Let's consider the function defined by the infinite product:
$$f_p(x)=\prod_{k=0}^\infty \frac{1}{2} (1+x^{1/p^k})$$
$$p \geq 2$$
The only closed form I know is for $p=2$.
$$f_2(x)=\frac{x^2-1}{2 \ln x}$$
However, numerically, it turns out that this function is very close to linear, and the larger $p$, the better the approximation.
For my purposes, it's enough to consider:
$$0 \leq x \leq 1$$
For small $x$ and small $p$ the linear approximation doesn't fit. See for $p=3$:
However, for $p=15$ linear approximation fits much better:
Thus, my question is:
Can we obtain a highly accurate linear approximation $f_p(x)=A_p x+B_p$ without numerically computing the product for several points and using regression (i.e. least squares)?
Also, how can we prove that:
$$\lim_{p \to \infty} f_p(x)=\frac{1}{2}(1+x)$$
Motivation:
If we introduce a sequence:
$$a_{n+1}^p=a_n \frac{(a_n+b_n)^{p-1}}{2^{p-1}},\qquad b_{n+1}^p=b_n \frac{(a_n+b_n)^{p-1}}{2^{p-1}}$$
$$a_0>b_0, \qquad x=\frac{b_0}{a_0}$$
Then:
$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=(f_p(x))^{\frac{p-1}{p}} a_0$$
As $x \to 1$, $p \to \infty$ or $k \to \infty$, I think the following are right
$$ \begin{align} x^{1/p^k} &= \exp( p^{-k} \log x) \\&= \exp\left( p^{-k} (x-1) + O(p^{-k} (x-1)^2 ) \right) \\&= 1 + p^{-k} (x-1) + O\left(p^{-k} (x-1)^2 \right) \end{align}$$
$$ \frac{1}{2} \left(1 + x^{1/p^k} \right) = 1 + \frac{x-1}{2p^k} + O(p^{-k} (x-1)^2) $$
$$\begin{align} \log \prod_{k=0}^{\infty} \frac{1}{2} \left(1 + x^{1/p^k} \right) &= \sum_{k=0}^{\infty} \frac{x-1}{2p^k} + O\left(p^{-k}(x-1)^2 \right) \\&= \frac{p(x-1)}{2(p - 1)} + O((x-1)^2) \end{align}$$
$$ \prod_{k=0}^{\infty} \frac{1}{2} \left(1 + x^{1/p^k} \right) = 1 + \frac{p(x-1)}{2(p-1)} + O((x-1)^2) $$