0

I stuck at part (a) after 1 have done the partial fraction, can anyone give me some hints?

for part (b) I express $$f_1(z)=(-1)^n-1(n-1)!f_n(z)$$ What I suppose to do next?

user111605
  • 17
  • 1
    You said you did a partial fraction decomposition? Will you provide more details in your question please so we can see where you are stuck and give appropriate hints? Thanks. – Tom Nov 26 '13 at 14:11
  • I have found its three pole $$ \frac1{z-1}+\frac{i}{\sqrt{3}(z+\frac12-\frac{\sqrt{3}i}{2})} +\frac{i}{\sqrt{3}(z+\frac12+\frac{\sqrt{3}i}{2})} $$ But I don't know how can I turn it into series by $$ \frac{1}{1-c}=1+c+c^2+c^3+....... $$ where |c|<1 – user111605 Nov 26 '13 at 17:18

2 Answers2

1

Hint: From the question, it seems that you are trying to do the Laurent expansion about $z=1$. So, you should result in something that looks like $$ C + \sum_{r=1}^{\infty} \left\{ \frac{A_r}{(z-1)^r} + B_r (z-1)^r \right\} $$

By your partial fraction decomposition, you are left with something of the form $$ \frac{1}{z-1} + \frac{c_1}{z-a}+\frac{c_2}{z-b} $$ A potential good next step is to rewrite this as $$ \frac{1}{z-1} - \frac{c_1}{(a-1)-(z-1)} - \frac{c_2}{(b-1)-(z-1)} \\ = \frac{1}{z-1} - \left(\frac{c_1}{a-1}\right)\left(\frac{1}{1 - \big(\frac{z-1}{a-1}\big)}\right) - \left(\frac{c_1}{b-1}\right)\left( \frac{1}{1 - \big(\frac{z-1}{b-1}\big)}\right). $$

Now, you will notice that if $|z-1|<\sqrt{3}$, then $\left| \frac{z-1}{a-1} \right| < 1$ and $\left| \frac{z-1}{b-1} \right| < 1$, so you can apply the geometric series formula $$\frac{1}{1-c} = \sum_{n=0}^{\infty}c^n$$ with $c = \frac{z-1}{a-1}$ in the middle term and $c = \frac{z-1}{b-1}$ in the last term.

Tom
  • 9,978
  • Thank you! But the expression should be z-1/a-1 and z-1/b-1 respectively, the G.S formula will still hold at this situation BTW, for (b) what should I do after expressing $$ f_1^{(n-1)}=(-1)^{n-1}(n-1)!f_n(z) $$ I am stuck again – user111605 Nov 26 '13 at 19:02
  • @user111605 You're right! I made a typo with $a\pm 1$ and $b \pm 1$. No worries, I fixed it.. though the technique remains the same! – Tom Nov 26 '13 at 20:02
1

Hint: Let $z=w+1$, then $$ \begin{align} \frac{z^2+2z}{z^3-1} &=\frac{w^2+4w+3}{w^3+3w^2+3w}\\ &=\frac1w\frac{w^2+4w+3}{w^2+3w+3}\\ &=\frac1w+\frac1w\frac{w}{w^2+3w+3}\\ &=\frac1w+\frac1{w^2+3w+3} \end{align} $$ Show that $$ \frac1{w^2+3w+3}=\sum_{k=0}^\infty\left(-\frac13\right)^ka_kw^k $$ where $a_0=\frac13$, $a_1=1$, and $a_k=3(a_{k-1}-a_{k-2})$.

robjohn
  • 345,667