We have a new vector defined as \( x(0)=0 \), \( x(1)=0 \), \( x(2)=\beta_0 \), \( x(3)=\beta_1 \), \( x(4)=\beta_2 \), \( x(5)=\beta_3 \), \( x(6)=0 \), and \( x(7)=0 \).
We have added four new elements, which are all zero. The benefit is that we can rewrite the equation for \( \boldsymbol{y} \), with \( i=0,1,\dots,5 \),
$$ y(i) = \sum_{k=0}^{k=m-1}w(k)x(i+(m-1)-k). $$As an example, we have
$$ y(4)=x(6)w(0)+x(5)w(1)+x(4)w(2)=0\times \alpha_0+\beta_3\alpha_1+\beta_2\alpha_2, $$as before except that we have an additional term \( x(6)w(0) \), which is zero.
Similarly, for the fifth-order term we have
$$ y(5)=x(7)w(0)+x(6)w(1)+x(5)w(2)=0\times \alpha_0+0\times\alpha_1+\beta_3\alpha_2. $$The zeroth-order term is
$$ y(0)=x(2)w(0)+x(1)w(1)+x(0)w(2)=\beta_0 \alpha_0+0\times\alpha_1+0\times\alpha_2=\alpha_0\beta_0. $$