[uc_mc_semigroups]6_correct_errors_in_solution4

[shlff]

Hi @jstac , this PR corrects errors in the following math expression of lecture uc_mc_semigroups's Solution 4:

• K^{m+1}(i, j) = \sum_n K(i, n) K^m (n, j) = \sum_n K(i, n) \mathbb 1\{j = i + m\} = K(i, m-j) -->> K^{m+1}(i, j) = \sum_n K(i, n) K^m (n, j) = \sum_n K(i, n) \mathbb 1\{j = n + m\} = K(i, j-m)

[jstac]

Great job, thanks @shlff .