Simplifying An Infinite Radical Expression

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts) and my first video in short form! Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#ChallengingMathProblems #Radicals
via @KZbin @Apple @Desmos @NotabilityApp
@googledocs @canva
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus

[@SyberMath]

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts) and my first video in short form! Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡 www.youtube.com/@SyberMathShorts

[@DucLE-zv2lh]

\\âbio

[@DucLE-zv2lh]

\\âbio

[@randomjin9392]

A bit more formal way for the sum: we set f(x) = ∑sx⁻ᔆ, x > 1; then ∫(f(x)/x)dx = -∑x⁻ᔆ = -(1/x)/(1-1/x) = -1/(x-1), differentiating back gets us f(x) = x/(x-1)² and our sum is simply f(2)

I'm seem to be the only one doing the infinite product directly -- without using series. I don't want to brag -- it could be a dumb thing to do. But this far I have always got it right. On the other hand multiplication is of course based on addition. Many of these KZbin mathematics channels -- this one in particular -- is very entertaining and give very useful POV's even if you like me are really only interested in unsolved problems -- besides the valuable didactics tips.

[@suryatejapopuri9610]

How did you get to type sigma in KZbin.

[@rotten-Z]

@@suryatejapopuri9610 ∆≠∞

[@HoSza1]

@@suryatejapopuri9610 Magic of utf8 tables... and a bit of copy/paste glue that holds the entire internet together!

[@HoSza1]

@ With all due respect sir, saying that still counts as bragging unless you show your solution, that is.

[@harshkhare7084]

Another way: NOTE: the dotted lines mean that the sequence IN THE BRACKETS continue till infinity in a set fashion. A single dot is same as the multiplication (*) sign. Let: x =√2(√2²√2³.....). Note that in the ABSENCE of brackets, the √ covers the successive terms as well. Square both sides to get: 2(√2²(√2³....)) = x² √2²(√2³(√2⁴......)) = x²/2. This can be re written as: √2.2(√2.2²(√2.2³.......)) = x²/2 Note that the obtained equation is SOMEWHAT similar to the original equation, except for the fact that instead of 2, we have 2√2 in each term. If we manage to filter out √2 from each of these terms, we can then get a characteristic equation in terms of x. After filtering the √2 terms and expressing them as filtered-out individual terms, we can re-write the equation as: (√2)(√√2)(√√√2).... * (√2(√2(√2.....))) = x²/2 Notice the brackets. They separate the terms √ is distributed over from the ones it is not. Note that the last term is same as x. And the terms before them result in 2^y where y is a Geometric Progression (GP). We again re-write the equation as: 2^((1/2)+(1/4)+(1/8)+.....) * x = x²/2 Solving the GP in the power, we get: 2x = x²/2 x² - 4x = 0 x.(x-4) = 0 x = 0,4. Since x is increasing function, the value of x is 4.

[@SyberMath]

Nice!

[@Micro-Moo]

@@SyberMath Hey, you up-vote people admiring your video and ignore most of the comments showing much better solutions. Think what people should think of you.

[@pwmiles56]

Nice one. I got bored doing it the video way so I found this. Trickle through a factor of 2, which becomes a 4 under the next square root: x = sqrt(2 sqrt(4 sqrt(8 sqrt( ...)))) 2x = sqrt(2.2.2 sqrt(4 sqrt(8 sqrt( ...)))) 2x = sqrt(2.2 sqrt(2.4.2 sqrt(8 sqrt( ...)))) 2x = sqrt(2.2 sqrt(2.4.sqrt(2.8.2 sqrt( ...)))) 2x = sqrt(4 sqrt(8.sqrt(16 sqrt( ...)))) 4x = 2 sqrt(4 sqrt(8.sqrt(16 sqrt( ...)))) sqrt(4x) = x 4x = x^2 x = 4

[@pwmiles56]

@Nigelfarij Fair point. Maybe demonstrate the sequence is increasing, then find an upper bound?

[@ufukkoyuncu3408]

Very nice solutıon

[@artham2676]

@Nigel 0:22

[@РичиЯркий]

You lost value 0, and autor of video too)

[@serulu3490]

x also equals to 0 lmao

[@akifbaysal9141]

I noticed that there is a shortcut solution by squaring and pulling 4 from all levels beyond 2 to one earlier level, which gives L^2 = 4 L (where L is the limit or solutiom). I saw few people have already mentioned this quicker solution..

