Is there a real sequence $0 \leq a_n $ such that for $x > 1 $ we have :
$$\ln(x) = a_0 + \sum_{n=1}^{\infty} (-1)^n \cdot a_n \cdot x^{1/n}$$
Or
$$\ln(x) = a_0 + \sum_{n=1}^{\infty} (-1)^{1+n} \cdot a_n \cdot x^{1/n}$$
where we take the principle logarithm for $\ln(x)$
Is that sequence $a_n$ unique ? Does that sequence $a_n$ have any closed forms ?
For clarity $x$ is real.
For the nonreal case it is proven to be false :
Let $z=re^{i\theta}$.
On left hand side, we know: $$\ln(z)=\ln(re^{i\theta})=\ln(r)+i\theta$$
On right hand side : $$a_0 + \sum_{n=1}^{\infty} a_n \cdot z^{1/n}$$ $$=a_0 + \sum_{n=1}^{\infty} a_n \cdot r^{1/n}e^{i\theta/n}$$ $$=\sum_{n=1}^{\infty} a_n \cdot r^{1/n}\cos\left(\frac{\theta}{n}\right)+i\sum_{n=1}^{\infty} a_n \cdot r^{1/n}\sin\left(\frac{\theta}{n}\right)$$
By comparing real and imaginary parts respectively, we obtain:
$$\ln(r)=a_0 + \sum_{n=1}^{\infty} a_n \cdot r^{1/n}\cos\left(\frac{\theta}{n}\right)$$ $$\theta = \sum_{n=1}^{\infty} a_n \cdot r^{1/n}\sin\left(\frac{\theta}{n}\right)$$
However, fundamentally, we know that $r$ and $\theta$ in general is independent of each other. Therefore, $a_n=0$ for $n \ge 1$ to remove dependence. Then we have $$\ln(r)=a_0 \rightarrow r=e^{a_0}$$ and $$\theta=0$$. But $r$ and $\theta$ are variables, not constants!
Thus, we conclude it is not valid for nonreal $z$.
I had the idea to try and show that for very large $x$ the LHS and the RHS would not be asymptotic.
That would be Nice stronger result of a weaker statement. Perhaps l’hopital can be used then ?
As follows l’hopital seems to work ??
$$(1/x)/(RHS ‘ (x)) $$
Which results in the claim that a sum of $n$ th roots ( $ x RHS ‘ (x) $ ) can have the constant $1$ as an asymptotic. That seems even weirder than the original question.
Another idea was too use lin algebra.
But in both cases I am stuck.
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EDIT
I would like to comment that
$$ ln(z^2) = 2 \cdot ln(z) $$
Both for real or complex $z$ shows that there is No uniqueness for the potential representation !!
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Also $$ ln(x^n) = n \cdot ln(z) $$ and the nonuniqueness lead to the idea of a system of equations ( $n$ equations ) and hence lin algebra. Kinda.
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