• October 3rd 2008, 05:36 PM
Consider the equation $x^2-x-c=0, \ \ \ x>0$
Define the sequence $\{ x_n \}$ by $x_{n+1} = \sqrt {c+x_n} \ \ \ x_1 >0$

So that the sequence converges monotonically to the solution of the equation.

Proof so far.

I have $x_2 = \sqrt {c +x_1 }$ and $x_3 = \sqrt {c+x_2} = \sqrt {c + \sqrt {c+x_1 } } > \sqrt {c+x_1} = x_2$

So this sequence is monotonically increasing, but do I need to do induction to be more convincing?

I want to show that this sequence is bounded, so it does converge. Then suppose it converges to L. But I'm stuck here, guess I didn't start this right.

Any hints? Thanks.
• October 3rd 2008, 05:58 PM
Jhevon
Quote:

Consider the equation $x^2-x-c=0, \ \ \ x>0$
Define the sequence $\{ x_n \}$ by $x_{n+1} = \sqrt {c+x_n} \ \ \ x_1 >0$

So that the sequence converges monotonically to the solution of the equation.

Proof so far.

I have $x_2 = \sqrt {c +x_1 }$ and $x_3 = \sqrt {c+x_2} = \sqrt {c + \sqrt {c+x_1 } } > \sqrt {c+x_1} = x_2$

So this sequence is monotonically increasing, but do I need to do induction to be more convincing?

yes, use induction. can you take it from there?

Quote:

I want to show that this sequence is bounded, so it does converge. Then suppose it converges to L. But I'm stuck here, guess I didn't start this right.

Any hints? Thanks.
yes, say the limit is L, then what do we know? $\lim x_{n + 1} = \lim x_n$. so that, from our equation we have, $L = \sqrt{c + L}$. now solve for L. this would complete the proof after you prove convergence
• October 4th 2008, 12:22 PM
First, is $\sqrt {c+ \sqrt {c+x_1 } } > \sqrt {c+x_1}$? It really depends on what c is.
Second, I have $L = \sqrt { c+L }$, then obtains $L^2 - L - c = 0$, now how do I solve for L?
$L = \frac{{1 \pm \sqrt {1 + 4c} }}{2}$