[@iabervon]

When I saw that sum, I was wondering how you were going to evaluate it without power series and calculus, which seemed like a lot more advanced math to introduce for this problem.

[@jennifertate4397]

Love it! Mathematics is so cool! Thanks syber dude!

[@SyberMath]

Happy to help!

[@angelmendez-rivera351]

If we want to be rigorous, then we need to talk about sequences and limits of sequences. Here, it seems as though the sequence of interest goes as 1, sqrt(2), sqrt(2·sqrt(4)), sqrt(2·sqrt(4·sqrt(8))), etc. This can be rewritten as 1, 1·sqrt(2), 1·sqrt(2)·sqrt(sqrt(4)), 1·sqrt(2)·sqrt(sqrt(4))·sqrt(sqrt(sqrt(8))), etc. Let f : [0, ∞) -> [0, ∞) such that f(x) = sqrt(x) everywhere, and let f^m denote the mth iterate of f, such that f°f^m = f^(m + 1), and f^0 = id, the identity function. The sequence can further be rewritten as (f^0)(2^0), (f^0)(2^0)·(f^1)(2^1), (f^0)(2^0)·(f^1)(2^1)·(f^2)(2^2), etc. It is now clearly visible that this is a partial product sequence. Thus, let g : N -> (0, ∞) be defined by the recursion g(0) = (f^0)(2^0) = 1, and g(m + 1) = g(m)·(f^(m + 1))(2^(m + 1)). We have now formally defined the sequence we are studying, but how are we supposed to proceed with the analysis? Notice that 2^m = exp(ln(2)·m). Also, notice that f(exp(ln(2)·m)) = exp(ln(2)·m/2), and more generally, (f^n)(2^m) = exp(ln(2)·m/2^n). If this is not obvious, then you can prove it by induction. Therefore, (f^m)(2^m) = exp(ln(2)·m/2^m). Also, g(m) = exp(ln(g(m))). Thus, we have exp(ln(g(m + 1)) = exp(ln(g(m)))·exp(ln(2)·m/2^m) = exp(ln(g(m)) + ln(2)·m/2^m), and this is equivalent to ln(g(m + 1)) = ln(g(m)) + ln(2)·m/2^m. Let h(m) = ln(g(m)), so that h(0) = 0. Hence, h(m + 1) - h(m) = ln(2)·m/2^m. Let j(m) = m/2^m. The idea is to find the sequence of partial sums of j, multiply it by ln(2), set that equal to h, and find lim h, since lim h = ln(lim g). Hence, lim g = exp(lim h). Consider k : (-1, 1) -> (-∞, 1/2) such that k(x) = 1/(1 - x) everywhere. k has a Maclaurin series expansion, where the mth order Maclaurin term is equal to M(k, m, x) = x^m. The derivative of k, k', also has a Maclaurin series expansion, where the mth order Maclaurin term is M(k', m, x) = m·M(k, m - 1, x) = m·x^(m - 1), which you obtain by differentiation. Let λ(x) := x·k'(x) = x/(1 - x)^2. λ thus has mth order Maclaurin term M(λ, m, x) = m·x^m. Notice this: j(m) = M(λ, m, 1/2), and therefore, lim h = ln(2)·λ(1/2). Therefore, lim g = exp(lim h) = exp(ln(2)·λ(1/2)) = exp(ln(2)·(1/2)/(1 - 1/2)^2) = exp(ln(2)·(1/2)/(1/4)) = exp(2·ln(2)) = 2^2 = 4.

[@leif1075]

But hiw can you write a function f(×)= sqrt of x when x doesn't take every possible value between o and infinity it only.tales power of 2 values so it's not a continuous function so not sure how that is accurate or what you mean sorry?

[@angelmendez-rivera351]

@@leif1075 You do not need x to take on every value possible.

[@prashantgujar9159]

Well illustrated!

[@SyberMath]

Thanks!

[@thevectorspaceofmath]

A joy to watch. Very inspiring :)

[@SyberMath]

Glad you enjoyed it! 💖

[@septembrinol1]

I liked the idea of breaking the series in pieces!

