2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Does a complex Fourier series count? Here is an example $\endgroup$ Jun 11, 2023 at 21:12
  • $\begingroup$ 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. $\endgroup$
    – Poseidaan
    Jun 11, 2023 at 21:21
  • 1
    $\begingroup$ The RHS of your last formula is a power series in $e^x - 1$, not $e^x$. $\endgroup$
    – Rob Arthan
    Jun 11, 2023 at 21:24
  • $\begingroup$ 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 $e^x$ $\endgroup$ Jun 11, 2023 at 21:48
  • 2
    $\begingroup$ 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 $yd/(dy) (\sum a_ky^k)=d/(dx)(\sum a_k e^{kx})$ $\endgroup$
    – Conrad
    Jun 11, 2023 at 22:55

0

You must log in to answer this question.

Browse other questions tagged .