- #1
Caeder
- 13
- 0
Given the following sequence:
[tex]x_0 = 1, \quad x_1 = \sqrt{3+1}, \quad x_2 = \sqrt{3+\sqrt{4}}, \quad x_3 = \sqrt{3+\sqrt{5}},[/tex]
[tex]x_4 = \sqrt{3+\sqrt{3+\sqrt{5}}}, \quad x_5 = \sqrt{3+\sqrt{3+\sqrt{3+\sqrt{5}}}} \ldots[/tex]
prove the above sequence converges and determine the limit.
......
So from [tex]n=3[/tex] onwards, I notice that the sequence is recursively defined:
[tex]n \geq 4,\;x_n = \sqrt {3 + x_{n - 1} }[/tex]
To prove convergence, I'd stimply have to show sequence is bounded above and that it's increasing.
I'm not quite sure how to do this with a recursive function.
To find the limit, I realized that
[tex]x_n = \sqrt {3 + x_{n - 1} }[/tex] is at it's "equilibrium point" when [tex]x = 3 + \sqrt{x}[/tex]. I solved for [tex]\sqrt{x}[/tex] and found the limit to be:
[tex]\frac{1\pm\sqrt{13}}{2}[/tex]. I then see that we go towards:
[tex]\frac{1+\sqrt{13}}{2}[/tex]
So I know the limit will be the above if the sequence converges.
Any help on the proof?
[tex]x_0 = 1, \quad x_1 = \sqrt{3+1}, \quad x_2 = \sqrt{3+\sqrt{4}}, \quad x_3 = \sqrt{3+\sqrt{5}},[/tex]
[tex]x_4 = \sqrt{3+\sqrt{3+\sqrt{5}}}, \quad x_5 = \sqrt{3+\sqrt{3+\sqrt{3+\sqrt{5}}}} \ldots[/tex]
prove the above sequence converges and determine the limit.
......
So from [tex]n=3[/tex] onwards, I notice that the sequence is recursively defined:
[tex]n \geq 4,\;x_n = \sqrt {3 + x_{n - 1} }[/tex]
To prove convergence, I'd stimply have to show sequence is bounded above and that it's increasing.
I'm not quite sure how to do this with a recursive function.
To find the limit, I realized that
[tex]x_n = \sqrt {3 + x_{n - 1} }[/tex] is at it's "equilibrium point" when [tex]x = 3 + \sqrt{x}[/tex]. I solved for [tex]\sqrt{x}[/tex] and found the limit to be:
[tex]\frac{1\pm\sqrt{13}}{2}[/tex]. I then see that we go towards:
[tex]\frac{1+\sqrt{13}}{2}[/tex]
So I know the limit will be the above if the sequence converges.
Any help on the proof?