# Calculate coefficients of power series

Calculate the coefficients of the power series expansion of $f(z)=\frac{2}{\sqrt{1-3z}}+\frac{1}{(1-z)(1-2z)}$

Could you check if I understood the task and calculated it correctly?

\begin{align*} f(z)&=\frac{2}{\sqrt{1-3z}}+\frac{1}{(1-z)(1-2z)}\\ &=2 \cdot \frac{1}{(1-3z)^{1/2}} - \frac{1}{1-z} + 2 \cdot \frac{1}{1-2z}\\ &=2 \sum \binom{n-\frac{1}{2}}{-\frac{1}{2}}(3z)^n-\sum z^n + 2 \sum (2z)^n\\ &=\sum \Big( 2 \cdot 3^n \cdot \binom{n-\frac{1}{2}}{-\frac{1}{2}} - 1 + 2^{n+1} \Big) z^n \end{align*}

I've calculated the power series. Now, what is the answer? Which are the coefficients?

• You did not expand the first series correctly! – Mhenni Benghorbal Feb 27 '16 at 20:14
• I see and I've updated it. Can you check if it is correct now? – dash Feb 27 '16 at 20:24
• You got an answer! – Mhenni Benghorbal Feb 27 '16 at 20:29

We have \begin{align} f(z) &= \frac2{(1-3z)^{\frac12}} + \frac1{(1-z)(1-2z)}\\ &= \frac2{(1-3z)^{\frac12}} - \frac1{1-z} + \frac2{1-2z}\\ &= 2\sum_{n=0}^\infty \binom{-\frac12}n(-3)^nz^n -\sum_{n=0}^\infty z^n + 2\sum_{n=0}^\infty 2^nz^n\\ &= \sum_{n=0}^\infty \left[2\binom{-\frac12}n(-3)^n - 1 + 2^{n+1} \right]z^n, \end{align} so the $n^{\mathrm{th}}$ coefficient is $$2\binom{-\frac12}n(-3)^n - 1 + 2^{n+1}.$$

• Thanks. Shouldn't the first fraction be expanded as $\mathbf{2} \cdot \sum \binom{-\frac{1}{2}}{n}(-3)^nz^n$ ? – dash Feb 27 '16 at 20:33
• Ah, left out the $2$ in my computations, thanks. – Math1000 Feb 27 '16 at 20:42