rigorous way to do this is to define the double-sequence S : N^2 -> R such that S(m, m) = sqrt(1/2^(2^m)) everywhere and S(m, n) = sqrt(1/2^(2^n) + S(m, n + 1)) everywhere. To get a sense of what is happening, notice that S(0, 0) = sqrt(1/2), so S(0, 0) = sqrt(1/2 + S(0, 1)) = sqrt(1/2 + sqrt(1/4 + S(0, 2))), etc. S(1, 1) =sqrt(1/4) = 1/2, so S(1, 1) = sqrt(1/4 + S(1, 2)) = sqrt(1/4 + sqrt(1/16 + S(1, 3))), etc. S(1, 0) = sqrt(1/2 + S(1, 1)) = sqrt(1/2 + sqrt(1/4)), so S(2, 0) = sqrt(1/2 + sqrt(1/4 + sqrt(1/16))), etc. Now, it is more apparent what we should do: we want is to let f : N -> R, such that f(m) = S(m, 0) everywhere. The task is to find lim(f), if it exists. To try to make some progress, let us go to the equation S(m, n) = sqrt(1/2^(2^n) + S(m, n + 1)). Let m |-> m + n + 1, so that the equation reads S(m + n + 1, n) = sqrt(1/2^(2^n) + S(m + n + 1, n + 1)). Now, let m |-> 0, giving us that S(n + 1, n) = sqrt(1/2^(2^n) + S(n + 1, n + 1)) = sqrt(1/2^(2^n) + sqrt(1/2^(2^(n + 1))) = sqrt(1/2^(2^n) + 1/2^(2^n)) = sqrt(2/2^(2^n)) = sqrt(2)•S(n, n). Now we go back to S(m, n) = sqrt(1/2^(2^n) + S(m, n + 1)), and let m |-> m + n + 2, hence S(m + n + 2, n) = sqrt(1/2^(2^n) + S(m + n + 2, n + 1)). Let m |-> 0, so S(n + 2, n) = sqrt(1/2^(2^n) + S(n + 2, n + 1)) = sqrt(1/2^(2^n) + sqrt(2)•S(n + 1, n + 1)) = sqrt(1/2^(2^n) + sqrt(2)•1/2^(2^n)) = sqrt(1 + sqrt(2))•sqrt(S(n, n)). Go back, and instead, let m |-> m + n + 3. This yields S(m + n + 3, n) = sqrt(1/2^(2^n) + S(m + n + 3, n + 1)), and m |-> 0 implies S(n + 3, n) = sqrt(1 + sqrt(1 + sqrt(2))•sqrt(S(n, n)). This motivates the following conjecture: S(n + p, n) = (f^p)(1)•S(n, n), where f : [0, ∞) -> [0, ∞) and f(x) = sqrt(1 + x) everywhere. The base case was already proven, so we proceed to use induction. So, in the equation S(m, n) = sqrt(1/2^(2^n) + S(m, n + 1)), let m |-> m + n + p + 1, so S(m + n + p + 1, n) = sqrt(1/2^(2^n) + S(m + n + p + 1, n + 1)), and m |-> 0 implies S(n + p + 1, n) = sqrt(1/2^(2^n) + S(n + p + 1, n + 1)) = sqrt;(1/2^(2^n) + (f^p)(1)•S(n + 1, n + 1)) = sqrt(1/2^(2^n) + (f^p)(1)•sqrt(1/2^(2^(n + 1)))) = sqrt(1/2^(2^n) + (f^p)(1)•1/2^(2^n)) = sqrt(1 + (f^p)(1))•sqrt(1/2^(2^n)) = (f^(p + 1))(1)•S(n, n). This completes the proof of the conjecture by induction, and so, we can conclude for all p in N that S(n + p, n) = (f^p)(1)•sqrt(1/2^(2^n)). Hence, n |-> 0 means S(p, 0) = h(p) = (f^p)(1)•sqrt(1/2) everywhere. From here, it is easy to recognize that lim(h) = φ/sqrt(2) = (1 + sqrt(5))/(2•sqrt(2)). Q. E. D.

[@SyberMath]

Great, as always!

[@leif1075]

@@SyberMath I'm curious whybdidnt you use the infinite series Sqrt 1/2 + sqrtswrt1/2 instead of using 1 as the term I the test series..since we have 1/2...and youndidnt make it clear what is x..x is the original infinite series I presume right?

[@metehan9185]

Holy shi.....

[@Qermaq]

This one is like watching a really complicated movie where at the end all the threads are tied together in a bow. :D

[@SyberMath]

Good to hear!

[@wannabeactuary01]

I like this approach - partial sequences or series are fine but this is better and with some mental effort clearer - thank you.

[@AdityaKumar-gv4dj]

Nice video with a amazing solution

[@SyberMath]

Thank you!

[@MrLidless]

Similar to how you started, you can see that x² - 1/2 = x / √ 2, and it drops out from there.

