Convergence of sequence $x_{n+1}=\sqrt{2x_n+3}$

Question

Let $f:\Bbb R \rightarrow \Bbb R$ be the function $f(x)=\sqrt{2x+3}$.

(a) Show that for all $x\in[1,3], f(x)\geq x$. You may use the intermediate value theorem if you like.

Let $x_0=1$ and recursively define $x_{n+1}=\sqrt{2x_n+3}$.

(b) Show that the sequence $<x_n>$ converges.

(c) Find $\lim x_n$.

I have found the sequence $x_n =\frac{1+5^{1/2}}4\cdot3^n+\frac{3-5^{1/2}}4\cdot(-1^n)$ using characteristic equation. I don't understand how come it converges?

Please help!

Answer 1

Hint:

The characteristic equation is used for linear recurrence equations, here it's not linear.

However, you may to study "directly" the variation of $x_n$ : since $f(x)\geq x$, the sequence is increasing.

Show that it's bounded, then use continuity to find the limit with the equation $f(x)=x$, which amounts to a trinomial.

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Notice that of the two main arguments for convergence ($f(x)\geq x$ and $x_n\leq3$) one is proved, and the other is suggested, in question (a). Usually, questions are asked in some order for a good reason, and you can take advange of this.

Answer 2

First show that $x_n$ is upper-bounded, by induction. Clearly, $x_1 < 3$. Assume that $x_n < 3$; then $x_{n+1} = \sqrt {2x_n + 3} < \sqrt {2 \cdot 3 + 3} = 3$. So, $x_n < 3 \space \forall n$.

Next, show that $x_n$ is lower-bounded: by how they are defined, all $x_n$ are positive, so $x_{n+1} = \sqrt {2 x_n + 3} > \sqrt 3, \space \forall n>1$.

So, $(x_n)_{n \geq 1}$ is a bounded sequence. Let us also show that it is monotonous.

$x_{n+1} - x_n = \frac {2 x_n + 3 - x_n ^2} {\sqrt {2 x_n + 3} + x_n}$, so the sign of $x_{n+1} - x_n$ is the sign of $2 x_n + 3 - x_n ^2$ because the denominator is positive. Therefore, let us study the sign of the polynomial $-x^2 + 2x +3$. Its roots are $-1$ and $3$. It takes positive values between the roots, and negative outside of $[-1,3]$. But we have shown that $x_n \in [\sqrt 3, 3]$, so the polynomial is positive in $x_n$, so $x_{n+1} - x_n \geq 0$, so $(x_n)_{n \geq 1}$ is increasing.

Now, you have a bounded, monotonous sequence; by Weierstrass's theorem it must be convergent. Let $l$ be its limit; passing to the limit in the recurrence formula, you get $l = \sqrt {2l + 3}$, and this equation has the roots $-1$ and $3$. Since $x_n > 0$, the limit can only be $3$.

Answer 3

Obviously, for $x>2$ one has $\sqrt{2x+3}>\sqrt{7}>2$. On the interval $(2,+\infty)$ $$ \left|\sqrt{2x+3}-\sqrt{2y+3}\right|=\frac{|(2x+3)-(2y+3)|}{\sqrt{2x+3}+\sqrt{2y+3}}<\frac{2}{2\sqrt{7}}·|x-y| $$ gives a contraction constant $\frac1{\sqrt7}<\frac12<1$ for the fixed point iteration for the function $f(x)=\sqrt{2x+3}$. By the Banach fixed-point theorem, exactly one fixed point exists in $(2,+∞)$.