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I am working with generating functions and am required to prove that the generating function for the sequence $\{a_k\}$ where $a_k = (-8)^k$ for all integers $k\geq0$ is $\cfrac{1}{1+8x}$ and I have gotten it to:

$$g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$$

I can see that I want to take $(-8)^{n+1}x^{n+1}\to0$, but I don't know how I can say this is alright.

It works for $0\lt x\lt\lt\lt1$ I suppose, but I don't see how I can justify this.

This has been shown to me. It is a very similar question, and they have gotten to teh stage where $g(x)-xg(x)=1$, this is what I want, but this ugly one above can only reduce if $x\lt\frac18$. The current only answer says that I can't assume this, and I feel as though that is true. But his formula assumes what he is saying I can't.

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Wow you have a good name
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2 Answers2

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When using generating functions we treat the series $$\sum_{n=0}^{\infty}a_nx^n$$ as formal expression, and we dont care about convergence rate and stuff like that.

The notation $\frac{1}{1+8x}$ is nothing but a simple notation, and we dont actually care about subtituting x with anything, just with the cofficient of $x^k$ in the series.

For your series the generating function is $$f(x) = \sum_{n=0}^{\infty}(-8)^nx^n = \sum_{n=0}^{\infty}(-8x)^n = \frac{1}{1+8x}$$

Where the last part is that notation that is convinient since its the sum of the geometric series

Snufsan
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  • Yeah it is indeterminate, but can I assume that it is significantly less than $1$, so the limit goes to $1-0$ allowing me to divide by $1+8x$ giving me $g(x)=\frac{1}{1+8x}$ – Wow you have a good name Sep 11 '14 at 12:49
  • That last step is what I am trying to prove haha, you have just assumed it – Wow you have a good name Sep 11 '14 at 12:54
  • Its just a notation that is convenient because of the formula for sum of geometric series, but it has no meaning mathematically – Snufsan Sep 11 '14 at 12:56
  • I meant the step from $\sum \limits_{n=0}^\infty (-8x)^n$ to $\frac{1}{1+8x}$ how is this step achieved? I know we take $g(x)-(-8x)g(x)=g(x)[1+8x]$, but we get a limit as shown above, which only equals what I want for $x\lt \frac18$ – Wow you have a good name Sep 11 '14 at 13:01
  • Your formula for the geometric series actually uses the assumption $x\lt\frac18$ within it I believe. – Wow you have a good name Sep 11 '14 at 13:12
  • But i'm not applying the formula, I have no use in the fraction, just a simpler notation for the series..

    The fact that we write it is just writing it, we dont care about it's mathematic meaning.

     – Snufsan Sep 11 '14 at 13:20
  • Well obviously it is derived from something, e.g. http://mathworld.wolfram.com/GeometricSeries.html

    which is specifically my question, why won't this method work here.

     – Wow you have a good name Sep 11 '14 at 13:30
  • Do you understand whats the meaning of a generating function is? Why would you mind if the series converges or not? – Snufsan Sep 11 '14 at 13:31
  • That last step isn't true by definition if $x$ isn't less than $\frac18$ that is what I am not understanding, you can't just say it is true without proving it. I am not trying to be rude for the record, I know that text probably makes it seem like I am. – Wow you have a good name Sep 11 '14 at 13:32
  • There is not meaning of truth and false if $x$ never gets a value. – Snufsan Sep 11 '14 at 13:34
  • If we wrote instead of that fraction, $D(-8x)$ - will it be cool with you? – Snufsan Sep 11 '14 at 13:36
  • My problem is I have $g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$

    If you ignore the $x$ term, this diverges to infinity, which is not the case. For it to be a true statement$^1$, $x$ does have to have some range of value, I know it is indeterminate, but this doesn't give it a value, it just constrains it

    $ ; ;\dots ;;$ 1) True in that your sum above does equal $\frac1{1+8x}$

     – Wow you have a good name Sep 11 '14 at 13:39
  • But the limit has no meaning here... You need to understand that this series is a 'formal series' and not acctually a series with convergence and divergence – Snufsan Sep 11 '14 at 13:47
  • @Snufsan on a technical level, there is a notion of convergence and divergence in the ring of formal power series - it's based on the $(x)$-adic topology. In this topology, $8^{100}$ and $8^{-100}$ are equally close to $0$, and $x^k\to0$ as $k\to\infty$. – anon Sep 12 '14 at 00:29
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Generating functions are often expressed in closed form (rather than as a series), by some expression involving operations defined for formal power series. These expressions in terms of the indeterminate x may involve arithmetic operations, differentiation with respect to x and composition with (i.e., substitution into) other generating functions; since these operations are also defined for functions, the result looks like a function of x. Indeed, the closed form expression can often be interpreted as a function that can be evaluated at (sufficiently small) concrete values of x, and which has the formal power series as its Taylor series; this explains the designation "generating functions".

However such interpretation is not required to be possible, because formal power series are not required to give a convergent series when a nonzero numeric value is substituted for x. Also, not all expressions that are meaningful as functions of x are meaningful as expressions designating formal power series; negative and fractional powers of x are examples of this.

http://en.wikipedia.org/wiki/Generating_function

Snufsan
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