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The Gaussian is infinitely flat at infinity. Equivalently, the
Maclaurin expansion (Taylor expansion about
) of
![$\displaystyle f(t) = e^{-\frac{1}{t^2}}$](img2722.png) |
(D.3) |
is zero for all orders. Thus, even though
is
differentiable of all orders at
, its series expansion fails to
approach the function. This happens because
has an
essential singularity at
(also called a
``non-removable singularity''). One can think of an essential
singularity as an infinite number of poles piled up at the same
point (
for
). Equivalently,
above has an
infinite number of zeros at
, leading to the problem with
Maclaurin series expansion. To prove this, one can show
![$\displaystyle \lim_{t\to 0} \frac{1}{t^k} f(t) = 0$](img2725.png) |
(D.4) |
for all
. This follows from the fact that exponential
growth or decay is faster than polynomial growth or decay. An
exponential can in fact be viewed as an infinite-order polynomial,
since
![$\displaystyle e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots.$](img2727.png) |
(D.5) |
We may call
infinitely flat at
in the ``Padé sense'':
Another interesting mathematical property of essential singularities is
that near an essential singular point
the
inequality
![$\displaystyle \left\vert f(z)-c\right\vert<\epsilon$](img2730.png) |
(D.6) |
is satisfied at some point
in every neighborhood of
, however small. In other words,
comes arbitrarily close
to every possible value in any neighborhood about an
essential singular point. This was first proved by Weierstrass
[42, p. 270].
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