I am trying to calculate the inverse Laplace transform of a probability distribution, and while I don't believe I can get a closed form expression, I would like to get an idea of the general shape of the distribution. The Laplace transform is a nice rational function, however, the roots of the denominator are the inverse (I mean $1/z$) of the following $$\frac{z^n-1}{z-1}=b$$

where $b>1$ and real. What can be said about the zeros of the polynomial? Are they all real, what is their degeneracy? Are there any $b$ derived bounds on the roots (i.e. do they all fall in a circle of radius $2b^{1/n}$ for example-- I know it is not true, just giving an example)?


1 Answer 1


The expression $$\frac{z^{n+1}-1}{z-1}-b = 1-b+z+z^2+\cdots+z^n$$ can be thought of as the $n^{\text{th}}$ partial sum of the power series for $\frac{1}{1-z}-b$, whose circle of convergence is the unit circle. By Jentzsch's theorem, every point on the unit circle will thus be a limit point of the zeros of these partial sums.

Here is a plot of all of the zeros for $n=3,4,\ldots,70$ with $b = 4$. The unit circle is shown in blue.


It appears that there are no limit points outside of the circle of convergence, which I believe follows from the fact that the reciprocal polynomials of the partial sums at the top of this answer converge to $1/(1-z)$ inside the unit circle, which has no zeros.

Also, there is one limit point inside of the circle of convergence at $z=(b-1)/b$, which is the lone zero of the limit function of the partial sums which lies within the circle of convergence.

(The above should be read in the context of Hurwitz's theorem, which essentially says that any zero of the limit function which is inside the circle of convergence will be a limit point of the partial sums.)

Lastly, if I had to guess the rate of convergence of the zeros to the unit circle, I would say they approach at about $O(1/n)$ based on working with similar problems.

  • 1
    $\begingroup$ I wish I could +more than 1 for this $\endgroup$ Mar 22, 2012 at 4:46

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