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I was trying to figure out the expansion of the product:

$$Q= (1+x)(1+x^2)(1+x^4)..(1+x^{2^n})$$

Multiplying and dividing by $1-x$

$$ Q = \frac{1}{1-x} [ 1-x^{2^{n+1}}]$$

Expanding the denominator as geometric series:

$$ Q = - [ x^{2^{n+1}} - 1 ] [ \sum_{j=0}^{\infty} x^j] = - [ \sum_{j=0}^{\infty} x^{2^{n+1}+j} - x^{j}]$$

Now,

$$ \sum_{j=0}^{\infty} x^j = \sum_{j=0}^{2^{n+1}} x^j + \sum_{j=2^{n+1}}^{\infty} x^{j} = \sum_{j=0}^{2^{n+1}} x^j + \sum_{j=0}^{\infty} x^{2^n+j}$$

Using the above on $Q$,

$$ Q = \sum_{j=0}^{2^{n+1}} x^j$$

Or,

$$ (1+x)(1+x^2)(1+x^4)..(1+x^{2^n})=\sum_{j=0}^{2^{n+1}} x^j$$

My question: $ \frac{1}{1-x} = \sum_{j=0}^{\infty} x^j$ only holds if $ |x|<1$ so this chain of logical implications? But it is weird to me because I had a finite polynomial end ended in a finite polynomial so I am pretty convinced that my final polynomial is actually the expansion of first.. so, how do I justify this gives me the equation which holds for all $ x \in R$

Clemens Bartholdy
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  • how about now? I think I caught everything – Clemens Bartholdy Sep 28 '20 at 05:46
  • The argument is fine and can be justified in many ways. Two polynomials are equal iff they're equal on infinitely many points so if they're equal when $|x| < 1$ they're equal identically. You can also appeal to the identity theorem from complex analysis, or work with formal power series. – Qiaochu Yuan Sep 28 '20 at 05:54
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    Actually, $Q = \sum_{j=0}^{2^{n+1}-1} x^j$ – Angina Seng Sep 28 '20 at 05:55
  • All the answers so far are nice, but you could also, of course, recognize $\frac{1-x^{2n+1}}{1-x}$ as a finite geometric sum. No need to involve geometric series. – Will Orrick Sep 28 '20 at 20:01
  • ...but even there you need an argument that it holds when $x=1$. – Will Orrick Sep 28 '20 at 20:28

3 Answers3

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If we start with a polynomial and end with a polynomial, as long as there's infinitely many values of $x$ at which everything we did is allowed, then we know the conclusion is valid. Two polynomials that agree at infinitely many values must be equal. (Even having more values of $x$ than the degree of the polynomials would be enough.)

In particular: since the difference of your starting and ending polynomial is $0$ at every value of $x$ with $|x|<1$, it must be the zero polynomial, because any nonzero polynomial has only finitely many roots.

Misha Lavrov
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  • Well then I'd have to do trial and error and check if they equal ... which is gonna be difficult. Is there any general proof without trial and erroring at many points? – Clemens Bartholdy Sep 28 '20 at 06:02
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    Why trial and error? You don't have to check if they're equal. For every value of $x$ between $-1$ and $1$, you've already done the work to know they're equal, because all your operations with infinite series are valid for such $x$. – Misha Lavrov Sep 28 '20 at 06:03
  • @Buraian The last paragraph of Misha's answer tells you that you don't need to check anymore. You already know that your expansion is correct for all $|x|<1$. This is an infinite set (and hence has more elements than the degree), and hence you are done. – Yuzuriha Inori Sep 28 '20 at 06:04
  • Hmm I mean the whole real set is much larger than that interval.. I"m not sure how holding on interval implies that it holds everywhere – Clemens Bartholdy Sep 28 '20 at 06:40
  • @Buraian: prove that if two polynomials are equal at infinitely many points then they're identical. Actually you can prove something a bit stronger: if two polynomials of degree at most $n$ are equal at $n+1$ points then they're identical. – Qiaochu Yuan Sep 28 '20 at 06:43
  • I'm thinking of something with linear equations but I'm not sure how to formulate it in a very precise fashion – Clemens Bartholdy Sep 28 '20 at 09:30
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The calculations used by OP to show the identity is valid are based on algebraic manipulations but no analytical means, no limiting processes are used. In such cases we can switch the point of view, consider polynomials resp. series as formal power series and calculate in the ring $\mathbb{R}[[x]]$ of formal power series.

In this ring of formal power series questions of convergence are non-existent. We do not consider any evaluation of the variable $x$ but instead treat the series as objects which are used to work with and manipulate the coefficients of the series.

Here OP has shown the identity \begin{align*} (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})=\sum_{j=0}^{2^{n+1}} x^j\tag{1} \end{align*}

In fact he has shown the validity of (1) in the ring of formal power series by showing that $[x^n]$, the coefficients of $x^n$ are equal on both sides of (1) for $n\geq 0$. Evaluation of $x$ or convergence considerations were not needed at all.

After having shown the validity in the ring of formal power series we can switch back and consider LHS and RHS of (1) as two real-valued polynomials. We know from the formal derivation above that the coefficients are the same and we conclude by analytical means that these polynomials are identical as real valued functions.

Note: A great introduction into formal power series is presented in chapter 2 of H. Wilf's Generatingfunctionology.

Markus Scheuer
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Actually, you do not need to introduce the $1/(1-x)$ factor, since the equality between the two polynomials can be established in a direct elegant way.

Consider the development of $$ \eqalign{ & \left( {1 + x} \right)\left( {1 + x^{\,2} } \right)\left( {1 + x^{\,4} } \right) = \cr & = \left( {x^{\,0} + x} \right)\left( {x^{\,0} + x^{\,2} } \right)\left( {x^{\,0} + x^{\,4} } \right) = \cr & = x^{\,0} \cdot x^{\,0} \cdot x^{\,0} + \cr & + x^{\,1} \cdot x^{\,0} \cdot x^{\,0} + \cr & + x^{\,0} \cdot x^{\,2} \cdot x^{\,0} + \cr & + x^{\,1} \cdot x^{\,2} \cdot x^{\,0} + \cr & + x^{\,0} \cdot x^{\,0} \cdot x^{\,4} + \cr & + \cdots + \cr & + x^{\,1} \cdot x^{\,2} \cdot x^{\,4} \cr} $$

That, is from each of the three binomials you take (weight 1) or don't take (weight 0) each of the exponents $1,2,4$ , thereby obtaining $2^3$ triplets where the exponents are the $3$ -digit binary representation of the numbers $0$ to $7$.

Since the binary representation of a natural number is unique, then ...

Refer to this interesting paper.

Will Orrick
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G Cab
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