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Given these condition, I am seeking a proof:

Define a sequence of real numbers by $x_1 = 3$ and then, for $n \geq 2$,

$x_n = \sqrt{2 \; x_{n-1} + 1 }$

Prove that for all positive integers $n$, $x_n \geq x_{n+1}$.

I began a proof by induction, but ran out of steam. I tested the base case for $n=2$ , but I could not seem to get anywhere after that. I feel like there is not enough information (ie recursion, sequence, etc) to prove by induction. Is an induction proof an efficient way to proceed? Are there easier methods of proof?

Aryabhata
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Dominick Gerard
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3 Answers3

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For the induction, assume that $x_n\leq x_{n-1}$. Then

$$ \begin{align*} x_n &\leq x_{n-1}\\ 2x_n &\leq 2x_{n-1}\\ 2x_n+1 &\leq 2x_{n-1}+1\\ \sqrt{2x_n+1} &\leq \sqrt{2x_{n-1}+1}\\ \end{align*} $$

J126
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$ x_n \geq x_{n+1} $ if and only if $x_n \geq \sqrt{2x_n+1}$ and this is true if and only if ${x_n}^2 \geq 2x_n+1$ which is same as ${x_n}^2 -2x_n-1 \geq 0$, and this expression can be written as $(x_n+\sqrt{2}-1)(x_n-\sqrt{2}-1) \geq 0$

The signs in first expression was flipped, just fixed it (Patrick was right)

So try to show that expression is always positive for every $n$

Now how do we show $x_n \geq \sqrt{2}+1$ and $x_n \geq 1-\sqrt{2}$ for every $n$ by induction?

I think Joe Johnson has better proof, follow that.

Kirthi Raman
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  • The roots of the equation are $\frac 12 \left(-(-2) \pm \sqrt{(-2)^2 - 4(-1)(1)} \right) = \frac 12 \left( 2 \pm \sqrt 8 \right) = 1 \pm \sqrt 2$. So the factorization should be $$ x_n^2 - 2x_n -1 = (x_n - (1+\sqrt 2))(x_n - (1-\sqrt 2)) $$ which is not the same as your expression. But you got the idea so I upvoted anyway. (I noticed this because the roots should be $a \pm \sqrt b$ something, so the sign should change on the $\sqrt{b}$, not on the other term.) – Patrick Da Silva Mar 15 '12 at 22:53
  • I understand what you guys have done here, but do you have any hints on how to proceed to prove that $x_n \geq (1+\sqrt{2})$ and $x_n \geq (1-\sqrt{2})$? – Dominick Gerard Mar 15 '12 at 22:59
  • @Patrick I think it is the same expression without the parenthesis, i.e. $(x_n-\sqrt{2}+1)(x_n-\sqrt{2}-1)=(x_n-(\sqrt{2}-1))(x_n-(\sqrt{2}+1))$ – Jeremy Carlos Mar 15 '12 at 23:50
  • @Jeremy : it cant be the same, the roots are not$ \sqrt 2 \pm 1$, theyre $1\pm 2$. Recall that a polynomial with two distinct roots a and b is always of the form$ (x-a)(x-b)$ or expand it correctly, youll see i am right – Patrick Da Silva Mar 16 '12 at 04:23
  • Hm at first i thought you said i was wrong. But yes the parenthesis.dont matter, i was putting emphasis on the roots ; it is a good mathematician reflex, since a polynomial with as many distinct roots as its degree is a product of linear factors of the form $x-a_i$, where $a_i$'s are the roots. Therefore since my roots are $1\pm \sqrt 2$ i dont want to remove my parenthesis to show the roots better. – Patrick Da Silva Mar 16 '12 at 04:27
  • But my expression is different from kirthi's thats for sure. – Patrick Da Silva Mar 16 '12 at 04:29
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Calculate $x_2$, and notice that $x_{n+1}-x_n$ has the same sign as $x_{n}-x_{n-1}$, and inductively, the same sign as $x_2-x_1$, which you just calculated.

Beni Bogosel
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