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I was told that it is not possible to write $x$ as a power series in $e^x$ i.e. $$x = \sum_{k = 0}^\infty a_k e^{kx}.$$

The proof given stated that if such a power series did exist, then one could take the derivative of both sides to obtain $$1 = \sum_{k=1}^\infty ka_k e^{kx},$$ which contradicts the linear independence of the $e^{kx}$. However, I wonder if this proof is really correct as one is only allowed to exchange derivative and sum unless the sum of the derivatives converges uniformly. Since we do not know what the coefficients $a_k$ are, we cannot know whether that sum will converge uniformly.

That being said, it is possible to write $x$ as a pointwise-converging series in $e^x$? I know this is possible on a bounded domain since $$x = \log (e^x) = (e^x-1) + \frac{1}{2}(e^x-1)^2 + \cdots$$ but this series for the logarithm only converges for $e^x$ between 0 and 1. I wonder if it is possible to write this series so that it does converge on all positive real numbers.

Poseidaan
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  • Does a complex Fourier series count? Here is an example – Тyma Gaidash Jun 11 '23 at 21:12
  • I'd say it doesn't count as the series is supposed to run over the real numbers; nonetheless an interesting take on the problem. – Poseidaan Jun 11 '23 at 21:21
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    The RHS of your last formula is a power series in $e^x - 1$, not $e^x$. – Rob Arthan Jun 11 '23 at 21:24
  • Use $-\ln(\frac1x)=\ln(x)$ to extend convergence. Also, a Binomial theorem on $(e^x-1)^m$ technically gives a double series in terms of $e^x$ – Тyma Gaidash Jun 11 '23 at 21:48
  • @Tyma the Fourier series of $x$ (as periodic function on say $[-\pi, \pi]$ and extended by periodicity then) is not a power series in $e^{kx}$ but a Laurent series in $e^{ikx}$; similarly in the power series of the logarithm above, one cannot really expand the binomials and group as the series obtained that way is not convergent anymore (eg $e^x$ appears in all terms with coefficients $\pm 1$ depending on parity so grouped together one would get $e^x (1-1+1-1....)$ which is not convergent) – Conrad Jun 11 '23 at 22:49
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    The proof of impossibility in the Op is correct since with $e^x=y$, convergence of the series $\sum a_ky^k$ at a single non zero point (and any finite $x$ will do as it implies $y =e^x \ne 0$) implies uniform and absolute convergence on a disc in $y$ hence on a domain (including a real interval) in $x$ too; then differentiating term by term is allowed (in $y$ first but then in $x$ too as there is a simple rule that connects the two derivatives $yd/(dy) (\sum a_ky^k)=d/(dx)(\sum a_k e^{kx})$ – Conrad Jun 11 '23 at 22:55

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