limit of a sequence -> possibly squeeze theorem

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August 21st 2012, 06:54 AM
Angela11

limit of a sequence -> possibly squeeze theorem

http://img96.imageshack.us/img96/8606/qqqqbj.jpg

so what i'm thinking is that you let

x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1

then show that the first term can eventually go to the last term

squeezing the term into the middle

but the problem with this is i dont see where the 1/3 comes from

any ideas?
August 21st 2012, 08:05 AM
emakarov

Re: limit of a sequence -> possibly squeeze theorem

Quote:

Originally Posted by Angela11

so what i'm thinking is that you let

x_n+3 + x_n+2<=x_n+2 + x_n+1 <= x_2 + x_1

What do you mean by "let"? One uses the word "let" during a definition: e.g., let w = 6. Why do you think these inequalities are true?

Quote:

Originally Posted by Angela11

any ideas?

One way is to find an explicit formula for $x_n$ . It is convenient to define $a_n$ and $b_n$ so that xₙ = (aₙx₁ + bₙx₂) / 2ⁿ⁻². Then it is possible to write recurrence relations and initial conditions on aₙ and bₙ, solve them and find the limits of aₙ / 2ⁿ⁻² and bₙ / 2ⁿ⁻².
September 18th 2012, 02:53 AM
johnsomeone

Re: limit of a sequence -> possibly squeeze theorem

The way to do simple linear recursions is to try $x_n = ar^n$ . This is very much analogous to ODEs, where you'd try $y(x) = Ae^{f(x)}$ , seeing what f must be, and then using superposition to add together the different possibilities for f.

Solve: $x_{n+2} = \frac{x_{n+1} + x_n}{2}, n \in \mathbb{N}$ .

Solution: Try $x_n = ar^n$ for some constants $a, r \in \mathbb{C}$

From $x_{n+2} = \frac{x_{n+1} + x_n}{2}$ get $\ ar^{n+2} = \frac{ar^{n+1} + ar^n}{2}$ , so $2r^2 = r + 1$ .

Solving that gives $r = 1$ or $r = -1/2$ . Now just like with ODEs, put them together to get the general form of the solution:

$x_n = a_0(1)^n + a_1(\frac{-1}{2})^n = a_0 + \frac{a_1}{(-2)^n}$ .

Note the required 2 unknown constants for a degree 2 linear recursion.

When you plug that in to the recursion equation, it works. It's just algebra - I won't work it out here. Thus we have the generic solution to that recursion.

Just like with ODEs, we use the initial conditions to get the specific solution:

$x_1 = a_0 + (\frac{a_1}{(-2)^1}) = a_0 - a_1/2$ , and $x_2 = a_0 + (\frac{a_1}{(-2)^2}) = a_0 + a_1/4$ . Thus

$2x_1 = 2a_0 - a_1$ , and $4x_2 = 4a_0 + a_1$ , which add to give $2x_1 + 4x_2 = 6a_0$ , so $a_0 = \frac{x_1 + 2x_2}{3}$ .

To solve the problem, we don't even need to bother to find $a_1$ , though it would be simple to do.

$\underset{n \rightarrow \infty}{\lim} x_n = \underset{n \rightarrow \infty}{\lim} \left(a_0 + \frac{a_1}{(-2)^n}\right) = a_0 = \frac{x_1 + 2x_2}{3}$