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Given the geometric series:

$1 + x^2 + x^4 + x^6 + x^8 + \cdots$

We can recast it as:

$S = 1 + x^2 \, (1 + x^2 + x^4 + x^6 + x^8 + \cdots)$, where $S = 1 + x^2 + x^4 + x^6 + x^8 + \cdots$.

This recasting is possible only because there is an infinite number of terms in $S$.

Exactly how is this mathematically possible?

(Related, but not identical, question: General question on relation between infinite series and complex numbers).

Community
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UGPhysics
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    Are you asking why $\frac{1}{1-x^2} = 1 + x^2 \frac{1}{1-x^2}$? Note that the identification of the formal series $S$ with the function $\frac{1}{1-x^2}$ is valid for $|x| \lt 1$, otherwise it's just a game you can play with formal power series. – t.b. Oct 05 '11 at 12:31
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    +1 For realizing that it is essential for there to be an infinite number of terms. Infinite sums usually make sense only under some kind of convergence (in a ring of formal power series the convergence is of a very different kind). But yeah, it is the same phenomenon that is underlying the following. Let $S=0.9999\ldots$. Then $$S=0.9+0.09999\ldots=0.9+0.1\cdot S,$$ Therefore $0.9=S-0.1\cdot S=0.9 \cdot S$, so $S=0.9/0.9=1$. – Jyrki Lahtonen Oct 05 '11 at 12:40
  • Well, I'm basically trying to understand why we can do this recasting. Normally - (I'm assuming) - we cannot recast a finite series. But the above recasting is allowed since the terms go to infinity. ... What is the exact mathematical (i.e. rigorous) basis which allows this manipulation? – UGPhysics Oct 05 '11 at 12:47
  • You can take out a common (non-zero) factor of a finite sum. If the sequence of finite sums converges, then you can take that common factor out of the limit, too. If the only difference between two sequences is that the other is one step behind, then the 'lagging' sequence is a subsequence of its faster brother, and thus converges to the same limit. IOW the argument of Joe Johnson's answer. – Jyrki Lahtonen Oct 05 '11 at 13:04
  • @Jyrki: This would only apply if the series converged. | However, for this series, the recasting is valid regardless of convergence or divergence of the series, (i.e. the issue revolves around the specific infinite number of terms for the series - which allows recasting). ... My question is: how (in a rigorous mathematical sense) is this mathematically meaningful / (possible)? – UGPhysics Oct 05 '11 at 13:11
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    You cannot even define the sum of infinitely many elements without some notion of a convergence. In the ring of formal power series (that many of us want to use here!) the convergence is based on the so called $x$-adic topology. Meaning that an infinite sum $\sum_{i=0}^\infty p_i(x)$ makes sense, if and only if the degrees of the lowest degree terms of the power series $p_i(x)$ tend to infinity. Here $p_i(x)=x^{2i}$ is its lowest degree term, and convergence thus follows from the fact that $2i\to\infty$ as $i\to\infty$. – Jyrki Lahtonen Oct 05 '11 at 13:18
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    IOW those three dots $\ldots$ hide a lot of reasoning. And 'no', the recasting does not make sense without some kind of a convergence allowing infinite sums to be formed in whatever structure we feel is most appropriate (here the complex numbers or the ring of formal power series). – Jyrki Lahtonen Oct 05 '11 at 13:22
  • @Jyrki: Ok. Thanks for the response. -- May I courteously request a formal answer to the above question? (Use whatever mathematics you like, as long as your answer is completely rigorous.) - cheers – UGPhysics Oct 05 '11 at 13:33
  • @t.b.: Within the radius of convergence the series is identified with / [or possibly mathematically equivalent to(?)] $\frac{1}{1-x^2}$. My question is: [preliminarily, and from receiving various responses: can we recast it the way we have if $S$ is not convergent?, and] (if we can recast it regardless or not the series is convergent or divergent [which I now believe the answer is a 'no' to]) how is it mathematically justified? .. [Part of my question I think has already been answered since I don't believe you can recast it as above if the series isn't convergent [or outside its domain... – UGPhysics Oct 05 '11 at 16:53
  • ... of convergence.] – UGPhysics Oct 05 '11 at 16:54
  • @t.b. Such formal "games" often yield great power, e.g. see this famous problem of Halmos. – Bill Dubuque Oct 05 '11 at 17:03
  • @Bill: Sure, I didn't say or mean to say otherwise and thanks for the nice link.@UGPhysics: You seem to insist that a formula has an intrinsic sense, without further explanations. That's simply not true. You can write down this identity, but asking whether it is valid means that you need a formal context in which you can make sense out of it. Two interpretations were given and they happen both to involve a notion of convergence (which always is around when dealing with infinitary operations). If you interpret the series as a convergent power series then it represents the same function on – t.b. Oct 05 '11 at 17:41
  • the open unit disk as the function $1/(1-x^2)$. So yes, in this sense they are mathematically equivalent. – t.b. Oct 05 '11 at 17:43

