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How to simplify the expression:

$$\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\cdots}}}}.$$

If I could at least know what kind of reference there is that would explain these type of expressions that would be very helpful.

Thank you.

J. M. ain't a mathematician
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hyg17
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  • Related: http://math.stackexchange.com/questions/186652/how-can-one-solve-for-x-scriptsize-x-sqrt2-sqrt2-sqrt2-sqrt2-sq/186669#186669 – timur May 31 '13 at 13:20

4 Answers4

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We should really make the problem precise, and prove convergence. But this is the GRE, we manipulate. Let $x$ be the number. Then $x^2-5=2x$. Our number is the positive root of the quadratic.

Adam
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André Nicolas
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    ...and the reason for choosing the positive root is that the negative root doesn't satisfy the original equation $x = \sqrt{5+2x}$, given the usual definition of $\sqrt{z}$ as the principal square root of $z$. – Ilmari Karonen Mar 30 '13 at 15:08
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    It made perfect sense. Thank you very much ! – hyg17 Mar 31 '13 at 03:08
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Let $x = 2\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{...}}}}$. Then (if this converges) $x = 2\sqrt{5+x}$. Solving, $x = 2(1+\sqrt6)$, so the answer to your original question is $1+\sqrt{6}$

Euler....IS_ALIVE
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Put $$x_0:=0,\quad x_1:=\sqrt{5},\quad x_2:=\sqrt{5+2\sqrt{5}},\quad x_3:=\sqrt{5+2\sqrt{5+2\sqrt{5}}}\ ,$$ and so on, which amounts to $$x_0:=0,\qquad x_{n+1}:=\sqrt{5+2x_n}\quad(n\geq0)\ .$$ Then $$x_{n+1}-x_n=\sqrt{5+2x_n}-\sqrt{5+2x_{n-1}}={2(x_n-x_{n-1}) \over \sqrt{5+2x_n}+\sqrt{5+2x_{n-1}}}\ .$$ As $x_1-x_0>0$ this shows that the sequence $(x_n)_{n\geq0}$ is momotonically increasing.

Furthermore $0\leq x_0<4$, and for any $n\geq0$ the statement $0\leq x_n< 4$ implies $$0\leq x_{n+1}<\sqrt{5+2\cdot 4}<4\ .$$ This shows that our sequence is as well bounded, so it has a limit $\xi\in[0,4]\ $. This limit satisfies the equation $x=\sqrt{5+2x}$ and is therefore given by $\xi=1+\sqrt{6}\doteq 3.45$.

Christian Blatter
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Let $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$.
Then $x^2=5+2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$.
So, $x^2-5=2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$
Remember that $x=\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$
So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}$.
So, $2\sqrt{5+2\sqrt{5+\sqrt{5+2\sqrt{\dots}}}}=2x$
This automatically makes our equation to be $x^2-5=2x$. Just solve the quadratic equation, and take the positive root. Why only positive? Think about it: how can $\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}$ equal a negative number? It must be positive! So that's why we only take the positive root. $$x^2-2x-5=0$$ $$x=\dfrac{2\pm \sqrt{(-2)^2-4(1)(-5)}}{2(1)}$$ $$x=\dfrac{2\pm \sqrt{4+20}}{2}$$ $$x=\dfrac{2\pm \sqrt{24}}{2}$$ $$x=\dfrac{2\pm 2\sqrt{6}}{2}$$ $$x=1\pm \sqrt{6}$$ Since $1-\sqrt{6}$ equals a negative number, we reject that root. So, $x=1+\sqrt{6}$, which is also the answer to the original problem.

Answer: $$\displaystyle \boxed{\sqrt{5+2\sqrt{5+2\sqrt{5+2\sqrt{\dots}}}}=1+\sqrt{6}}$$

Anonymous Computer
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