# Thread: help me to simplify this ...

1. ## help me to simplify this ...

2. let $x_1 = \sqrt{2}$
and $x_{n+1} = \sqrt{2x_n}$

Then $x_{n+1}^2=2x_n$

If the limiting value of $x_n$ is $x$, then the limiting value of $x_{n+1}$ is also $x$.

So $x^2=2x$

$x^2-2x=0$
$x(x-2)=0$

So we see that $x=2$

Note that the given sequence is increasing and bounded above (you can show easily by induction on n that each member of the sequence is bounded by 2). Thus the sequence is guaranteed to converge.

3. Hello, rcs!

$\text{Evaluate: }\;2\sqrt{2\sqrt{2\sqrt{2\hdots}}}$

Let: . $x \;=\;2\sqrt{2\sqrt{2\sqrt{2\hdots}}}$

$\text{Square: }\;x^2 \;=\;2\cdot\underbrace{2\sqrt{2\sqrt{2\sqrt{2\hdot s }}}}_{\text{This is }x}$

We have: . $x^2 \:=\:2x \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0$

Therefore: . $x \:=\:2$

4. Soroban - I found a little mistake in each of our arguments. The answer is actually 4.

In my argument, I misread the problem and left off the constant 2 in front. Since constants can be pulled outside limits this is no big deal, and you can just multiply my answer of 2 by 2 to get 4.

In your argument, the right hand side of where you wrote "Square" should actually start with a 4 (since 2^2 = 4).

5. Here is my corrected solution:

Let $x_1 = 2\sqrt{2}$
and $x_{n+1} = 2\sqrt{x_n}$

Then $x_{n+1}^2=4x_n$

If the limiting value of $x_n$ is $x$, then the limiting value of $x_{n+1}$ is also $x$.

So $x^2=4x$

$x^2-4x=0$
$x(x-4)=0$

So we see that $x=4$

Note that the given sequence is increasing and bounded above (you can show easily by induction on n that each member of the sequence is bounded by 4). Thus the sequence is guaranteed to converge.

6. thank you so so much ... u have been so good guys. i was actually trying to solve it add i was stuck i couldn't finish it. thanks a lot