[@danik0011]

5:00 At this point, you just calculate those geometric series, 1/2+1/4+1/8+1/6...=1, there is a fully visual proof for that. 1/4+1/8+1/16+1/32...=1/2 and so on so we get 1+1/2+1/4+1/8+1/16... which goes to 2, and that is the exponent.

[@giuseppemalaguti435]

Se t=prodotti dei radicali, lnt=sommatoria dei logaritmi, e applicando la derivata delle serie risulta lnt=ln2*2=ln4 per cui t=4

[@theothetorch8016]

My brain is screaming because all the expansion forces are starting to take a toll. Love it.

[@SyberMath]

😁

[@leif1075]

@@SyberMath isn't that a dirty trick and not fair at 6:50 to keep writing them in the form with 1/2 in the denominator since no one ever does that and you would only know to do that if you knew the answer beforehand right? So shouldn't thst not count?

[@oahuhawaii2141]

This is an infinite product of diminishing terms. Each term is a successive power of 2 (2, 4, 8, 16, 32, ..., 2^i) under an increasing number of square roots applied to it to get the 2^i root. Thus, the expression is the square root of 2, times the 4th root of 4, times the 8th root of 8, times the 16th root of 16, times the 32nd root of 32, ..., ad infinitum. We write the continued product, P: P = √2*√√4*√√√8*√√√√16*√√√√√32*... We square this P: P^2 = 2*√4*√√8*√√√16*√√√√32*√√√√√64*... We divide this P^2 by P, by aligning terms with the same number of radicals: P^2/P = P = 2*√2*√√2*√√√2*√√√√2*√√√√√2*... This is a new form of P. We square this P: P^2 = 4*2*√2*√√2*√√√2*√√√√2*√√√√√2*... We divide this P^2 by P, by aligning terms with the same number of radicals: P^2/P = P = 4

[@aguyontheinternet8436]

This seems to be a unique solution. So, I will assume that this value converges, and does not oscillate or converge to infinity, and I will set it equal to x x=2^{1/2}2^{2/4}2^{3/8}2^{4/16}2^{5/32}... I will then take the square root x^{1/2}=2^{1/4}2^{2/8}2^{3/16}2^{4/32}... To further simplify this, I will also find the value of another infinite series, y, which I will make the same assumptions as with x y=2^{1/4}2^{1/8}2^{1/16}2^{1/32}2^{1/64}... y^{1/2}=2^{1/8}2^{1/16}2^{1/32}2^{1/64}2^{1/128}... y*y^{-1/2}=2^{1/4} y=2^{1/2} Now to continue, using my newly discovered y to my advantage y*x^{1/2}=2^{1/4}2^{1/4}2^{2/8}2^{1/8}2^{3/16}2^{1/16}2^{4/32}2^{1/32}... y*x^{1/2}=2^{2/4}2^{3/8}2^{4/16}2^{5/32}... 2^{1/2}*x^{1/2}=2^{2/4}2^{3/8}2^{4/16}2^{5/32}... (2x)^{1/2}=2^{2/4}2^{3/8}2^{4/16}2^{5/32}... x*(2x)^{-1/2}=2^{1/2} x=(2x)^{1/2}*2^{1/2} x=(2x*2)^{1/2} x=(2^{2}x)^{1/2} x=2x^{1/2} x^{1/2}=2 x=4 and there we go. This would look a lot nicer if youtube had a math function or something

[@imonkalyanbarua]

Beautiful! 😍👏👏👏

[@SyberMath]

Thank you! 😊

[@shacharh5470]

I used another method, a method I picked up when I learned about generating functions.. S = sum from 1 to infinity of n/2^n. then 2 times S = sum from 1 to inifity of n/2^(n-1) = sum from 0 to infinity of (n+1)/2^n. S = 2S - S = the 2nd sum minus the first sum. So you get 1 + sum from 1 to infinity of 1/2^n = 1 + 1 = 2

[@TomatoFryEgg]

I like the solution a lot! by using sum of geometric sequence and sum of geometric sequence of the each sum of geometric sequence it is a fantastic solution.

[@nasrullahhusnan2289]

The same technique can be used to find 1²+2²+3²+4²+...+n²=n(n+1)(2n+1)

[@mathematics67]

Which software are U using?

[@SyberMath]

Notability

[@arkajitbhowmik7]

Was there any way that we could Integrate n/(2^n) from 1 to infinity?

