Conjecture about Continued Square Roots

Let \((s_n)_{n = 0}^{\infty}\) be a sequence of the form \(s_0 = \sqrt{a}, s_1 = \sqrt{a + \sqrt{a}}, s_2 = \sqrt{a + \sqrt{a + \sqrt{a}}}, ...\)

As nn approaches infinity, the sequence approaches a "continued square root", a number of the following form:

x=a+a+a+...x = \sqrt{a + \sqrt{a + \sqrt{a + ...}}}

I'll leave it as an exercise to show that if a>0a > 0, then this continued square root always converges, and the result is the positive root of the equation a=x2xa = x^2 - x.

Let's take a=2a = 2 as an example. By the formula I just mentioned, we have 2+2+2+...=2\sqrt{2+\sqrt{2+\sqrt{2+...}}} = 2.

One question I asked myself is, how fast do these continued square roots converge to 22?

To put it more precisely, let's consider the sequence 22,22+2,22+2+22 - \sqrt{2}, 2 - \sqrt{2 + \sqrt{2}}, 2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}, and so on. The decimal approximations for these numbers are 0.5858, 0.1522, 0.03843, 0.0096305, etc. Do you notice the pattern? The numbers seem to be getting about 44 times smaller each time.

Let's make this more rigorous. In the original sequence, we have sn+1=2+sns_{n+1} = \sqrt{2 + s_n}. Therefore, limn2sn+12sn=limn22+sn2sn=limn(22+sn)(2+2+sn)(2sn)(2+2+sn)\lim_{n \to \infty} \dfrac{2 - s_{n+1}}{2 - s_n} =\lim_{n \to \infty} \dfrac{2 - \sqrt{2 + s_n}}{2 - s_n} = \lim_{n \to \infty} \dfrac{(2 - \sqrt{2 + s_n})(2 + \sqrt{2 + s_n})}{(2 - s_n)(2 + \sqrt{2 + s_n})}=limn4(2+sn)(2sn)(2+2+sn)=limn12+2+sn=14 = \lim_{n \to \infty} \dfrac{4 - (2 + s_n)}{(2 - s_n)(2 + \sqrt{2 + s_n})} = \lim_{n \to \infty} \dfrac{1}{2 + \sqrt{2 + s_n}} = \dfrac{1}{4}

So, the differences between the continued square roots and their limit (2) do, in fact, approach a geometric sequence with 14\dfrac{1}{4} as the common ratio. Hence, sns_n can be asymptotically approximated by a function sn2b4ns_n \approx 2 - \dfrac{b}{4^n}.

Now, the million dollar question: What is this constant bb?

In other words, what is limn4n(2sn)\displaystyle\lim_{n\to\infty} 4^n(2 - s_n)?

I won't tell you the answer in case you want to solve it on your own. If you succeed, I'd love to see how you solved it so feel free to share it in the comments. If you'd like to see my solution, you can find it here.

Now let's generalize the previous result for the real number aa. Let's replace 22 with x2xx^2 - x, so that our sequence looks like this: sn={x2x,(x2x)+(x2x),(x2x)+(x2x)+x2x...}s_n = \left\{\sqrt{x^2 - x}, \sqrt{(x^2 - x) + \sqrt{(x^2 - x)}}, \sqrt{(x^2 - x) + \sqrt{(x^2 - x)+\sqrt{x^2 - x}}} ...\right\} You can show (using a similar proof to mine above) that limnxsn+1xsn=12x\lim_{n \to \infty} \dfrac{x - s_{n+1}}{x - s_n} = \dfrac{1}{2x}

Hence, the generalized sequence can be approximated by the function snxLx(2x)ns_n \approx x - \dfrac{L_x}{(2x)^n} where LxL_x is the following limit: Lx=limn(2x)n(xsn)L_x = \lim_{n \to \infty} (2x)^n (x - s_n)

Now, I have a conjecture for you. I haven't been able to prove/disprove it yet; see if you can:

Let sns_n be a sequence defined recursively by s0=x2xs_0 = \sqrt{x^2 - x} and sn+1=(x2x)+sns_{n+1} = \sqrt{(x^2 - x) + s_n} for nNn \in \mathbb{N}, for some x>1x > 1. Let Lx=limn(2x)n(xsn)L_x = \lim_{n \to \infty} (2x)^n (x - s_n). Then,

limxLx=12 \lim_{x \to \infty} L_x = \dfrac{1}{2}

#Algebra #Sequences #Limits #SquareRoot #ContinuedSquareRoots

Note by Ariel Gershon
5 years, 9 months ago

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Comments

Just to confirm, is b=π216b=\dfrac{\pi^2}{16} ?

