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.

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