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This is the recursive formula: $$\begin{cases} a_1=1/2,\\ a_n= \sqrt {\frac {a_{n-1}+1}{2}} \end{cases}$$

I have calculated the first 4 terms to be $\frac12, \frac {\sqrt3}2, \frac {\sqrt{\sqrt3+2}}2, \frac {\sqrt{\sqrt{\sqrt3+2}+2}}2$

How can I find the explicit formula?

Looking at the first 4 terms, I notice that the denominator remains at 2, while the numerator appears to also be recursive. I will call this numerator $b_n$ $$\begin{cases} b_1=1,\\ b_n= \sqrt {b_{n-1}+2} \end{cases}$$

I am thinking that if I find the explicit formula for $b_n$, perhaps I could use the formula $\frac {b_n}2=a_n$ and solve.

I also figure out that in the recursive formula where $c_1=x$, and $c_{n}=\sqrt{c_{n-1}}$, the explicit formula would be: $$c_n=x^\frac1{2^{(n-1)}}$$

I am currently struggling on how to solve for the case $b_n=\sqrt{b_{n-1}+2}$

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  • $\begingroup$ Can you elaborate more please? $\endgroup$ Commented Aug 11, 2020 at 20:03
  • $\begingroup$ Have you tried to compute the first values ? What does that give ? $\endgroup$ Commented Aug 11, 2020 at 20:40
  • $\begingroup$ If you want to sqare the sequence, it would go like this: $b_n = a_n^2 = \frac{a_{n-1} + 1}{2} = \frac{\sqrt{b_{n-1}} + 1}{2}$. It's not enough to square the recurrence formula. That's why your graph doesn't line up. $\endgroup$ Commented Aug 11, 2020 at 21:33

1 Answer 1

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Such non-linear recurrences very rarely have explicit solutions.

You can go for a limit, though: If the limit is $A$, it must be that:

$\begin{align*} A &= \sqrt{\frac{A + 1}{2}} \\ A^2 &= \frac{A + 1}{2} \\ A &= \frac{1 \pm \sqrt{1 + 4 \cdot 2}}{4} \\ &= 1 \text{ or } -\frac{1}{2} \end{align*}$

The negative value makes no sense, so the limit (if it exists) is 1. To prove the limit exists, see that $a_n$ is strictly increasing, limited from above by 1.

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