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So I came across this silly video "When mathematicians get bored (ep1)" on bprp fast's YouTube channel, and it mentions the following recursive sequence:

For all natural $n \geq 1$, \begin{align}f_0(x) &= \sqrt{x}, \\ f_n(x) &= \sqrt{x+f_{n-1}(x)} \end{align}

In the video, it is claimed that $\lim_{n \to \infty}{f_n{(20)}}=5$, but I couldn't actually prove this when I tried to!

I don't know whether I'm forgetting something obvious or not, but can someone sketch a proof for me or at least point me in the direction of a method of proof?

Much appreciated!

Blue
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jason
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    First, you must proof the existence of the limit (Hint: increasing bounded sequence). Then, apply limit on both sides on the recursive relationship and solve for $L$. – Jorge Jun 29 '21 at 13:48
  • It's wrong equation – jasmine Jun 29 '21 at 13:49
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    The map $t \mapsto g(t) = \sqrt{x + t}$ is a continuous increasing function in $t$ from $[0,\infty) \to [0,\infty)$. In general, if you generate a sequence by repeat iteration of this type of function, the sequence will convergence to the fixed point immediately above or below the seed value $x_0$ (above if $x_1 = g(x_0) > x_0$ and below if $x_1 < x_0$) or diverge to $+\infty$ if $x_1 > x_0 >$ last fixed point. The proof is essentially the same for all $g$. See this for an example. – achille hui Jun 29 '21 at 14:02
  • @achillehui Thank you! The example you linked to is very nice and I can see how this all works now! :) – jason Jun 29 '21 at 14:09
  • I'm sure I have seen this before, it turn out it is one of the abstract duplicated question. see this which is essentially the same question,... – achille hui Jun 29 '21 at 14:12

2 Answers2

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Use monotone convergence to show that the sequence converges, and then use $$P=\sqrt{20+\sqrt{20+\sqrt{20+\dots}}}$$ which implies $$P=\sqrt{20+P}\implies P^2=20+P\implies P=5$$

Does that help?

Sayan Dutta
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First, you can forget the $x$ and simply define:

$$u_0=\sqrt{20}$$ $$u_{n+1}=\sqrt{20+u_n}=f(u_n)$$

where $f(x)=\sqrt{20+x}$; and show that $\lim_{n\to\infty}u_n=5$.

Since $f$ is an increasing function, and since $$u_1 = \sqrt{20+\sqrt{20}}>u_0,$$ the sequence $(u_n)$ is increasing by induction.

Also, since $u_0<5$, by induction, you can prove that $u_n<5$ for any $n$.

Therefore the sequence converges, and can converge only to the fixed point of $f$, that is 5 since $\sqrt{20+5}=5$.

Voilà!

Florian Ingels
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