# Another Limit Problem

I have been discussing the problem of antique equations of the form

$$\sqrt{12+\sqrt{12+\sqrt{12 + \ldots}}} = x$$

previously. This case can be studied by the recursive sequence

$$x_{n+1} = \sqrt{12+x_n}$$

It turns out that the sequence converges no matter where we start. So we can say that we can stop the infinite process of taking square roots at some point

$$\sqrt{12+\sqrt{12+\sqrt{12 + a}}} = x$$

and get very close to the value 4. It should stated clearly that, for a fixed starting value a, this will converge to 4 only if we make the sequence of square roots longer and longer. On the other hand we could study how close the value x will be to 4 for a fixed number of square roots depending on the starting point. These are two different but interesting aspects of the problem. Both are related to the fixed point of the mapping

$$g(x) = \sqrt{12+x}$$

which happens to be x=4 (i.e., g(4)=4). But this is only a necessary condition for the convergence of the sequence above, not a sufficient one.

Here is another problem which is attributed to Ramanujan. Probably, it is not his, because the problem looks too easy for the mastery of this mathematician. I could not find any reference anyway. We study the process

$$\sqrt{1+2 \sqrt{1+3 \sqrt{1 + \ldots}}}$$

The idea to transfer this into an iteration is no longer as easy as before. The result depends on the starting point and the number of steps as before, but the iterations are not identical steps anymore. We can define

$$g_n(x) = n \sqrt{1+x}.$$

Then

$$f_n(a) = g_1(g_2(g_3(\ldots g_n(a)))).$$

The starting value is again a and we take n square roots. It happens that

$$g_n((n+1)(n+3)) = n(n+2).$$

So we get

$$\sqrt{1+2 \sqrt{1+3 \sqrt{1 + \ldots + n \sqrt{1+(n+1)(n+3)}}}} = 3.$$

Now it is interesting to ask if the value x=3 is in some way the value of the infinite sequence independent of the starting point a. Let us do some experiments in Euler Math Toolbox.

>function map f(a,n) ...
$for i=n to 1 step -1;$     a=i*sqrt(a+1);
$end;$  return a;
\$  endfunction
>plot2d("f(x,20)-3",0,10000000,xl="a",yl="x"): The function f(a,n) computes n iterations starting with a. We see from the plot that the values gather very closely around 3 (closer than 0.00002). Why does this happen?

We discover the reason immediately if we look at an individual sequence of iterations.

$$s_n = a, \quad s_{n-1} = g_{n-1}(s_n), \quad \ldots, \quad s_1 = g_1(s_2) = f_n(a).$$

>n=40; s=ones(1,n);
>for i=n-1 to 1 step -1; s[i]=i*sqrt(s[i+1]+1); end;
>plot2d(s,>points):
>plot2d("x*(x+2)",>add): We started with the value a=1 and took n=40. The vector s contains the elements of the sequence and it is plotted point by point from right to left. I have already added the clearly visible „attractive function“ f which has the property that f(1)=3. To see that the function attracts the values of the sequence no matter where we start, involves the following facts.

$$g_n(f(n+1)) = f(n), \quad g’_n(x) = \dfrac{n}{2 \sqrt{1+x}}.$$

Using the analysis of contractions, it is clear that the sequence gets closer and closer to the function, if we start with a value larger than f(n). In the example, we start below the function at 1. I leave it to the reader to find an argument for the fact that the sequence gets into the region of attraction fast enough to see the very good results from our first plot.

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