I want to calculate the limit of this sum :

$$\lim\limits_{x \to 1} {\left(x - x^2 + x^4 - x^8 + {x^{16}}-\dotsb\right)}$$

My efforts to solve the problem are described in the self-answer below.

  • 6
    $\begingroup$ The limit is obviously $\dfrac12$ $\endgroup$ – Lucian Dec 25 '14 at 17:25
  • 1
    $\begingroup$ That's not a polynomial. $\endgroup$ – Daniel Hast Dec 25 '14 at 17:43
  • $\begingroup$ @Lucian it isn't I have corrected my ansewr $\endgroup$ – RE60K Dec 25 '14 at 18:57
  • 5
    $\begingroup$ @Lucian : I agree that that is obvious, but apparently it's not true. $\endgroup$ – Michael Hardy Dec 25 '14 at 21:06
  • 1
    $\begingroup$ @MichaelHardy: Appearances can be deceiving. ;-$)$ $\endgroup$ – Lucian Dec 26 '14 at 5:22

From here is the amazing solution:

Since $S$ satisfies the functional equation $$S(x) = x − S(x^2)$$ it is clear that if $S(x)$ has a limit as $x$ approaches 1 then that limit must be $1/2$. One might guess that $S(x)$ in fact approaches $1/2$, and numerical computation supports this guess — at first. But once $x$ increases past $0.9$ or so, the approach to $1/2$ gets more and more erratic, and eventually we find that $S(0.995) = 0.50088\ldots > 1/2$. Iterating the functional equation, we find $S(x) = x − x^2 + S(x^4) > S(x^4)$. Therefore the fourth root, 16th root, 64th root, … of $0.995$ are all values of x for which $S(x) > S(0.995) > 1/2$. Since these roots approach $1$, we conclude that in fact $S(x)$ cannot tend to $1/2$ as $x$ approaches $1$, and thus has no limit at all! So what does $S(x)$ do as $x$ approaches $1$? It oscillates infinitely many times, each oscillation about 4 times quicker than the previous one; If we change variables from x to $\log_4(\log(1/x))$, we get in the limit an odd periodic oscillation of period 1 that's almost but not quite sinusoidal, with an amplitude of approximately $0.00275$. Remarkably, the Fourier coefficients can be obtained exactly, but only in terms of the Gamma function evaluated at the pure imaginary number $\pi i / \ln(2)$ and its odd multiples!

| cite | improve this answer | |
  • 2
    $\begingroup$ This is if the limit does exist. No? $\endgroup$ – Olivier Oloa Dec 25 '14 at 17:34
  • 1
    $\begingroup$ it is kind of trigonometric function when we see its curve im62.gulfup.com/TRaAc7.png $\endgroup$ – Abdou Abdou Dec 25 '14 at 17:40
  • 2
    $\begingroup$ The equation above is only true if you have infinitely many terms, and in this case $f$ is not a polynomial -- the existence of the limit is not obvious. $\endgroup$ – Clement C. Dec 25 '14 at 17:40
  • 1
    $\begingroup$ Your final argument does not work. For example $g(x) = \sum_{n=0}(-x)^n = \frac{1}{1+x}$ so by the "continuity of a polynomial function" $g(1) = 1/2$. But the series diverges for $x=1$. $\endgroup$ – Winther Dec 25 '14 at 17:59
  • 1
    $\begingroup$ @Winther yes you are correct, w8 i am currently thinking on the problem, and this solution is temporary, will work out a correct solution!thanks for all your support! $\endgroup$ – RE60K Dec 25 '14 at 18:01

Based on a paper "Summability of alternating gap series" by J.P. Keating and J.B.Reade in year $2000$, (an online copy can be found here) , one can use Poisson summation formula to show $$ S(x) =\sum_{n=0}^\infty (-1)^n x^{2^n} = \frac12 x + \frac{2}{\log 2} \Re\sum_{n=0}^\infty\left( \frac{\Gamma(\alpha_n i)}{\lambda^{\alpha_n i}} - \sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{\lambda^k}{\alpha_n i + k} \right)$$ where $\alpha_n = \frac{(2n+1)\pi}{\log 2}$ and $x = e^{-\lambda}$.

As $x \to 1^{-}$, $S(x) - \frac12 x$ will be dominated by the first term ( the term for $\alpha_0$ ) which oscillate with amplitude

$$\frac{2}{\log 2}\left|\Gamma\left(\frac{\pi i}{\log 2}\right)\right| = \frac{2}{\sqrt{\log 2\sinh(\pi^2/\log 2)}} \sim 0.00275 $$ and periodic in $\log_2 \lambda = \frac{\log\log\frac1x}{\log 2}$ with period $2$.

