Consider an LTI system whose impulse response is $$h[n]=\frac{1}{2^n}u[n]+\frac{1}{3^n}u[n]$$ The input signal to this system is $x[n]$ and is null for $n<0$ but may or may not be null for $n=0$.
I found the difference equation, which turned out to be $$y[n]-\frac{5}{6}y[n-1]+\frac{1}{6}y[n-2]=2x[n]-\frac{5}{6}x[n-1]$$
Then the instruction goes as follows:
It is desired to compute the output $y[n]$ for $n=0,1,...,1000$ with the input signal being the same $x[n]$. Determine the impulse response $\hat{h}[n]$ of the minimum-sized FIR system whose output $\hat{y}[n]$ equals $y[n]$ for $n=0,1,...,1000$ and find the difference equation of that FIR filter.
I don't know for certain how to deal with this. My first attempt consisted of establishing that
$$\hat{h}[n]=\left\{ \begin{array}{ll} h[n] & \mbox{if } 0\leq n\leq 1000 \\ 0 & \mbox{otherwise } \end{array} \right.$$
I don't know if that is correct. I just thought that, due to the causality of both $h[n]$ and $x[n]$, the convolution between them would be exactly the same for the first 1001 values if $h[n]$ were cut off at $n=1000$. I named $\hat{h}[n]$ that cut-off version of the first impulse response. Nevertheless, I'm really doubtful about this reasoning. On top of it, I don't even know if that response is the shortest possible as I am asked to find in the exercise.
Assuming that the expression above for $\hat{h}[n]$ is okay, it can be expressed as $$\hat{h}[n]=\frac{1}{2^n}(u[n]-u[n-1001])+\frac{1}{3^n}(u[n]-u[n-1001])$$
After some algebra, that equation becomes impossible to deal with. The $\hat{H}(z)$ has a 1001st grade polynomial in its numerator and a 2nd grade one in its denominator. Because the filter is a FIR one, those poles should be cancelled by two of the zeros of the numerator, but it's just impossible to find the zeros of a 1001st grade polynomial, so I really don't think that this is the way to solve this problem.
Any ideas?
Thanks for your time!