A Radical That Does Not Stop

[@Blaqjaqshellaq]

Nice one! (At first glance, I wasn't even sure it converged...)

[@lukasjetu9776]

same

[@matt-fitzpatrick]

6:14 the geometric series of sums of geometric series was chefs kiss

[@brendanward2991]

x = given radical. It's fairly easy to show that x^2 = 2*2*x. x^2-4x =0, therefore x=0 or x=4. x=0 is obviously not applicable, so x=4.

[@reizinhodojogo3956]

x^2 = 4x?

[@brendanward2991]

@@reizinhodojogo3956 Yes.

[@akeebplazy1124]

How can u show that x^2 = 4x???

[@brendanward2991]

@@akeebplazy1124 If you can't see yourself, I don't know if you'll be able to follow my method without proper notation, but here it is: x = √2√4√8√16√32√64... x^2 = 2√4√8√16√32√64... = 2*2√√8√16√32√64... =4*√√8√16√32√64... =4*√2√2√16√32√64... =4*√2√2*2√4√32√64... =4*√2√4√4√32√64... =4*√2√4√4*2√8√64... =4*√2√4√8√8√64... =4*√2√4√8√8√64... =4*√2√4√8√8*2√16... =4*√2√4√8√16√16... and so on

[@OKchromia]

​@@akeebplazy1124 by solving x^2=4x

[@MineKukele]

Very intriguing. Thank you for sharing!

[@albertoambram8025]

Very nice one. Thanks for sharing.

[@pietergeerkens6324]

Nice problem and solution! Less rigorously, I set the expression as x; and then observed: 2x = 2 √[ 2 √[ 4 √[ 8 √[ 16 √[ 32 √[ ... = √[ 4 * 2 √[ 4 √[ 8 √[ 16 √[ 32 √[ ... = √[ 4 √[ 8 * 2 √[ 8 √[ 16 √[ 32 √[ ... = √[ 4 √[ 8 √[ 16 * 2 √[ 16 √[ 32 √[ ... : = √[ 4 √[ 8 √[ 16 √[ 32 √[ ... and x² = 2 √[ 4 √[ 8 √[ 16 √[ 32 √[ ... = 2 * 2x = 4x. Hence 0 = x² - 4x = x(x - 4) with x = 0 clearly being a failing spurious case from the squaring at step 2.

[@danielobeng1587]

Informative and helpful

[@michaeledwards2251]

Beautifully clear, at first I didn't think it was possible for a finite result : an infinite number of terms greater than 1.

[@irhzuf]

None of the terms are greater than 1. They are greater than 0, but the known 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... = 1 has also it's terms all greater than 0. (if all the terms would be greater than 1 then a normal sum wouldn't exist as you would have to have a number for summing 1's any number of times.

[@Sugarman96]

The sum of x^1+x^2+x^3+...=x/(1-x) for |x|

[@OnDaTrainToMath]

hi

[@ShauyanOfficial13]

your voice is back, nice

[@tunneloflight]

How do you solve it. Easy, you look it up in Jan Tuma's amazing reference text of such formulas.

[@Twi_543]

This is a nice problem

[@andirijal9033]

The sum of the powers is, deferentiate and multiply again by the ratio

[@edwardsteen3058]

Cool!

[@AliKhavashi]

The Greatest Math-Professor of all time........

[@cytos]

What an incredibly long solution. This is how I solved it: Notice y=√2√4√8√16… can be written as y=√2ay because the sequence after the first √2 is the same as y multiplied by some constant a. a will be equal to √2√2√2… (multiplying each consecutive entry by 2). a=√2√2√2… can be written as a=√2. After performing basic algebra, we get a=2. (a can not equal 0) We now know that a=2, so y=√2ay can now be written as y=√4y. After performing more basic algebra, we get y, the variable we are looking for, is equal to 4. This is a somewhat lengthy explanation, but if you know the tricks, you can solve it in under a minute like this.

[@irhzuf]

This constant doesn't exist as it is a nested radical not just sqrt(2)*sqrt(4)*..., but sqrt(2*sqrt(4...)). Actual answer is 2.

[@cameronspalding9792]

Personally I would have used the equation 1/(1-x)= 1+x+x^2+x^3+… for |x|

[@Sugarman96]

Due to the indexing, I'd start from x/(1-x), but otherwise that's the approach I'd take too

[@matthewfeig5624]

@@Sugarman96 You can start the index at n=1 or n=0 since 0/2^0 = 0. Also x/(1-x) + 1 = x/(1-x) + (1-x)/(1-x) = 1/(1-x), so the two functions x/(1-x) and 1/(1-x) differ by the constant 1.

[@Ezechiel-pk6ku]

Nevermind I answered that myself by looking at the first term, it technically being 1.414, no 1/2 of 2, I'm stupid

[@aczajka74]

I'm a simple man. I see infinite product, I take log

[@TinyFoxTom]

7:33 Wow, -1/12 really does try to pop up where infinite sums abound.