Please look at the paper mentioned above for more details.

| cite | improve this answer | |

We can define a rotation point when the curve switches its direction from increasing ordinates to decreasing values as x advances.


So when we try to locate those points in our polynomial, the derivative of this function must be 0 in that abscissa.

$$fn'(x)={1-2x+4x^3-......(-|+)2^n x^{2^n-1}}$$

This polynomial has (n-1) factors, which means that $fn'(x)=0$ has $(n-1)$ solutions which implies that there are $(n-1)$ deviation points in our graph.

When $n$ tends to infinity we have limitless deviations as shown in this graph:

enter image description here

Well, consequently, our limit is hovering endlessly between $0.5+λ$ and $0.5-λ$ with $λ$ extremely small -- it may even be zero!

Finally, we can say that the function is diverging, and the limit doesn't exist.

| cite | improve this answer | |
  • After consulting @achille hui's answer, looking on this paper he/she brought, I found an approbation to my answer posted awhile ago at the bottom of the page -97- proven by Hardy in this book.Well, I came little bit late :p.

-"for $k$ even. Therefore, the limit as $x->\infty$ does not exist: $y_n$ has maxima whose heights tend to $\frac{2}{3}$, and minima whose heights tend to $\frac{1}{3}$. Hence the series is not Cesaro summable. It then follows from the second of the results quoted above that neither can it be Abel summable (because the partial sums $s_m$ are bounded). Hardy gave a direct proof of this in 2. The question we address here is: what is the asymptotic form of the gap series (1.1) in this case as $i -> 1$"

Well as this graph illustrates that f(x) oscillates infinitly between $0.5+\delta$ and $0.5-\delta$ . now , need to prove it analytically since an algebric way would be tooo long.

so when f(x) changes direction as x tends to 1 means .

I tried to zoom in the behavior of the graph at the nearest right of x=1 using variable substitution :

x -> 1

$sin(( \frac{x-1}{a}+ \frac{1}{2}) \Pi)$ -> 1 with closer values

The graph i received looks like this :

enter image description here

Matlab code :

syms x;syms k;for i=1 : 50
f=@(x) sin((1/(i)*(x-1)+1/2)*pi);
y= f(x)+symsum(((-1)^k)*(f(x)^(2^k)),k,1,10+i);

Now lets find out the upper and lower limits:

Lets prove firstly that the curve is changing direction endlessly as x gets closer to 1.

$f'(x)= 1-2x+4x^3-...+(-1)^{k+1}x^{2^k+1}$

$lim_{x\rightarrow 1} f'(x)= \pm \inf$

the limit sways between $+inf$ and $-inf$ it means it crosses x-axis repeatedly infinite number of times, this can be proved by reccurence but lets do it easy way since it is not the aim of this question.

to prove that $f'(x)$ changes diection infinitely it suffices to prove that $lim_{x\rightarrow 1} f'(x) \neq f'(1) $




f'(x) has an average limit between $+\inf$ and $-\inf$ which is not one of those, that is a blatant proof that $0.5$ isnt the real limit of $f(x)$.

Now lets find the mysterious limit.

$$xf'(x) = x-2x²+4x^4 ...$$

$$x²f'(x²) = x²-2x^4 +4x^8..$$


$$x^n f'(x^n) = x^n-2x^{2n}+4x^{4n}...$$


$$f(x) = xf'(x) + x²f'(x²)-x^4f'(x^4)...$$

$$f(x) = xf'(x) + \frac{x(f'(x)-1)}{(-2)} - \frac{x(f'(x)-1+2X)}{2²}+\frac{x(f'(x)-1+2X-4x^3)}{-2^3} ... $$

the derived $f'(x)$ is nil when $f(x)$ is at the level of either its lower or higher summit, lets denote $x_0$ the group of abscissas where the curve of $f(x)$ changes direction from $l+\delta$ to $l-\delta$.

