# Quod Erat Demonstrandum

## 2009/03/04

### boring discussion on limit of sequence

Filed under: Additional / Applied Mathematics,HKALE,Pure Mathematics — johnmayhk @ 3:30 下午

Here is just a typical, basic, level-zero question about the limit of sequence for formative assessment.

Let $x_1 = 3$, $x_{n + 1} = \frac{2x_n^3 + 8}{3x_n^2}$ for $n \in \mathbb{N}$.

(a) Given that $x_n > 2 (\forall n \in \mathbb{N})$, show that $x_{n + 1} - 2 < \frac{2}{3}(x_n - 2) (\forall n \in \mathbb{N})$.

(b) By squeezing principle, show that $\lim_{n \rightarrow \infty} x_n = 2$.

Actually, $x_n > 2$ can be shown easily (no need to be given) by M.I. or something else, students you may try.

To solve part (a), we simply consider $x_{n + 1} - 2 - \frac{2}{3}(x_n - 2) = \frac{8 - 2x_n^2}{3x_n^2} < 0$ ($\because x_n > 2 (\forall n \in \mathbb{N})$).

To set the question in a ‘better’ way, we may ask, show that $x_{n + 1} - 2 < k(x_n - 2)$ for some constant $k$ such that $0 < k < 1$.

Part (b) requires a common technique: recurrence. That is, we may apply the result $x_{n + 1} - 2 < \frac{2}{3}(x_n - 2)$ again and again.

$0 < x_n - 2$
$< \frac{2}{3}(x_{n - 1} - 2)$ (by (a))
$< \frac{2}{3} \frac{2}{3}(x_{n - 2} - 2)$ (by (a) again)
$< \frac{2}{3} \frac{2}{3} \frac{2}{3}(x_{n - 3} - 2)$ (by (a) again and again)
…inductively…
$< (\frac{2}{3})^{n - 1}(x_1 - 2)$

Hence we made a ‘sandwich’ with lower bound 0 and upper bound $(\frac{2}{3})^{n - 1}(x_1 - 2)$. Together with the fact that $\lim_{n \rightarrow \infty} (\frac{2}{3})^{n - 1}(x_1 - 2) = 0$, hence the ‘content’ in the sandwich, $(x_n - 2)$, will also tend to zero, that is, $\lim_{n \rightarrow \infty}(x_n - 2) = 0$ and the result follows.

As I said at the beginning, it is only for formative assessment. To show the convergence of the limit, we may simply check that {$x_n$} is monotonic decreasing and bounded from below by 2. By monotonic sequence theorem (MST), {$x_n$} converges. And taking limit from $x_{n + 1} = \frac{2x_n^3 + 8}{3x_n^2}$, it is easy to have $\lim_{n \rightarrow \infty}x_n = 2$, try.

How to create this question? It is a piece of cake.

Let $f(x) = x^3 - 8$.

It is easy to know $x = 2$ is a root of $f(x) = 0$, right?

Then I use Newton’s method to solve it numerically.

Hence, we may obtain the recurrence relation trivially, i.e.

$x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{x_n^3 - 8}{3x_n^2} = \frac{2x_n^3 + 8}{3x_n^2}$

See, it is exactly the formula given in the original question. For a suitably choice of an initial guess, take $x_1 = 3$ (say), we may generate a sequence which converges to 2.

In general, we may generate a sequence which converges to $\sqrt[k]{\alpha}$ (where $\alpha$ is a positive integer) by taking $f(x) = x^k - \alpha$. It is easy to see that, once the initial guess (i.e. $x_1$) is rational, the numbers in the sequence generated by Newton’s method are all rational. Hence, for irrational number $\sqrt[k]\alpha$, we have at least one way (e.g. by Newton’s method) to generate a sequence of rational numbers $x_1, x_2, \dots$ which converges to $\sqrt[k]\alpha$.

Let’s see a concrete example $f(x) = x^2 - 2$, yes, the irrational number $\sqrt{2}$ is a root of $f(x) = 0$. Applying Newton’s formula, yield $x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)} = \frac{x^2 + 2}{2x}$, take $x_1 = 1$ (suitably chosen), then, we have

$x_1 = \frac{3}{2} = 1.5$
$x_2 = \frac{17}{12} = 1.416666667$
$x_3 = \frac{577}{408} = 1.414215686$

The rational numbers in the sequence tend to the irrational number $\sqrt{2}$ (i.e. $\lim_{n \rightarrow \infty}x_n = \sqrt{2}$), see?

In the discussion above, we consider irrational number in the form of $\sqrt[k]{\alpha}$. How about irrational numbers in other form? As for example, $\pi$, can we still find a sequence of rational numbers which converges to $\pi$?

Urm, it seems to be very easy, because

$\pi \approx 3.14159...$

Then, we may create a sequence

$x_1 = 3$
$x_2 = 3.1$
$x_3 = 3.14$
$x_4 = 3.141$
$x_5 = 3.1415$
$x_6 = 3.14159$

Actually, there are many ways to create sequences converging to $\pi$, just give an easy one

$x_1 = 1$ and $x_n = x_{n - 1} + \frac{(-1)^{n - 1}}{2n - 1}$ ($n > 1$), then $\lim_{n \rightarrow \infty}x_n = \pi$.