I have a hard time finding the analytical solution to the following non-linear equation: ( 1 + x ) 1 − p + p p 1 − p x ( p − 1 ) − 1 = 0where p ∈ ] 0 , 1 [ and x &gt; 1. I would like to have a solution x in terms of p for each fixed p in the specified interval. Of course it's possible that no analytical form of the solution exists. If so, I'd also be happy to hear an argument why that is the case. However, for some values of p the solution is 'nice', for example for p = 1 2 it's easy to compute that x = 8 and for p = 1 3 I obtained x = 9 16 ( 5 3 + 11 + 64 3 )Any kind of help is greatly appreciated!

Question

I have a hard time finding the analytical solution to the following non-linear equation:  (  1  +  x      )          1      −      p        +      p                  p                  1          −          p                      x  (  p  −  1  )  −  1  =  0where   p  ∈  ]  0  ,  1  [ and   x  &amp;gt;  1. I would like to have a solution   x in terms of   p for each fixed   p in the specified interval. Of course it&#039;s possible that no analytical form of the solution exists. If so, I&#039;d also be happy to hear an argument why that is the case. However, for some values of p the solution is &#039;nice&#039;, for example for   p  =      1    2   it&#039;s easy to compute that   x  =  8 and for   p  =      1    3   I obtained   x  =      9    16    (  5      3    +      11    +    64          3        )Any kind of help is greatly appreciated!

lutzantsca885 · Accepted Answer

This a highly transcendental equation and, for general p, you will need some numerical method.It effectively happens that for some particular values of p, there are explicit solutions (this is the case for   p  =      1    5    ,      2    5    ,      3    5    ,      4    5    ,      1    4    ,      2    4    ,      3    4    ,      1    3    ,      2    3   but I do not think that we could do anything for, say,   p  =      1    6  .Let   k  =  (  p  −  1  )      p                  p                  1          −          p                     (  k varies between −1 and 0) and   p  =      m    n   (  m and   n being integers,   m  &amp;lt;  n). The equation write  (  x  +  1      )          1      −              m        n              =  1  −  k  x    ⟹    (  x  +  1      )          n      −      m        =  (  1  −  k  x      )    n  which makes that we face a polynomial of degree   n in   x. In fact   x  =  0 is a solution; so the polynomial of degree   (  n  −  1  ) and we do not know to solve analytically quintic polynomials. Then ... ?In French, we have an expression which says &quot;This, Sir, is the cause of your daughter&#039;s being dumb&quot;We can have approximations; for example, when   p is close to zero, a Taylor expansion gives  p  (  x  −  x  log  ⁡  (  p  )  −  x  log  ⁡  (  x  +  1  )  −  log  ⁡  (  x  +  1  )  )  +  O      (          p      2        )  If we neglect the term   log  ⁡  (  x  +  1  ) we have   x  ∼      e    p    −  1 which could be a good starting point.Similarly, when   p is close to 1, we have  (  p  −  1  )      (          x      e        −    log    ⁡    (    x    +    1    )    )    +  O      (    (    p    −    1          )      2        )  which gives  x  =  −  e      W          −      1            (    −          e              −        1        −                  1          e                      )    −  1  ≈  4.76which is probably very close to the limit.Between these two limits, it seems that the function looks like an hyperbola. A quick and dirty nonlinear regression gives as an estimate      x    0    =      1    p        (          373      181              p              2                  /                3              +          1882      697        )  which seems to be quite good (as shown below).We can generate a first iterate of Newton method and get      x    1    =            (              x        0            +      1              )        p            −      p              x        0            −      1              (      p      −      1      )              (                  p                                    p                              1                −                p                                                    (                  x          0                +        1                  )          p                −        1        )                  (                            p                                      x            0                                                x            1                                    solution                                      0.05                          59.5967                          59.4965                          59.4964                                      0.10                          31.4412                          31.4184                          31.4184                                      0.15                          21.8795                          21.8791                          21.8791                                      0.20                          17.0246                          17.0316                          17.0316                                      0.25                          14.0718                          14.0808                          14.0808                                      0.30                          12.0789                          12.0876                          12.0876                                      0.35                          10.6389                          10.6465                          10.6465                                      0.40                          9.54726                          9.55343                          9.55343                                      0.45                          8.68954                          8.69423                          8.69423                                      0.50                          7.99670                          8.00000                          8.00000                                      0.55                          7.42457                          7.42663                          7.42663                                      0.60                          6.94356                          6.94455                          6.94455                                      0.65                          6.53306                          6.53316                          6.53316                                      0.70                          6.17829                          6.17768                          6.17768                                      0.75                          5.86837                          5.86720                          5.86720                                      0.80                          5.59508                          5.59353                          5.59353                                      0.85                          5.35213                          5.35033                          5.35033                                      0.90                          5.13459                          5.13268                          5.13268                                      0.95                          4.93857                          4.93666                          4.93666                      )  Starting with this estimate, Newton method will probably converge in a couple of iterations.Trying for   p  =  0.23456      (                            n                                      x            n                                                0                          14.853058                                      1                          14.861762                                      2                          14.861760                      )

I have a hard time finding the analytical solution to the following non-linear equation: ( 1

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