Let us give another expression to the cost function at instant K.
We have found: with and and Wiener solution of .
The cost can be put in the form:
Let us state (the origin of the axes is at present ); it becomes:
and easily: : the factor K representing the instant where we are considering the gradient.
Let us simplify the preceding expressions to find simple geometric interpretations by changing the base.
Matrix R being symmetric, we say that it is diagonalizable ...