# Power series expansion of $\frac{z}{(z^3+1)^2}$ around $z=1$

I want to expand $f(z)=\frac{z}{(z^3+1)^2}$ around $z=1$. That is, I want to find the coefficients $c_n$ such that $f(z) = \sum_{n=0}^\infty c_n (z-1)^n$.

So far, my first strategy was using long division after I expanded both the numerator and the denominator about $z=1$. To this end, we first apply a substitution $u:= z-1$, after we'll expand around $u=0$.

For the denominator we simply get $$1 + u$$.

For the numerator we have

$$\frac{1}{\big((u+1)^3+1\big)^2} = \frac{1}{2^2} \frac{1}{\big(1 + \frac{u^3+3u(u+1)}{2}\big)^2} \quad \mbox{put x:= \frac{u^3+3u(u+1)}{2}}\\ =-\frac{1}{2^2}\frac{\partial d}{\partial x}\frac{1}{x+1} \\ = -\frac{1}{2^2}(-1+2x-3x^2 +\ \ldots\ ) \\ =-\frac{1}{2^2}\big(1+6u-21u^2-52u^3+\ \ldots \ \big)$$

After using long division, we could try to see a pattern in the coefficients of $u^n$ but this doesn't seem to bear any quick results.

My second attempt was using the binomial theorem to expand the power series of $$f(z) = \sum_{n=0}^\infty a_n z^{3n+1} \quad \mbox{with} \quad a_n = (n+1)(-1)^n$$ about $z=1$.

To this end, we need to reorganize the $u^n$ in the following sum

$$\sum_{n=0}^\infty (u+1)^n = \sum_{n=0}^\infty \sum_{k=0}^n \binom nk u^k .$$

This seems fairly laborious. Below we have the terms for increasing $n$.

$$1 \\ 1 + u \\ 1 + 2u + u^2 \\ 1 + 3u + 3u^2 + u^3 \\ \vdots \\ 1 + mu + \binom m2 + \ldots + mu^{m-1} + u^m \\ 1 + (m+1)u + \binom{m+1}{2}u^2 + \ldots + (m+1)u^m + u^{m+1} \\ 1 + (m+2)u + \binom{m+2}{2}u^2 + \ldots + (m+2)u^{m+1} + u^{m+2} \\ \vdots$$

If we look at the terms of $1$, $u$ and $u^2$, their coefficients $b_n$ in the sum over $n$ suggest the following coefficients for $1$, $u$ and $u^2$ respectively.

$$b_0 = \sum_{n=0}^\infty 1 \\ b_1 =\sum_{n=0}^\infty n \\ b_2 = \sum_{n=0}^\infty \binom n2 \ .$$

Expanding on this, we arrive at

$$\sum_{i=0}^\infty b_i u^i = \sum_{i=0}^\infty \bigg( \sum_{k=0}^\infty \binom ki \bigg)u^i$$ , which seems very incorrect, since we have a double infinite sum.

• Okay, I think I've done it. Jan 14 '17 at 16:17
• See my new comment; your second approach only gives rise to an expansion in $v:=(z^3-1)$, as opposed to $u^3 := (z-1)^3$. Jan 14 '17 at 17:49
• @BenjaminDickman How did you arrive at $(n+1)(n^2+n+1)$? Jan 14 '17 at 17:59
• Alternatively, can you start with a substitution $x = -z$? Then you seek a power series for $-x/(x^3 - 1)^2$, which can be decomposed using partial fractions (e.g., Mathematica / WA: Apart[-x/(x^3 - 1)^2, x]). Jan 14 '17 at 18:10
• @BenjaminDickman Never mind my last comment; PFD seems the way to go. Jan 14 '17 at 19:19

