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Recently I came across this question. Compute $$\sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}.$$

I know how to do it. By equating the above expression to $x$ (i.e., $x = \sqrt{10+\sqrt{100+\sqrt{10,000 +\dots}}}$) and $x = \sqrt{10+x}$ and so on. But how is this justified? I mean if we take the expression $\sqrt{10+x}$, literally it is equal to $\sqrt{10+\sqrt{10+\sqrt{100 +\dots}}}$ which alters the original expression. Can someone clarify my doubt?

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I'm not quite sure I understand precisely the value you're computing. Is it: $$\sqrt{10+\sqrt{100+\sqrt{1000+\cdots}}}$$ – Drew Christianson Sep 22 '12 at 16:22
the third term under the square root is 10,000. – Ranjan Yajurvedi Sep 22 '12 at 16:25
the 4th term is 100000000 – Ranjan Yajurvedi Sep 22 '12 at 16:32
It might be easier if you'd written $$\sqrt{10+\sqrt{10^2+\sqrt{10^4+\cdots}}}$$ instead. – Cameron Buie Sep 22 '12 at 17:31
@CameronBuie : Not just easier, but also clear. The original question is not clear what is meant to be. – Arjang Sep 23 '12 at 12:29
up vote 3 down vote accepted

In this case, it isn't so simple as putting $x=\sqrt{10+x}$--as you've noted, this changes the expression. Now, if you were looking at $\sqrt{10+\sqrt{10+\sqrt{10+\cdots}}}$, that's precisely what you'd do. The reason we can do that, there, is that we're dealing with a sequence defined recursively by $x_1=\sqrt{10}$, $x_{n+1}=\sqrt{10+x_n}$, and then determining the limit of this sequence. That sequence does converge (as it is increasing and bounded above, for example by $10$), so letting $x$ be the limit, we take $n\to\infty$ in the recursion equation $x_{n+1}=\sqrt{10+x_n}$, yielding $x=\sqrt{10+x}$, which we can then use to determine the limit.

Let's consider the sequence given by $y_1=\sqrt{10}$, $y_{n+1}=\sqrt{10+y_n\sqrt{10}}$. Then $$y_2=\sqrt{10+10}=\sqrt{10+\sqrt{100}},$$ $$y_3=\sqrt{10+\sqrt{10}\sqrt{10+\sqrt{100}}}=\sqrt{10+\sqrt{100+10\sqrt{100}}}=\sqrt{10+\sqrt{100+\sqrt{10,000}}},$$ and so on. Now, if this sequence converges (say to $y$), then we take $n\to\infty$ in the equation $y_{n+1}=\sqrt{10+y_n\sqrt{10}}$, yielding $y=\sqrt{10+y\sqrt{10}}$, which we may solve to determine $y$. The trickier part (in this case) is determining an upper bound, so you can prove convergence. From the sounds of it, though, you're supposed to take for granted that it converges, so all you've got to do is the procedure above.

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Thank You Very Much – Ranjan Yajurvedi Sep 23 '12 at 12:10


$\implies \frac x{\sqrt{10}}=\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ (This is the well-known Golden Ratio)

so, $\frac x{\sqrt{10}}=\frac{ 1+\sqrt 5}{2}$

Alternatively, $ \frac x{\sqrt{10}}=\sqrt{1+\frac x{\sqrt{10}}}$

Squaring we get, $\frac {x^2}{10}=1+\frac x{\sqrt{10}}\implies x^2-\sqrt{10}x-10=0$

$\implies x=\frac{\sqrt 10\pm \sqrt {50}}{2}=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$

or $y=\sqrt{1+y}$ putting $\frac x{\sqrt{10}}=y$

$\implies y^2=1+y\implies y=\frac{1\pm\sqrt5}2\implies x=\frac{\sqrt{10}(1\pm\sqrt 5)}{2}$

But $x>0,$ which means $x=\frac{\sqrt{10}(1+\sqrt 5)}{2}$

For the converge, one may check for "Geometric Infinite Surd" here.

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I think we cannot take 10 common from the first term i.e., 10 – Ranjan Yajurvedi Sep 22 '12 at 16:37
@RanjanYajurvedi, thanks for your observation, we need to take $\sqrt {10}$ common. – lab bhattacharjee Sep 22 '12 at 16:39
Thanks for the help. But my doubt is, normally the procedure to calculate such expressions is to equate the original expression to x and replacing from second term with to infinity with x. That is x=SQRT(10+x). Doesn't it deviate from the original question? – Ranjan Yajurvedi Sep 22 '12 at 16:49
You need $\sqrt{2.5}+\sqrt{12.5}$ about 5.1166727, a solution to $\frac x{\sqrt{10}}=\sqrt{1+\frac x{\sqrt{10}}}$ – Henry Sep 22 '12 at 16:51
This is helpful, but it shows only that IF IT CONVERGES, then this is its value. – GEdgar Sep 22 '12 at 16:52

Since the OP expressed some doubts about the procedure not the solution, here are some elements of reassurance. Let $a=10$, or, more generally, any positive real number.

In any interpretation of the exercise, the value of $x$, if it exists, should correspond to the limit of a sequence $(x_n)_{n\geqslant0}$ such that $x_{n+1}=\sqrt{a+\sqrt{a}x_n}$, for every $n\geqslant0$. Here is a fact:

Let $(x_n)_{n\geqslant0}$ denote any sequence defined as above. For every $x_0\geqslant-\sqrt{a}$, $x_n\to\ell_a$, where $\ell_a=\frac12(\sqrt5+1)\sqrt{a}$.

Thus, about any reasonable procedure used to define $x$, first, will succeed, and second, will yield $x=\ell_a$. We happy. (When $a=10$, $\ell_a=\frac{5+\sqrt5}{\sqrt{2}}$.)

To prove the fact stated above, consider the function $u_a$ defined by $u_a(t)=\sqrt{a+\sqrt{a}t}$, for every $t\geqslant-\sqrt{a}$. Then $u_a$ is continuous, increasing, such that $t\lt u_a(t)\lt\ell_a$ for every $-\sqrt{a}\leqslant t\lt\ell_a$, $u_a(\ell_a)=\ell_a$, and $\ell_a\lt u_a(t)\lt t$ for every $t\gt\ell_a$.

As a consequence, $x_n=\ell_a$ for every $n\geqslant0$ if $x_0=\ell_a$, $(x_n)_{n\geqslant0}$ is increasing and bounded above by $\ell_a$ and converges to $\ell_a$ for every $-\sqrt{a}\leqslant x_0\lt\ell_a$, and $(x_n)_{n\geqslant0}$ is decreasing and bounded below by $\ell_a$ and converges to $\ell_a$ for every $x_0\gt\ell_a$. In every case, $x_n\to\ell_a$.

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Thanks for the Help. – Ranjan Yajurvedi Sep 23 '12 at 12:10

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