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Let $a_k$ be a sequence of fixed positive integers, $k \in [1,n]$. Consider the periodic nested radical:

$$x=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n+x}}}$$

We can transform this nested radical into the polynomial equation:

$$\left( \cdots \left(\left(x^2-a_1 \right)^2-a_2\right)^2-\cdots-a_{n-1}\right)^2-x-a_n=0$$

This equation has order $2^n$, so it has exactly $2^n$ roots.

I observed that the nested radical converges to the largest positive real root of this equation (for small $n$ at least).

How to prove that this kind of equation will always have a positive real root, and that the nested radical converges to the largest of them? (If this is true)

It's obvious that the nested radical should converge to a real and positive value.


Edit

An example. Consider the periodic nested radical, satisfying:

$$x=\sqrt{3+\sqrt{5+\sqrt{2+\sqrt{7+x}}}}$$

It's value can be found by iterations of the above expression:

$$x \approx 2.38592$$

Now, consider the polynomial equation, obtained from this radical:

$$\left(\left(\left(x^2-3 \right)^2-5\right)^2-2\right)^2-x-7=0$$

It has $16$ roots, exactly $8$ real roots and $8$ complex roots.

Of the real roots there are $4$ positive roots and $4$ negative roots.

Four positive roots are:

$$x_1 \approx 0.566147$$

$$x_2 \approx 1.15226$$

$$x_3 \approx 2.15887$$

$$x_4 \approx 2.38592$$

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Here is a partial solution ; I can show that the limit always exists, but I cannot show (yet) that the limit is independent of the initial value (this would in turn immediately yield the fact that the limit is the largest root).

Let $f(x)=\sqrt{a_1+\sqrt{a_2+\cdots+\sqrt{a_n+x}}}$ and $P(x)=((x^2-a_1)^2-a_2)^2-\cdots-a_{n-1})^2-a_n$. Then $P$ is a polynomial of degree $2^n$, $f$ and $P$ are (formally at least) inverses of each other, so that $f(x)=y$ forces $P(y)=x$, though the converse is not true. In fact, unnesting inequalities as in the OP and taking care of signs , we have

$$ f(x)=y \Leftrightarrow \Big( P(y)=x, \text{and} \ y \geq f(0) \Big) \tag{1} $$

In particular,

$$ f(x)=x \Leftrightarrow \Big( P(x)=x, \text{and} \ x \geq f(0) \Big) \tag{2} $$

Since the square root function and all the $x\mapsto x+a$ function are increasing, $f$ which is a big composite of all such functions is increasing also.

Consider the sequence $u=(u_p)$ defined by $u_0=0$, $u_{p+1}=f(u_p)$. Then $u_0<u_1$, so $f(u_0)<f(u_1)$, i.e. $u_1<u_2$ and by induction we see that $u$ is increasing.

It follows that $u$ either converges to a finite limit or diverges to $+\infty$. But if it diverged to $+\infty$, $u_p=P(u_{p+1})$ would be asymptotically equivalent to $u_{p+1}^{2^n}$ contradicting $u_p<u_{p+1}$ for large enough $p$.

So $u$ must converge to a finite limit $l$. By (2) above, $l$ must be a root of $P$. Let $m$ be any value $>l$. Then $m>u_k$ for every $k$, hence $f(m)>f(u_k)=u_{k+1}$. Passing to the limit, we have $f(m)\geq l$.

Consider the sequence $v=(v_p)$ defined by $v_0=m$, $v_{p+1}=f(v_p)$. By induction, we have $v_p\geq l$ for every $p$. Also, $v_p=P(v_{p+1})$ by (1). By arguments already used above, $v$ is monotonic and cannot diverge to $+\infty$. So $v$ must have a finite limit.

(... Proof to be finished ; looks promising from here ...)

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I know this is not a formal answer but consider a polynomial $x^2 -c = 0$. This has 2 solutions $\sqrt{c}$ and $-\sqrt{c}$.

When you convert this polynomial equation into a fixed point equation by taking square roots, you will obtain the following equation

$x = \sqrt{c}$.

This second equation has only one solution, which is the positive one because of the square root, $\sqrt{c}$.

Again, now consider the equation $x = \sqrt{c + x}$. (Eq 1)

Take squares of both sides, we have $x^2 = c + x \implies x^2 - x - c = 0$.

$\Delta = 1 + 4c$

$x_1 = \frac{1 + \sqrt{1 + 4c}}{2}$ and $x_2 = \frac{1 - \sqrt{1 + 4c}}{2}$.

We know $\sqrt{1+4c} > 1$ thats why the $x_2$ root of the polynomial is not the solution of the original equation (Eq 1).

Thus, when you take squares of the original $x = \sqrt{a_1 + \sqrt{a_2 + ...\sqrt{a_n + x}}}$ equation, each time you will double the number of solutions with contributions of corresponding (Eq 1) omitted $-$ signed root, while the original solution was the bigger one, which is obtained by $+$ sign in discriminant.

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  • $\begingroup$ One problem with this however is that not all roots are real, so your analogy fails for complex roots $\endgroup$ – Ewan Delanoy Apr 6 '16 at 16:58
  • $\begingroup$ @EwanDelanoy I think $a_i$'s being positive leads all the roots to be real numbers. $\endgroup$ – crbah Apr 7 '16 at 6:11
  • $\begingroup$ @corbah, all the roots can't be real (see the example I provided). But thank you for the answer, now I understand this better $\endgroup$ – Yuriy S Apr 7 '16 at 7:30

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