Transcendental Equation with Quadratic Part (Can it be solved via Lambert W function)?There I hope to minimize an optimization problem: min x ∈ R + f ( x ) = x 2 − y e − x 2 + r ( x − d ) 2 ,where y , d ∈ R and r ∈ R + .For this equation, the most natural idea is to find the critical point x 0 where f ′ ( x 0 ) = 0. That is, ( r + 1 ) x + y x e − x 2 = r.But this form is not similar with Lambert W function since quadratic term is involved. Is there any method to find the analytic/ numerical solution of this equation?

Question

Transcendental Equation with Quadratic Part (Can it be solved via Lambert W function)?There I hope to minimize an optimization problem:      min          x      ∈                        R                +              f  (  x  )  =      x    2    −  y      e          −              x        2              +  r  (  x  −  d      )    2    ,where   y  ,  d  ∈      R   and   r  ∈            R        +  .For this equation, the most natural idea is to find the critical point       x    0   where       f    ′    (      x    0    )  =  0. That is,   (  r  +  1  )  x  +  y  x      e          −              x        2              =  r.But this form is not similar with Lambert W function since quadratic term is involved. Is there any method to find the analytic/ numerical solution of this equation?

Miah Scott · Accepted Answer

Step 1Considering that you look for the zero of function  g  (  x  )  =  (  r  +  1  )  x  +  y  x      e          −              x        2              −  rIts derivative      g    ′    (  x  )  =  (  r  +  1  )  +  y        e          −              x        2                  (    1    −    2          x      2        )  cancels at      x    ∗    =      1          2            1    −    2    W          (      −                                    e                    (          r          +          1          )                          2          y                    )      Step 2(since you look for       x    ∈                  R            +      ). Now, two possibilities :- x∗ does not exist in             R        +   and then g′(x) does not change sign and Newton method would not face any trouble (more or less iterations depending on       x    0  ). What we know is that   g  (  0  )  =  −  r,       g    ′    (  0  )  =  1  +  r  +  y,       g    ″    (  0  )  =  0. There is a solution only if (  (  1  +  r  +  y  &amp;gt;  0; if not, no solution. If yes, start with       x    0    =  0 or better       x    0    =      r          r      +      y      +      1       which is the result of the first iteration of Newton method.- x∗ does exist in             R        +  . Now, we can build a Taylor series around x∗ which would give as initial estimate      x    0    =      x    ∗    +      −    2                  g        (                  x          ∗                )                              g          ″                (                  x          ∗                )            with  g  (      x    ∗    )  =                    2            (      r      +      1      )              x        ∗        3            −      2      r              x        ∗        2            +      r              2              x        ∗        2            −      1          and        g    ″    (      x    ∗    )  =      2    (  r  +  1  )      x    ∗        (          2              1        −        2                  x          ∗          2                      +    1    )

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