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

Additional questions (added dec 16)

Consider the similar case for $z$ being real or having a small imaginary part:

$$ g(z) = \frac{1}{z} + \frac{1}{\exp(z)} + \frac{1}{\exp(\exp(z))} + \frac{1}{\exp(\exp(\exp(z)))} + \dotsb$$ As such $g(z)$ converges for real $z$ but diverges for nonreals.

So we try Some sort of continuation:

$$ g(z) = 1/z + 1/\exp + \dotsb$$ $$g(\exp) = 1/\exp + \dotsb $$

Thus $ g(z) - g(\exp(z) ) = 1/z $

$$ \exp(z) = g^{[-1]} ( g(z) - 1/z ) $$

Call that solution $g_1(z)$.

Assume differentiability and take the derivative at both sides. (Notice we can repeat to get infinite many equations!)

$$ \exp(z) = \frac{d}{dz} g^{[-1]} ( g(z) - 1/z) $$ [ ofcourse we can simplify the RHS by applying the chain rule and the rule for the derivative of the functional inverse - name ? - ]

And in General the equations

$$ exp(z) = \frac{d^m}{dz^m} g^{[-1]} ( g(z) - 1/z ) $$

Point is: are these equations getting analytic solutions or not?

Also we could combine the equations to get new ones.

Like this for example :

$$ g^{[-1]} ( g(z) - 1/z ) = \frac{d}{dz} g^{[-1]} ( g(z) - 1/z ) $$

And we could continue by taking any positive integer number of $a$ th derivative on the LHS and any positive integer number of $b$ th derivative on the RHS !

So are all these equations nowhere analytic ?? Or Some ? Or all of them ? And when they are analytic is it possible to do analytic continuation ? Are there natural boundaries ??

Many questions.

In fact ; not even sure how to solve these equations , neither with expressions nor numerical. - in terms of complex Numbers ofcourse otherwise i simply use the Sum from the beginning -.

< ps i considered using the functional inverse of $ g $ with notation $G$ so the simplifications of the derivatives take a different form , yet this makes no essential difference i guess >

  • $\begingroup$ @Steven Nice tex edit. $\endgroup$
    – mick
    Commented Jan 28, 2013 at 22:58
  • $\begingroup$ The alternative question should be: does it converge to its Taylor series anywhere? I believe the answer is no .... $\endgroup$
    – Sheldon L
    Commented Jan 29, 2013 at 19:08
  • $\begingroup$ @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. $\endgroup$
    – mick
    Commented Jan 29, 2013 at 22:12
  • $\begingroup$ 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. $\endgroup$
    – Sheldon L
    Commented Jan 30, 2013 at 12:58
  • 2
    $\begingroup$ 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. $\endgroup$
    – Sheldon L
    Commented Jan 30, 2013 at 13:17

2 Answers 2


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$

  • $\begingroup$ But $1/z -1/z + 1/z - ... = 0$ ? $\endgroup$
    – mick
    Commented Jan 28, 2013 at 23:03
  • 5
    $\begingroup$ No, $1/z - 1/z + 1/z -\ldots$ is a sequence that diverges (the partial sums alternate between $1/z$ and $0$). $\endgroup$ Commented Jan 28, 2013 at 23:04
  • 1
    $\begingroup$ The periodic points of exponentiation are a dense subset of $\mathbb{C}$, so actually this function is no-where analytic $\endgroup$ Commented Jan 21, 2020 at 17:36
  • 1
    $\begingroup$ @DarkMalthorp Do you have a reference for that assertion? $\endgroup$ Commented Jan 21, 2020 at 18:09
  • 1
    $\begingroup$ Yes, sorry. Was looking for it and forgot to update my comment. cambridge.org/core/journals/… $\endgroup$ Commented Jan 21, 2020 at 18:10

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


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
  • $\begingroup$ Is the defining equation for $\delta$ correct? It seems to approach $z$ instead of zero as we advance without bound on the positive real axis. $\endgroup$ Commented Dec 31, 2016 at 0:22

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