Closed shlff closed 4 years ago
Hi @jstac , this PR corrects errors in the following math expression of lecture uc_mc_semigroups's Solution 4:
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)
Great job, thanks @shlff .
jstac
commented
4 years ago jstac
commented
4 years ago
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)