Compute filter output with given values of $n$ \begin{align}
x[n] &amp;=\begin{cases}
 1&amp; \text{for}\quad n \ge 0\\
0&amp; \text{otherwise}
\end{cases}\\
h[n] &amp;= \delta[n] - \delta[n-1]\\
y[n] &amp;= 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&amp;x[n]&amp;d[n]&amp;d[n-1]&amp;h[n]&amp;y[n]\\\hline
-1&amp; 0 &amp; 0 &amp; 0 &amp; 0&amp;0\\\hline
0&amp;1&amp;1&amp;0&amp;1&amp;1\\\hline
1&amp; 1 &amp; 0 &amp; 1 &amp; -1&amp;-1\\\hline
2&amp; 1 &amp; 0 &amp; 0 &amp; 0&amp;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

$$y[n]=\left\{
 \begin{array}{ll}
  1  &amp; \mbox{if } n = 0 \\
  0 &amp; \mbox{otherwise }
 \end{array}
\right. \implies y[n]=\delta[n]$$