[@vladimironoprienko7177]

I guess the hard part is to prove that integral over continuous range equals discrete series sum.

[@michaelpurtell4741]

Black pen had a video on n/2^n 3 years ago with a slightly different approach

[@virentanti16]

i was thinking power series convergence and all but taking infinite number of gps and then taking sum of infinite gps was a pretty cool method! made my day

[@subhashkumarsinha8252]

Beautiful

[@SyberMath]

Thank you!

[@yoav613]

Nice one!😀💯

[@tbg-brawlstars]

Did in my mind in around 25 seconds My ans = 4 Edit : I am right and my approach is WAYYYYYYYYY shorter than yours Just call the expression x, then multiply both sides by (*2) , you'll get x² = 4x, x≠0, x = 4 *Note : 2 = √2(√2(√2(√2...

[@knutthompson7879]

Need to be careful. Your method assumes the limit exists. To be rigorous, you would need to show that.

[@tbg-brawlstars]

@@knutthompson7879 Umm, √2(√2√.. = X X² = 2X X≠0, X=2 Where's the problem?

[@thomaspickin9376]

@@tbg-brawlstars The problem is the first line when you say the thing = X. You're assuming there is a defined value (which there may not be). I could use similar logic and say: 1/0 = X Take reciprocals and say: 1/X = 0/1 = 0. That isn't true though as X never existed in the first place. See what I mean? Or to make it clearer you could take an infinite sum say +1-1+1-1... Which has no defined value. Assign that to X, your calculations after wouldn't be valid as X actually doesn't have a value.

[@knutthompson7879]

@@tbg-brawlstars As I say, there is an assumption that you skipped over. When you say “the value is x”, you assume there IS such a value, which has not been established. You can claim this is a stupid detail, but if you skip this step, you can prove a lot of incorrect things.

[@tbg-brawlstars]

@@knutthompson7879 O lol so true Some months ago , I proved that the sum 2 + 2² + 2³ + ...∞ is equal to -2 😂 Jokes aside, how do I prove it ?

[@qwang3118]

Let P = the expression. Then P = sqrt(4P). P^2 = 4P. P =4 or P = 1 (out). So the final solution is P = 4.

[@MDSaad_IIT_BHU]

Sir you just blowed my mind ❤️🧡💛 Keep updating by these types

[@SyberMath]

So nice of you

[@dzigerica666]

Nice one!

[@captainteach007]

Nicely done!

[@SyberMath]

Thanks!

[@SuperMath111]

@@SyberMath Can you tell me, which software are using

[@iRReligious]

That was dope! 👍

[@SyberMath]

Glad you liked it!

[@mantubera9073]

Let, the given function is x Then, squaring both side we have, X²=2.2x x²-4x=0, x(x-4)=0, x is not equal to zero, Hence, x=4, the value of the given function is 4

[@meehow72]

This literally just fried my brain

[@SyberMath]

Sorry 😁

[@ygalel]

Waaaahhh that's awesome😂

[@samarthchohan106]

That series is agp series and it's sum using the agp infinite sum formula is 3 so the answer is 8 not 4

[@pavankushwaha9848]

Awesome bro

[@SyberMath]

Thanks

[@billliu1441]

没必要解这么复杂！ 令原式=A 则A*A=2*A*root(2root2(root2…)))) 于是A=2*root(2root2(root2…)))) 则A*A=4*2*root(2root2(root2…)))) 则A*A=4*A 结果 A=4

[@corkdiamond663]

It's 2 power sum to infinity of agp

[@dusanlazic7422]

KAKVU PRAKTICNU ULOGU IMA OVA TRAUMA MOZGA ?

[@comfycomfy2003]

The calculator said that this task equals 3,0844.. so on, please explain me so wrong? Task: 2^(1÷2)×4^(1÷4)×8^(1÷8)×16^(1÷16)

[@mubashirraza1608]

Can anyone tell me, which software he is using?

[@SyberMath]

Notability

[@Тюлень-о1м]

Nice video, good luck

[@social6332]

good question.

[@SyberMath]

Thanks

[@ahmetarifcan7798]

Ercüment senin yapacağın işi

[@mintusaren895]

Root of straight line

[@travisporco]

fascinating

[@SyberMath]

😊

[@braydentaylor4639]

Beautiful problem

[@SyberMath]

Thank you!

