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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?

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    What does this tell you about the symmetry of $\Phi(s)$? Make a [plot](http://tinyurl.com/cclnshw)...2012-07-31
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    Could you elaborate?2012-07-31
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    Compare the symmetry of $x^{2n}$ resp. $x^{2n+1}$, when you substitute $-x$ for $x$. What do you get?2012-07-31
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    I don't think it is related. Why symmetry is need to consider here?2012-07-31
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    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 $s^{2k-1}$.2012-07-31
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    @did Could you provide a wiki link about the method you are using?2012-07-31
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    Google `Binomial series`.2012-07-31
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    The series representation of $\Phi(s)$ shares the same symmetry as the original function \begin{equation*} \Phi(s) = \frac{1 - \sqrt{1 - 4pqs^2}}{2qs} = \sum_{n=0}^\infty \phi_ns^n. \end{equation*} Since $\Phi(s)=- \Phi(-s)$, coefficients belonging to even powers are $0$ since $x^{2n}=\color{red}{+}(-x)^{2n}$.2012-07-31
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    From the title of the post you seem to wonder how you get only odd coefficients. You can test whether this is an odd function. You can also note that you have a binomial expansion of a term (the square root) in $s^2$. This will give you only terms of even degree $(s^2)^r = s^{2r}$. Then you divide by $s$ which gives you only terms of odd degree.2012-07-31
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    I have edited my question a little bit to make it clear.2012-07-31
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    Thanks did. I have my answer ready.2012-07-31

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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.

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    thx for reminding2012-08-21