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Let $z$ be a complex number. Is the alternating infinite series $ f(z) = \frac{1}{\exp(z)} - \frac{1}{\exp(\exp(z))} + \frac{1}{\exp(\exp(\exp(z)))} -\ldots$ an entire function ? Does it even converge everywhere ?

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@Steven Nice tex edit. –  mick Jan 28 '13 at 22:58
    
The alternative question should be: does it converge to its Taylor series anywhere? I believe the answer is no .... –  Sheldon L Jan 29 '13 at 19:08
    
@SheldonL I think you are correct. The chaotic behaviour of iterations of exp AND the growth rate of the nth derivative suggests this. I believe analytic continuation does not even help. Since we need a nonzero radius somewhere first. I will come back to these ideas later since it relates to tetration as you probably know. –  mick Jan 29 '13 at 22:12
    
Hey Mick, the interesting thing is the derivatives at f(0) behave as though the radius of convergence were around 0.45, although technically, the approximation with n=4 terms is analytic and entire. With n=5 terms, it acts the same for the first 650,000 or so derivatives ... then switches to acting like the radius of convergence is 0.038 when n=5 takes over. –  Sheldon L Jan 30 '13 at 12:58
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by the way, thanks for posting this example! I think this is the easiest to define nowhere analytic function I've ever seen! Although, I would define it without the alternating signs, so as to focus simply on the non analytic taylor series convergence. I don't know why I hadn't come up with it myself, although I've looked at the similarly behaving nowhere analytic base change function and tommysexp for tetration. –  Sheldon L Jan 30 '13 at 13:17
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2 Answers

up vote 12 down vote accepted

No. There are infinitely many $z$ for which $e^z = z$, namely the branches of $-\text{LambertW}(-1)$: approximately $ 0.3181315052 \pm 1.337235701\,i, 2.062277730 \pm 7.588631178\,i, 2.653191974 \pm 13.94920833\,i, \ldots$

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+1 I was typing too slowly –  mrf Jan 28 '13 at 23:03
    
But $1/z -1/z + 1/z - ... = 0$ ? –  mick Jan 28 '13 at 23:03
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No, $1/z - 1/z + 1/z -\ldots$ is a sequence that diverges (the partial sums alternate between $1/z$ and $0$). –  Robert Israel Jan 28 '13 at 23:04
    
@mrf: I'm not sure what you mean. $\exp(-W(1)) = W(1)$, but $\exp(\exp(-W(1))) = 1/W(1)$, and from there further exponentials aren't particularly nice. –  Robert Israel Jan 28 '13 at 23:19
    
There are points such as (approximately) $1.66802405157611+5.03244706448615 i$ for which $\exp(z) = \overline{z}$, and then the series is $1/\overline{z} - 1/z + 1/\overline{z} - \ldots$ where each term's imaginary parts are equal (and nonzero). –  Robert Israel Jan 28 '13 at 23:28
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I don't even think its analytic! Here is a demonstration that f(z) is nowhere analytic at the real axis, but probably infinitely differentiable. $f(z)=\frac{+1}{\exp(z)}+\frac{-1}{\exp^{o2}(z)}+\frac{+1}{\exp^{o3}(z)}+\frac{-1}{\exp^{o4}(z)}....$

Iterated exponentials in the complex plane are really poorly behaved. For z at the real axis, f(z) seems to converge very nicely since $\exp^{on}(z)$ for real z quickly gets arbitrarily large meaning that $\frac{1}{\exp^{on}(z)}$ gets arbitrarily small. But what about for a complex value of z close to the real axis?

Consider a small complex delta, where we pick the value of delta such that: $\exp^{on}(z+\delta)=\exp^{on}(z)+\pi i$, where delta is $\delta=\log^{on}(\exp^{on}(z)+\pi i)$. Here we are saying that for an arbitrarily large real value, we can compute delta such that nearby z is an identical real value + $\pi i$. Then $\exp^{on+1}(z+\delta)=-\exp^{on+1}(z)$ And then $\exp^{on+2}(z+\delta)=\frac{1}{\exp^{on+2}(z)}$. Finally, $\frac{1}{\exp^{on+2}(z+\delta)}=\exp^{on+2}(z)$, which, somewhat unexpectedly, is an arbitrarily large real number, where we were expecting an arbitrarily small real number!

Lets take an example for $f(0+\delta)$, where we can easily calculate that

$f(0)=1-\frac{1}{^1e}+\frac{1}{^2e}-\frac{1}{^3e}+\frac{1}{^4e}\approx0.69810833$.

The 5th term is $\frac{1}{^4e}\approx4.289\times10^{-1656521}$, which is a very small number indeed! But consider $\delta=\log^{o3}(\exp^{o3}(0)+\pi i)\approx0.013141+0.073577i$. Then the 5th term switches to the reciprocal of the original 5th term, or $^4e\approx2.3315\times10^{1656520}$. If we were to do the same calculation for the sixth term, the delta value would be on the order of i/50000000, and yet the sixth term would become far too large to write down in exponential form.

But this same argument can be easily applied to any real value of z, where as n increases, $|\delta|$ gets arbitrarily small and $\exp^{on+2}(z+\delta)$ gets arbitrarily close to zero, so its reciprocal is arbitrarily large. So then, f(z) is only defined at the real axis, but not in the complex plane in the neighborhood of the real axis. I suspect similar arguments can be made for f(z) anywhere in the complex plane, since the fixed point of exp(z) is repelling, and then as n increases, then there is some arbitrarily small value of $|\delta|$ which leads to an arbitrarily large negative value for $\exp^{on+1}(z+\delta)$; the details would be messier than at the real axis.

I worked out how the Taylor series changes as the iteration count n grows, for a similar function involving tetration, which is a tricky computation! The weird thing is that this function is very nearly analytic. Below, I posted the first 21 terms of the Taylor series for f(z) centered at z=0, with n=4 approximation terms. We know that changing to n=5 adds $\frac{1}{^4e}$ to the constant term in the Taylor series, which of course is completely insignificant. It turns out that switching to n=5 also has an insignificant effect on the first 650,000 derivatives; where insignificant would mean at least the first 10 thousand digits in the mantissa are unchanged. But then suddenly this high frequency noise takes over, somewhere before the 700,000th derivative or so. The crossover point calculation is based on a previous similar example, not this particular case (but its probably correct).

The Taylor series terms below are for n=4 are from the current question, and its pretty accurate in the range from -0.3 to 0.3. As noted, switching to n=5 has virtually no effect whatsoever on any of these Taylor series terms. I think this is a fascinating infinitely differentiable nowhere analytic function.

{f=     0.698108332501131269345959
+x^ 1* -0.811483837735256689925511
+x^ 2*  0.564223607296152024212994
+x^ 3* -0.108713360197197249695425
+x^ 4*  0.0386545634563827924149361
+x^ 5* -0.0165764547202329162292425
+x^ 6* -0.136752426088883411516781
+x^ 7* -0.211624565156209581106389
+x^ 8*  0.756835546755967062843976
+x^ 9*  0.690325621331856063058907
+x^10* -5.14435389580661580579195
+x^11* -4.47947937782130852737809
+x^12*  32.5937752304652003488119
+x^13*  49.8669444598467536440491
+x^14* -167.436448420554441232387
+x^15* -496.152768269738246163963
+x^16*  443.431869230502056934185
+x^17*  3711.37945688838615014677
+x^18*  2906.96563801990937017428
+x^19* -17914.5512639999679052410
+x^20* -48878.6372819973401224653
}
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