Power series expansion of a function

Question

In this question we are asked to find the power series expansion of M(t) upto the t squared term. I understand how they've gone from step 1 to step 2, by dividing the top and bottom by p but I'm not able to understand at all how they have proceeded from step 2 to step three. Is there a general technique used to find the power series expansion for functions, how do we approach these types of questions from the start?

Any help would be much appreciated.

Answer 1

Note that we do not divide by $p$ but instead consider the reciprocal value and take from it the reciprocal again.

\begin{align*} M(t)&=\left(\frac{p}{1-qe^t}\right)^k=\left(\frac{1-qe^t}{p}\right)^{-k}\\ &=\left(\frac{1}{p}-\frac{q}{p}\left(1+t+\frac{t^2}{2}+\cdots\right)\right)^{-k}\\ &=\left(1-\frac{q}{p}\left(t+\frac{t^2}{2}+\cdots\right)\right)^{-k}\tag{1}\\ \end{align*}

Now we use the binomial series expansion with $\alpha=-k$ \begin{align*} (1+z)^{-k}&=\sum_{n=0}^\infty\binom{-k}{n}z^k=1+\binom{-k}{1}t+\binom{-k}{2}z^2+\cdots\\ &=1-kz+\frac{1}{2}(-k)(-k-1)z^2+\cdots \end{align*}

With \begin{align*} z:=-\frac{q}{p}\left(t+\frac{t^2}{2}+\cdots\right) \end{align*}

we obtain from (1) \begin{align*} M(t)&=1+k\cdot\frac{q}{p}\left(t+\frac{t^2}{2}+\cdots\right) +\frac{1}{2}(-k)(-k-1)\left(\frac{q}{p}\left(t+\frac{t^2}{2}+\cdots\right)\right)^2+\cdots\\ &=1+k\cdot\frac{q}{p}t+k\cdot\frac{q}{p}t^2/2+k(k+1)\left(\frac{q}{p}\right)^2t^2/2+\cdots \end{align*} and we obtain the wanted representation.

Two aspects which should be considered:

• Since each summand $\left(t+\frac{t^2}{2}+\cdots\right)$ starts with $t$ we do not need to consider higher powers than $2$ since then there is no longer any contribution to $t^2$ but only to higher powers of $t$.

• It is this clever transformation right at the beginning which enables us to easily find the powers up to $t^2$. If we would instead start with a $k$-th power of a geometric series expansion \begin{align*} M(t)&=\left(\frac{p}{1-qe^t}\right)^k\\ &=\left(p\left(1+qe^t+q^2e^{2t}+\cdots\right)\right)^k\\ &=p^k\left(1+q(1+t+t^2/2+\cdots)+q^2(1+t+t^2/2+\cdots)^2+\cdots\right)^k \end{align*} each of the inner summands contributes to $1,t$ and $t^2$ and this is not easily manageable.