Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In William Feller's 1st book p.272

It said the generating function $\Phi$ satisfies

\begin{equation*} qs\Phi^2(s) - \Phi(s) + ps = 0 \end{equation*}

so it has two roots. The first root is unbounded near $s = 0$.

So the generating function is given by the unique bounded solution

\begin{equation*} \Phi(s) = \frac{1 - \sqrt{1 - 4pqs^2}}{2qs}. \end{equation*}

Also, \begin{equation*} \Phi(s) = \sum_{n=0}^\infty \phi_ns^n. \end{equation*}

What I don't understand is how the coefficients \begin{equation*} \phi_{2k-1} = \frac{(-1)^{k-1}}{2q} \binom{1/2}{k} (4pq)^k, \quad \phi_{2k} = 0 \end{equation*} come up with binomial expansion of the unique root.

To be clear, I want to know how the unique root of $\Phi$ is converted to the form $(1 + t)^a$ and how and why the new form have the coefficients above?

share|cite|improve this question
What does this tell you about the symmetry of $\Phi(s)$? Make a plot... – draks ... Jul 31 '12 at 7:08
Could you elaborate? – RHS Jul 31 '12 at 7:12
Compare the symmetry of $x^{2n}$ resp. $x^{2n+1}$, when you substitute $-x$ for $x$. What do you get? – draks ... Jul 31 '12 at 7:15
The explicit formula for the coefficients $\phi_k$ follows from the expansion of $(1+x)^{1/2}$ for $x=-4pqs^2$. The coefficient of $x^k$ in $(1+x)^{1/2}$ is ${1/2\choose k}$, which yields the coefficient of $s^{2k-1}$. – Did Jul 31 '12 at 7:20
Google Binomial series. – Did Jul 31 '12 at 7:32
up vote 1 down vote accepted

Thanks for the magic word "Binomial series", now I solved it.

First, by Newton's binomial formula we have \begin{equation*} (1 - 4pqs^2)^{1/2} = \sum_{k=0}^\infty \binom{1/2}{k} (-4pqs^2)^k. \end{equation*} And its first term $\binom{1/2}{0} (4pqs^2)^0$ is $1$. So \begin{equation*} \Phi(s) = \frac{1 - \sqrt{1 - 4pqs^2}}{2qs} = - \frac{1}{2qs} \sum_{k=1}^\infty \binom{1/2}{k} (-4pqs^2)^k = \frac{(-1)^{k+1}}{2q} \sum_{k=1}^\infty \binom{1/2}{k} (4pq)^k s^{2k -1}. \end{equation*} Now we can see $\Phi$ only has odd coefficient, and we could get $\phi_{2k-1}$ and $\phi_{2k}$ immediately.

share|cite|improve this answer
Do not forget to accept this answer. – Did Aug 17 '12 at 0:01
thx for reminding – RHS Aug 21 '12 at 11:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.