How can I solve this system of equations for any $n$:
$$1.25a_1+0.5a_2=0.5\
0.5a_1+1.25a_2+0.5a_3=0\
0.5a_2+1.25a_3+0.5a_4=0\
0.5a_3+1.25a_4+0.5a_5=0\
...\
0.5a_{n-2}+1.25a_{n-1}+0.5a_n=0\
0.5a_{n-1}+1.25a_n=0$$

Question

How can I solve this system of equations for any $n$:
$$1.25a_1+0.5a_2=0.5\
0.5a_1+1.25a_2+0.5a_3=0\
0.5a_2+1.25a_3+0.5a_4=0\
0.5a_3+1.25a_4+0.5a_5=0\
...\
0.5a_{n-2}+1.25a_{n-1}+0.5a_n=0\
0.5a_{n-1}+1.25a_n=0$$

kirbykirby · Answer

The best I could do is get "some sort" of recursive formula:

$$a_1=0.4-0.4a_2\
a_2=1-2.5a_1\
a_k=-a_{k-2}-2.5a_{k-1} \text{    for    } 3 \le k \le n-1 \
a_n=-0.4a_{n-1}\ $$

But doesn't help too much since a1 depends on a2, and a2 depends on a1 :(

perl · Answer

interesting

ganeshie8 · Answer

n equations and n unknowns, so solvable

i may try something like below :- 
1) eliminate $a_1$ from first two equations : (2) - (1)/2.5 ----------(I)
2) eliminate $a_2$ from above resultant equation and third equation : (3) - (I) --------(II)
3) keep going, final two equations will have two unknowns : $a_{n-1}$ and $a_n$

ganeshie8 · Answer

you're looking for deriving a formula for finding last two equations is it ?

kirbykirby · Answer

well basically some sort of equation(s) that will provide a solution to $a_1$ to $a_n$

ganeshie8 · Answer

looks like bit involved task...

kirbykirby · Answer

Yeah :( I'm not sure how to do it exactly. and i can't really use software since I have to find it for any general n

kirbykirby · Answer

or can I? o_o I'm sorry I am so brain-dead. I have been up all night ;_;

ganeshie8 · Answer

another thing to try is (painful too) :-
1) setup a nxn matrix (u wil get a symmetric matrix here)
2) row reduce to triangular
3) setup equations

ganeshie8 · Answer

is this question from linear algebra... or just a general one ?

loser66 · Answer

for this $0.5 a_{n-2} + 1.25a_{n-1}+0.5a_n=0$ you can write down back ward and make it have a integer coefficient by time all by 4,
$$2a_{n +2}+ 5a_{n+1} +2a_n=0$$
characteristic equation is $2r^2+5r +2=0$
solve for it, you have 2 roots $r_1= -0.5$ and $r_2 = -2$
so $a_n = C_0(-0.5)^n + C_1(-2)^n$

loser66 · Answer

If you have 2 initial conditions for $a_0$ and $a_1$, you can solve for $C_0$ and $C_1$
If you don't have , just stop there. You are OK.
Is this from discrete math?

kirbykirby · Answer

actually this is a sub result I am getting from my stats problem

kirbykirby · Answer

I assume not many people here have done upper-year stats, but it's a forecasting problem and I need to solve these equations as I progress through the problem.

kirbykirby · Answer

the thing is I don't have any initial conditions so this was giving me a lot of grief

amistre64 · Answer

the initial conditions should result in arbitrary constants i believe

amistre64 · Answer

or at least one arbitrary constant with which to define when a specific initial condition is known

kirbykirby · Answer

thanks everyone for their responses!

loser66 · Answer

hope this helps

kirbykirby · Answer

Oh thanks very much^!