Page 29 - Foundations Of Differential Calculus
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12    1. On Finite Differences
        In a similar way we let
                                         2    3
                            −1     ω   ω    ω
                        ∆.x   = −    +    −    + ··· ,
                                  x 2  x 3  x 4
                            −2    2ω   3ω 2   4ω 3
                        ∆.x   = −    +     −     + ··· ,
                                  x 3   x 4   x 5

                            −3    3ω   6ω 2   10ω 3
                        ∆.x   = −    +     −      + ··· ,
                                  x 4   x 5    x 6
                                           2      3
                            −4    4ω   10ω     20ω
                        ∆.x   = −    +      −      + ··· .
                                  x 5    x 6    x 7
        We continue in the same way for the rest. For fractions we have

                          1/2    ω      ω 2     ω 3
                      ∆.x    =      −       +       − ··· ,
                               2x 1/2  8x 3/2  16x 5/2
                          1/3    ω      ω 2    5ω 3
                      ∆.x    =      −       +       − ··· ,
                               3x 2/3  9x 5/9  81x 8/3
                        −1/2      ω      3ω 2    5ω 3
                     ∆.x     = −      +      −        + ··· ,
                                 2x 3/2  8x 5/2  16x 7/2
                                           2     14ω 3
                                         2ω
                                  ω
                     ∆.x −1/3  = −    +      −         + ··· .
                                 3x 4/3  9x 7/3  81x 10/3
        18. It should be clear that if the exponent is not a positive integer, then
        these differences will progress without limit, that is, there will be an infinite
        number of terms. Nevertheless, these same differences can be expressed by
                                                    I
        a finite expression. If we let y = x −1  =1/x, then y =1/ (x + ω), so that
                                −1     1     1     1
                            ∆.x   =∆.    =       −  .
                                       x   x + ω   x
        Hence, if the fraction 1/ (x + ω) is expressed as a series, then we obtain the
        infinite expression we saw before. In a similar way we have

                             −2      1       1       1
                          ∆.x   =∆.    =        2  −  .
                                     x 2  (x + ω)   x 2
        Furthermore, for irrational expressions we have
                                 √    √        √
                               ∆. x =   x + ω −  x
        and
                                 1       1       1
                              ∆.√ = √        − √ .
                                  x     x + ω    x
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