Solved - Finding the function of $(1+x)(1+x^4)(1+x^{16})(1+x^{64})....$

Question

The question states: For $0 < x < 1$, let $f(x) = (1+x)(1+x^4)(1+x^{16})(1+x^{64})(1+x^{256}).... $
Find $f(x).$

$f(x)= \prod_{n=0}^{\infty}(1+x^{(4^n)})$

I've tried noticing that it looks similar to the telescoping series $(1+x)(1+x^2)(1+x^4)...$
but applying the same trick of multiplying by $(1-x)$ doesn't work here.

Can someone please guide me in the right direction to solving this problem? Thanks very much.

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Edit: The original problem was to find $f^{-1} \left ( \frac{8}{5f(\tfrac{3}{8})} \right)$, as can be seen in this thread.
The solution can be found by noticing that $f(x)f(x^2)=(1+x)(1+x^2)(1+x^4)(1+x^8)... = \frac{1}{1-x}$
and by plugging in $x = 3/8$ (credit to John Omelian)

we get $f(\frac{3}{8})f(\frac{9}{64}) = \frac{1}{1-\frac{3}{8}} = \frac{8}{5}$, in which then we can divide by $f(\frac{3}{8})$ and take the inverse of both sides to get $f^{-1} \left ( \frac{8}{5f(\tfrac{3}{8})} \right) = \boxed{\frac{9}{64}\:}$.

I apologize for any confusion that I've caused, but thanks to everyone for their support and feedback!

Answer 1

$$ (1+x)(1+x^2)(1+x^4) (1+x^8)\dots = \frac{1}{1-x} $$

$$ f(x) := (1+x) (1+x^4)(1+x^{16}) ( 1 + x^{64})\dots $$

$$ g(x) := (1+x^2)(1+x^8)(1+x^{32})(1+x^{128})\dots $$

First, $f(x) g(x) = \dfrac{1}{1-x} $

Next $$ g(\sqrt x) = f(x) $$

Answer 2

Just note that the coefficient of $x^k$ is the number of ways in which $k$ can be represented as distinct powers of $4$. We note that since the base 4 representation is unique, this coefficient is 1 or 0.

Let's find which coefficient is 1:
Clearly, the coefficient of $x^k$ is $1$ iff when $k$ is represented in base 4, the representation consists entirely of 1s and 0s. So, we can form a bijection with the binary numbers as in: $$0_2 \mapsto 0_4 = 0$$ $$1_2 \mapsto 1_4 = 1$$ $$(10)_2 \mapsto (10)_4 = 4$$ $$(11)_2 \mapsto (11)_4 = 5$$ and so on.
In this way, we observe that the coefficients of $x^k$ are 1 where $k$ is $0,1,4,5,16,17,20,21$ and so on. Not sure if this sequence has a closed form.

$$f(x) = 1+x+x^4+x^5+x^{16}+x^{17} \ldots $$

Edit: Found this on OEIS, which includes both mine and Will's ideas.

Answer 3

I am not sure if $f(z)$ has an elementary closed form, due to its lacunary behavior on the boundary $|z|=1$.

For example, the figure below demonstrates the graph of $|f(\frac{299}{300}e^{i\theta})|$ for $-\pi\leq \theta \leq\pi$, hinting an extremely wild behavior of $f(z)$ near the boundary $|z|=1$ that cannot be replicated by any elementary functions:

Let us analyze this behavior more closely.

Let $\alpha$ be a dyadic rational number, and let $m \in \mathbb{Z}_{\geq 0}$ be the smallest non-negative integer such that $2^{2m+1} \alpha \in \mathbb{Z}$. If we let $p = 2^{2m+1} \alpha$, then $p$ is either an odd integer, or $2$ times an odd integer. Now, for $0 < r < 1$ we have

\begin{align*} |f(re^{2\pi i \alpha})| &= \prod_{k=0}^{\infty} \left| 1 + r^{4^k} e^{(2^{2k+1}\alpha) \pi i} \right| \\ &= \left( \prod_{k=1}^{m} \left| 1 + r^{4^k} e^{p \pi i / 4^k} \right| \right) (1 + r^{4^k}e^{p\pi i}) \left( \prod_{k=m+1}^{\infty} ( 1 + r^{4^k} ) \right). \end{align*}

To estimate the behavior of this quantity as $r \to 1^-$, we note that

$$ \prod_{k=m+1}^{\infty} ( 1 + r^{4^k} ) \to \infty \quad \text{as } r \to 1^-, $$

whereas

\begin{align*} (1 - r^{4^m}) \prod_{k=m+1}^{\infty} ( 1 + r^{4^k} ) &= \frac{\prod_{k=m+1}^{\infty} ( 1 + r^{4^k} )}{\prod_{k=2m}^{\infty} ( 1 + r^{2^k} )} \\ &= \frac{1}{1+r^{4^m}} \frac{1}{\prod_{k=m}^{\infty} ( 1 + r^{2^{2k+1}} ) } \to 0 \quad \text{as } r \to 1^-. \end{align*}

Hence, it follows that

$$ \lim_{r \to 1^-} |f(re^{2\pi i \alpha})| = \begin{cases} 0, & \text{if $p$ is odd,} \\ \infty, & \text{if $p$ is even.} \end{cases} $$

This clearly demonstrates the lacunary behavior of $f$.