$d$ is a real number. In each integer $m\ge0$, we define $\{a_m(j)\}$ $(j=0, 1, 2, \cdots)$ as follows. $$a_m(0)=\frac{d}{2^m}, a_m(j+1)=\left(a_m(j)\right)^2+2a_m(j) \space\space(j\ge 0)$$ Then, evaluate $\lim_{n \to \infty}a_n(n)$.

It is known that the closed-form of the given sequence is $$a_m(j)+1 = (a_m(0)+1)^{2^{j-1}} \space\space\cdots(1)$$ It is not that hard to find $\lim_{n \to \infty}a_n(n)$ from here.

My question is that how the closed-form of the given sequence can be calculated like that?

For sure, it can be proven by mathematical induction. However, what I do want to know is how the closed-form is derived, not how the $eq.(1)$ is fit to the given recursion formula. Thanks for answering!

  • $\begingroup$ I think the way is to apply recurrence relation to guess the formula and then prove the formula by induction. You can read something here: en.wikipedia.org/wiki/Recurrence_relation $\endgroup$ – asv Feb 14 at 16:51

From the recurrence relation, you can get $a_m(j+1) + 1 = (a_m(j)+1)^2$. Then, you can get the closed-form by using this relation recursively.


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