[@reizinhodojogo3956]

-1/12 is a random value, i don't think it makes sense for it to try to go in infinite sums, unless its specifically constructed to get -1/12

[@irhzuf]

​@@reizinhodojogo3956it isn't a random value as it is an answer for the Ramanujan summation of 1 + 2 + 3 + ... which is a pretty known divergent infinite series, but yea when talking about actually converging infinite series then I think it could be considered pretty random

[@reizinhodojogo3956]

@@irhzuf how is 1+2+3+... -1/12?? that doesn't make any sense or logic to me, cause: 1. the sum is only natural numbers(positive integers), -1/12 is a negative fraction 2. all the numbers in the SUM are positive, and no negatives, -1/12 is negative 3. all the values only get bigger to the right of the sum, never smaller, and there are only plus(+) operators, which only add up to the total sum the sum is infinite as far as i know, 1+2+3+... = ∞

[@irhzuf]

@@reizinhodojogo3956 it isn't a normal sum basically Ramanujan summation is used as kind of characteristics of divergent infinite series (it is just a number that people sometimes place on divergent infinite series, because it can be useful)

[@irhzuf]

@@reizinhodojogo3956 wikipedia articles for more info: en.m.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF en.m.wikipedia.org/wiki/Ramanujan_summation

[@jf3518]

Just rewrite the equation as x = sqrt(2 * 2 * x) and solve with the quadratic formular

[@Starblazer-oc4nt]

I just turned that into sqrt(2x)=x solving for x

[@Ezechiel-pk6ku]

So it's 1?

[@aloi4]

Let x = 2√(2√(4√(8√...))) 2√(a) = √((2×a)×2) → 2√(2√(4√(8√...))) = √(4×2√(4√(8√...))) = √(4√(8×2√(8√...))) = √(4√(8√(16×2√...))) → 2x = √(4√(8√(16√...))) → x = √(2×2x) → x² = 4x → x=0 or x=4 However x≠0, therefore x=4

[@brendanward2991]

x = √2√4√8√16√32√64... x^2 = 2√4√8√16√32√64... = 2*2√√8√16√32√64... = 4*√√(4*2)√16√32√64... =4*√2√2√16√32√64... =4*√2√2√(4*4)√32√64... =4*√2√(2*2)√4√32√64... =4*√2√4√4√(4*8)√64... =4*√2√4√(4*2)√8√64... =4*√2√4√8√8√(4*16)... =4*√2√4√8√(8*2)√16... =4*√2√4√8√16√16... and so on.

[@black_eagle]

It's easier to use recurrence: Let x = sqrt(2*sqrt(4*sqrt(8 ... => x^2/2 = sqrt(4*sqrt(8*sqrt(16 ... = sqrt(2*2*sqrt(2*4*sqrt(2*8 ... = sqrt(2*sqrt(2*sqrt(2 ... * x = 2^1/2 * 2^1/4 * 2^1/8 * ... * x = 2^(1/2 + 1/4 + 1/8 + ...) * x = 2^1 * x => x^2/2 = 2x => x = 4.

[@yurenchu]

The last equation, (x^2)/2 = 2x actually has _three_ possible solutions: x = 0 , x = 4 , x = +infinity . How would you know that x = +infinity is not the actual answer to the question?

[@alexicon2006]

​@@yurenchuThe same way we dont put infinity as a possible solution for equations like x²-4x=0? Since when have we started getting infinity as ONE of the solutions. Infinity can ONLY be an answer when NO other valid solution is there. THEN it would limit to infinity. And we clearly know x isnt 0. I mean one look at the nested radical confirms that. These are called extraneous roots. They are completely normal in recurrence based polynomials. You just gotta have common sense to tell them apart from the actual solution.

[@aczajka74]

​@@alexicon2006infinity is a reasonable answer to consider here because the quantity is an infinite product that (a priori) very well might diverge to infinity. The point is that the Algebraic approach can be used to tell us the value of the expression IF IT CONVERGES, but it doesn't prove that the expression converges.

[@alexicon2006]

@aczajka74 Yeah I know. But that doesnt MAKE the Algebraic method itself WRONG. THATS my point. Proving convergence is a whole other process. The algebraic method is NOT responsible for that. That is what I meant.

[@robertveith6383]

Original poster, your sixth line is wrong, because each of your fractional exponents must be inside grouping symbols.

[@f5673-t1h]

The answer is 4 because the exponents are obvious and the resulting series is a well-known one summing to 2.

[@robertveith6383]

That is *not* sufficient work. Your post is *invalid.*

[@yurenchu]

_Answer_ : 4 _Calculation_ (from the thumbnail): X = = √(2 * √(4 * √(8 * √(16 * √(32 * √(...)))))) = (√2) * (√√4) * (√√√8) * (√√√√16) * (√√√√√32) * ... = 2^(1/2) * 4^(1/4) * 8^(1/8) * 16^(1/16) * 32^(1/32) * ... = 2^(1/2) * 2^(2/4) * 2^(3/8) * 2^(4/16) * 2^(5/32) * ... = 2^(1/2 + 2/4 + 3/8 + 4/16 + 5/32 + ...) = 2^W where W = { Σ k/(2^k) , from k=0 to k=∞ } Note that (1 - xⁿ⁺¹)/(1 - x) = 1 + x + x² + x³ + ... + xⁿ and hence for |x| < 1, 1/(1-x) = lim_{n--> ∞} (1 - xⁿ)/(1 - x) = 1 + x + x² + x³ + x⁴ + x⁵ + ... Let f(x) = 1/(1-x) , then f'(x) = 1/(1 - x)² = 0 + 1 + 2x + 3x² + 4x³ + 5x⁴ + ... Let g(x) = x*f'(x) ==> g(x) = x/(1 - x)² = x + 2x² + 3x³ + 4x⁴ + 5x⁵ + ... then W = ( ½ + 2*(½)² + 3*(½)³ + 4*(½)⁴ + 5*(½)⁵ + ... ) = g(½) = (½)/(1 - ½)² = (½)/(½)² = 1/(½) = 2 and therefore, X = 2^W = 2^2 = 4

[@Twi_543]

Cold