[@sinforpizero]

we can make an equation n=√4n n=4 n=0 and the answer clearly not 0.

[@lazymello6778]

i did it this way: sqrt(2(sqrt(2^2(sqrt2^3... ))))= x =sqrt(2(sqrt(2*2(sqrt(2*2^2(...)))))) (taking seperated 2's as sqrt(2) in the bigger/preceeding radicals) =sqrt(2sqrt(2)*sqrt(2sqrt(2)*sqrt(2^2sqrt(2)*...))) =sqrt(4sqrt(2sqrt(4sqrt(8sqrt(16)...)))) = sqrt(4x)=x therefore, 4x = x^2 x^2 - 4x = 0 x=4 [not 0 cuz... no] i had a stroke writing this lol

[@Serizon_]

nice video

[@SyberMath]

Thank you!

[@ikonkar1208]

Head spinning exercise

[@dusanlazic7422]

ZASTO BRISES KOMENTARE ?

[@vh73sy]

P(v,k) = Product( (v^k)^(v^-k) ), k=1 to inf = v^(v/(v-1)^2) Where v>1, v is integer When v=2 => P(2,k)= 4

[@imdatkoksal6187]

I didnt understand your solution

[@jamesstrickland833]

I generalized this technique for any polynomial numerator over a geometric term that converges. Even showed how it works for alternating sequences that absolutely convege kzbin.info/www/bejne/gGHKpIGNip56bdk

[@xukinei4323]

巧！

[@ВасилийДрагунов-н8т]

Why does noone care about convergence of the sequence...

[@астемиртлупов-х3д]

Незнаю как но я в уме решил

[@堀勇作-l5p]

答え　4

[@robertveith6383]

*SyberMath* -- You described this wrong right at the beginning. You said, "We have the square root of two times the square root of four times the square of eight ..." Then you corrected yourself as you explained this problem is nested square roots. Go back and listen to the very beginning, please.

[@rakenzarnsworld2]

Answer: 2

[@alfa2k3]

Niceee

[@SyberMath]

Thanks 🤗

[@sonaruo]

lol why you make your life hard?? 1/2+ 1/4 is 1 the other one will be 1/2 the other one will be 1/4 the other one 1/8 and if you add them again it will be a 2 so you have 2 raised on 2 so 4

[@temen1167]

sqrt(2*sqrt(4*sqrt(8*sqrt(16*....)))) = x; sqrt(2*2*sqrt(2*sqrt(4*sqrt(8*sqrt(16*....)))) = x; sqrt(4*x) = x; x^2 - 4x = 0; x(x - 4) = 0; so x = 4 and it brings us to the end of this comment

[@mathsknowledge6194]

Answer is 4

[@DeadJDona]

42.0

[@toanpham4110]

✅

[@speedcubesolver1195]

👍👍👍

[@letswaveabook3183]

I haven't worked it out on paper, but if we call the nested root a, we can take a sqrt(4) out of each root, then we find a=sqrt(4)*sqrt(a) from which follows that a=4

[@ssunahohlov]

Решается проще, в уме, за 1 минуту. Автор тут мути какой-то налил

[@李家生-j2v]

4

[@РичиЯркий]

U lost one value) Let this expression x, so x^2=4x x=4 and 0

[@vnagaraju6265]

32

[@redouanchab3an331]

22

[@Deepthivlogsa]

Jai Pawan Kalyan

[@ahmetarifcan7798]

Yaw bramınlar neyi yapamadınız hepsini 2 üzeri eşit kuvvette alın ne uğraşıyorsunuz 8/16 + 8/16 + 6/16 + 4/16 = 26/16 = 13/8 bu da 2 nin 1+5/8 inci kuvveti olduğunu söylüyor cevabın

[@Deepthivlogsa]

Jai Janasena

[@T.N.O.K-Century]

To evaluate Σ[k=1→∞] (2^(-k)), you can get the answer by thinking that you cut the 1×1 square in two halves continuously I reckon. Anyway, I love your math videos so hold up the good work😊

[@csongor-tiborpeli374]

Ans 3.8.....

[@ttrgan]

i used limit for the converging series and got ⁴√2

[@toanpham4110]

✅

[@vv7293]

4