3 Answers3

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There is a finite version of which the expression you have is the limit.

Suppose $S=1+x^2+x^4+x^6+x^8$, then we can put

$S+x^{10}=1+x^2(1+x^2+x^4+x^6+x^8)=1+x^2S$

And obviously this can be taken as far as you like, so you can replace 10 with 10,000 if you choose. If the absolute value of $x$ is less than 1, this extra term approaches zero as the exponent increases.

There is also a theory of formal power series, which does not depend on notions of convergence.

Mark Bennet
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  • If x is a matrix, nilpotent to some even power, can we then say, that this holds also for finite series? (Or, more generally, if we have an algebra with zero-divisors, can we formally relax the condition of infiniteness?) – Gottfried Helms Oct 05 '11 at 12:56
  • Hi @Mark, thanks for the response. | May I know the formal power series version of your response? (I think that's what I'm searching for.) – UGPhysics Oct 05 '11 at 13:21
  • lhf has put a link to some notes on formal power series, which covers more than any comment I could make. What I wanted to show was that, while it is not possible to recast a finite truncation of your series in the form you gave, it is possible to get close if you include a kind of error term. Roughly, in formal power series the powers of $x$ act as placeholders for the coefficients. Arithmetic operations work as you would expect, but you have to be careful that the calculation for each coefficient of the result is a finite calculation. – Mark Bennet Oct 05 '11 at 13:55
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The $n$th partial sum of your series is

$$ \begin{align*} S_n &= 1+x^2+x^4+\cdots +x^{2n}= 1+x^2(1+x^2+x^4+\cdots +x^{2n-2})\\ &= 1+x^2S_{n-1} \end{align*} $$

Assuming your series converges you get that $$ \lim_{n\to\infty}S_n=\lim_{n\to\infty}S_{n-1}=S. $$

Thus $S=1+x^2S$.

J126
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$S = 1 + x^2 \, S$ is true even in the ring of formal power series. No convergence is needed here.

lhf
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  • If I understand correctly, the series in the question can be recast only if it is convergent; otherwise it can't. [via @Jyrki | ... so, how then are "formal power series" related to the "usual" series above? ... – UGPhysics Oct 05 '11 at 16:36
  • @UGPhysics, if you want to talk about numerical series, where $x$ is a number, then of course you need to worry about convergence. The manipulation you mentioned in the question is really purely formal. That was my point. – lhf Oct 05 '11 at 17:58
  • So, -to clarify this whole debate-, in the space of real numbers I cannot write an expression like $S$, and claim that I can recast it [as $1 + x^2,S$] unless the expression is convergent, (or defined within a specified radius of convergence)? – UGPhysics Oct 05 '11 at 18:06
  • @UGPhysics, right. – lhf Oct 05 '11 at 19:44