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?

ghulamu51 2022-09-27 Answered
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?
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Karli Moreno
Answered 2022-09-28 Author has 7 answers
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 | < 1, the series (1) has the sum 1 / ( 1 + x ), which in this range agrees with f ( x ).
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions