Show that $f(x^a) = a f(x) $ and $f(x y) = f(x) + f(y)$

Question

Let $f(x) = (x-1) \large\prod_\limits{n=1}^{\infty} \dfrac{2}{x^{2^{-n}} + 1} $

For real $x > 0$ it is easy to show that $f(x^2) = 2 f(x)$.

Let $a$ be a real number.

Question 1

Show that for real $x>0$ and real $a$ We have

$f(x^a) = a f(x)$

Question 2

Show that for positive real $x,y$

$f(x y) = f(x) + f(y)$

Question 3

Show that $f(x)$ is $C^{\infty}$ for positive real $x$.

Question 4

Show that $f(2)= \ln(2)$.

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Let $z$ be a complex number.

Question 5

Where does $f(z)$ converge ?

And when it converges does it equal the Natural logarithm with the usual branches ?

It seems we have poles on the unit circle by the definition ( no analytic continuation ).

So it seems not to converge for $z$ with negative real part or close to the unit circle ??

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I found this product at the tetration forum.

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Edit : it might seem inevitable to some that a function satisfying $f(x^2) = 2 f(x) $ could not be the logarithm. But that is the case , hence these questions.

See for instance

http://mathworld.wolfram.com/FalseLogarithmicSeries.html

Answer 1

For real $x > 0$ and $a\in \mathbb{R}$ the convention that $x^a$ denotes $\exp(a\log x)$, where $\log \colon (0,+\infty) \to \mathbb{R}$ is the inverse of the exponential function $\exp \colon \mathbb{R}\to \mathbb{R}$ is so strong that any other interpretation needs to be explicitly announced. Hence we can assume that without further ado.

Looking at the partial products

$$f_N(x) = (x-1)\prod_{n = 1}^N \frac{2}{x^{2^{-n}}+1},\tag{1}$$

we see that $f_N(1) = 0$ for all $N$, and therefore $f(1) = 0$. For $x\neq 1$, we multiply the partial product with a helpful representation of $1$, namely $1 = \dfrac{x^{2^{-N}}-1}{x^{2^{-N}}-1}$. Noting that

$$\bigl(x^{2^{-k}}-1\bigr) \bigl(x^{2^{-k}}+1\bigr) = x^{2^{-(k-1)}}-1$$

for all $k\in \mathbb{Z}$, we see that the partial product telescopes,

\begin{align} f_N(x) &= (x-1)\prod_{n = 1}^N \frac{2}{x^{2^{-n}}+1}\\ &= \bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-N}}-1} \prod_{n = 1}^{N} \frac{2}{x^{2^{-n}}+1}\\ &= \bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-N}}-1}\frac{2}{x^{2^{-N}}+1} \prod_{n = 1}^{N-1} \frac{2}{x^{2^{-n}}+1}\\ &= 2\bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-(N-1)}}-1}\prod_{n = 1}^{N-1} \frac{2}{x^{2^{-n}}+1}\\ &= 2^2\bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-(N-1)}}-1}\frac{1}{x^{2^{-(N-1)}}+1} \prod_{n = 1}^{N-2} \frac{2}{x^{2^{-n}}+1}\\ &= 2^2\bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-(N-2)}}-1}\prod_{n = 1}^{N-2} \frac{2}{x^{2^{-n}}+1}\\ &= \dotsc\\ &= 2^k\bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-(N-k)}}-1}\prod_{n = 1}^{N-k} \frac{2}{x^{2^{-n}}+1}\\ &= \dotsc\\ &= 2^N \bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x^{2^{-(N-N)}}-1}\prod_{n = 1}^{N-N} \frac{2}{x^{2^{-n}}+1}\\ &= 2^N\bigl(x^{2^{-N}}-1\bigr)(x-1)\frac{1}{x-1}\\ &= 2^N\bigl(x^{2^{-N}}-1\bigr). \end{align}

Now writing $x^{2^{-N}} = \exp \bigl( 2^{-N}\log x\bigr)$ we find

$$\lim_{N\to \infty} f_N(x) = \log x$$

from familiar properties of the exponential function. Thus $f(x) = \log x$ for all $x \in (0,+\infty)$, and the answers to questions 1 to 4 are immediate from the known properties of the logarithm.

For complex arguments, not much changes, but we no longer have a canonical branch of the logarithm to interpret $x^a = \exp(a\log x)$, thus we must make a choice. If for $x \in \mathbb{C}\setminus \{0,1\}$ we always use the same logarithm of $x$ to compute $x^{2^{-n}}$, the same computation as above shows $f_N(x) \to \log x$ for the chosen logarithm of $x$, and hence every value of $\log x$ is a possible value of $f(x)$. If for different factors of the product we use different branches of the logarithm to compute $x^{2^{-n}}$, the product need not converge at all, or might converge to different values, not much can be said then. For $x = 1$, the factor $x-1$ ensures that $f_N(1) = 0$ for all $N$ and hence $f(1) = 0$, so there not all possible values of $\log 1$ can be obtained as the limit of the product, only the principal value. For $x = 0$, we can either refuse to accept $0^{2^{-n}}$ as a valid expression, or interpret it as $0$, in which case the product clearly diverges. In both cases, $f$ is not defined at $0$.

If, what is most reasonable, on a (sufficiently small) neighbourhood $U \subset \mathbb{C}\setminus \{0,1\}$ of $x_0$ we choose a continuous (and hence holomorphic) branch of the logarithm to compute $x^{2^{-n}}$ for all $x\in U$ (and all $n$), then we obtain that branch as $f = \lim f_N$ on $U$. For connected neighbourhoods of $1$, only choosing the principal branch of the logarithm yields a continuous limit function.

Answer 2

Partial stuff: Once you have solved Question 1, the rest is easy or greatly simplified:

Q2: If $x,y>0$, let $u=\ln x, v=\ln y$. Then $$\begin{align}f(xy)&=f(e^ue^v)\\&=f(e^{u+v})\\&=(u+v)f(e)\\&=uf(e)+vf(e)\\&=f(e^u)+f(e^v)\\&=f(x)+f(y)\end{align} $$

Q3: By the same method $f$ is proportional to logarithm $$f(x)=f(e^{\ln x})=\ln x\cdot f(e)$$

Q4: In the light of Q3, it suffices to show $f(e)=1$.