Stackexchange community,

i have a little trouble in understanding this condradtiction about singularities.

Lets consider the function $f(z)=\exp(-z)$.

This function can be expanded as a taylor series:


As this is "some sort of polynomial" it will never have a singularity on every finite domain of $z$.

But if I express $f(z)=\exp(-z)=\frac{1}{\exp(z)}$ and then apply the taylor series on $\exp(z)$ I get:


Now here comes my confusion, the new representation allows singularities of $f(z)$, as it is "some kind of polynomial" in the denominator. What is the problem here? Is it because of the radius of convergence?

EDIT: Ok it is because the fundamental theorem of calculus isn't true for infinite sums. Is there some intuitive way of understanding this phenomenon?

I would be glad if someone could solve this contradicion.

  • $\begingroup$ But what is the problem with second representation? $\endgroup$ – qwenty Nov 9 '15 at 17:22
  • $\begingroup$ The second representation allows singularities by the fundamental theorem of algebra. $\endgroup$ – MrYouMath Nov 9 '15 at 17:39
  • $\begingroup$ "As this is a polynomial" -- no it is not a polynomial. $\endgroup$ – zhw. Nov 9 '15 at 18:31
  • $\begingroup$ Ok it is power sum, but why exactly does the fundamental theorem of algebra not hold for infinite polynomials? $\endgroup$ – MrYouMath Nov 9 '15 at 19:56
  • $\begingroup$ See math.stackexchange.com/questions/109360/… for the behaviour of the roots of the partial sums of the exponential function. Also math.stackexchange.com/questions/51586/… $\endgroup$ – Lutz Lehmann Jul 3 '17 at 18:05

It is not "some sort of polynomial." It is a power series with an infinite number of terms.

Consider the polynomials $\sum_{k=0}^n x^k $ for any positive integer $k$ these are defined for all real (and complex) $x$.

But, if we consider the infinite power series $\sum_{k=0}^{\infty} x^k $, this diverges for $|x| > 1$ and converges for $|x| < 1$.

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the denominator is never 0. no singularities

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