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How can I solve the equation: $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}} = \frac{1+\sqrt{53}}{2}$$

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Is $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x}\sqrt{x...}}}}$ supposed to be $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}$? –  Andrew Salmon Mar 10 '13 at 1:08

1 Answer 1

up vote 8 down vote accepted

Set $y = \sqrt{x + \sqrt{x + \dots}}$.

$y^2 = x+\sqrt{x+\dots} = x + y$

So $y^2 - y = x$.

But $\displaystyle y = \frac{1 + \sqrt{53}}{2}$

Now just substitute.


Addendum for people who want to see something a bit more rigorous.

We will take the infinite square root expression to be defined as the limit of the sequence defined recursively:

$$s(1) = \sqrt{x}$$

$$s(n+1) = \sqrt{x + s(n)}$$

We want to take $\lim_{n \to \infty} s(n)$.

For $x > 0, s(n) \le s(n+1)$ by induction on $n$:

$s(1) \le s(2)$ because $s(2) = \sqrt{x + s(1)} > \sqrt{x}$ since $s(1) > 0$.

If $s(n) \le s(n+1)$, then $s(n+2) = \sqrt{x + s(n+1)} \ge \sqrt{x+s(n)} = s(n+1)$.

So $s$ is monotonic.

Moreover, $s$ is bounded above by $\max(x,2)$. This will also be proven by induction.

$s(1) = \sqrt{x} \le \max(x,1) \le \max(x,2)$

If $s(n) \le 2$, then $s(n+1) = \sqrt{x+2} \le \sqrt{4} \le 2$.

Similarly, if $2 < s(n) \le x$, then $s(n+1) = \sqrt{x+s(n)} \le \sqrt{2x} \le \sqrt{x^2} = x$

Since $s$ is bounded and monotonic, it has a limit as $n \to \infty$.

Since $f(y) = \sqrt{ x + y }$ is continuous, $$\lim_{n \to \infty} s(n) = \lim_{n \to \infty} s(n+1) = \lim_{n \to \infty} \sqrt{ x + s(n) } = \sqrt{ x + \lim_{n \to \infty} s(n)}$$

Justifying the reasoning above.

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This shows that if there is a solution to this equation it must satisfy the equation he derives, however it remains to show that each solution satisfies the equation given in the original post. This includes proving that the nested radical sequence given in the original post converges to some finite limit (e.g. by showing that it is bounded and monotone.) –  newToProgramming Mar 10 '13 at 1:11
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Can you please elaborate –  ChrisHeneghan Mar 10 '13 at 1:17
    
Let's assume this equation you gave has a solution. Set $y$ as I did. When we square $y$, the square root sign is removed, so we're left with $x$ plus a square root expression, which is precisely $y$. So $y^2 = x + y$. That should provide the rationale for the first expression given. –  Andrew Salmon Mar 10 '13 at 1:19
    
I think that writing that extra step as $y = \sqrt{x + \sqrt{x+\ldots}} = \sqrt{x + y} \Rightarrow y^2 = x + y$ would've made it clear –  Guest 86 Mar 10 '13 at 1:29
    
That doesn't seem right to me... That's like saying $1+\infty=x$, and $x-\infty=1$... –  agent154 Mar 10 '13 at 1:31

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