I want to approximate tanh by a low-degree rational function of form r ( x ) = p ( x ) q ( x ) = p 2 x 2 + p 1 x + p 0 q 1 x + q 0 such that the L 2 norm over the fixed interval [ x 0 , x 1 ] is small: I ( p , q ) = 1 2 ∫ x 0 x 1 ( r ( x ) − tanh ⁡ ( x ) ) 2 d x .That is, I would like to solve for the coefficients p 2 , p 1 , p 0 , q 1 , q 0 . Note that d e g ( p ) ≤ 2 and d e g ( q ) ≤ 1. I don't know where to begin, so I'm looking for suggestions on how to approach this problem; pointers to relevant numerical methods would also be greatly appreciated

Question

I want to approximate   tanh by a low-degree rational function of form  r  (  x  )  =            p      (      x      )              q      (      x      )        =                    p        2                    x        2            +              p        1            x      +              p        0                            q        1            x      +              q        0            such that the       L    2   norm over the fixed interval   [      x    0    ,      x    1    ] is small:  I  (  p  ,  q  )  =      1    2        ∫                  x        0                            x        1                        (      r      (      x      )      −      tanh      ⁡      (      x      )      )        2        d    x  .That is, I would like to solve for the coefficients       p    2    ,      p    1    ,      p    0    ,      q    1    ,      q    0  . Note that       d    e    g    (  p  )  ≤  2 and       d    e    g    (  q  )  ≤  1. I don&#039;t know where to begin, so I&#039;m looking for suggestions on how to approach this problem; pointers to relevant numerical methods would also be greatly appreciated

Camron Herrera · Accepted Answer

Not an answer but too long for a comment.An analytical expression of the       L    2   norm seems very difficult (not to say impossible).What I would suggest is to built the   [  2  ,  1  ] Padé approximant of   tanh  ⁡  (  x  ) around   a  =                    x        0            +              x        1              2  . This would give  tanh  ⁡  (  x  )  =            3      tanh      ⁡      (      a      )      +      2      (      x      −      a      )      +      (      tanh      ⁡      (      a      )      −      coth      ⁡      (      a      )      )      (      x      −      a              )        2                    3      +      (      3      tanh      ⁡      (      a      )      −      coth      ⁡      (      a      )      )      (      x      −      a      )        +  O      (    (    x    −    a          )      4        )  Trying for       x    0    =  1 and       x    1    =  2      ∫    1    2              [        tanh  ⁡  (  x  )  −  Padé                    ]              2      d  x  =  3.90537  ×      10          −      7      seems to be already decent. For the same problem, a full optimization would give   1.15644  ×      10          −      8            ∫    2    3              [        tanh  ⁡  (  x  )  −  Padé                    ]              2      d  x  =  8.80730  ×      10          −      9            ∫    3    4              [        tanh  ⁡  (  x  )  −  Padé                    ]              2      d  x  =  1.66518  ×      10          −      10        log  ⁡            [            ∫    n          n      +      1                  [        tanh  ⁡  (  x  )  −  Padé                    ]              2      d  x            ]        =  −  3.87288    n  −  11.2126May be, you could use the sum of these Padé approximants each of them being built for a short interval.Edit:The key problem is that we cannot have even an explicit solution for  ∫            (      tanh      ⁡      (      x      )      −                        a          x                          b          x          +          1                    )        2      d  xThere is one thing you could try taking into account the fact that the   [  2  ,  1  ] Padé approximant is   O      (    (    x    −    a          )      4        )  .Let  tanh  ⁡  (  x  )  =      ∑          n      =      0        3        b    n      (  x  −  a      )    n  and define the       L    2   norm  Φ  (      b    0    ,      b    1    ,      b    2    ,      b    3    )  =      ∫                  x        0                            x        1                        [        tanh  ⁡  (  x  )  −      ∑          n      =      0        3        b    n      (  x  −  a      )    n                      ]              2      d  xdeveloping the integrand, we have terms in   (  k  =  0  ,  ⋯  ,  6  ) (no problem with these) and a series of integrals      I    n    =  ∫      x    n      tanh  ⁡  (  x  )    d  x      k  =  0  ,  1  ,  2  ,  3They do not make much problems      I    0    =  log  ⁡  (  cosh  ⁡  (  x  )  )      I    1    =      1    2        (    x          (      x      +      2      log      ⁡              (                  e                      −            2            x                          +        1        )            )        −          Li      2              (      −              e                  −          2          x                    )        )        I    2    =  −  x      Li    2        (    −          e              −        2        x              )    −      1    2        Li    3        (    −          e              −        2        x              )    +            x      3        3    +      x    2    log  ⁡      (          e              −        2        x              +    1    )        I    3    =      1    4        (    −    6          x      2              Li      2              (      −              e                  −          2          x                    )        −    6    x          Li      3              (      −              e                  −          2          x                    )        −    3          Li      4              (      −              e                  −          2          x                    )        +          x      4        +    4          x      3        log    ⁡          (              e                  −          2          x                    +      1      )        )  and   ∫      tanh    2    ⁡  (  x  )    d  x  =  x  −  tanh  ⁡  (  x  )So, we have all the elements. Now solve            ∂      Φ              ∂              b        0              =            ∂      Φ              ∂              b        1              =            ∂      Φ              ∂              b        2              =            ∂      Φ              ∂              b        3              =  0which makes a linear system of four equations for four unknowns.So, we have the series expansion that we could transform as a   [  2  ,  1  ] Padé approximant around the midpoint   a  =                    x        0            +              x        1              2  with      p    2    =      b    2    2    −      b    1        b    3          p    1    =      b    1        b    2    −      b    0        b    3          p    0    =      b    0        b    2          q    1    =  −      b    3          q    0    =      b    2  This procedure has been tried with       x    0    =  1,       x    1    =  3,   a  =  2.Using the original Padé approximant leads, for the       L    2   norm, to a value equal to   1.67  ×      10          −      4      . The optimization of the series gives   3.57  ×      10          −      6      . The new Padé approximant leads to   3.33  ×      10          −      4      .Much work for a more than bad result.

