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I'm having difficulty showing the following identity (on open unit disk):

$$\frac{1}{1-z} = \prod_{n=0}^\infty{1+z^{2^n}}$$

My first idea was to show that they agreed on a set with non isolated points and show that the product converges to an analytic function on the unit disk, so that by uniqueness, they are equal. However, aside from 0, I cannot think of more appropriate values to choose. As well, I'm not quite sure how to show the uniform convergence of the infinite sum of |z|^(2^n) on compact subsets of the unit disk (or uniform convergence on unit disk?) to be able to show that the product converges to an analytic function.

Thank you!

Mason
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kls3984
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2 Answers2

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Hint:

$$\frac{1}{1-z}=\prod_{n=0}^{\infty}(1+z^{2^n})$$

Recall that

$$\displaystyle \frac{1}{1-z}=\sum_{n=0}^{\infty}z^{n}$$

So, the equality

$$\prod_{n=0}^{\infty}(1+z^{2^n})=\frac{1}{1-z}=\sum_{n=0}^{\infty}z^{n}$$

boils down to the statement:

Every even natural number (including $0$) can be written uniquely as the sum of distinct powers of $2$.

Try proving this by induction.

You will then have shown that these two objects are formally equal, and since they both converge on the disc, they are equal as functions.

EDIT: I was being silly--see Joriki's comment below.

Mason
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Alex Youcis
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  • Hello. Thank you! I understand how to show that every even natural number can be written uniquely as the sum of distinct powers of two. But I still do not see how it boils down to that statement...? I appreciate your help. – kls3984 Mar 25 '13 at 23:12
  • I will as soon as I understand it :) I still do not see why the equality boils down to showing "Every even natural number...". – kls3984 Mar 25 '13 at 23:19
  • I was wondering if you could clarify please. – kls3984 Mar 25 '13 at 23:20
  • @kls3984 Oh, sorry! Try multiplying out the terms in $\displaystyle \prod(1+z^{2^n})$ to see that it really comes out as

    $$\sum_{n\in I}z^{n}$$

    where $I$ ranges over all the natural numbers (plus zero) which can be written as the sum of distinct powers of $2$!.

     – Alex Youcis Mar 25 '13 at 23:21
  • @kls3984 PS, note the implicit use of the fact that both the series and sum converge absolutely, so that order does not matter for the terms! – Alex Youcis Mar 25 '13 at 23:23
  • Oh great! Thanks. One last question: Why is it that we can treat an infinite product formally when showing the equality? I understand that the sum converges and that if it gets factored it yields the product (given the lemma: "Every even natural number..."). However, I'm still unsure about the "formal" treatment of the sum, i.e. breaking down the sum into infinite factors. – kls3984 Mar 25 '13 at 23:28
  • Sorry, I just saw your post-script. How does that still solve the problem of factoring infinitely many times? – kls3984 Mar 25 '13 at 23:29
  • @kls3984 Basically what you have shown is that when you expand your infinite product as an infinite sum (which is valid by expanding the finite partial products into finite partial sums and then passing to the limit) you get precisely the infinite product you wanted. The "formal" comment was really just the fact that this is more of a problem about generating functions then it is about complex analysis. – Alex Youcis Mar 25 '13 at 23:30
  • Why did you switch from $1-z$ and $n=0,\dotsc$ in the question to $1-z^2$ and $n=1,\dotsc$ in the answer? It seems like an unnecessary complication, since you then have to limit the argument to even numbers, whereas the equality in the question simply corresponds to the fact that every natural number can be written uniquely as a sum of distinct powers of $2$ (including $2^0=1$). – joriki Mar 26 '13 at 07:06
  • @joriki Yes, that was a silly mistake. Fixed! Thanks. – Alex Youcis Mar 26 '13 at 07:09
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Here's another take on this. The basic tools are the same what are in the comments of the answer above but also we will exploit

$$\prod_{n=0}^\infty(1+x_m)=\sum_{S\in\mathcal{P}_f(\mathbb{W})} \prod_{m\in S}{x_m}$$

Which seems true but I don't know the name of this supposed identity. I have asked about using this identity here.

Applying this to $\prod_{n=0}^\infty{1+z^{2^n}}$ we find

$$\prod_{n=0}^\infty{1+z^{2^n}}=\sum_{S\in\mathcal{P}_f(\mathbb{W})}{\prod_{n\in S}{z^{2^n}}}=\sum_{S\in\mathcal{P}_f(\mathbb{W})} z^{\sum_{s\in S} 2^{s_i}}=\sum_{n=0}^\infty {z^n}$$

The first equality is the identity we exploit. The second is just unpacking some of what a product is: we are using $z^az^b=z^{a+b}$. The last equality is the one of interest: It really holds because $g: \mathcal{P}_f(\mathbb{W})\to \mathbb{W}$ defined as $$g(\{s_1, \dots, s_k \})= \sum_{i}^k {2^{s_i}}$$ is in fact a bijection. Note that $g(\{\})=0$ and $g(\{0\})=1$. Indeed we are just formally saying "Every natural number can be written uniquely as the sum of distinct (non-negative) powers of $2$.

To finish the argument of the OP we will just say that the geometric sum identity is well known. And thereby we have established that

$$\prod_{n=0}^\infty{1+z^{2^n}}=\sum_{n=0}^\infty {z^n}=\frac{1}{1-z}$$

And we somehow have avoided induction but this might be tucked into the argument which establishes the product sum identity above. I am not sure.

Mason
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  • Apparently... no one but us farmland mathematicians use this $\mathbb{W}$ to mean $\mathbb{N} \cup {0}$. The post above can be improved by fixing this notation to something more accessible. I will get to this eventually I am sure. – Mason Jul 26 '18 at 14:37