# 1415 the derivative matrix of f with respect to the

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. . = 0 0 . . . (14.15) The derivative matrix of f with respect to the vector x is f ( x , a ) x vextendsingle vextendsingle vextendsingle vextendsingle x g , a g = 1 a g 1 0 0 0 0 . . . 0 0 1 a g 1 0 0 . . . 0 0 0 0 1 a g 1 . . . . . . , (14.16) always evaluated at the guessed values of the parameters (subscript g ). The derivative matrix of f with respect to the vector a is

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– 61 – f ( x , a ) a vextendsingle vextendsingle vextendsingle vextendsingle x g , a g = 1 t g 0 1 t g 1 1 t g 2 . . . , (14.17) again always evaluated at the guessed values of the parameters (subscript g ). 14.8. The Solution to the Lagrangian: Two Matrix Equations Jefferys does all this this 9 for us and, after some algebraic manipulation, provides the two following matrix equations (from the set of four in his equation 10): σ - 1 · δ x + f T ( x , a ) x vextendsingle vextendsingle vextendsingle vextendsingle x d , a * · λ = 0 (14.18a) This single matrix equation embodies MJ individual equations. Here we write f T x instead of f T ∂δ x for clarity, and use the fact that they are the negative of each other. f T ( x , a ) a vextendsingle vextendsingle vextendsingle vextendsingle x d , a * · λ = 0 (14.18b) This matrix equation embodies N individual equations. These two matrix equations embody MJ + N individual equations, which is equal to the number of unknowns, so we can solve for them! In both of these equations, the dimensions of the factors are ( Note the transposes! ): σ : ( MJ ) × ( MJ ) (14.19a) δ x : MJ × 1 (14.19b) f T : 1 × M (14.19c) f T ( x , a ) x : ( MJ ) × M (14.19d) 9 And more: he includes the possibility for additional constraints, the second line in his equation (7). This introduces additional complications and we ignore this possibility in the interest of pedagogical simplification. For example, his set of equations (10) has four equations, not two as we write here.
– 62 – λ : M × 1 (14.19e) f T ( x , a ) a : N × M (14.19f) Note that in 14.19d above the dimension is ( MJ ) × M : M elements in f , differentiated by ( MJ ) different variables x ; similarly for 14.19f above. 14.9. Solving Equations 14.18a and 14.18b Iteratively Generally, these equations are nonlinear and we solve them iteratively using guessed solutions. We denote the guessed values with subscript g , so we have a g = guessed values for a * (14.20a) x g = guessed values for x * (14.20b) ITERATION STEP 1: We define the difference between the measured data quantities and their currently-guessed counterparts Δx g x d x g (14.20c) Above, our Δx g is the negative of Jefferys’ ˆv . The nonlinear fit solves for corrections to these guesses, which we denote by Δx new (the negative of Jefferys’ ˆv new ) and Δa new (which is identical to Jefferys’ ˆ δ ).

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