The following expansion of $f(z)=\frac{z}{(1+z^3)^2}$ at $z=1$ is valid

We perform a partial fraction decomposition of $f(z)$. The zeros of the denominator $(1+z^3)^2$ are \begin{align*} \zeta_0&=-1\\ \zeta_1&=e^\frac{\pi i}{3}=\frac{1}{2}\left(1+i\sqrt{3}\right)\\ \zeta_2&=e^{-\frac{\pi i}{3}}=\frac{1}{2}\left(1-i\sqrt{3}\right)=\overline{\zeta_1} \end{align*} Since each of the zeros has muliplicity two we use the Ansatz: \begin{align*} \frac{z}{(1+z^3)^2}=\sum_{j=0}^2\frac{\alpha_j z+\beta_j}{(z-\zeta_j)^2}\tag{2} \end{align*}

Multiplication of the RHS of (2) and comparison of coefficients gives

\begin{align*} \frac{z}{(z^3+1)^2}=\frac{1}{9}\left(-\frac{z+2}{(z+1)^2}+ \frac{\overline{\zeta_1}z-2}{\left(z-\zeta_1\right)^2} +\frac{\zeta_1z-2}{\left(z-\overline{\zeta_1}\right)^2} \right)\tag{3} \end{align*}

Note, the second summand of the RHS of (3) is the complex conjugate of the third summand. This will help to simplify some calculations.

Next we want to expand the terms of the RHS of (3) at $z=1$. In order to do so it's convenient to do some preparatory work. We consider the general aspect of expanding a term around $z=z_0$.

We obtain \begin{align*} \frac{1}{(z-z_0)^n}&=\frac{1}{\left((z_1-z_0)+(z-z_1)\right)^n}\\ &=\frac{1}{(z_1-z_0)^n}\cdot\frac{1}{\left(1+\frac{z-z_1}{z_1-z_0}\right)^n}\\ &=\frac{1}{(z_1-z_0)^n}\sum_{j=0}^\infty \binom{-n}{j}\left(\frac{z-z_0}{z_1-z_0}\right)^j\\ &=\frac{1}{(z_1-z_0)^n}\sum_{j=0}^\infty \binom{n+j-1}{j}\left(\frac{z-z_0}{z_0-z_1}\right)^j\tag{4}\\ \end{align*}

Since we have to do a lot of calculations with the zeros $$\left\{-1,\frac{1}{2}\left(1+i\sqrt{3}\right),\frac{1}{2}\left(1-i\sqrt{3}\right)\right\}=\left\{-1,\zeta_1,\overline{\zeta_1}\right\}$$ its convenient to prepare some intermediate results \begin{array}{r|rrr|lll} \cdot&-1&\zeta_1&\overline{\zeta_1}\\ \hline -1&1&-\zeta_1&-\overline{\zeta_1}&\qquad \frac{1}{\zeta_0-1}=-\frac{1}{2}\\ \zeta_1&-\zeta_1&-\overline{\zeta_1}&1 &\qquad \frac{1}{\zeta_1-1}=-\zeta_1&\frac{1}{\left(\zeta_1-1\right)^2}=-\overline{\zeta_1} &\frac{1}{\left(\zeta_1-1\right)^3}=1\\ \overline{\zeta_1}&-\overline{\zeta_1}&1&-\zeta_1 &\qquad \frac{1}{\overline{\zeta_1}-1}=-\overline{\zeta_1}&\frac{1}{\left(\overline{\zeta_1}-1\right)^2}=-\zeta_1 &\frac{1}{\left(\overline{\zeta_1}-1\right)^3}=1\tag{5}\\ \end{array}

We are now ready and well prepared to continue.

Evaluation of (4) with $n=2$ and at $z_1=1$ gives \begin{align*} \frac{1}{\left(z-z_0\right)^2}=\frac{1}{\left(z_0-1\right)^2}\sum_{j=0}^\infty\frac{j+1}{(z_0-1)^j}(z-1)^j \end{align*} Since we use in (2) a numerator $\alpha z+\beta$ we respect it also in the expansion \begin{align*} \frac{\alpha z+\beta}{\left(z-z_0\right)^2} &=\frac{\alpha(z-1)+\alpha+\beta}{(z_0-1)^2}\sum_{j=0}^\infty\frac{j+1}{(z_0-1)^j}(z-1)^j\\ &=\sum_{j=0}^\infty\frac{(\alpha z_0+\beta)j+\alpha+\beta}{(z_0-1)^{j+2}}(z-1)^j\tag{6} \end{align*}

Now it's time to harvest. We take the three terms of (2) with $-1,\zeta_1,\overline{\zeta_1}$ and evaluate them using (6). We also note that according to (5) $\frac{1}{\left(\zeta_1-1\right)^3}=1$ so that we can split the sum according to the index $j$ modulo $3$.

