Background
Let us start with an equation:
$$ f(x) + f(x^2) + f(x^3) + f(x^4) + \dots= \sum_{r=1}^\infty f(x^r) = x$$
From the mobius inversion formula we have:
$$ f(x) = \sum_{r=1}^\infty \mu(r) x^r $$
where $\mu$ is the mobius function.
Standard Coordinate basis:
Let, $x=0 \implies f(0) = 0$
Now let us differentiate both sides:
$$ \sum_{r=1}^\infty r x^{r-1}f'(x^r) = 1$$
Let, $x=0 \implies f'(0) = 1 \implies \mu(1) = 1$
Again we can differentiate both sides:
$$ \sum_{r=1}^\infty r(r-1) x^{r-2} + r^2 x^{r-1} f'(x^r) = 0$$
Again let $x = 0 \implies f''(0) = -2! \implies \mu(2) = -1$
Different Coordinate Basis
Beginning with,
$$ \sum_{r=1}^\infty f(x^r) = x$$
Now, let us consider, $x \to x^k$ and divide LHS and RHS by $k!$
$$ \sum_{r=1}^\infty \frac{f(x^{kr})}{k!} = \frac{x^k}{k!}$$
Summing $k=0$ to $ \infty$:
$$ \sum_{k=1}^\infty \sum_{r=1}^\infty \frac{f(x^{kr})}{k!} = \sum_{k=1}^\infty \frac{x^k}{k!} $$
Writing the above explicitly:
$$ f(x) + f(x^2) + f(x^3) + f(x^4) + \dots = x$$ $$0 + \frac{f(x^2)}{2!} + 0 + \frac{f(x^4)}{2!} + \dots= \frac{x^2}{2!}$$ $$ 0 + 0 + \frac{f(x^3)}{3!} + 0 + \dots= \frac{f(x^3)}{3!} = \frac{x^3}{3!}$$ $$\vdots $$
And adding vertically:
$$ f(x) +(1+ \frac{1}{2!})f(x^2) + \dots = e^x-1$$
Defining the coefficients of $f(x^r)$ as $A(r)$:
$$ \sum_{k=1}^\infty \sum_{r=1}^\infty \frac{f(x^{kr})}{k!} = \sum_{k=1}^\infty \frac{x^k}{k!} \implies \sum_{r=1}^\infty A(r) f(x^r) = e^x-1$$
where $A(r)$ is the sum of the reciprocal factorials of $r$ divisible which are $r$.
Now, we replace, $x \to e^x$: $$ \implies \sum_{r=1}^\infty A(r) f(e^{xr}) = e^{e^x}-1$$
Now let, $x \to - \infty \implies f(0) = 0$
Differentiating and dividing by $e^{x}$:
$$ \implies \sum_{r=1}^\infty r A(r) e^{x(r-1)} f'(e^{xr}) = e^{e^x}$$
Let $ x \to - \infty \implies f'(0) =1 \implies \mu(1) = 1$
Differentiating and dividing by $e^{x}$:
$$ \implies \sum_{r=1}^\infty A(r) r^2 e^{2x(r-1)} f''(e^{rx})+ r(r-1) A(r) e^{x(r-1)} f'(e^{xr}) = e^{e^x}$$
Again, let us put $x \to - \infty \implies f''(0) = -2 \implies \mu(2) = -1 $
Question
How does one find the co-ordinates basis such that computing $\mu(r)$ is simplest (Given one must use the differentiation trick)?