# $\mathcal{Z}$-transform of $\frac{1}{n^2}$

This is a Question asked in IISC ( Indian Institute of Science,Bangalore,India) interview for MS admission.

What is the $\mathcal{Z}$-transform of $\dfrac 1{n^2}$ ?

• I just couldn't get any idea. Some similar problems can be solved by series expansion etc. But this one is terrifying. Jul 20 '16 at 10:00
• do you look for a closed form explicit expression? btw if $n=0$ allowed in the domain of $x[n]=\frac{1}{n^2}$ then I guess $X(z)$ won't exist for any finite $z$ either... Jul 20 '16 at 12:08

The problem is not sufficiently specified, because the range of admissible values of $n$ is missing. Here I make the assumption that we consider $n>0$. With this assumption we have

$$X(z)=\sum_{n=1}^{\infty}x[n]z^{-n}=\sum_{n=1}^{\infty}\frac{z^{-n}}{n^2}\tag{1}$$

And that's the point where we might get stuck, if we didn't have a list of mathematical series, or if we didn't know about the polylogarithm, which is defined by

$$\text{Li}_{s}(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^s},\qquad |z|<1\tag{2}$$

where $s$ is an arbitrary complex number.

In your case, $s=2$ and the corresponding function is called the dilogarithm or Spence's function.

Comparing $(1)$ and $(2)$ we get for the $\mathcal{Z}$-transform of $1/n^2$ for $n>0$

$$X(z)=\text{Li}_2\left(\frac{1}{z}\right),\qquad |z|>1\tag{3}$$

Another way to arrive at the solution is to use the differentiation property of the $\mathcal{Z}$-transform:

$$nx[n]\Longleftrightarrow -z\frac{dX(z)}{dz}\tag{4}$$

Applying $(4)$ twice will give you the result. You need the correspondence

$$u[n-1]\Longleftrightarrow \frac{1}{z-1},\qquad |z|>1\tag{5}$$

In a first step you'll arrive at the transform of $1/n$, $n>0$, and in a second step you'll arrive at the transform of $1/n^2$, $n>0$.

In this case you will be using the integral representation of the dilogarithm:

$$\text{Li}_2(z)=-\int_0^z\frac{\ln(1-u)}{u}du\tag{6}$$

• Correct me if I'm wrong but the final solution according to the differentiation property (which seems a preferrable choice to me) yields a differential equation in the unknown function $X(z)$ of the given signal $\frac{1}{n^2} u[n-1]$ I found something like this (not sure though) $X(z)'' + \frac{1}{z} X(z)' = \frac{z^{-2}}{1-z}$ Jul 20 '16 at 13:06
• @Fat32: Yes, but it's easier if you do it step by step, i.e., first compute the transform of $1/n$, and from that compute the transform of $1/n^2$; by doing so, you'll avoid the DE (you'll just have a trivial one). Jul 20 '16 at 13:11
• oh yes, of course solving the first then moving to the second ! thanks... Jul 20 '16 at 13:18
• @MattL, Is there any way to get the answer using duality? Aug 5 '16 at 9:11
• @spectre: What exactly do you mean? Aug 5 '16 at 9:18