0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ You did not expand the first series correctly! $\endgroup$ – Mhenni Benghorbal Feb 27 '16 at 20:14
  • $\begingroup$ I see and I've updated it. Can you check if it is correct now? $\endgroup$ – dash Feb 27 '16 at 20:24
  • $\begingroup$ You got an answer! $\endgroup$ – Mhenni Benghorbal Feb 27 '16 at 20:29
0
$\begingroup$

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 $$ 2\binom{-\frac12}n(-3)^n - 1 + 2^{n+1}.$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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