We obtain using the relations in (5) \begin{align*} \frac{\overline{\zeta_1}z-2}{\left(z-\zeta_1\right)^2} &=\sum_{j=0}^\infty\frac{\left(\overline{\zeta_1}\zeta_1-1\right)j+\overline{\zeta_1}-2}{\left(\zeta_1-1\right)^{j+2}}(z-1)^j\\ &=\sum_{j=0}^\infty\frac{\left(-\overline{\zeta_1}\right)\left(-j+\overline{\zeta_1}-2\right)}{\left(\zeta_1-1\right)^j}(z-1)^j\\ &=\sum_{j=0}^\infty\frac{\overline{\zeta_1}(j+2)+\zeta_1}{\left(\zeta_1-1\right)^j}(z-1)^j\\ &=\sum_{j=0}^\infty\left[(3j+2)\overline{\zeta_1}+\zeta_1\right](z-1)^{3j}\\ &\qquad+\sum_{j=0}^\infty\left[(3j+3)\overline{\zeta_1}+\zeta_1\right]\left(-\zeta_1\right)(z-1)^{3j+1}\\ &\qquad+\sum_{j=0}^\infty\left[(3j+4)\overline{\zeta_1}+\zeta_1\right]\left(-\overline{\zeta_1}\right)(z-1)^{3j+2}\\ &=\sum_{j=0}^\infty\left[(3j+2)\overline{\zeta_1}+\zeta_1\right](z-1)^{3j}\\ &\qquad+\sum_{j=0}^\infty\left[(3j+3)(-1)+\overline{\zeta_1}\right](z-1)^{3j+1}\\ &\qquad+\sum_{j=0}^\infty\left[(3j+4)\overline{\zeta_1}-1\right](z-1)^{3j+2}\\ \end{align*} We can now earn that first and second term are complex conjugates and obtain \begin{align*} \frac{\overline{\zeta_1}z-2}{\left(z-\zeta_1\right)^2}+\frac{\zeta_1z-2}{\left(z-\overline{\zeta_1}\right)^2} &=\sum_{j=0}^\infty\left[3j+3\right](z-1)^{3j}\\ &\qquad+\sum_{j=0}^\infty\left[-6j-5\right](z-1)^{3j+1}\\ &\qquad+\sum_{j=0}^\infty\left[3j+2\right](z-1)^{3j+2}\tag{7}\\ \end{align*}

Last but not least the final term of (2)

Combining (7) and (8) and the claim (1) follows.

Note: The nearest singularities of $z=1$ are $\zeta_1$ and $\overline{\zeta_1}$, both having the same distance to $1$. This shows the radius of convergence is \begin{align*} \left|1-\zeta_1\right|&=\left|1-\overline{\zeta_1}\right|=\left|1-\frac{1}{2}\left(1-i\sqrt{3}\right)\right|\\ &=\frac{1}{2}\left|1+i\sqrt{3}\right|\\ &=1 \end{align*}

OPs was asking how to obtain the coefficients $\alpha_j,\beta_j$, $j=0,1,2$ in the partial fraction decomposition (2)

\begin{align*} \frac{z}{(1+z^3)^2}=\sum_{j=0}^2\frac{\alpha_j z+\beta_j}{(z-\zeta_j)^2} \end{align*}

One method is to multiply both sides with the denominator and perform a coefficient comparison for $[z^k]$, the coefficient of $z^k$ with $0\leq k\leq 5$ and obtain this way six equations for the six unknowns.

But its laborious to obtain six equations for all coefficients this way. We prefer a two step approach which reduces somewhat the calculations. Nevertheless it is somewhat tedious.

Step 1: Coefficients of highest power two.

