!!!!!!!!!!!!
>at (e,1)
so n=1
>looking for a[t] = na = 1a = a
So a[t] = correct a?
>At that value, d[t] = na+x
So d[t] = 1a + x like always.
>x[t] = x
I'm confused this would just mean we're looking for the right x I'm not sure how much this other stuff helps unless I'm completly missing something