Solution to Problem 12

The best solution went as follows:

The sequence of weights is given by the recursive formula: $a_0=\sqrt{2}$and then $a_n=\sqrt{2-a_{n-1}}$ for $n\geq 1$.

If we assume that the limit exists (big assumption), then we can let $\lim\limits_{n\to\infty} a_n=L$. Now it will also be the case that $\lim\limits_{n\to\infty} a_{n-1}=L$. So

$$L=\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sqrt{2-a_{n-1}}=\sqrt{2-\lim_{n \to\infty} a_{n-1}}=\sqrt{2-L}.$$

But then $L=\sqrt{2-L}$, so L²+L-2=0 and (L+2)(L-1)=0. So L=1.

By the same logic: Consider the sequence defined by the recursive formula:

$$b_0=3, \mbox{ and then } b_n=\frac{1}{2}(b^2_{n-1}+1)\mbox{ for }n\geq 1.$$

A quick check on your calculator shows that this sequence diverges. Or you may simply show that b_n>2b_n-1, so the sequence diverges.

Let us assume that the limit exists (bad assumption), and let $\lim\limits_{n\to\infty} b_n=M$. So $\lim\limits_{n\to\infty} b_{n-1}=M$. Then

$$M=\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{1}{2}(b_{n-1}^2... ... \frac{1}{2}(\lim_{n\to\infty} b^2_{n-1}+1)=\frac{1}{2}(M^2+1).$$

So $M=\frac{1}{2}(M^2+1)$, M²-2M+1=0, and (M-1)²=0. So M=1 which is obviously not true.

So what is wrong with the first argument?

We never checked that the sequence converges!

A few people observed (based upon calculator/computer simulations?) that the terms with even subscripts decrease monotonically to 1, while the terms with odd subscripts increased monotonically to 1.

Can we prove this?

The subsequence with even subscripts is given by the recursive formula:

$$a_2=\sqrt{2},\quad\mbox{ and }a_n=\sqrt{2-\sqrt{2-a_{n-2}}}\mbox{ for }n\geq 2,$$

while the subsequence with odd subscripts is given by the recursive formula:

$$a_1=\sqrt{2-\sqrt{2}},\quad\mbox{ and }a_n=\sqrt{2-\sqrt{2-a_{n-2}}}\mbox{ for }n\geq 3.$$

In any case, we are interested in the formula a_n-2=2-(a_n²-2)². In particular we would like to show that $a_{n-2}\leq a_n$ for odd n, while $a_{n-2}\geq a_n$ for even n (the odd subsequence is increasing while the even subsequence is decreasing). In other words, we would like to show that

$$2-(x^2-2)^2\leq x\quad \mbox{ for }\sqrt{2-\sqrt{2}}\leq x\leq 1,$$

while

$$2-(x^2-2)^2\geq x\quad \mbox{ for }1\leq x\leq \sqrt{2}.$$

In other words we would like to show that

f(x)=x⁴-4x²+x+2

is greater than 0 for $\sqrt{2-\sqrt{2}}\leq x\leq 1$, while f(x) is less than 0 for $1\leq x\leq \sqrt{2}$. This is a pre-calculus problem. Note that

$$f(x)=(x+2)\cdot(x-1)\cdot\Big(x-\frac{1+\sqrt{5}}{2}\Big)\cdot \Big(x- \frac{1-\sqrt{5}}{2}\Big).$$

So $f(x)\geq 0$ for $-0.61\leq x\leq 1$, while $f(x)\leq 0$ for $1\leq x\leq 1.61$. All this is much more than we need, but at least it tells us that odd subsequence is increasing while the even subsequence is decreasing.

To know that they converge we prove that the subsequences are bounded. To do this we need to show that the odd terms are smaller than the even terms. Start with a_n-1=2-a_n². We need to show that a_n-1>a_n if $\sqrt{2-\sqrt{2}}\leq a_n\leq 1$, while a_n-1<a_n if $1\leq a_n\leq \sqrt{2}$. In other words, we need to show that

$$2-x^2>x\quad\mbox{ if }\sqrt{2-\sqrt{2}}\leq x\leq 1,$$

while

$$2-x^2<x\quad\mbox{ if }1\leq x\leq \sqrt{2}.$$

Alternatively, we need to show that

g(x)=x²+x-2=(x+2)(x-1)

is positive for $1\leq x\leq \sqrt{2}$ and negative for $\sqrt{2-\sqrt{2}}\leq x\leq 1$. But using pre-calculus, this is obvious. So each odd term is smaller than the next term (necessarily even). Even terms are smaller than $\sqrt{2}$, so the odd subsequence is increasing, bounded and so converges. each even term is larger than the next term (necessarily odd). Odd terms are larger than $\sqrt{2-\sqrt{2}}$. So the even subsequence is decreasing, bounded and so converges.

Just for fun, let N be the limit of the even terms. So for n be even, $\lim\limits_{n\to\infty} a_n=N$, and also $\lim\limits_{n\to\infty} a_{n-2} =N$. Using $a_n=\sqrt{2-\sqrt{2-a_{n-2}}}$, we find that

$$N=\sqrt{2-\sqrt{2-N}}.$$

So f(N)=0. We determined the roots of f(x) above, and the only root in $\sqrt{2-\sqrt{2}}\leq x\leq\sqrt{2}$ is 1. So the even terms converge to 1. Similarly the odd terms also converge to 1.

Uncle ?

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Nov 17 12:53:39 CST 2000