$f(x_0)=l \pm \delta $ means $f'(x_0)=0$

as $f'(x_0)=0$ at any $x_0$ where $f(x)$ reaches its upper or lower summit $l+\delta$ or $l-\delta$,

$$f(x_0) = x+ \frac{x_0(-1)}{(-2)} - \frac{x_0(-1+2x_0)}{2²}+\frac{x_0(-1+2x_0-4x_0^3)}{-2^3} ... $$

and since $x_0\rightarrow 1$,

$$\lim_{x_0\rightarrow 1} f(x_0)=(\frac{1}{2}+\frac{1-2}{2²}+\frac{1-2+4}{2^3}+...)=\frac{1}{2}-\frac{1}{4}+\frac{3}{8}-\frac{5}{15}+\frac{11}{32}-...=0.333 or 0.666$$

the mysterious limits are:

$l=\frac{1}{2} \pm (\frac{1}{6}) $

if we take deeper look (using last trigonometric magnifier) at functions $f(x)$ and $f(x_0)$ we should notice remarkable matching

enter image description here

$f(x)=x+\sum_k {(-1)^k x^{2^k}}$

$f(x_0)=x_0.\sum_k (\frac{1}{2^k} \sum_l ((-1)^l 2^l x_0^{2^l-1})$

syms k;syms l;x=-1:0.1:1;
for i=1 : 1000
f=@(x) sin((1/i.*(x-1)+1./2).*pi);
y=@(x,n) f(x)+symsum(((-1).^k).*(f(x).^(2.^k)),k,1,n);
yy=@(x,n) f(x).* symsum((1./2.^k).*symsum((-1).^l.*2.^l.*f(x).^(2.^l-1),l,0,k-1),k,1,n);
a=plot(x,y(x,i),'Color', [0.5, 1.0, 0.0], 'LineStyle', '--');
hold on;
b=plot(x,yy(x,i+1),'Color', [1.0, 0.0, 0.5]);
set(gca, 'XLim',[-1 1], 'YLim',[-1 2]);

similitude of two curves:

syms x;syms k;syms l;
for i=1 : 1000
y=@(x,n) (x)+symsum(((-1)^k)*((x)^(2^k)),k,1,n);
yy=@(x,n) (x)* symsum((1/2^k)*symsum((-1)^l*2^l*(x)^(2^l-1),l,0,k-1),k,1,n);
set(gca, 'colororder', [1, 0.5, 0.753;0.5, 0.5, 1;1, 1, 0.753]);
hold on;
set(gca, 'colororder', [0.1, 0.3, 0.53;0.35, 0.85, 1;1, 0.1, 0.73]);
legend({'y' 'yy'}, 'Location','NorthWest')

snapshots: ($f_n(x)$,$f_{n+1}(x_0)$)


enter image description here


enter image description here


enter image description here


enter image description here

  • note:

-I know it has a high probability of being wrong but, that is well argumented, and any critics should be also well argumented. thanks.

| cite | improve this answer | |
  • $\begingroup$ Dear Abdou, please see also my graphics in the (same )question at MO from february: mathoverflow.net/questions/198665/… $\endgroup$ – Gottfried Helms Jun 23 '15 at 16:03
  • $\begingroup$ I gave it a +1 although it seems that the result contradicts other (and my own) results because of the lot of serious work in this answer... $\endgroup$ – Gottfried Helms Jun 23 '15 at 17:54
  • $\begingroup$ Didn't you see the formula 3.18 at page 100 of your link? They say the following . Amplitude is about (pg 100 formula 3.18): $ 2/\log(2) \cdot | (\Gamma(\pi \cdot I/ \log(2)))| \approx 0.00274922168400 $ and this is alternating around $0.5$. So there is nothing with $0.333...$ and $0.6666...$ in that article... $\endgroup$ – Gottfried Helms Jun 23 '15 at 18:13
  • $\begingroup$ Mr @GottfriedHelms thanks for giving interest to my work, i posed this problem last december 2014 and I'm still working on it, trying to enclosing it from all sides, I just don't know yet how to integrate this summation using theta function maybe it would lead me somewhere? $\endgroup$ – Abdou Abdou Jun 23 '15 at 22:17
  • $\begingroup$ I do follow your researches on MO and Im amazed by your advancement in solution, $\endgroup$ – Abdou Abdou Jun 23 '15 at 22:20

This is a copy of my answer in mathoverflow which I've linked to in my OP's comment but it might really be overlooked

This is more an extended comment than an answer.

