# Compute filter output with given values of $n$

\begin{align} x[n] &=\begin{cases} 1& \text{for}\quad n \ge 0\\ 0& \text{otherwise} \end{cases}\\ h[n] &= \delta[n] - \delta[n-1]\\ y[n] &= x[n]*h[n] \end{align}

Find the values of $y[-1]$, $y[0]$, $y[1]$, $y[2]$.

I thought $\delta[0] = 1$ and $\delta[n] = 0$ otherwise, and that $y[n] = x[k]h[n-k]$:

\begin{array}{|c|c|c|c|c|c|}\hline n&x[n]&d[n]&d[n-1]&h[n]&y[n]\\\hline -1& 0 & 0 & 0 & 0&0\\\hline 0&1&1&0&1&1\\\hline 1& 1 & 0 & 1 & -1&-1\\\hline 2& 1 & 0 & 0 & 0&0\\\hline \end{array}

So I put $\text{ 0 1 -1 0 }$ as my answer but apparently that's wrong. What did I mess up? Did I misinterpret the meaning of $\delta[n]$?

Your impulse response is $h[n]=\delta [n] - \delta [n-1]$, so it basically takes a copy of the input and then substracts to it a delayed version of the former. When $x[n]=u[n]$, the output signal is:
$$y[n] = u[n] - u[n-1]$$
This means that we have a step function (i.e. it is equal to $1$ for $n \geq 0$ and $0$ otherwise), and we substract to it a shifted step (i.e. it is equal to $1$ for $n \geq 1$ and $0$ otherwise). Therefore, $\forall n \geq 1$ we are cancelling out the two signals ($1-1=0$). This leaves us with only one not-zero value, which happens to be at the origin. So the output turns out to be
$$y[n]=\left\{ \begin{array}{ll} 1 & \mbox{if } n = 0 \\ 0 & \mbox{otherwise } \end{array} \right. \implies y[n]=\delta[n]$$