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Hi anassinator, great work and thanks for the contribution. I am wondering if this can handle problems with with one step cost being L(X; u) = (X-X*)TQ(X-X*)+ (u-u*)TR(u-u*), where X* and u* are my target trajectory? @anassinator
Hi anassinator, great work and thanks for the contribution. I am wondering if this can handle problems with with one step cost being L(X; u) = (X-X*)TQ(X-X*)+ (u-u*)TR(u-u*), where X* and u* are my target trajectory? @anassinator