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?

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 $\displaystyle x_n$. It is convenient to define $\displaystyle a_n$ and $\displaystyle b_n$ 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.

Re: limit of a sequence -> possibly squeeze theorem

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

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

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

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

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

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

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:

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

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

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

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