How can the endless square root problem be solved?

[Chirag B]

I'm having a little bit of trouble figuring out how exactly to do this.

Prove that \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}}} = \frac{1\pm\sqrt{4n+1}}{2}.

How exactly does one go about doing this? I mean, I understand it goes on infinitely, but doesn't that create an infinitely large number? Why would the number be so precise?

Also, it appears to be the solution of a quadratic equation (with the \pm and all). Is that an approach? If so, how does one derive such a quadratic equation?

I tried working backwards, but when I do, all I get is that \sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\cdots}}}} = \frac{1\pm\sqrt{4n+1}}{2}.

There's one less n in the square root, but I still end up getting the same equation. Can this process by repeated infinitely until I get \sqrt{n} = \frac{1\pm\sqrt{4n+1}}{2}? But this isn't true! Can someone help me?

Any help would be gladly appreciated.

 

[micromass]

Here's the trick:

Let x=\sqrt{n+\sqrt{n+\sqrt{n+...}}}

Then x^2-n=\sqrt{n+\sqrt{n+\sqrt{n+...}}}

So x^2-n=x.

This is a quadratic equation which you can solve.

Note, I did not take into consideration any convergence issues here.

 

[Chirag B]

Ah, that's a neat trick. So you would have to think a little outside of the square root, if you get what I mean.

Thanks!

 

[Office_Shredder]

If you want to solve the problem carefully, you have to prove that the sequence
\sqrt{n},\ \sqrt{n+\sqrt{n}},\ \sqrt{n+\sqrt{n+\sqrt{n}}},... converges before you can start doing algebraic manipulations

Otherwise you could have something like x=n+n+n+..., and get x=x+n. Since n is arbitrary we get that n=0 always, which isn't true, and the mistake was assuming that we had a number to begin with

 

[lugita15]

If you want to solve the problem carefully, you have to prove that the sequence
\sqrt{n},\ \sqrt{n+\sqrt{n}},\ \sqrt{n+\sqrt{n+\sqrt{n}}},... converges before you can start doing algebraic manipulations

Otherwise you could have something like x=n+n+n+..., and get x=x+n. Since n is arbitrary we get that n=0 always, which isn't true, and the mistake was assuming that we had a number to begin with

I think an appropriate fixed point theorem is what's needed to sort out the convergence issues.

 

[Chirag B]

I think an appropriate fixed point theorem is what's needed to sort out the convergence issues.

Unfortunately, I'm not familiar with Fixed Point Theorems. However, I am familiar with rudimentary sequences/series. Would I have to prove the convergence of the sequence as Office_Shredder has suggested, and only then use the formula that micromass gets? Thank you in advance.

 

[Dickfore]

Try this one instead:
<br /> \sqrt{1 + \sqrt{2 + \sqrt{3 + \ldots}}} = ?<br />

 

[A. Bahat]

How devious! That expression has no known closed-form.

 

[Office_Shredder]

Unfortunately, I'm not familiar with Fixed Point Theorems. However, I am familiar with rudimentary sequences/series. Would I have to prove the convergence of the sequence as Office_Shredder has suggested, and only then use the formula that micromass gets? Thank you in advance.

That's the standard procedure. Most examples where you get a simple polynomial for the limit wool converge because that's the point, but I have once or twice in real life scenarios seen sequences where you can find a" limit" but the sequence doesn't actually converge.

Typically you prove these converge by showing they're cauchy (hard), or by proving they're monotone and bounded (easier to prove if it's true)

 

[Chirag B]

Typically you prove these converge by showing they're cauchy (hard), or by proving they're monotone and bounded (easier to prove if it's true)

So I would have to use the Monotonic Sequence Theorem, I take it?

 

[lugita15]

So I would have to use the Monotonic Sequence Theorem, I take it?

Yes, here you can use the fact that for real numbers, a monotone bounded sequence is convergent. But for general problems of this type, your sequence may not be monotone, so it's better to read up on fixed point theorems to prove convergence.