There is a known identity:

$$\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)=\frac{1}{1-x}, ~~~~~|x|<1$$

It's easy to derive it by converting it to a telescoping product as shown in this answer.

However, we can't use the same method here.

$$\left( 1-x^{2^k} \right) \left( 1+x^{2^k} \right)=\left( 1-x^{2^{k+1}} \right)$$

$$\left( 1-x^{2^k} \right) =\frac{\left( 1-x^{2^{k+1}} \right)}{ \left( 1+x^{2^k} \right)}$$

This product will not telescope. We can't even use this to find something new about it:

$$p(x)=\prod_{k=0}^{\infty} \left( 1-x^{2^k} \right)=\frac{\prod_{k=0}^{\infty} \left( 1-x^{2^{k+1}} \right)}{\prod_{k=0}^{\infty} \left( 1+x^{2^k} \right)}$$

We only get the obvious recurrence relation:


Mathematica gives this plot (for 25 terms).

enter image description here

Does this product have a closed form?

  • 6
    $\begingroup$ expanding the product $\prod_{k=0}^\infty (1-x^{2^k}) = \sum_{n=1}^\infty x^n (-1)^{b(n)}$ where $b(n)$ is the number of 1's in the binary representation of $n$ $\endgroup$ – reuns Mar 10 '16 at 22:58
  • $\begingroup$ So does it prove that there is no way to get a closed form? Since $b(n)$ is not regular? $\endgroup$ – Yuriy S Mar 11 '16 at 20:09
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    $\begingroup$ Doing some combinatorics with $b(n)$ gives the series $\frac1{1-x}\left(1-2\sum_{n\geq0}\frac{(-1)^nx^{2^n}}{1+x^{2^n}}\right)$, by summing the $x^k$ whose $n$th binary digit is $1$, and putting $(-1)^n$ to compensate double counting according to whether the number of $1$s is odd or even (this $n$ is the same as in the series). $\endgroup$ – Bart Michels Jan 2 '18 at 17:48
  • $\begingroup$ @barto, this is really nice, could you possibly post this as an answer? $\endgroup$ – Yuriy S Feb 5 '18 at 21:58
  • 1
    $\begingroup$ @YuriyS It's quite far from a closed form, but I'll put it on my todo list. $\endgroup$ – Bart Michels Feb 5 '18 at 22:00

Not a closed form, but we have the following series representations:

$$\begin{align*} p(x) = \prod_{k=0}^{\infty} \left( 1-x^{2^k} \right) &= \frac1{1-x}\left(1+2\sum_{k\geq0}\frac{(-1)^kx^{2^k}}{1+x^{2^k}}\right) \\ &= \frac{1}{1-x} - \frac{4x}{(1-x)^2} + \frac6{1-x} \sum_{k=0}^\infty \frac{x^{2^{2k}}}{1-x^{2^{2k+1}}} \end{align*}$$ where the last sum is $$\sum_{\nu_2(n)\text{ even}}x^n$$

Proof. We have: (on the level of formal series; although everything converges absolutely for $|x|<1$)

$$p(x)-1 = \sum_{n=1}^\infty x^n (-1)^{b(n)}$$ where $b(n)$ is the number of $1$'s in the binary expansion of $n$. We can write: $$(-1)^{b(n)} = 1+2\sum_{\substack{k \text{ for which the }\\k\text{th digit is 1,}\\\text{starting from }k=0}}(-1)^k$$ plug this in the sum and change the order of summation $n \leftrightarrow k$:

$$\begin{align*}p(x)-1 &= \sum_{n=1}^\infty x^n + 2 \sum_{n=1}^\infty x^n\sum_{\substack{k \text{ for which the }\\k\text{th digit in }n\text{ is 1}}}(-1)^k \\ &= \frac x{1-x} + 2 \sum_{k=0}^\infty (-1)^k x^{2^k} \prod_{j \neq k}(1+x^{2^j}) \\ &= \frac x{1-x} + 2 \sum_{k=0}^\infty (-1)^k \frac{x^{2^k}}{(1-x)(1+x^{2^k})} \\ \end{align*}$$

For the series $S = \sum_{k=0}^\infty (-1)^k \frac{x^{2^k}}{1+x^{2^k}}$, we can write each term as a geometric series: $$S = -\sum_{k=0}^\infty \sum_{n=1}^\infty (-1)^{k+n}x^{2^kn}$$ Grouping the terms with the same power of $x$ gives: $$\sum_{n=1}^\infty c_nx^n$$ where $c_n=1$ if $\nu_2(n)$ is even, and $c_n=-2$ if $\nu_2(n)$ is odd.

We get $$\begin{align*}S &= -2\frac x{1-x} + 3\sum_{\nu_2(n)\text{ even}}x^n \\ &= -2\frac x{1-x} + 3\sum_{k=0}^\infty \left( (x^{4^k})^1 + (x^{4^k})^3 + (x^{4^k})^5 + \cdots \right)\\ &= -2\frac x{1-x} + 3\sum_{k=0}^\infty \frac{x^{2^{2k}}}{1-x^{2^{2k+1}}} \end{align*}$$

  • $\begingroup$ It's probably worth noting that $a_k=\frac{x^{\left(2^{2k}\right)}}{1-x^{\left(2^{2k}\right)}}$ rapidly approaches $a_k=\begin{cases}0&|x|<1\\-1&|x|>1\end{cases}$, as $k$ increases. $\endgroup$ – Jam Apr 26 '18 at 13:28

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