Iterative mapping of A (n+1) = F[a (n)] type recursive sequence (play calculator playing a problem)

Put any positive number of square plus \ (1\), the result is also open square plus \ (1\), continue to calculate, eventually always get \ (\frac{3+\sqrt{5}}{2} \approx 2.61804 \), namely:

\ (\sqrt{\cdots \sqrt{\sqrt{\sqrt{x}+1}+1}+1}+1 = \frac{3+\sqrt{5}}{2} \;\;\;\; (x>0) \)

This is a classmate when playing calculators found, very interesting. No matter what number is given at the beginning, the end will stop at this magical number, which is the square of the Golden ratio \ (\frac{\sqrt{5}+1}{2} \).

such as substituting \ (x=2\), established:

Substituting \ (x=20\), also established:

To change the formula from \ (\sqrt{x}+1 \) to \ (\sqrt{x}+2 \) also has similar conclusion:

In other words, the recursive sequence

\ (A_{n+1} = \sqrt{a_{n}}+1 \;\;\;\; (a_{1}>0) \)

At \ (n \rightarrow +\infty \), converge to \ (\frac{3+\sqrt{5}}{2} \). replacing \ (+1\) with a positive number also has a similar conclusion, except that the values converge to a different value.

So, what is this for? Why Convergence? Why the limit is this number, and is not related to \ (A_{1} \)?

After I asked the question online, I got the answer:

You see, if this sequence converges at \ (n \rightarrow +\infty \), then when \ (n\) is "very big", \ (A_{n+1} and A_{n} \) is a thing, namely:

\ (\lim_{n \rightarrow +\infty} a_{n} = \lim_{n \rightarrow +\infty} a_{n+1} \)

The recursive surrogate is available:

\ (\lim_{n \rightarrow +\infty} a_{n} = \lim_{n \rightarrow +\infty} (\sqrt{a_{n}}+1) \)

\ (\lim_{n \rightarrow +\infty} a_{n} = \left (\lim_{n \rightarrow +\infty} \sqrt{a_{n}} \right) +1 \)

\ (\lim_{n \rightarrow +\infty} a_{n} = \sqrt{\lim_{n \rightarrow +\infty} a_{n}}+1 \)

Order \ (x = \lim_{n \rightarrow +\infty}} \), get:

\ (x = \sqrt{x}+1 \)

Solution to

\ (x = \frac{3+\sqrt{5}}{2} \)

This answer is well-reasoned and explains why the limit is irrelevant to \ (A_{1} \) and the limit is solved. But the question of "Why Convergence" is still unresolved.

Later I did not know where to see a drawing method, just can solve the problem. Although not strictly proven, but very intuitive.

, the Blue Line (set to \ (l\)) is \ (y = x \), the red curve (set to \ (m\)) is \ (y = \sqrt{x}+1 \), their intersection (i\) of the horizontal axis is the equation \ (x = \sqrt{x}+1 \) solution, that is, the limit of the sequence.

On the \ (x\) axis fetch point \ (A_{1} (a_{1},\; 0) \), over \ (a_{1}\) for vertical straight line (m\) in \ (p_{1}\). At this point \ (p_{1}\) has an ordinate of \ (\sqrt{a_{1}}+1\), which is \ (a_{2}\). This is equivalent to completing the first generation.

In order to do the second generation, in order to out \ (a_{3}\), we can cross \ (p_{1}\) for the horizontal line (l\) in \ (r_{1}\), \ (r_{1}\) on the \ (x\) axis of the projection set to \ (a_{2}\), its horizontal axis is \ (a_{2}\).

Then. We can just like the \ (a_{1}\), over \ (a_{2}\) for vertical straight line (m\) in \ (p_{2}\), find out \ (a_{3}\) ... Repeat this process, you will leave a bunch of dots on the \ (x\) axis (a_{1}\; a_{2}\; A_{3}\cdots \), their horizontal axis is the sequence of the various \ (A_{1}\;a_{2}\;a_{3}\cdots \).

A ladder-shaped figure was left in the middle of the two curves. It is easy to see that this "ladder" is not worn (i\) point. This shows that the series \ (\{a_{n}\}\) converges to \ (i_{x} = \frac{3+\sqrt{5}}{2}\).

This method of plotting applies to various recursive sequences of \ (A_{n+1} = f (a_{n}) \), such as the following \ (a_{n+1} = \cos (A_{n}) \):

But this is not a ladder-shaped, but spiral-shaped, a circle of an enclosure to go in.

At this point, finally ask why, playing with the calculator to play the problem solved.

Iterative mapping of A (n+1) = F[a (n)] type recursive sequence (play calculator playing a problem)