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What is the interval where the function converges to this fixed point?

So we have a function $$g(x) = \frac{x^2+a}{2x}$$

We want to know the interval where this function for sure converges to fixed point $\sqrt{a}$.

As far as I know this is true for $|\frac{d}{dx}g(x)| < 1$, so we get:

$$-1<\frac{x^2-a}{2x^2}<1$$

and we get $$3x^2 >a>-x^2$$

However If we look at for what $x$ this is true we get: $x < - \sqrt{a/3}$ or $x>\sqrt{a/3}$.

So I would say that the interval is $x < - \sqrt{a/3}$ or $x>\sqrt{a/3}$. However in solutions the interval is

$$(\sqrt{a} - \sqrt{a/3},\sqrt{a} + \sqrt{a/3} )$$

Why is that so ?

VLC
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  • You have $|g'|\lt 1$ implies $|g| \lt 1$, why? – herb steinberg Dec 14 '21 at 17:42
  • Well, the negative values converge to the other fixed point, $-\sqrt a.$ But not sure why the narrow range. – Thomas Andrews Dec 14 '21 at 17:43
  • It seems like you are using a particular technique for finding the bounds on $x$ such that $x_0=x, x_{n+1}=g(x_n)$ converges to a fixed point of $g.$ The sequence definitely converges to $\sqrt a$ for all $x_0>0,$ but can’t converge to a positive value when $x_0<0,$ because $g(x)<0$ when $x<0.$ So your “correct” answer is also wrong, unless you are deducing a minimal interval using some specific theorem. – Thomas Andrews Dec 14 '21 at 18:04
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    Maybe you can state what you mean by “where the function converges to this fixed point” means. Does it mean convergence repeated application of $g?$ That is, convergence of $x_n$ where $x_0=x$ and $x_{n+1}=g(x_n)?$ Because both your answer and the “correct answer” both seem wrong to me, if that is what you mean. – Thomas Andrews Dec 14 '21 at 18:17
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    See https://math.stackexchange.com/questions/2289703/finding-the-square-root-newton-raphson-convergence-proof/2291600#2291600 and similar answers inspired by the proof of the Newton-Kantorovich theorem. – Lutz Lehmann Dec 15 '21 at 08:11

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If you start by noticing that, when $x_0 <\sqrt{a}$, you have $x_1>\sqrt{a}$, you can, without loss of generality, study the convergence on the interval $[\sqrt{a}, +\infty)$. In this interval, since the iteration function $g(x)=\frac x2 +\frac{a}{2x}$ is invariant and contractive, the fixed point theorem guarantees that the sequence will converge to $\sqrt{a}$, regardless of the choice of $x_0\in [\sqrt{a},+\infty)$. The initial observation implies that conv ergence will in fact occur for $x_0 \in (0,+\infty)$.

PierreCarre
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