Recursive sequence of functions

Question

Recursive sequence of functions: $f_{n+1}= \sqrt{x+f_n}$, $f_1(x)= \sqrt{x}$.

this sequence is monotonic, but what is bounding it? thanks

Answer 1

So you have defined $$f_n(x) = \underbrace {\sqrt{x + \sqrt {x + \dots \sqrt x}}}_{\text {n square roots}}$$ and you say that it is "monotonic", suggesting some comparison like $$f_{n+1} \ge f_n$$ which, correct me if I'm wrong, I'm going to assume means $\forall x\, f_{n+1}(x) > f_n(x)$.

A bound (..but not a least upper bound...) then would imply that you are looking for a function $g(x)$ such that $$\forall n,\,x,\, g(x) \ge f_n(x)$$

This is a bit more than I would expect in a regular infinitessimal calculus class. The upper bound obviously has to be larger than $g(x) = \sqrt x$, but it doesn't seem like it has to be much more. Let's try $g(x) = \sqrt {2x}$. To use induction though, we have to prove a stronger result:

$$h_n(x) = \sqrt{\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^{n-1}}\right)\cdot x}$$ $$h_n(x) \ge f_n(x)$$

Base case: $$\sqrt{x} \ge \sqrt{x}$$

Recursive assumptions: $$h_n(x) \ge f_n(x)$$

Recursive cave: $$h_{n+1}(x) \ge f_{n+1}(x)$$

$$\sqrt{\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^n}\right)\cdot x} \ge \sqrt{x + f_n(x)}$$

$$\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^n}\right)\cdot x \ge x + f_n(x)$$

$$\left(1 + \frac 1 2 + \frac 1 4 ... \frac 1 {2^{n - 1}}\right)\cdot x + \frac x {2^n} \ge x + f_n(x)$$

$$h_n(x)^2 + \frac x {2^n} \ge f_n(x) + x $$

$$\text{Is implied by}$$

$$f_n(x)^2 + \frac x {2^n} \ge f_n(x) + x $$

$$f_n(x)^2 - f_n(x) \ge x - \frac x {2^n} $$

This last statement....sorry...we have to prove with induction over n.

Base case: $$x - \sqrt{x} > x - \frac x 1$$ For $x\ge 1$ this holds.

Recursive assumption: $$f_n(x)^2 - f_n(x) \ge x - \frac x {2^n} $$

Recursive case: $$f_{n+1}(x)^2 - f_{n+1}(x) \ge x - \frac x {2^{n+1}} $$

$$x+f_n(x) - \sqrt{x+f_n(x)} \ge x - \frac x {2^{n+1}} $$

$$f_n(x) - \sqrt{x+f_n(x)} \ge \frac x {2^{n+1}} $$

$$f_n(x) + \frac x {2^{n+1}} \ge \sqrt{x+f_n(x)} $$

$$f_n(x)^2 + 2f_n(x)\frac x {2^{n+1}} + \frac {x^2} {2^{n+2}} \ge x+f_n(x) $$

$$\text{Is implied by: (using the recursive assumption)}$$

$$x - \frac x {2^n} + 2f_n(x)\frac x {2^{n+1}} + \frac {x^2} {2^{n+2}} \ge x$$

$$x^2 - 4 x + 4f_n(x)\ge 0$$

Which holds for $x \ge 1$.

Since $\forall n\, g(x) \ge h_n(x)$, $g$ is an upper bound for all $f_n$, for $x \ge 1$.

Answer 2

Let $g_{x}(a) = \sqrt{x+a}$. Observe that, $g$ is increasing on $a$. Hence, if there exists some $a_{0}$ such that $g_{x}(a_{0}) = a_{0}$, then

$$\text{Fact 1: for every }a \leq a_{0}, g_{x}(a) \leq a_{0}$$

Solving for $g_{x}(a_{0}) = a_{0}$ yields $a_{0} = \frac{1+\sqrt{1+4x}}{2}$. Hence, for every $x \geq 0$ there exists such an $a_{0}$.

Observe that $f_{1}(x) = \sqrt{x} \leq a_{0}$. Hence, it follows from Fact 1 that $f_{2}(x) = g_{x}(f_{1}(x)) \leq a_{0}$. Conclude inductively that, for every $i \in \mathbb{N}$, $f_{i} \leq a_{0}$. This gives you a bound.

But how sharp is this bound?

Let $h_{x}(a) = \sqrt{a+x}-a$. Observe that, if $a \leq a_{0}-\epsilon$,

\begin{align*} h_{x}(a) &= \sqrt{a+x}-a \\ &= \frac{a+x-a^2}{\sqrt{a+x} + a} \\ &\geq \frac{\epsilon(a_{0}-1)-\epsilon^{2}}{\sqrt{a_{0}+x} + a_{0}} \end{align*}

The last inequality follows from $a \leq a_{0}$ and $a_{0}+x-a_{0}^{2} = 0$.

Observe that, if $a \leq a_{0}-\epsilon$, then $a \leq a_{0}-\delta$, for every $\delta \leq \epsilon$. Hence,

$$h_{x}(a) \geq \frac{\delta(a_{0}-1)-\delta^{2}}{\sqrt{a_{0}+x} + a_{0}}$$.

Since $a_{0} > 1$, there exists an appropriately small $\delta(\epsilon)$ such that one obtains that, for all $a \leq a_{0}-\epsilon$, $h_{x}(a) \geq k$, for some $k > 0$.

Observe that since $f_{n}$ is bounded and increasing, it converges. Assume that $f_{n}$ converges to a number $a_{0}-\epsilon$. Hence, for every $n$,

$$f_{n+1}-f_{n} = h_{x}(f_{n}) > k$$

A contradiction. Conclude that $f_{n}$ converges to $a_{0}$.