6.6. Geometric interpretation
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:
or 
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 ...
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