Convergence Proof

Say that you have a recurring sequence such that: $a_{0} = 1/2$
$a_{n+1} = \sqrt{1-a_n}$

How do you prove that this sequence converges, given that it does. If it doesn't, how do you prove it diverges?

#Calculus

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let's see.

Supposing we start off with a number $a_0$ .

$a_1 = \sqrt{ 1 - a_0}$

$a_2 = \sqrt {1 - a_1 }$

$a_2 = \sqrt { 1 - \sqrt{1 - a_0} }$

$a_3 = \sqrt { 1 - a_2}$

$a_3 = \sqrt { 1 - \sqrt{1 - a_1 }}$

$a_3 = \sqrt { 1 - \sqrt{1 - \sqrt{1- a_0}}}$

As you can notice, $a_n$ seemingly becomes a nested radical starting from $a_0$ ..

If we extend this toward infinity, we will know if $a_n$ approaches a specific value.

$a_n = \sqrt{1 - \sqrt{ 1 - \sqrt { 1 - \sqrt { 1 - \sqrt{...}}}}}$

We can condense this into

$a_n = \sqrt { 1 - a_n }$

$a_n^2 + a_n - 1 = 0$

We are now left with an equation in terms of $a_n$ , which surprisingly means that no matter what $a_0$ we begin with, it reaches this value of $a_n$ .

Coincidentally, this $a_n$ , when solved, gives the reciprocal of the golden ratio $\frac {-1 + \sqrt{5}}{2}$ . Well the other value is discarded as it is extraneous.

Efren Medallo - 4 years, 10 months ago

Very elegant solution. That's the exact process I used to get the solution, given that it does converge. I just wonder if this is enough to prove that the sequence certainly does converge.

Oli Hohman - 4 years, 10 months ago

I think since it was assumed that $lim_{n \to \infty} a_n = y$ , so we reached for a particular value of $y$ , which verifies that indeed the limit exists.

Efren Medallo - 4 years, 10 months ago

@Efren Medallo – No, that's not a sufficient condition. Consider the following counter-example of a sequence $\{a_n\}_{n\geq 0}$ defined by $a_{n+1}=2a_n~\forall~n\geq 0$ and the base case $a_0=1$

If we assume that the sequence converges, i.e., assume that $\lim\limits_{n\to\infty} a_n=y$ exists and is finite, then we get using the recurrence relation that $y=2y\implies y=0$ which is finite. But the sequence is actually $a_n=2^n~\forall~n\geq 0$ which is divergent.

Prasun Biswas - 4 years, 10 months ago

One possible way to show the convergence is to consider the two subsequences $\{a_{2n}\}_{n\geq 0}$ and $\{a_{2n+1}\}_{n\geq 0}$ and show that they are monotone and bounded. The first subsequence is monotonically increasing and strictly bounded above by $1$ and below by $1/2$ and the second subsequence is monotonically decreasing and strictly bounded below by $0$ and above by $\sqrt{1/2}$ . Since monotone bounded sequences have a finite limit, denote the limits of the two subsequences by $l_1$ and $l_2$ respectively. Then, both $l_1$ and $l_2$ are solutions of the equation below:

$k=\sqrt{1-\sqrt{1-k}}\implies k^2=1-\sqrt{1-k}\implies (k^2-1)^2=1-k\implies (k-1)(k^2+k-1)=0$

Now, to show that the sequence $\{a_n\}_{n\geq 0}$ is convergent, we need to show that $l_1=l_2$ , i.e., we need to show that there exists a unique real root $k_0$ of the above equation such that $1/2\leq k_0\leq\sqrt{1/2}$ . Solving the equation shows that there does exist a unique real root $k_0$ which is equal to $1/\phi$ and hence the sequence $\{a_n\}_{n\geq 0}$ is convergent with the limit being $1/\phi$ .

Another way to show the convergence would be to show that the map $x\mapsto\sqrt{1-x}$ is a contraction mapping in the metric space $[0,1]$ with the usual absolute value metric and then invoke the Banach fixed point theorem to show that the sequence $\{a_n\}_{n\geq 0}$ is convergent with the limit being the fixed point of the mentioned contraction mapping, which is $1/\phi$ .

Prasun Biswas - 4 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$