# Where is the mistake in the convolution?

Let : $$x[n]=\begin{cases} 1&\text{if 0\leq n\leq 4}\\ \\ 0&\text{if otherwise} \end{cases} \qquad \text{and} \qquad h[n]=\begin{cases} \alpha^{n}&\text{if 0\leq n\leq 6}\\ \\ 0&\text{if otherwise} \end{cases}$$ We must compute the convolution : $$(x*h)[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]$$ Observe that $$x[n]=u[n]$$ for $$n\in[0,4]$$ and $$0$$ otherwise. Furthermore, $$h[n]=\alpha^{n}u[n]$$ for $$n\in[0,6]$$. Thus : $$\begin{array}{c | c c c c c c c } \mathbf{x} & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ \hline 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 1 & 1 & \alpha & \alpha^{2} & \alpha^{3} &\alpha^{4} &\alpha^{5} &\alpha^{6}\\ 0 & 0 & 0 & 0& 0& 0 & 0 & 0 \\ 0 & 0 & 0 & 0& 0& 0 & 0 & 0 \end{array}$$ \begin{align*} \displaystyle (x*h)[n]&=\left[\sum_{k=0}^{0}\alpha^{k}\;,\;\sum_{k=0}^{1}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=0}^{4}\alpha^{k}\;,\;\sum_{k=1}^{5}\alpha^{k}\;,\;\sum_{k=2}^{6}\alpha^{k}\;,\;\cdots\;,\;\sum_{k=5}^{6}\alpha^{k}\;,\;\sum_{k=6}^{6}\alpha^{k}\right]\\ \\ &=\begin{cases} \displaystyle\sum_{k=0}^{n}\alpha^{k}&\text{if 0\leq n\leq 4}\\ \displaystyle\sum_{k=1}^{n}\alpha^{k}&\text{if n=5}\\ \displaystyle\sum_{k=n+2}^{6}\alpha^{k}&\text{if 0\leq n\leq4} \end{cases} \end{align*} I wish to know where my mistake is and how to fix it. Thank you :)

$$3^{rd}$$ term in your final answer should have $$6 \le n \le 10$$, instead of $$0 \le n \le 4$$. And, the limit of summation will go from $$n-4$$ to $$6$$.

$$\sum^{6}_{k = n-4} \alpha^k \ , 6 \le n \le 10$$

$$0 \le n \le 4$$ case has already been covered by the $$1^{st}$$ term of final answer.