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With $a_0 =1$, $a_n = 8a_{n-1} + 10^{n-1}$

Let a generating function with it, $G(x) = \sum a_k x^k$ = $a_0 + \sum (8a_{k-1} +10^{k-1}) x^k =a_0 + \sum 8a_{k-1}x^k + \sum10^{k-1} x^k = a_0 + 8x\sum a_kx^k + x\sum10^{k} x^k $

Now I think that, $\sum 10^k x^k \neq\frac{1}{1-10x}$, since it has no limitation on $x$, such as $|x|<1$ (If so, $10^nx^n$ cannot be less than 1, too).

What should I do? How can I deal with $\sum 10^k x^k$ ??

Or is it fine to do just like that?

JAEMTO
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1 Answers1

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The generating function uses formal power series. As long as there is some radius of convergence, the result will apply.

We have $$\begin{align} G(x) &= \sum_{k=0}^\infty a_k x^k \\ &= a_0 + \sum_{k=1}^\infty (8a_{k-1} + 10^{k-1})x^k \\ &= a_0 + 8x \sum_{k=0}^\infty a_k x^k + x \sum_{k=0}^\infty (10x)^k \\ &= 1 + 8x G(x) + \frac{x}{1 - 10x} \\ &= 8x G(x) + \frac{1 - 9x}{1 - 10x}. \end{align}$$

Therefore, $$G(x) = \frac{1 - 9x}{(1 - 8x)(1 - 10x)} = \frac{1}{2} \left(\frac{1}{1-8x} + \frac{1}{1-10x}\right) = \frac{1}{2} \sum_{k=0}^\infty (8^k + 10^k) x^k$$ and $$a_n = \frac{8^n + 10^n}{2}$$ and you can check that this works.

heropup
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  • Well then, we have an assumption that $|10x| < 1$? I am confusing at that point... – JAEMTO May 02 '22 at 07:37
  • Also, the form of $a_n$ is somewhat different from Marin's answer, [https://math.stackexchange.com/questions/3937669/how-to-solve-recurrence-relation-a-k-ba-k%e2%88%921-crk-assuming-b-neq-r], – JAEMTO May 02 '22 at 07:38
  • At first, I did like what you did, but I think it is wrong, because of $10x$, so is it fine to do? It is just an assumption? – JAEMTO May 02 '22 at 07:40
  • @JAEMTO You can assume that $|x|<\frac{1}{10}$, or $|x|<\frac{1}{100}$, or $|x|<\frac{1}{1000}$. I mean, why couldn't you? – Gary May 02 '22 at 07:47
  • @Gary Yes in that case, I understand it, but I am wondering that if I can think of for any $x$, like $x=2$... is it just an assumption for recurrence relation for what you said? – JAEMTO May 02 '22 at 07:52
  • @JAEMTO No, you can't. But from the viewpoint of the problem it does not matter. What matters is that the series converge in some neighbourhood of the origin. The size of it is totally irrelevant. I feel that you have not embraced the concept of a generating function fully yet. – Gary May 02 '22 at 07:54
  • @Gary Yeah I have studied generating function these days... Then what you mean is that $x$ should be in some neighborhood for convergence? That is an assumption because if $x$ is not in some neighborhood for convergence, then it just diverges and it has no meaning, is that right? – JAEMTO May 02 '22 at 08:03
  • @Gary Oh... generating function is important, because we can get $a_n $ easily, if there is a convergence radius, without too much calculation, is that right? – JAEMTO May 02 '22 at 08:22
  • @JAEMTO Yes, you can obtain a closed form of the sequence using a generating function if you are lucky. The convergence is nedeed so that you can write down a closed form for the function represented by the power series. – Gary May 02 '22 at 08:52