Is ma solution correct to this equation?

Question

$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=3$$

My take: $$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=3\implies {x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}}=9$$ Thus: $$x+3=9\implies x=6$$

Did I do it right?

Yes, you did... assuming the sequence of infinite nested square roots converges finitely. Can you prove this? — DonAntonio, Mar 23 '14 at 20:21

score 3 · Answer 1 · answered Mar 23 '14 at 20:21

3

You got the right answer. This is a recursive question. Take: $$u=\sqrt{x+u}$$ We know this is true because, if you evaluate it, it will come out to be: $u=\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}$ expanded to $\infty$.

We are given what $u$ is. It is $3$ according to the first equation. Thus, plugging it back in: $$3=\sqrt{x+3}$$ $$x=6$$

answered Mar 23 '14 at 20:21

Shahar

3,302

Why the down vote? – Shahar Mar 23 '14 at 20:42
Haters loooolol ^^ – user137465 Mar 23 '14 at 20:47

score 2 · Accepted Answer · answered Mar 23 '14 at 20:30

2

One way to compute this limit is the following:

Let $x_1 = x $, and $x_n = \sqrt{x + x_n}$, for $n \geq 2$. Then prove that the limit of the $x_n$'s exists by showing that the sequence is monotonic and bounded.

Then, we have that $L = \sqrt{x + L}$ but $L =3$ so you get that $x = 6$.

answered Mar 23 '14 at 20:30

Mustafa Said

4,137

Thank you man and good luck in your graduate studies :))))) – user137465 Mar 23 '14 at 20:51

Is ma solution correct to this equation?

2 Answers2