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Define a sequence {$\ x_n$} recursively by

$$ x_{n+1} = \sqrt{2 x_n -1}, \ and \ x_0=a \ where \ a>1 $$ Prove that {$\ x_n$} is strictly decreasing. I'm not sure where to start.

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    $\begingroup$ Welcome to Mathematics Stack Exchange! I would suggest you to explain a little bit how you tried to solve it (at least your thoughts), so other people could help you better. Good luck! $\endgroup$
    – iadvd
    Commented May 29, 2015 at 0:07

5 Answers 5

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We need to show that $x_{n+1}-x_n<0$ for all $n\ge 1$ whenever x_0=a>1$

First note that if $x_n>1$, then $x_{n+1}=\sqrt{2x_n-1}>1$ also. And since $x_0>1$, then we have that $x_n>1$ for all $n$.

Next, to show $x_{n+1}-x_n<0$, all we need to show is that $2x_n-1<x_n^2$. But this is trivially the same as $x_n^2-2x_n+1=(x_n-1)^2>0$. And since we know $x_n>1$, then we are done!

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One can view it as a function $a_n = f(a_{n-1})$ whereas $f(x) = \sqrt{2x-1}$, then the monotonicity of $a_n$ depends on $f'(x)$. And $f'(x) = \dfrac{1}{\sqrt{2x-1}} > 0$, this means the sequence is increasing as stated.

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Hint:

It is enough to show the graph of $y=\sqrt{2x-1}$ is under the straight line $y=x$. For an increasing sequence it should be over this line.

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Use induction. Your base case is $a > \sqrt{2a - 1} \Leftrightarrow a^2 > 2a - 1 \Leftrightarrow (a-1)^2 > 0$, which is true since $a > 1$. Furthermore, notice that $x_1 = \sqrt{2a - 1} > 1$.

Inductive step: assume that $x_k < x_{k-1}$ and $x_k > 1$.

We wish to show that $x_{k+1} < x_k$. Well, we have $x_{k+1} = \sqrt{2 x_k - 1}$, so

$$\sqrt{2 x_k - 1} < x_k$$

$$0 < (x_k - 1)^2$$

Finally, this last statement is true since we assumed $x_k > 1$. Thus, we get that $x_{k+1} < x_k$ and that $x_{k+1} > 1$, so the induction is proven.

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  • $\begingroup$ Careful with the logic in your base case. You have shown your desired conclusion implies something True, which shows nothing. It can be fixed by changing all your $\Rightarrow$ into $\Leftrightarrow$. $\endgroup$
    – Simon S
    Commented May 29, 2015 at 0:17
  • $\begingroup$ @SimonS Yeah, it was a typo. Sorry $\endgroup$
    – MT_
    Commented May 29, 2015 at 4:47
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The condition to prove is that $x_{n+1}<x_n$, that is, $$ \sqrt{2x_n-1}<x_n $$ that's equivalent to $$ 2x_n-1<x_n^2 $$ or $$ x_n^2-2x_n+1>0 $$ that's true provided you prove in advance that $x_n\ne1$, for all $n$, and moreover that $x_n\ge1/2$, for all $n$, in order that the expressions are meaningful.

Try proving, instead, that $x_n>1$, by induction on $n$.

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  • $\begingroup$ Why do you say that $x_n > 1/2$ must be true? Isn't it just that $x_n \neq 1$? $\endgroup$
    – MT_
    Commented May 29, 2015 at 0:13
  • $\begingroup$ @Soke If $x_n<1/2$, then $2x_n-1<0$ and you can't define $x_{n+1}$. $\endgroup$
    – egreg
    Commented May 29, 2015 at 8:20
  • $\begingroup$ Okay, but you're glossing over the fact that $x \neq 1$ is necessary for the inequality to hold. $\endgroup$
    – MT_
    Commented May 29, 2015 at 15:29
  • $\begingroup$ @Soke You're right, I added the condition. $\endgroup$
    – egreg
    Commented May 29, 2015 at 15:50

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