[@TechyMage]

After wasting some time here is what i did x=√(1\2+√(1/4+......)) Take 1/√2 common we get: x=√(1/2+x/√2) Squaring both sides we get: 2x²-√2x-1=0 Use quadratic formula and u will end up with the same answer(don't forget to reject the negative value)

[@jmart474]

Nice solution! Really difficult to see the result of extracting the common factor.👍

[@zunaidparker]

Interesting radical! I would like to see the actual convergence proof itself. As it is now, it feels like we just applied a clever trick to "get" to the answer in a predetermined way. But the true value of this question lies in learning how to prove convergence and learning about how such radicals behave and their properties. Maybe something for a follow up video?

[@elrichardo1337]

to start maybe look up “herschfeld’s convergence theorem”? it’s a generalization of convergence conditions to any infinitely nested radical

[@SyberMath]

Thank you for the feedback! I like the 'doing' part not the 'proving' part so I'm not rigorous in that sense. 😜😁🤓 Check out the pinned comment by Angel Mendez-Rivera, though!

[@zunaidparker]

@@SyberMath thanks will do! There's nothing wrong with the doing part either, only challenge is that it feels sometimes very arbitrary how you get to the solution. It can come across that you either already had the answer and reverse engineered the question, or you looked up the solution and are just going through the motions to get there without applying any thought out method yourself while skipping over the important details. Especially when it's some sort of clever trick or substitution like in this case, it doesn't teach us anything to just watch you plug and play without any deeper thought process or explaining interesting properties of the question. I could just ask Wolfram Alpha if I wanted the answer in that case. The beauty of having a human do the problem on a KZbin video is that they can explain the method and WHY you would use such a method in each case to solve a similar class of problems. Then we actually learn something from the video and there's value in you doing it live and narrating to us. Hope this helps.

If one really knows/understands the distribution on the left hand is the correct one (it's given if you see that it continues the pattern) then since the left hand side and the right hand side is equal in this case. The only question is about the radical equal to the Golden Section. And of course that you know about it and see the connection. The tremendous note by Angel Mendez-Rivera is a fine example of my saying that laziness don't take you far in a field. Besides being a good exercise in being formal and presenting a proof -- of what it states -- even though this isn't needed. (For this specific case, provided you know what you are doing to the left hand side gives the same thing numerically on the right hand side. For all l.h.s's a corresponding r.h.s.). It's maybe even an "overkill". Good how this question makes it so clear that algebra is so powerful. But also that algebra is very subtle and hard.

[@anshnamdev4630]

Nice sir.I like your explanation

[@SyberMath]

Thanks for liking

[@paragogosolokliroma6448]

Very interesting!Very good idea.

[@SyberMath]

Thank you very much!

[@ngothuyh]

nice vid with a great solution!

[@SyberMath]

Glad you think so!

[@ItsSurgeee]

was a bit confused about the x development But great vid

[@BOBPERIO2]

Excellent video! Thank you very much. 👍🏻

[@SyberMath]

You are welcome!

[@skwbusaidi]

For what value of x , the equation you show converge

[@davidr482]

Beautiful

[@SyberMath]

Thank you

[@sachs6]

How can one prove it converges?

[@SyberMath]

Check the pinned comment by Angel Mendez-Rivera .

[@-basicmaths862]

Substitute 1/2=a^2 & solved .

[@emanuellandeholm5657]

Happy holidays SyberMath! :)

[@SyberMath]

Thank you! To you, too!!! 🥰🎉

[@notlin1976]

Merry Christmas, Professor! 🧑🏽‍🎄🧑🏽‍🎄🧑🏽‍🎄🧑🏽‍🎄🧑🏽‍🎄🧑🏽‍🎄🧑🏽‍🎄

[@SyberMath]

Thank you! Merry Christmas and Happy Holidays!!! 🥰🎉🥳

[@stephenlesliebrown5959]

Great problem and very thorough discussion of the solution. You are an exceptionally talented professor of mathematics! Have a happy new year 🙂

[@SyberMath]

Aww, thank you for the kind words! Happy Holidays and Happy New Year! 🥰💖

[@John-Spartan03]

number syndrome or language syndrome!, I have number rejection syndrome

[@RAG981]

Clever.

[@robertveith6383]

*@ SyberMath* You have a typo at the end of the video in green. Instead of 1/6, it needs to be 1/16.

[@SyberMath]

oopsies. Thanks for letting me know

[@please.visit.wmscog.watv.]

Could you please help me to solve this: square root (1+ 2 square root (1 +3 square root (1 + 4 square root (1+ ...))))

[@please.visit.wmscog.watv.]

Thank youuu!

[@SyberMath]

kzbin.info/www/bejne/lX-ummuLo7CkfLM

[@-basicmaths862]

Answer=1?

[@barakathaider6333]

👍

[@andirijal9033]

My answere differen metod

[@angelamusiema]

1,2⭐

[@michaelempeigne3519]

cheap method but good.

[@SyberMath]

Why cheap? 🤓😁😜

[@michaelempeigne3519]

@@SyberMath it converges ?

[@Drk950]

Let "x" be the seeked value and "r" the Golden ratio, I found: x^2 = (1/2)*(1+r) ----> x = (1/2)*sqrt(3+sqrt(5)). It's numerically equal to the answer expressed by @SyberMath, but looks uglier >•