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Let $g(x)$ be infinite formal power series and $$g(x) = (1 + x)(1 + x^2)\cdots(1 + x^{2^k})\cdots$$ Show that $(1 - x) g(x) = 1$. My book gives following proof:

Using a fact that $(1 - x^k)(1 + x^k)=1-x^{2^k}$

$$ (1 - x) g(x) = (1 - x)(1 + x)(1 + x^2)\cdots(1 + x^{2^k})\cdots = (1 - x^2)(1 + x^2)\cdots(1 + x^{2^k})\cdots =$$ $$(1 - x^4)(1 + x^4)(1 + x^8)\cdots = (1 - x^8)(1 + x^8)(1 + x^{16})\cdots = 1 $$

Each step of reasoning shows that every term is "eaten" by multiplication. I do not see how it equates to 1 and I am not even close to accept it as a formal enough proof. If it was regular power series (numeric not formal) then I see how this proof could be easily turned in formal one by using limit but I do not see why formal power series of this form equates to $1$.

I have added logic tag because I do not understand proof technique used here. It is not any "inifitiary" proof technique which is widely used like limits or induction.

Trismegistos
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  • Why not just think about how $g(x)=\frac{1}{1-x}$ and recall that the binary representations of numbers are unique? – Bulberage Mar 22 '14 at 20:37
  • This item may be of interest. – Michael Hardy Mar 22 '14 at 20:37
  • @Bulberage This is supposed to be a proof that binary representation of numbers is unique so I can not use this fact. Besides I am more interested in technique of this proof. – Trismegistos Mar 22 '14 at 20:41
  • Here's the whole paper: http://www.obm.org.br/export/sites/default/semana_olimpica/docs/2011/semananivel3.2_samuel.pdf – Michael Hardy Mar 22 '14 at 20:44
  • @MichaelHardy This is some general book/paper about formal power series. I am asking questions about single specific proof. Besides I skimmed over this paper and I know most facts comprising this book. – Trismegistos Mar 22 '14 at 20:46
  • If all you want is a reason to understand why these sorts of formal manipulations are valid, then consider that these manipulations are valid for the first all of the coefficients of $x^i$ in $g(x)$ where $i<n$ for any fixed $n$. – Bulberage Mar 22 '14 at 20:49
  • @Bulberage I understand why these terms are "eaten" by multiplication I just do not see how the result of this procedure is $1$. This is just illogical for me because repeating this procedure would make every term vanish. I can understand procedure for finite sum because it has to stop and yield some number but I do not understand the proof for infinitely many terms and why in the end you get $1$. – Trismegistos Mar 22 '14 at 20:52
  • You know what a power series is. You know this infinite product has no powers of $x$ greater than $1$ in its power series. The only thing left is the constant term. The constant term is obviously $1$. What's there left to say? – anon Mar 22 '14 at 20:54
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    By using the "eating" process you can show that the coefficient of each $x^i$ with $i>0$ in the product is $0$. By repeating the process you discover all of the coefficients of the formal power series, so you know what the formal power series is. – Bulberage Mar 22 '14 at 20:56
  • @seaturtles There is a lot left to it. This proof does not use any accepted proof technique. Normally for infinite procedures like this you use induction or limits but this proof has some kind of infinite process which is not induction not any know by my proof scheme. – Trismegistos Mar 22 '14 at 20:58
  • @Trismegistos would a limit satisfy you? – Guy Mar 22 '14 at 20:59
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    @Trismegistos This proof does use induction: you use this to show that for every $i>0$, the coefficient of $x^i$ in the resulting power series is $0$. We also know the constant term is $1$. Do you disagree with the statement "since we know every coefficient of the power series, we know the power series"? You can also use limits if you want, by invoking the $(X)$-adic topology. – anon Mar 22 '14 at 21:02
  • @Bulberage now after what you said I understand why every coefficient of $x^i$ where $i > 0$ is 0. If you post your comment as an answer and explain briefly why a coefficient isn't 0 for $x^0$ I will accept your answer. – Trismegistos Mar 22 '14 at 21:02
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    Since every partial product has the same constant term, the resulting formal power series must have that same constant term. Namely, $1$. Also, why would it be $0$ for $x^0$? Why did you pick $0$ for a coefficient, as opposed to any other number? Do you really wonder why $1$ doesn't get eaten? How would it? – anon Mar 22 '14 at 21:03
  • @Sabyasachi I do not see how you would use limit for formal power series. I know how to do this for numerical power series. Can you post it in form of answer (comment is probably bit to short for that)? – Trismegistos Mar 22 '14 at 21:05
  • Thanks @seaturtles now I understand why constant term is $1$ and I understand why every other coefficient is 0. – Trismegistos Mar 22 '14 at 21:09
  • See this answer for the appropriate notion of convergence. – Bill Dubuque May 02 '16 at 20:47

2 Answers2

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What we really want to understand is why these sorts of formal manipulations are valid. To do so it will suffice to convince ourselves that for all of the coefficients of $x^i$ in $(1-x)g(x)$ where $i<n$ for any fixed $n$ we have a coefficient of $0$. By using the "eating" process you can show that the coefficient of each $x^i$ with $i>0$ in this product is $0$. By inductively repeating the process, you discover all of the coefficients of the formal power series, so you know what the formal power series is. Then, note that every time we "ate" the previous term, the constant term, $1$, is preserved, so, inductively, we see that the constant term will always be $1$, and thus that $(1-x)g(x)=1$.

Bulberage
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In the formal power series ring $R[[X]]$, what does it mean for a sequence of elements to converge?

This is the question you should be asking yourself before you even tried to look at the proof. The answer is that it means for every $i\ge0$, the sequence of $X^i$ coefficients of $f_1(X),f_2(X),f_3(X),\cdots$ will eventually be constant, say $a_i$. The series it converges to is then $\sum_{i\ge0}a_iX^i$.

The partial product is $(1-X)(1+X)(1+X^2)\cdots(1+X^{2^k})=1-X^{2^{k+1}}$. Clearly then, if $i>0$ the coefficient of $X^i$ in the partial products will eventually be constantly $0$ (for $i>2^{k+1}$) and if $i=0$ the term is constantly $1$. Hence the infinite product is $1+0X+0X^2+\cdots=1$.

Induction is implicit in the simplification of the partial products. Limits are implicit in how we define convergence of sequences of formal power series. There is even a topology lurking underneath all of this called the $(X)$-adic topology. Here, a neighborhood basis of the identity $0\in R[[X]]$ is given by the nested decreasing sequence of ideals $(X)\supset(X^2)\supset(X^3)\supset\cdots$. This construction is a special case of more general ones, that of completions and more generally inverse limits of rings and groups. It is also called the Krull topology (for example in the definition of an absolute Galois group of a field). And a different generalization is through valuation rings.

anon
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