For positive $p$
$$\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}}}=x$$
I want to show that this is convergent by using contractive mapping.
Can anyone help me?
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Martin Sleziak
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jessie
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1See also these posts: Proof $\lim\limits_{n \rightarrow \infty} {\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}=2$ using Banach's Fixed Point and $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ – Martin Sleziak Oct 03 '15 at 07:17
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You have $x=\sqrt{p+\text{something}}$, and "something" is equal to $x$ itself. So
$$ x = \sqrt{p + x}. \tag 1 $$
Thus $x$ is a fixed point of the mapping $$w\mapsto \sqrt{p+w}. \tag 2$$ We can ask whether that mapping is a contraction.
$$ \frac d {dw} \sqrt{p+w} = \frac{ 1 }{2 \sqrt{p+w}}. $$
If the absolute value of the derivative remains less than some number $c<1$, then we can use the mean value theorem to show that this is a contraction. The derivative is less than $c$ iff $2\sqrt{p+w} > 1/c$, and that means $p+w > 1/(4c^2)$. Thus on the interval
$$ \left( \frac 1 {4c^2} - p, \infty \right) \tag 3 $$
the mapping $(2)$ is a contraction, and thus must have exactly one fixed point in that interval.
If we square both sides of $(1)$ we get a quadratic equation $$ x^2 - x - p = 0. $$ The solutions are $$ x = \frac{1 \pm\sqrt{ 1+4p }} 2. $$
Is exactly one of those inside the interval $(3)$?
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Hi Thank you for your answer. Would you mind explaining more about why derivative less than 1 implies contractive mapping? – jessie Oct 03 '15 at 00:59
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The way my professor defined Contraction is that IF(x)-F(y)I is less than or equal to cIx-yI for all x, y on given interval. Is there any relationship between this definition and derivative less than 1? – jessie Oct 03 '15 at 01:02
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A derivative everywhere less than $1$ does not imply a mapping is a contraction, but a derivative everywhere less than or equal to some particular number less than $1$ does imply that. The mapping $x \mapsto \sqrt{1+x^2}$ has a derivative whose absolute value is everywhere less than $1$, but there is no number $c<1$ for which the absolute value of the derivative is everywhere less than or equal to $c$. That is because the derivative approaches $1$ as $x\to\infty$. And that mapping is not a contraction and has no fixed point. [continued below] ${}\qquad{}$ – Michael Hardy Oct 03 '15 at 03:29
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Suppose $|f'(x)| \le c < 1$ everywhere. Then the mean value theorem implies $\dfrac{f(x)-f(y)}{x-y} = f'(\cdots)$ and the absolute value of the derivative is $\le c$. That imples $|f(x) - f(y)| \le c|x-y|$ for all $x,y$. That is the definition of a contraction. ${}\qquad{}$ – Michael Hardy Oct 03 '15 at 03:31
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Does it matter whether starting point of the sequence is not contained in the interval you found? for instance let starting point of the sequence be 0. However, 0 is not contained in the interval. – jessie Oct 03 '15 at 23:06
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The reason why I am asking this is that in my book, "Let C be closed interval and F is contractive mapping of C, then F has a unique fixed point. Moreover, this fixed point is the limit of every sequence with a starting point contained in C" – jessie Oct 03 '15 at 23:07