I refer to your function $$ f(x) = \sum_{k=0}^\infty (-1)^k x^{2^k} \qquad \qquad 0 \lt x \lt 1 \tag 1$$ and as generalization $$ f_b(x)= \sum_{k=0}^\infty (-1)^k x^{b^k} \qquad \qquad 1 \lt b \tag 2$$

[update 8'2016] The same values as for the function $g(x)$ as described below one gets seemingly simply by the completing function of $f(x)$ -the sum where the index goes to negative infinity, such that, using Cesarosummation $\mathfrak C$ for the alternating divergent series in $h(x)$ , we have that $$h(x) \underset{\mathfrak C}=\sum_{k=1}^\infty (-1)^k x^{2^{-k}} \qquad \qquad \underset{\text{apparently } }{=} -g(x) \tag{2.1} $$ (Note, the index starts at $1$) .
The difference-curve $d(x)$ as shown in the pictures of my original answer below occurs then similarly by $$d(x)= h(x)+f(x) \underset{\mathfrak C}= \sum_{k=-\infty}^\infty (-1)^k x^{2^k} \tag {2.2}$$ Using the derivation provided in @Zurab Silagadze's answer I get the alternative description (using the constants $l_2=\log(2)$ and $\rho = \pi i/l_2$ ) $$ d^*(x) {\underset{ \small \begin{smallmatrix}\lambda=\log(1/x) \\ \tau = \lambda^\rho \end{smallmatrix}}{=}} \quad {2\over l_2 } \sum_{\begin{matrix}k=0 \\ j=2k+1 \end{matrix}}^\infty \Re\left[{\Gamma( \rho \cdot j ) \over \tau^j}\right] \tag {2.3}$$ which agrees numerically very well with the direct computation of $d(x)$.
(I don't have yet a guess about the relevance of this, though, and whether $d(x)$ has also some other, especially closed form, representation)
[end update]

To get possibly more insight into the nature of the "wobbling" I also looked at the (naive) expansion into double-series. By writing $u=\ln(x)$ and expansion of the powers of $x$ into exponential-series we get the following table: $$ \small \begin{array} {r|rr|rrrrrrrrrrrrrr} +x & & +\exp(u) &&= +(& 1&+u&+u^2/2!&+u^3/3!&+...&) \\ -x^2 & & -\exp(2u)& &= -(& 1&+2u&+2^2u^2/2!&+2^3u^3/3!&+...&) \\ +x^4 & & +\exp(4u)& &= +(& 1&+4u&+4^2u^2/2!&+4^3u^3/3!&+...&) \\ -x^8 & & -\exp(8u)& &= -(& 1&+8u&+8^2u^2/2!&+8^3u^3/3!&+...&) \\ \vdots & & \vdots & \\ \hline f(x) & & ??? &g(x) &=( & 1/2 & +{1 \over1+2} u &+{1 \over1+2^2} {u^2 \over 2!} &+{1 \over1+2^3} {u^3 \over 3!} & +... &) \end{array}$$ where $g(x)$ is computed using the closed forms of the alternating geometric series along the columns.
The naive expectation is, that $f(x)=g(x)$ but which is not true. However, $$g(x)=\sum_{k=0}^\infty {(\ln x)^k\over (1+2^k) k! } \tag 3 $$ is still a meaningful construct: it defines somehow a monotonuous increasing "core-function" for the wobbly function $f(x)$ , perhaps so to say a functional "center of gravity".
The difference $$d(x)=f(x)-g(x) \tag 4$$ captures then perfectly the oscillatory aspect of $f(x)$. Its most interesting property is perhaps, that it seems to have perfectly constant amplitude ($a \approx 0.00274922$) . The wavelength however is variable and well described by the transformed value $x=1-4^{-y}$ as already stated by others. With this transformation the wavelength approximates $1$ very fast and very well for variable $y$.

For the generalizations $f_b(x)$ with $b \ne 2$ the amplitude increases with $b$, for instance for $b=4$ we get the amplitude $A \approx 0.068$ and for $b=2^{0.25}$ is $a \approx 3e-12$

Here are pictures of the function $f(x)$, $g(x)$ and $d(x)= f(x)-g(x)$ :

The blue line and the magenta line are nicely overlaid over the whole range $0<x<1$ and the amplitude of the red error-curve (the $y$-scale is at the right side) seems to be constant.
In the next picture I rescaled also the x-axis logarithmically (I've used the hint from Robert Israel's anwer to apply the exponentials to base $4$):

Having the x-axis a logarithmic scale, the curve of the $d(x)$ looks like a perfect sine-wave (with a slight shift in wave-length). If this is true, then because $g(0)=1/2$ the non-vanishing oscillation of $f(x)$, focused in the OP, when $x \to 1$ is obvious because it's just the oscillation of the $d(x)$-curve...
For the more critical range near $y=0.5$ I've a zoomed picture for this:

But from here my expertise is exhausted and I've no tools to proceed. First thing would be to look at the Fourier-decomposition of $d(x)$ which might be simpler than that of $f(x)$. Possibly we have here something like in the Ramanujan-summation of divergent series where we have to add some integral to complete the divergent sums, but I really don't know.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.