How To Solve This Crazy Equation. Ramanujan's Radical Brain Teaser

In 1911, Srinivasa Ramanujan posed this problem involving an infinite nested radical (a never-ending expression that has square roots inside of square roots). Can you figure out how to simplify the equation? The video presents an intuitive solution. Technically it is necessary to prove the equation has a solution (the nested radical converges) because crazy things can happen with infinite expressions. The technical proof is included in a link below.

Blog post (text/image explanation). I also justify why the pattern continues for the first solution:

Technical proof

StackExchange discussion

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[nikolaytsvetkov]

I think you should modify this question to "can you prove that this equation equal 3?".

[alexismandelias]

You didn't solve it. You knew the solution and proved it. Solving is a completely different thing to do

[Schindlabua]

Once you realize that n² = 1 + (n-1)(n+1) you're golden. Really cool puzzle!

[albinkx]

sir. Ramanujan the great mathematician, , a magician of maths

[helloitsme]

i think the real question is: can you prove no other value than 3 can do this?

[VirtualMedic]

Ramanujan was a genius, but this explanation is bad, because u do not solve equation, but start from solution and prove expression. That is not a way to solve any equation, because u have to know solution to approach to equation like this....

[AdityaSharma-qinu]

As a math professor, I can see that this solution has some serious issues.
While the answer posted is correct by chance, the solution offered is incomplete. Mathematics is not about the correct answers it's about the correct methods to reach them!

The mistake is that you did not prove that the nested radical is actually convergent to 3.This can lead to some serious confusion... For example one could start with 4 and through a slightly more complicated process get the same nested radical. This is also shown in a Mathologer video.

[tuerda]

Q: Did you figure it out?
A: Hell no!
This one was really hard! Just how smart do you expect us to be?

[AoSCow]

Seems a bit backwards, how would you work it out not knowing what the sequence approaches?

[Jerry_licious]

So here is the fun thing:
4 = √(16)
=√(1 + 2 * 15/2)
=√(1 + 2√(56.25))
=√(1 + 2√(1 + 55.25))
=√(1 + 2√(1 + 3*(221/12)))
=√(1 + 2√(1 + 3√(48841/144)))
...
As long as you do not have the last number, we can push anything into this fraction.

[LoboSolitario-ubsb]

1:42 --> To make it easier to understand the equation, just make a numerical sequence at the end. Like this: 9+7=16, 16+9=25, 25+11=36, 36+13=49, ....

[GreenDayxRock]

Nicee. You did the proof by induction without even scaring people by mentioning it. Keep the awesome videos coming!

[GuzmanTierno]

Actually there's nothing to solve because x is already alone on one side of the equation. On the other side we have an expression to evaluate after proving that it converges.

[maximumprobability]

Ok here is the complete rigorous proof it is equal to 3 and the bonus result for other radicals. Bare with me due to length but it is worth it to clean this proof for good in my opinion. It doesnt ever require to know the answer to produce it. Even beyond any wikipedia references on nested radicals in my opinion that do not go that deep to secure the solution.

You can consider the function (E1) This function satisfies the functional equation f(x)^2-1=x*f(x+1) (E2) as can be seen by taking the square of the infinite radical subtracting 1 and noticing the rest is x*f(x+1). This functional equation has as easy solution f(x)=x+1 (E3)

(you can see it if you try a linear guess or even a polynomial guess f(x)=ax+b with higher terms vanishing). Notice that for x=2 we have the original Ramanujan problem and the result is f(2)=3. Of course also f(0)=1 as expected. However to be thorough one needs to show this x+1 solution is the only solution of the radical not that it's just a solution of the functional equation the radical obeys because a functional equation can have many solutions that are different. So here is how to do that.

Imagine instead that there is a solution other than x+1 for instance a solution like f(x)=x+1+h(x) (E4) where h(x) is another nonzero function to make the solution different from x+1. We will show now that h(x) has to be zero in all x>0.

First of all notice that the functional equation f(x)^2-1=x*f(x+1) when f(x)=1+x+h(x) leads to a functional equation for h(x) as well.

This is basically

(E5)


Notice that from the initial value problem (E1) at x=0 f(0)=1 and h(0)=0 also from (E4).

