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I have to prove that a generating function for the sequence $\{a_k\}$ where $a_k = (-8)^k$ for all integers $k\geq 0$ is $\cfrac{1}{1+8x}$. But I don't have any information on what $x$ is. Nor is there a $\sum$ in front of $a_k$ or the fraction $\dfrac{1}{1+8x}$.

Is this $x$ meant to be $k$? Meaning typographical error by the lecturer?

My question: Is this problem solvable? Please don't show me how to solve it if it is, as it is an assignment question. Some explanation would be nice though, and definitely a yes/no it is solvable.

JMP
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Display Name
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  • $x$ is a purely formal variable in this context; it exists to make certain operations easier to use, and doesn't have a direct meaning in and of itself. All you need to show is that the coefficients of the function $\dfrac{1}{1+8x}$ are the same as the $a_k$ you're given. – Semiclassical Sep 07 '14 at 04:23
  • @Semiclassical so my $x$ is the indeterminate, and this is solved by the geometric subtraction $g(t) - tg(t)$(for example)? Sorry, I just saw that edit, I will think about this. – Display Name Sep 07 '14 at 04:25
  • @Semiclassical Thank you for your comment I appreciate it. I understand now, based on the above and heropup's answer below. – Display Name Sep 07 '14 at 04:31

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An ordinary generating function $f(x)$ for a sequence $\{a_k\}_{k=0}^\infty$ satisfies the relationship $$f(x) = \sum_{k=0}^\infty a_k x^k.$$ Thus, substitute $a_k = (-8)^k$ and evaluate the resulting sum, which is a geometric series with common ratio $-8x$.

heropup
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  • Thank you, that makes sense. I understand now. – Display Name Sep 07 '14 at 04:31
  • One question. I can see when we look at the $\sum_0^n$ we can use geometric series subtraction, and then we can take the limits as $n\to\infty$, but this only seems to work if our indeterminate $|x|\lt1$, is this assumed? If so, why can it be? – Display Name Sep 07 '14 at 04:41