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series expansion, namely,
1 = 1-x2 + x4 - ... + (-ltx2n + ...
1 +x^2
Chapter 8converges only for lxl < 1, i.e., for x in the interval (-1, 1). This fact can
be hardly explained in terms of real analysis in view of the continuity of
1/(1 + x^2 ) on the real line. The extension of this function to the complex
domain, namely, 1/(1 + z^2 ) has the Maclaurin expansion
_· _l_ = 1 - z2 + z4 - ••• + ( -1 r z2n + ...
1 +z^2valid for lzl < 1, which is now clear since 1/(1 + z^2 ) has poles at z = ±i,
so R = 1. Hence for real values of z the corresponding series also has 1 as
a radius of convergence. We see that the complex case has repercussions
on the real one, thus throwing light on an apparent paradox.
II. To see that a removable singularity has no effect on the radius of
convergence of the Taylor expansion of a function, consider the simple
example
f(z) = z2 -1
z-1(z:fl)
This function is analytic in a neighborhood of the ongm and has a
removable singularity at z = 1. About z = 0 f(z) has the series expansion
1 + z + Oz^2 + Oz^3 + · · ·
which converges for all z. In fact, the series represents the extended
f(z) with analyticity restored according to Riemann theorem by setting
f(l) = 2.
We may even expand a function about a removable singularity. For
instance, we have
sin z l ( z
3
) z
2
g(z)=-=- z--+··· =1--+···
z z 3! 3!for z :f 0. However, the series on the right represents the function for
all values of z if the definition of g(z) is extended by setting g(O) = 1, as
required by Riemann theorem (see Section 7.26).
Thus, as a matter of convenience, the definition of an analytic func-
tion is always supposed to be extended so as to eliminate any removable
singularities. This process will be called analytic extension of the
function.