We consider \begin{align*} \frac{z}{(z^3+1)^2}= \frac{\alpha_0z+\beta_0}{\left(z+1\right)^2} +\frac{\alpha_1z+\beta_1}{\left(z-\zeta_1\right)^2} +\frac{\alpha_2z+\beta_2}{\left(z-\overline{\zeta_1}\right)^2}\tag{10} \end{align*}

We now multiply both sides of (10) with $(z+1)^2$ and evaluate the resulting equation at $z=-1$. \begin{align*} \frac{z}{(z^3+1)^2}\cdot(z+1)^2&=\left(\alpha_0z+\beta_0\right) +\left(\frac{\alpha_1z+\beta_1}{\left(z-\zeta_1\right)^2}+\frac{\alpha_2z+\beta_2}{\left(z-\overline{\zeta_1}\right)^2}\right) \cdot\left(z+1\right)^2 \end{align*} Note that at the left hand side $-1$ is a removable singularity and evaluating both sides at $z=-1$ gives \begin{align*} \color{blue}{-\frac{1}{9}=-\alpha_0+\beta_0} \end{align*}

Multiplication of (9) with $(z-\zeta_1)^2$ and with $(z-\overline{\zeta_1})^2$ and evaluating at $z=\zeta_1$ and $z=\overline{\zeta_1}$ gives two more equations \begin{align*} \color{blue}{-\frac{1}{9}}&\color{blue}{=-\alpha_1\zeta_1+\beta_1}\\ \color{blue}{-\frac{1}{9}}&\color{blue}{=-\alpha_2\overline{\zeta_1}+\beta_2} \end{align*}

We put the results of these three equations into (9) and obtain

Since we have three unknowns $\alpha_0,\alpha_1,\alpha_2$ left, we need three more equations for determination. We next consider coefficient comparison for $z^0,z^1$ and $z^2$. Observe that each of the three summands in (11) contain five factors. So for $[z^0]$ we derive from each summand $\binom{5}{0}=1$ terms, for $[z^1]$ we derive $\binom{5}{1}=5$ terms and for $[z^2]$ we derive from each summand $\binom{5}{2}=10$ terms.

We will also consequently use the formulas from (5) to simplify expressions.

Step 2: Coefficient comparison for $[z^0],[z^1]$ and $[z^2]$.

Coefficient comparison $[z^0]$:

We obtain from (11) \begin{align*} 0&=\left[\left(\alpha_0-\frac{1}{9}\right)\zeta_1^2\overline{\zeta_1}^2\right] +\left[-\left(\alpha_1+\frac{1}{9}\right)\overline{\zeta_1}^2\right] +\left[-\left(\alpha_2+\frac{1}{9}\right)\zeta_1^2\right]\\ &=\left[\alpha_0-\frac{1}{9}\right]+\left[-\alpha_1\overline{\zeta_1}+\frac{1}{9}\zeta_1\right] +\left[-\alpha_2\zeta_1+\frac{1}{9}\overline{\zeta_1}\right]\\ \color{blue}{0}&\color{blue}{=\alpha_1-\alpha_1\overline{\zeta_1}-\alpha_2\zeta_1} \end{align*}

Coefficient comparison $[z^1]$:

Coefficient comparison $[z^2]$:

Now it's time to harvest.

Step 3: Linear equations for $\alpha_0,\alpha_1,\alpha_2$

We obtain via comparison of coefficients the three equations \begin{align*} \alpha_1-\alpha_1\overline{\zeta_1}-\alpha_2\zeta_1&=0\\ \alpha_0-\alpha_1\zeta_1-\alpha_2\overline{\zeta_1}&=0\\ \alpha_0+\alpha_1+\alpha_2&=0 \end{align*} with solutions \begin{align*} \alpha_0=-\frac{1}{9},\quad\alpha_1=\frac{1}{9}\overline{\zeta_1},\quad\alpha_2=\frac{1}{9}\zeta_1 \end{align*} We obtain by putting these values in the equations of step 1 \begin{align*} \beta_0&=\alpha_0-\frac{1}{9}=-\frac{2}{9}\\ \beta_1&=\alpha_1\zeta_1-\frac{1}{9}=\frac{1}{9}\overline{\zeta_1}\zeta_1=\frac{1}{9}\\ \beta_2&=\alpha_2\overline{\zeta_1}-\frac{1}{9}=\frac{1}{9}\zeta_1\overline{\zeta_1}=\frac{1}{9} \end{align*}