Imagine now that that for some x>0 eg x=1, h(1) is not 0 and it takes instead some value d.

If d>0 then (E5) tells you that h(2)>2*h(1). Also h(3)>2*h(2) etc . Basically h(x) increases exponentially (positive) with x for large x.

What about d<0? What does recursive equation (E5) say?

First of all lets establish some facts about h(x) and f(x) in general.

One can easily see that for x>0 >

From (E4) we can see then that h(x)+1+x>x or h(x)>-1 (C1)

We could ask now;

Under what starting values for h(x) does it prove true that |h(x+1)|>g*|h(x)| for some g>1 as a result of the recursive equation (E5) and condition (C1)?

You must have or (2*(x+1)+h(x))^2>g^2*x^2 which leads to h(x)<-gx-2x-2 or h(x)>(g-2)*x-2
The first inequality is never true due to (C1) but the second is always true for some g small enough a bit larger than 1 and smaller than 2 for eg x>=1.

So we have established that if h(1) is some d other than 0 consistent with the constraints for h(x) then the recursive equation that gives h(2), h(3) etc will be in terms of absolute magnitude bounded by a geometric sequence with ratio g>1. In other words for x large enough |h(x)| becomes larger than some exponential function of x with exponent g>1.

But that cannot be true. Because we can show for example that f(x) <a*x for some positive constant a>2 and the exponential growth of h(x) for large enough x would require this to not be true.

Here is how to see that f(x)<a*x for large enough x;

Notice that eg for x>1 < (x+1)^(1/2)*(x+2)^(1/4)*...(x+n)^(1/2^n)....<(2*x)^(1/2)*(3*x)^(1/4)*...*x*n)^(1/2^n)...)<x^(1/2+1/4+1/8+...)*Product((n+1)^(1/2^n)), {n, 1, Infinity}]= x*Product((n+1)^(1/2^n)), {n, 1, Infinity}]

(basically i claimed that the general product term that was say (x+n)^(1/2^n) is smaller than (x*n)^(1/2^n) turning the sum into product inside to bound it easier).

That last infinite product is just a number, some constant, as can be seen by taking the logarithm of it, turning it into an infinite sum and then bounding that sum with the

constant *Integral(log(z)/e^z, z=1 to Infinity) which is yet smaller than some Integral (e^-(z/2), {z, 1, Infinity}] = some constant because it is always true that log(z)<e^(z/2) eg for z>1 as can be seen by taking the derivative of the function e^(z/2)-log(z) and noticing it is positive for z>1 and checking the value of the function at z=1 etc securing the inequality for all z of interest and the bounding by the last integral that is some new constant.

Basically all this lengthy bounding effort served as purpose to show that

f(x)<Constant*x for some large enough constant.

But this is all we needed to show. Because now we have come to a contradiction because we showed that f is bounded by a linear function of x while |h(x)| increases exponentially if it has some starting value other than 0 at some point like x=1 or any other point x>0 for that matter.

It is therefore impossible to have h(x) anything other than 0 at any point x>0 because we reach contradictions based on the definition of f as the infinite nested radical.


This is how we know for sure the solution is f(x)=x+1 here. And so f(2)=3 for the original problem.

[Tehom]

But you also ought to show that the sequence converges. For an example of how it might not, consider 17 instead of 3:

17 = sqrt(289) = sqrt(1 + 288) = sqrt(1 + 2 * sqrt(72))

So there's a sequence that starts the same way but clearly is not equal to 3. The 17 sequence peters out after two iterations, but it still makes the point that you cannot judge the sequence by its first few components.

[alemorita]

Any way to solve it as a recurrence equation? I have (x_n)^2 = 1 + n x_(n-1), which in principle should be solvable.

[utpalsavaliya]

He staggered at 2:41
when he said
"Thanks for .... watching this video."

[IlTrojo]

I think the major issue here is not to see what the solution might be, but rather how to properly define the radical. In other words, you have to well-define the radical before the problem even makes sense. This is what I still find myself unable to do.

[saxbend]

James Grime posted this one too a few days ago. It's a good one though so I won't spoil it.

[ShoVo]

You can start with any value and by carefully factoring, it will equal the samething