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

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

Proof so far.

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

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.

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

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

Proof so far.

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

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?

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? $\displaystyle \lim x_{n + 1} = \lim x_n$. so that, from our equation we have, $\displaystyle L = \sqrt{c + L}$. now solve for L. this would complete the proof after you prove convergence

3. I think I'm completely lost on this one.

First, is $\displaystyle \sqrt {c+ \sqrt {c+x_1 } } > \sqrt {c+x_1}$? It really depends on what c is.

Second, I have $\displaystyle L = \sqrt { c+L }$, then obtains $\displaystyle L^2 - L - c = 0$, now how do I solve for L?

Thanks.

4. $\displaystyle L = \frac{{1 \pm \sqrt {1 + 4c} }}{2}$
Be careful to check for domain problems.