Expert: Paul Klarreich - 10/7/2008

Question
in the maclaurin's series expansion of log e(1+x), why is this valid only for -1< x ≤1 ?

Answer
Questioner:   aarthi
Category:  Calculus
Private:  no
 
Subject:  maclaurin's series
Question:  in the maclaurin's series expansion of log e(1+x), why is this valid only for -1< x ≤1 ?
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Hi, Aarthi,

When  x = -1,  log e(1 + x) = log e(1 + (-1)) = log e(0) = log 0, which is undefined.

Since you are expanding a Taylor series about  x = 0 (that what Maclaurin series means) you cannot expand it beyond a circle (in the complex plane) whose center is at  z = (0,0) going up to any singular point.  That's the 'circle of convergence.'  

That singular point is at (-1,0), i.e.  x = -1, one unit away from your center.    So it can't go beyond one unit in ANY direction, therefore it can't go beyond x = +1.

Does this help?

(P.S. The expert you sent this to moved it to the question pool.  That's where I got it.)