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Given the following Newton iteration: $x_{n+1}=2x_n-x^{2}_{n}y$
Identify the function being iterated.

The solution was:
$$x_{n+1}=2x_n-x^{2}_{n}y= x_n-(x^{2}_{n}y-x_n)=x_n-\frac{y -\frac{1}{x_n}}{\frac{1}{x^{2}_{n}}}$$

which gives $f(x)=y - \frac{1}{x}$

I understand that we need to put $$(x^{2}_{n}y-x_n)$$ into the form $$\frac{f(x)}{f'(x)}$$

But I don't understand how you "derive" $f(x)$ and $f'(x)$ from $(x^{2}_{n}y-x_n)$
I'd like to know what the derivation method was.

Jean Marie
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user137481
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  • We have $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ so if you are given than $x_{n+1} = 2x_n - x_n^2y$ comes from Newton's method then we have $2x_n - x_n^2y = x_n - \frac{f(x_n)}{f'(x_n)}$ for some function $f$ which gives us the ODE $2x - x^2y = x - \frac{f(x)}{f'(x)}$ or $\frac{f'(x)}{f(x)} = -\frac{1}{x-x^2y}$. Solve this (integrate both sides) to find your function. – Winther May 07 '17 at 21:17
  • @Winther Thanks. I've got to brush up on my ODE since it's been a long time... – user137481 May 07 '17 at 21:20
  • Also, if $r=1-xy$, then $1-r=xy$ and $1=y·x(1+r+r^2+r^3+…)$ which to second order error $O(r^2)$ gives the improved inverse $x_+=x(1+r)=x(2-xy)$, which is the given recursion. – Lutz Lehmann May 07 '17 at 22:19
  • @Winther In the answers of my numerical analysis book, it comes out that it is the $1/y$ function but I do not know why, could someone help me please? – Nash Mar 04 '18 at 01:57

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