rzfansubs87 · Accepted Answer

Here is a (bad!) attempt. First, suppose without loss of generality that       q    1    =  1, and define      Q          (      k      )        (      q    0    )  =      ∫                  x        0                            x        1                                x        k            tanh      ⁡      (      x      )              x      +              q        0                  d    x  ,then expanding the optimality conditions       ∂                  p                  (          ⋅          )                      I  =  0 we obtain                              ∂                                    p              0                                      I                            =                  ∫                                    x              0                                                          x              1                                      (        r        (        x        )        −        tanh        ⁡        (        x        )        )                  1                      x            +                          q              0                                                d                x        =                  ∫                                    x              0                                                          x              1                                                            r            (            x            )                                x            +                          q              0                                                d                x        −                  Q                      (            0            )                          (                  q          0                )        =        0                                      ∂                                    p              1                                      I                            =                  ∫                                    x              0                                                          x              1                                      (        r        (        x        )        −        tanh        ⁡        (        x        )        )                  x                      x            +                          q              0                                                d                x        =                  ∫                                    x              0                                                          x              1                                                            x            r            (            x            )                                x            +                          q              0                                                d                x        −                  Q                      (            1            )                          (                  q          0                )        =        0                                      ∂                                    p              2                                      I                            =                  ∫                                    x              0                                                          x              1                                      (        r        (        x        )        −        tanh        ⁡        (        x        )        )                              x            2                                x            +                          q              0                                                d                x        =                  ∫                                    x              0                                                          x              1                                                                          x              2                        r            (            x            )                                x            +                          q              0                                                d                x        −                  Q                      (            2            )                          (                  q          0                )        =        0            where for a fixed value of       q    0   such that   −      q    0    ∉  [      x    0    ,      x    1    ], the above becomes a system of 3 equations in       p    2    ,      p    1    ,      p    0  . If we can precompute a discretization of       Q          (      k      )        (      q    0    ) for different values of       q    0  , then we may search over the combination of coefficients that approximately minimizes the norm.

I want to approximate tanh by a low-degree rational function of form r ( x )

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