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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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  • $\begingroup$ It's not clear to me what your independence argument is in the complex case - given that both $r$ and $\theta$ appear on both sides of their individual equations I can see no reason to expect $a_n=0$. $\endgroup$ – Steven Stadnicki Feb 21 '18 at 16:38
  • $\begingroup$ The complex case can probably also be proved by considering riemann surfaces of the log and the roots. For instance the range of log is smaller than of root sums ( finite) $\endgroup$ – mick Feb 21 '18 at 16:54
  • $\begingroup$ But room for improvement exists yes $\endgroup$ – mick Feb 21 '18 at 16:55
  • $\begingroup$ Stone-Weirstrass might relate. But Maybe that is not necc. Rather avoid it If possible imho $\endgroup$ – mick Feb 21 '18 at 16:57
  • $\begingroup$ $\ln z^n \neq n \ln z$ in general $\endgroup$ – Antonio Vargas Feb 21 '18 at 23:48

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