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Hi this contains the question that I am trying to solve.

The first line of the last paragraph is what I am having trouble with. I don't see immediately how is x(n) > 0

Thank you for any help.

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    If $0 – 2017-02-10
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    Thank you for your comment. – 2017-02-10

2 Answers 2

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Considering $y_n=\frac1{x_n^2}$ one gets the recursion \begin{align} y_{n+1}=\frac1{x_n^2(1-x_n^2/2)^2} &=\frac{1}{x_n^2}\Bigl(1+x_n^2+\frac34x_n^4+…+\frac{k+1}{2^k}x_n^{2k}+… \Bigr)\\ &=y_n+1+x_n^2\Bigl(\frac34+\frac12 x_n^2+\frac{5}{16}x_n^4+…\Bigr) \end{align} by the square of the geometric series. This results in, for $04·10^8$ or $n>1.5·10^8$ which indeed is larger than 100 million. The actual index to reach that value will be closer to the crossing of the upper bound, or around $n=300·10^6$

Or perhaps without power series $$ y_{n+1}=\frac{y_n^2}{y_n-\frac12}=y_n+1+\frac1{4y_n-2} $$ which again immediately gives, assuming $y_0\ge1$, $y_n\ge y_0+n$ and using that $$ y_0+n\le y_n\le y_0+n+\frac{n}{4y_0-2} $$ which will give a tighter lower bound for the $n$ in the example.

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    Sorry am I missing something here, how does the iteration change to x^3 from x^2. From the screenshot where it states x(n+1).. to the second line? – 2017-02-11
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    You have $x_{n+1}=x_n-x_n^3/2=x_n(1-x_n^2)$ as used in the text and other answer. Then $x_{n+1}^2=x_n^2(1-x_n^2)^2$ so that everything is in squares and the rest follows. Essentially this replicates the substitution used in Bernoulli ODE for difference equations. – 2017-02-11
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The easy seeing happens in the next sentence: We have $x_{n+1}=x_n(1-x_n^2/2)$, so if we know $0

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    Sorry am I missing something here, how does the iteration change to x^3 from x^2. From the screenshot where it states x(n+1).. to the second line? – 2017-02-11
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    That is because $x^3=x·x^2$, the examination of the second quatratic factor explores the contraction behavior. – 2017-02-11
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    Quite possibly my last post but I have looked at this question for the better part of today. Your end argument makes perfect sense, however are you letting xn being bounded by sqrt2 or is this factual, how do you gather the bracket is bounded by 1, and finally what how can x(n+1) be represented differently as in is it equivalent to something similar to x*x(n) Thank you. – 2017-02-13