# Two functions in two variables

1. Mar 4, 2006

### jeffceth

Sorry, I don't know what category to put this in.

Here's my problem:

I have some corresponding values for two variables a and b.

I also know that
$$a_n = 2a_{n-1} + b_{n-1}$$
and
$$b_n = b_{n-1} + a_n$$
for all a and b.

I need to find functions $$f_1$$ and $$f_2$$ in terms of n so that $$f_1(n) = a_n$$ and $$f_2(n) = b_n$$

I don't really know where to start. I tried recombining the first two equations in myriad ways but I didn't get anywhere there. I have figured out that this would be pretty tough without the sample values to generate the others, and that the answers will be in general form because n = 1 could refer to any of the pairs of values, although all of them can be generated from a single pair. What do I do?

2. Mar 4, 2006

### 0rthodontist

Why not start by telling us your sample values?

One way to do this problem if you know linear algebra is to express the equations a_n and b_n each in terms of a_(n-1) and b_(n-1). This is then a simple Markov chain problem. Luckily the matrix you get is diagonalizable, so its powers will yield an easy algebraic expression.

Last edited: Mar 4, 2006
3. Mar 4, 2006

### jeffceth

Sorry, should have included that. my sample values are:
a = 1 when b =1 and a = 3 when b = 4(I know the second is unneccessary). The entire sequence can be generated from these values, though this point is not neccessarily n = 1. This is why the answer should be in general form.

4. Mar 4, 2006

### AlphaNumeric

1) $$a_{n} = 2a_{n-1}+b_{n-1}$$
2) $$b_{n} = b_{n-1} + a_{n}$$

2) => $$a_{n} = b_{n}-b_{n-1}$$

Put that into 1 to get

$$b_{n}-b_{n-1} = 2a_{n-1}+b_{n-1}$$
3) $$b_{n} = 2(a_{n-1}+b_{n-1})$$

2) => $$a_{n-1} = b_{n-1}-b_{n-2}$$

3) => $$b_{n} = 2(b_{n-1}-b_{n-2}+b_{n-1}) = 4b_{n-1}-2b_{n-2}$$

You need 2 initial values for b, one you're given, the other you generate from your initial condition on a. Now you can just use the equation for b and equation 2 to get a's equation, which will also need 2 initial values, one you're given, the other you get from b's initial value.

5. Mar 4, 2006

### 0rthodontist

The general equation looks irritating to work out by hand, involving a lot of sqrt 2 and large expressions, unless my calculator is missing a major simplification somewhere.

Last edited: Mar 4, 2006
6. Mar 4, 2006

### jeffceth

Thanks for your help!(I really should have seen that. It must be getting too late) Unfortunately the nature of the answer shows this series is not what I am looking for. I'll keep trying!

7. Mar 4, 2006

### 0rthodontist

What do you mean?