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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.

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    $\begingroup$ The limit is obviously $\dfrac12$ $\endgroup$ – Lucian Dec 25 '14 at 17:25
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    $\begingroup$ 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
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    $\begingroup$ @Lucian : I agree that that is obvious, but apparently it's not true. $\endgroup$ – Michael Hardy Dec 25 '14 at 21:06
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    $\begingroup$ @MichaelHardy: Appearances can be deceiving. ;-$)$ $\endgroup$ – Lucian Dec 26 '14 at 5:22
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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!

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    $\begingroup$ This is if the limit does exist. No? $\endgroup$ – Olivier Oloa Dec 25 '14 at 17:34
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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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.

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

curve

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.

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  • 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);
a=ezplot(y,-1,1);
grid;
pause(0.1);
delete(a);
end

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) $

$f'(x=1)=f'(x²=1)=1-2(x=1)f'(x²=1)$

$f'(1)=1-2f'(1)$

$f'(1)=\frac{1}{3}$

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.

$$xf'(x) = x-2x²+4x^4 ...$$

$$x²f'(x²) = x²-2x^4 +4x^8..$$

$$....$$

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

So,

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

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

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]);
grid;
pause(0.1);
delete(a);
delete(b);
grid;
end

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);
a=ezplot(y(x,i));
set(gca, 'colororder', [1, 0.5, 0.753;0.5, 0.5, 1;1, 1, 0.753]);
hold on;
b=ezplot(yy(x,i+1));
set(gca, 'colororder', [0.1, 0.3, 0.53;0.35, 0.85, 1;1, 0.1, 0.73]);
legend({'y' 'yy'}, 'Location','NorthWest')
grid;
pause(0.1);
delete(a);
delete(b);
grid;
end

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

n=1

enter image description here

n=3

enter image description here

n=6

enter image description here

n=9

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.

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  • $\begingroup$ 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

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