I was recently reading this wikipedia page: https://en.wikipedia.org/wiki/Lacunary_function and found atleast the example they are giving must be wrong because I have kind of managed to analytically get an asymptotic result:

The example he gives seems to be a special version of this series:

$$ S(n,x) = e^{n\ln(x)} + e^{n^2\ln(x)} + e^{n^3 ln(x)} + ... $$

However this series also satisfies the partial differential equation:

$$ x \frac{\partial S}{\partial x} - n x^2 \frac{\partial^2 S}{\partial x^2} = nS $$

(This was found by me after alot of brute force)

Dividing the PDE by $ x^2 $

$$ \frac{\partial S}{x \partial x} - n \frac{\partial^2 S}{\partial x^2} = \frac{nS}{x^2} $$

Taking limit $ x $ tends to infinity:

$$ \lim_{x \to \infty} \frac{\partial S}{x \partial x} -\frac{nS}{x^2} = n \frac{\partial^2 S}{\partial x^2} $$

Now, we have options depending on assuming $ \lim_{x \to \infty} S=\infty $ or $ \lim_{x \to \infty} S'=\infty $

With alot of guesswork we get (the other options when solved the assumption is not justified):

$$ -\frac{nS}{x^2} = n \frac{\partial^2 S}{\partial x^2} $$

We take the solution:

$$ S \sim C \sqrt(x) \sin (\frac{\sqrt{3}}{2} \log(x)) $$

P.S: In general I find it hard to believe there can exist series which cannot be analytically continued.


Have I done something wrong? Is wikipedia wrong? Is this a known method of converting a function into a PDE and then taking limits to guess the asymptotic relationship known?

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    $\begingroup$ I think he means in this particular case. $\endgroup$ – Jorge Fernández Hidalgo Sep 12 '15 at 17:00
  • $\begingroup$ @LeeMosher I wasn't sure because at the same time I find it difficult to find a flaw in their argument ... So I'm very confused who is wrong over here ... $\endgroup$ – drewdles Sep 12 '15 at 17:01
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    $\begingroup$ So you seem to deny the very existence of lacunary functions? That would be quite a bold claim to make. Now, what's the behavior of your series in the close vicinity of 0? Kind of $S(x)\approx x$, apparently. And what about your analytical solution? $\endgroup$ – Ivan Neretin Sep 12 '15 at 17:41
  • $\begingroup$ @IvanNeretin ... If you follow my method taking limit $x$ to $0$ ... We get $S \sim x $ $\endgroup$ – drewdles Sep 12 '15 at 17:50
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    $\begingroup$ Are you sure that your derivation of the PDE is correct? I plugged the series directly and it looks like the LHS doesn't match the RHS. $\endgroup$ – pregunton Sep 12 '15 at 18:28

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