13
$\begingroup$

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:

$$p(x)=(1-x)~p(x^2)$$

Mathematica gives this plot (for 25 terms).

enter image description here

Does this product have a closed form?

$\endgroup$
  • 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
  • 3
    $\begingroup$ 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). $\endgroup$ – punctured dusk 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. $\endgroup$ – punctured dusk Feb 5 '18 at 22:00
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.