Binomial expansion on quadratic solutionpγ2+(−2mdc)pγ+2mdBd=0&nbsp;(1)

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Binomial expansion on quadratic solutionpγ2+(−2mdc)pγ+2mdBd=0&amp;nbsp;(1)

Pentyrch9ci · Accepted Answer

Let&#039;s simplify the notation by dropping subscripts, and turn this into an arithmetic problem by making everything dimensionless:pmc=1±1−2Bmc2We can treat the square root as a binomial and expand. The binomial expansion is(1+ϵ)n=1+nϵ+n(n-1)2!ϵ2+⋯+nmϵm+...and the trick is to neglect the higher-order terms; the trick works better when&amp;nbsp;nϵ≪1. Taking the negative discriminant in your solution, the first-order approximation gives                                           p                          m              c                                                          ≈          1          −          (          1          −                      1            2                                              2              B                                      m                              c                2                                              )          =                      B                          m                              c                2                                                        To get the correction you&#039;re looking for, retain the second-order term,&amp;nbsp;−18(2Bmc2)2.

Binomial expansion on quadratic solution \(\displaystyle{{p}_{{\gamma}}^{{2}}}+{\left(-{2}{m}_{{d}}{c}\right)}{p}_{{\gamma}}+{2}{m}_{{d}}{B}_{{d}}={0}{t}{a}{g}{1}\)

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