Ishan Singh - 5 years, 9 months ago

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Yes, very good!

Ariel Gershon - 5 years, 9 months ago

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If we can solve the functional equation g(x)=ag2(kxa)1g(x)=a g^2\left(\dfrac{kx}{a}\right) - 1 where aa is a fixed constant and kk is arbitarary, we are almost done solving the conjecture.

Ishan Singh - 5 years, 9 months ago

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@Ishan Singh I'm sorry, how would this help?

Ariel Gershon - 5 years, 9 months ago

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@Ariel Gershon Let fn(y)=a+a++a+yn timesf_{n}(y)=\underbrace{\sqrt{a+\sqrt{a+\ldots+\sqrt{a+y}}}}_{n \ \text{times}}, substitute y=ag(x)y=a\cdot g(x) to simplify fn(y)f_{n}(y). We can then replace aa by t2tt^2-t and calculate LtL_{t} by letting nn \to \infty. Then we can find the conjectured limit by letting tt \to \infty. For example if we let a=2a=2, k=1k=1 we will get g(x)=2g2(x2)1g(x)=2g^2(\frac{x}{2})-1. One solution is g(x)=cos(x)g(x)= \cos (x). From this we can calculate b=π216b=\dfrac{\pi^2}{16}.

Ishan Singh - 5 years, 9 months ago

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@Ishan Singh I'm missing quite a few steps in your logic.... What are g(x)g(x) and kk?

Ariel Gershon - 5 years, 9 months ago

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@Ariel Gershon g(x)g(x) is an unknown function which satisfies g(x)=ag2(kxa)1g(x)=a g^2\left(\dfrac{kx}{a}\right) - 1. If we substitute y=ag(x)y=a\cdot g(x) in fn(y)=a+a++a+yn timesf_{n}(y)=\underbrace{\sqrt{a+\sqrt{a+\ldots+\sqrt{a+y}}}}_{n \ \text{times}}, we have fn(y)=ag(knxan)f_{n} (y) = a \cdot g\left(\dfrac{k^nx}{a^n}\right). And if we replace aa by t2tt^2-t, we can then calculate LtL_{t} as a function of tt. This will simplify the conjectured limit LtL_{t}. If my reasoning is missing any steps, you can ask and I'll try to fix it.

Ishan Singh - 5 years, 9 months ago

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@Ishan Singh OK, it makes more sense now as I was able to finish your argument. This is a very interesting approach! If you don't mind though, I'm going to simplify the equation for gg to make it g(x)=ag2(cx)1g(x) = a g^2\left(cx\right) - 1 (since kk is an arbitrary constant, let c=kac = \dfrac{k}{a}).

There are two more stipulations for gg and cc which you didn't mention, but they are important. The proof by induction of this approach requires that the following be true:

(1) there must exist a real number zz such that g(z)=aag(z) = \dfrac{\sqrt{a}}{a}

(2) g(cnz)0g(c^n z) \ge 0 for all nNn \in \mathbb{N}

Also, we can write fn(a)=snf_n(a) = s_n. Then by induction we can show that sn=ag(cnz)s_n = a * g(c^n z). If we replace aa with t2tt^2 - t, then LtL_t will have the following formula: Lt=limn(2t)n(t(t2t)g(cnz))L_t = \lim_{n\to\infty} (2t)^n \left(t - (t^2 - t) g(c^n z)\right)

In the case of t=2t = 2, we can have c=12c = \dfrac{1}{2} and g(x)=cos(x)g(x) = \cos(x), since, as you mentioned, cos(x)=2cos2(x2)1\cos(x) = 2\cos^2\left(\dfrac{x}{2}\right) - 1. We have z=π4z = \dfrac{\pi}{4} since cos(π4)=22\cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}. Then we get L2=limn4n(22cos(π42n))L_2 = \lim_{n\to\infty} 4^n \left(2 - 2\cos\left(\dfrac{\pi}{4*2^n}\right)\right) This limit does, in fact, evaluate to π216\dfrac{\pi^2}{16}.

Interesting as this is, though, I think it won't be as easy to find g,c,zg, c, z for tt in general.

Ariel Gershon - 5 years, 9 months ago

I want to point out that it only applies for a>0a > 0. (For a=0a = 0, it converges to the lower root of a=x2xa = x^2-x, namely x=0x = 0 instead of x=1x = 1.)

Ivan Koswara - 5 years, 9 months ago

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Fair point. I'll fix it

Ariel Gershon - 5 years, 9 months ago
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