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I'm having hard time figuring out the sequence generated by each of these generating functions.

  1. $f(x) = {x^4\over (1-x)}$

  2. $f(x) = {1\over (3-x)}$

  3. $f(x) = {1\over (1-x)+3x^7-11}$

  4. $f(x) = {1\over (1+3x)}$

for #1, since $1\over (1-x)$ is $1 + x + x^2 + x^3+ X^4...$ and so on, I have $x^4+x^8+x^12+x^16$ and so on.. but how do I get the sequence(like 0,0,1,1 and so on)?

Any help would be very much appreciated thankyou

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As you observe, #1 is just \begin{equation*} \frac{x^4}{1-x} = \sum_{i=0}^\infty x^{4+i} = \sum_{i=0}^\infty a_ix^i, \end{equation*} where $a_i = 1$ if $i\ge 4$, and $a_i = 0$ otherwise. For #2, rewrite $\frac{1}{3-x}$ as \begin{equation*} \frac{1}{3-x} = \frac{1/3}{1-(x/3)} = \frac{1}{3}\cdot\frac{1}{1-(x/3)}. \end{equation*} Then use the power series expansion of $\frac{1}{1-x}$ with $\frac{x}{3}$ in place of $x$.

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  • $\begingroup$ thankyou, so for #1 is it just 0,0,0,1,0,0,0,1,0,0,0,1 and so on? $\endgroup$ – ivanmike Feb 18 '15 at 4:36
  • $\begingroup$ That's right. And for #2, if you think of the power series for $1/(1-x)$ as having $1$'s as each coefficient, then do the arithmetic I suggested, you'll be able to figure out #2 as well. $\endgroup$ – rogerl Feb 18 '15 at 4:37
  • $\begingroup$ Actually, you need to be a bit careful. The first number in your list should correspond to $i=0$, right? So for #1 it should be $0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, \dots$ (four leading zeros, not three). $\endgroup$ – rogerl Feb 18 '15 at 4:38
  • $\begingroup$ The first equality is not correct. $\sum_{i=1}^\infty x^{4i}=\frac 1{1-x^4}$, not $\frac {x^4}{1-x}$ $\endgroup$ – Ross Millikan Feb 18 '15 at 4:38
  • $\begingroup$ Ack. Fixed, thanks. $\endgroup$ – rogerl Feb 18 '15 at 4:41
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For 1, you are correct that $\frac 1{1-x}=1+x+x^2+x^3+\dots $. You should multiply it by $x^4$, so there are terms in $x^5$ and $x^6$, etc. rogerl has dealt with 2. For 3 you just update the affected terms in $\frac 1{1-x}$ For 4, define $y=-3x$, write the generating function in $y$ and substitute back in for $x$

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  • $\begingroup$ sorry about that, I just edited number 4 $\endgroup$ – ivanmike Feb 18 '15 at 4:39
  • $\begingroup$ Updated in light of the edit. $\endgroup$ – Ross Millikan Feb 18 '15 at 4:40
  • $\begingroup$ Are you sure that you don’t want to suggest $y=-3x$ in the last one? $\endgroup$ – Brian M. Scott Feb 18 '15 at 4:42
  • $\begingroup$ @BrianM.Scott yes, that is better. $\endgroup$ – Ross Millikan Feb 18 '15 at 4:50

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