Power series of $\frac{(1-qx)^k}{(1-qx)^{k+1}-(px)^{k+1}}$

Question

I need to expand the function $$ f(x)=\frac{(1-qx)^k}{(1-qx)^{k+1}-(px)^{k+1}},\quad p>0,\;q>0,\;p+q=1 $$ into a power series $$ \sum_{n=0}^{\infty} r_n x^n. $$ $k$ is an integer $>1$.

The only thing that comes to my mind is the Taylor series expansion, but here it does not inspire enthusiasm. Maybe there is some nice way?

Edit: Following the Ted Shifrin's hint, I got the product of the sums of geometric series $$ \frac{(1-qx)^k}{(1-qx)^{k+1}-(px)^{k+1}}=\frac1{1-qx}\cdot \frac1{1-\left( \frac{px}{1-qx} \right)^{k+1}}; $$ Then, $$ \frac1{1-qx}= 1+qx+q^2x^2+\ldots $$ $$ \frac1{1-\left( \frac{px}{1-qx} \right)^{k+1}}= 1+\left( \frac{px}{1-qx} \right)^{k+1}+ \left( \frac{px}{1-qx} \right)^{2(k+1)}+\ldots $$ $$ \left( \frac{px}{1-qx} \right)^{m(k+1)}= px^{m(k+1)}\frac1{\left( 1-qx \right)^{m(k+1)}}= px^{m(k+1)} \sum_{n=0}^\infty \binom{-m(k+1)}{n} q^n x^n $$ This is great progress, but it seems to me that this is not the end. It’s not for nothing that it’s given $p+q=1$. Does this somehow make things easier?

Answer 1

$$f(x)=\frac{(1-q x)^k}{(1-q x)^{k+1}-(p x)^{k+1}}$$ $$\frac 1{f(x)}=1-qx-p x \left(\frac {px}{1-qx}\right)^k$$

$$\left(\frac {px}{1-qx}\right)^k=(px)^k\sum_{n=0}^\infty (-1)^n \binom{-k}{n}q^n \,x^n$$ So, the reciprocal of $f(x)$ is simple.

Now, have fun with the long division.

If you use it for $$g_k(x)=\frac{(1-q x)^k}{(1-q x)^{k+1}-(p x)^{k+1}}$$ and you expand them to $O(x^{k+3})$, you have the simple $$g_k(x)=\sum_{n=0}^{k} (qx)^n+(p^{k+1}+q^{k+1})\,x^{k+1}+q\big((k+2)p^{k+1}+q^{k+1}\big)\,x^{k+2}+O(x^{k+3})$$ The next term would be more than nasty.

Answer 2

We show the following power series expansions are valid for non-negative integer $k\geq 0$: \begin{align*} \color{blue}{\frac{(1-qx)^k}{(1-qx)^{k+1}-(px)^{k+1}}} &\color{blue}{=\sum_{n=0}^{\infty}\left(\frac{1}{k+1}\sum_{j=0}^{k}\left(e^{\frac{2\pi i j }{k+1}}p+q\right)^n\right)x^n}\\ &\color{blue}{=\sum_{n=0}^{\infty}\left(\sum_{l=0}^{\left\lfloor\frac{n}{k+1}\right\rfloor} \binom{n}{(k+1)l}p^{(k+1)l}q^{n-(k+1)l}\right)x^n}\tag{1} \end{align*} The key to get (1) is a partial fraction decomposition.

We obtain \begin{align*} \color{blue}{f_k(x)}&\color{blue}{=\frac{(1-qx)^k}{(1-qx)^{k+1}-(px)^{k+1}}}\\ &=\frac{1}{k+1}\sum_{j=0}^{k}\frac{1}{(1-qx)-e^{\frac{2\pi i j}{k+1}}px}\tag{2}\\ &=\frac{1}{k+1}\sum_{j=0}^{k}\frac{1}{1-\left(e^{\frac{2\pi i j}{k+1}}p+q\right)x}\\ &\color{blue}{=\frac{1}{k+1}\sum_{j=0}^{k}\sum_{n=0}^{\infty}\left(e^{\frac{2\pi i j }{k+1}}p+q\right)^nx^n}\tag{3}\\ &=\frac{1}{k+1}\sum_{j=0}^{k}\sum_{n=0}^{\infty}\sum_{l=0}^n \binom{n}{l}e^{\frac{2\pi i j l}{k+1}}p^lq^{n-l}x^n\tag{4}\\ &=\sum_{n=0}^{\infty}\left(\sum_{l=0}^n\binom{n}{l}p^lq^{n-l} \frac{1}{k+1}\sum_{j=0}^k\left(e^{\frac{2\pi i l}{k+1}}\right)^j\right)x^n\tag{5}\\ &=\sum_{n=0}^{\infty}\left(\sum_{{l=0}\atop{k+1|l}}^n\binom{n}{l}p^lq^{n-l}\right)x^n\tag{6}\\ &\,\,\color{blue}{=\sum_{n=0}^{\infty}\left(\sum_{l=0}^{\left\lfloor\frac{n}{k+1}\right\rfloor} \binom{n}{(k+1)l}p^{(k+1)l}q^{n-(k+1)l}\right)x^n}\tag{7} \end{align*} and the claim (1) follows from (3) and (7).

Comment:

• In (2) we perform a partial fraction decomposition. Some details: We set \begin{align*} z:=1-qx,\qquad z_0:=px,\qquad \omega_j:=e^{\frac{2\pi i j }{k+1}}\quad 0\leq j\leq k \end{align*} we can then write $f_k(x)$ as \begin{align*} \frac{z^k}{z^{k+1}-z_0^{k+1}} =\frac{c_0}{z-z_0}+\frac{c_1}{z-\omega_1 z_0}+\cdots +\frac{c_k}{z-\omega_k z_0} \end{align*} with unknowns $c_j, 0\leq j\leq k$. We use the decomposition in the form: \begin{align*} \frac{P(z)}{Q(z)}=\sum_{j=0}^k\frac{P\left(\omega_j z_0\right)}{Q^{\prime}\left(\omega_j z_0\right)}\,\frac{1}{z-\omega_j z_0} \end{align*} and we obtain \begin{align*} \color{blue}{\frac{z^k}{z^{k+1}-z_0^{k+1}}} &=\sum_{j=0}^{k}\frac{\left(\omega_j z_0\right)^k}{(k+1)\left(\omega_j z_0\right)^k}\,\frac{1}{z-\omega_j z_0}\\ &\,\,\color{blue}{=\frac{1}{k+1}\sum_{j=0}^{k}\frac{1}{z-\omega_j z_0}} \end{align*}

• In (3) we make a geometric series expansion.

• In (4) we apply the binomial theorem.

• In (5) we make some rearrangements.

• In (6) we make a simplification by noting that \begin{align*} \frac{1}{k+1}\sum_{j=0}^k\left(e^{\frac{2\pi i l}{k+1}}\right)^j =\begin{cases} 1\qquad& k+1|l\\ 0\qquad&\mathrm{otherwise} \end{cases} \end{align*}

• In (7) we take from the inner sum only summands with indices $l$ being a multiple of $k+1$, since other summands do not contribute.