# Interpretation of the final_cost field in Ceres.

01 Jul 2022I love using the Ceres solver, but couldnâ€™t figure out why when I computed the value of the cost function using my code, it was different from `final_cost`

field of a `Solver::Summary`

object.

Finally, I figured out that I constructed a cost function as

\[\sum_i \| f_i(x_i, \dots , x_{ik}) \|^2 \label{eqone}\]and in Ceres, the cost function is assumed to have this form, Ceres link:

\[\frac{1}{2}\sum_i \| f_i(x_i, \dots , x_{ik}) \|^2 \label{eqtwo}\]The only difference is that \(\frac{1}{2}\) in the Ceres version. To convert the final `final_cost`

to my cost function, I need to multiply `final_cost`

by two.

This was such a simple thing, but I spent more time than I would have liked figuring it out, so here this is as a TIL.