- #1

Poopsilon

- 294

- 1

So for those of you who don't have the book the problem goes like this:

Fix a > 1, take x_1 > sqrt(a), and define:

[tex]x_{n+1}=\frac{a+x_n}{1+x_n}=x_n+\frac{a-x_n^{2}}{1+x_n}[/tex]

a) Prove that x_1 > x_3 > x_5 > ...

b) Prove that x_2 < x_4 < x_6 < ...

Basically my strategy is to show that the sequence r_n=x_n+2 - x_n alternates sign, with it giving negative terms for n odd and positive terms for n even. My proof so far is as follows:

[tex]x_{n+1} - x_n = \frac{a - x_n^{2}}{1+x_n}[/tex] and [tex]x_{n+2} - x_{n+1} = \frac{a - x_{n+1}^{2}}{1+x_{n+1}}[/tex]

adding these two equations and setting r_n = x_n+2 - x_n produces:

[tex]r_n = x_{n+2} - x_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - x_{n+1}^{2}}{1+x_{n+1}}[/tex]

now substituting in [tex]\frac{a+x_n}{1+x_n}[/tex] for [tex]x_{n+1}[/tex] gives:

[tex]r_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - (\frac{a+x_n}{1+x_n})^2}{1+(\frac{a+x_n}{1+x_n})}[/tex]

from here I simplify the right side until obtaining:

[tex]r_n = \frac{2x_n(a-x_n^{2})+2a}{(1+x_n)(2x_n+a+1)}[/tex]

I have checked my work multiple times on this simplification and I cannot find any error. Yet if we set n=1 than clearly the denominator is positive, but if one chooses x_1 to be just a tiny bit greater than sqrt(a) than the numerator can be made positive as well, which contradicts the desired conclusion for part a).

I'm fairly certain my strategy is viable, hopefully someone can do the simplification and see if they are getting the same result. Its certainly possible that I'm missing something.

Fix a > 1, take x_1 > sqrt(a), and define:

[tex]x_{n+1}=\frac{a+x_n}{1+x_n}=x_n+\frac{a-x_n^{2}}{1+x_n}[/tex]

a) Prove that x_1 > x_3 > x_5 > ...

b) Prove that x_2 < x_4 < x_6 < ...

Basically my strategy is to show that the sequence r_n=x_n+2 - x_n alternates sign, with it giving negative terms for n odd and positive terms for n even. My proof so far is as follows:

[tex]x_{n+1} - x_n = \frac{a - x_n^{2}}{1+x_n}[/tex] and [tex]x_{n+2} - x_{n+1} = \frac{a - x_{n+1}^{2}}{1+x_{n+1}}[/tex]

adding these two equations and setting r_n = x_n+2 - x_n produces:

[tex]r_n = x_{n+2} - x_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - x_{n+1}^{2}}{1+x_{n+1}}[/tex]

now substituting in [tex]\frac{a+x_n}{1+x_n}[/tex] for [tex]x_{n+1}[/tex] gives:

[tex]r_n = \frac{a - x_n^{2}}{1+x_n} + \frac{a - (\frac{a+x_n}{1+x_n})^2}{1+(\frac{a+x_n}{1+x_n})}[/tex]

from here I simplify the right side until obtaining:

[tex]r_n = \frac{2x_n(a-x_n^{2})+2a}{(1+x_n)(2x_n+a+1)}[/tex]

I have checked my work multiple times on this simplification and I cannot find any error. Yet if we set n=1 than clearly the denominator is positive, but if one chooses x_1 to be just a tiny bit greater than sqrt(a) than the numerator can be made positive as well, which contradicts the desired conclusion for part a).

I'm fairly certain my strategy is viable, hopefully someone can do the simplification and see if they are getting the same result. Its certainly possible that I'm missing something.

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