and we finally conclude \begin{align*} \frac{z}{(1+z^3)^2}&=\sum_{j=0}^2\frac{\alpha_j z+\beta_j}{(z-\zeta_j)^2}\\ &=\frac{1}{9}\left( -\frac{z+2}{(z+1)^2} +\frac{\overline{\zeta_1}z-2}{\left(z-\zeta_1\right)^2} +\frac{\zeta_1z-2}{\left(z-\overline{\zeta_1}\right)^2} \right) \end{align*}

• Would it have made the problem noticeably simpler [or structurally different] to begin with the substitution suggested in my initial comment, i.e., $z = -x$ so that the denominator factors over $\mathbb{R}$? Jan 14 '17 at 19:28
• The form of $(1)$ seems to suggest three different coefficients for the powers of order $3j$, $3j+1$ and $3j+2$, respectively. What makes this the case; i.e. what exactly causes this? Jan 14 '17 at 19:29
• @BenjaminDickman: I think, since the zeros of the denominator after substitution $z\rightarrow -z$ are $1,-\zeta_1,-\overline{\zeta_1}$ any structurally simpler approach should be possible in both variants equally well due to symmetry. Jan 14 '17 at 19:39
• @MusséRedi: Yes, I think this is possible. Do you like to find it by yourself? Jan 14 '17 at 19:40
• Yes, I would, indeed. Jan 14 '17 at 19:44

Notice that

$$\frac z{(1+z^3)^2}=-z\frac{\partial}{\partial z^3}\frac1{1+z^3}=-z\frac\partial{\partial z^3}\sum_{n=0}^\infty(-1)^n(z^3)^n=\sum_{n=0}^\infty n(-1)^{n+1}z^{3n-2}$$

This expansion works for $|z|<1$, and since we know the expansion exists at $z=1$, we may apply the method described in this answer to get the expansion at $z=1$.

$$=\sum_{j=0}^\infty \left(\sum_{k=j}^\infty \binom kj a_k \right)(z-1)^j$$

where $a_{3n+1}=(-1)^n(n+1)$ for $n\in\mathbb N$, else $a_k=0$.

However, note that this method is usually only viable when the center for the radius of convergence was within the radius of convergence of the original series. For example, plugging $z=1$ into our "expansion" yields

$$\frac14\stackrel?=1-2+3-4+5-6+\dots$$

However, an interesting case happens that we may regularize our divergent series to equal our original function by applying an Euler sum to it:

$$f(z)=\sum_{j=0}^\infty\sum_{k=\lfloor j/3\rfloor}^\infty \binom{3k+1}j (-1)^k(k+1)(z-1)^j\\=\sum_{j=0}^\infty\sum_{i=\lfloor j/3\rfloor}^\infty\frac1{2^i}\sum_{k=\lfloor j/3\rfloor}^i\binom ik\binom{3k+1}j (-1)^k(k+1)(z-1)^j$$

which now converges as desired.

• Hmm, that's a very useful remark. Jan 9 '17 at 15:23
• @MusséRedi Yes, it really is :D Jan 9 '17 at 15:23
• @SimpleArt: The proof of the referred answer is based upon the fact that $D(c_2; r-|c_2|)$, the disc of convergence at $z=c_2$ with radius $r-|c_2|$ is completely contained in $D(c_1=0; r)$. This is not the case here. Jan 10 '17 at 14:03
• @MarkusScheuer yes, I contemplated if that should be a problem but couldn't manage much. Since it has a power series expansion at the point in question and it lay on the edge of convergence, I thought it might be fine here. Jan 10 '17 at 14:08
• @SimpleArt: I've added an answer based upon partial fraction decomposition. Jan 14 '17 at 19:07