Is there an analytical solution to the equation below? A α k e ( k − α ) t + B β k e ( k − β ) t = 1 where α and β are roots of z 2 − ( a + b + c ) z + a c = 0

Question

Is there an analytical solution to the equation below?                    A                  α          k                          e                      (            k            −            α            )            t                          +        B                  β          k                          e                      (            k            −            β            )            t                          =        1            where   α and   β are roots of                              z          2                −        (        a        +        b        +        c        )        z        +        a        c        =        0

pelankp · Accepted Answer

Step 1As already said in comments and answers, finding the zero of the function  f  (  t  )  =  A    α        e          −      α      t        +  B    β        e          −      β      t        −  k        e          −      k      t      (as Justpassingby wrote it) will require, for the most general case, numerical methods (such as Newton).Assuming that A, B,   α ,   β , k are all positive real numbers, it will probably be a good idea to search for the zero of  g  (  x  )  =  log  ⁡            (        A    α        e          −      α      t        +  B    β        e          −      β      t                  )        −  log  ⁡            (        k        e          −      k      t                  )      which should look almost as a straight line.Starting iterations at       t    0    =  0 would provide as a first iterate      t    1    =  (  α  A  +  β  B  )            log      ⁡      (      k      )      −      log      ⁡      (      α      A      +      β      B      )              α      A      (      k      −      α      )      +      β      B      (      k      −      β      )

Hailey Newton · Accepted Answer

Step 1If we ignore the title then there is no actual constraint on   α and   β  .. For arbitrary   α and   β  . you can set   a  =  α  ,,   b  =  0 and   c  =  β and the quadratic equation is satisfied.The equation can be re-written as  A  α  exp  ⁡  (  −  α  t  )  +  B  β  exp  ⁡  (  −  β  t  )  =  k  exp  ⁡  (  −  k  t  )  . .The left hand side is a linear combination of exponentials. In the special case where   α  ,  β and k are positive we are looking at the intersections between, on the one hand, a linear combination of density functions of two exponential distributions, and on the other hand, the density function of a third exponential distribution. If   A  +  B  =  1 (this is not given) then the LHS is a classical example of a hyperexponential distribution. Telecom engineers are not happy with such a distribution.This already illustrates the kind of difficulty we are up against: depending on clever choices of the parameters there may be zero, one or two solutions - and not in the obvious way of quadratic equations and discriminants.

Is there an analytical solution to the equation below? <mtable columnalign="right left right le

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