Maclaurin series for 1 | 1 + x | However, would it be appropriate for me to refer 1 | 1 + x | as 'not a smooth' curve (discontinuity at x = 1) hence the non-existence of a Maclaurin or in any case a Taylor Series?

Question

Maclaurin series for       1                  |            1      +      x              |            However, would it be appropriate for me to refer       1                  |            1      +      x              |             as &#039;not a smooth&#039; curve (discontinuity at   x  =  1) hence the non-existence of a Maclaurin or in any case a Taylor Series?

Karli Moreno · Accepted Answer

In order to write down a Taylor series for   f centered at   a, we only need   f to have derivatives of all orders at   a. (This, in turn, requires that for every   n the derivative       f          (      n      )       be defined in some neighborhood of   a, so that       f          (      n      +      1      )        (  a  ) makes sense.) The function   f  (  x  )  =  1      /        |    1  +  x      |   meets this requirement at every point   a  ≠  −  1.For example, with   a  =  0 the series is                    (1)                              ∑                      n            =            0                                ∞                          (        −        1                  )          n                          x          n                    which, not coincidentally, is the same series as for 1/(1+x). I have not yet considered the question of whether the series actually converges to   f. This can be addressed using the geometric series formula: when       |    x      |    &amp;lt;  1, the series (1) has the sum   1      /    (  1  +  x  ), which in this range agrees with   f  (  x  ).

Maclaurin Series for 1/|1+x| as the latter is not differentiable at x=−1. However, would it be appropriate for me to refer 1/|1+x| as 'not a smooth' curve (discontinuity at x=1) hence the non-existence of a Maclaurin or in any case a Taylor Series?

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