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If $S = x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

Find S.

Note:This is not a GP series.The powers are in GP.

My Attempts so far:

1)If $S(x)=x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$

Then $$S(x)-S(x^{2})=x$$

2)I tried finding $S^{2}$ and higher powers of S to find some kind of recursive relation.

3)When all failed I even tried differentiating and integrating S.Obviously that was of no good either.

Could anyone give me a hint to solve this?Thanks!

Shaswata
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    What kind of a closed form are you looking for? For $x=1/2$, [wolfram alpha has no idea](http://www.wolframalpha.com/input/?i=sum+%281%2F2%29%5E%7B2%5En%7D+for+n%3D0+to+infinity), so it's looking pretty bleak. – Bruno Joyal May 28 '13 at 05:57
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    I think that this is a function with an essential singularity on the unit circle. – marty cohen May 28 '13 at 06:06
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    @shaswata: Bruno's link was interpreted correctly in my browser. "Maybe if you type the commands properly it will give you an answer." Have *you* tried this? "...because other questions weren't that tough." What other questions? – Jonas Meyer May 28 '13 at 06:09
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    On my computer, I don't have a problem with the input as I wrote it. WA is made to take a beating with notation and I enjoy giving it one. Have you tried typing the commands properly yourself, and seeing if it gives an answer? Do you know that this has a closed form? If not, there is no reason to expect that it does. Inquisitiveness is a good quality, but rudeness and presumptuousness are not. – Bruno Joyal May 28 '13 at 06:11
  • @JonasMeyer:Yeah you're right.Something was wrong with my browser. – Shaswata May 28 '13 at 06:11
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    @Bruno:Sorry!My bad.Actually the place where I got this question,other questions were not that tough. – Shaswata May 28 '13 at 06:14
  • @shaswata : No worries. :) You might want to get your problems from some other place! – Bruno Joyal May 28 '13 at 06:15
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    What do you expect from [Lacunary functions](http://en.wikipedia.org/wiki/Lacunary_function)? – Sangchul Lee May 28 '13 at 06:38
  • sos440:Didn't know about that.Thanks! – Shaswata May 28 '13 at 06:40
  • @marty cohen: It actually has a *natural boundary* on the unit circle, meaning a "wall" that it cannot be analytically continued through. – The_Sympathizer May 28 '13 at 06:51

1 Answers1

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I have worked on this series before.

There is no simple closed form as a geometric series has.

(such as $x+x^2+x^3+x^4+...=\frac{x}{1-x}$ where $|x|<1$).

You can see more information in the link about Lacunary function.

The series can be expressed in closed form of double integral. I shared my result below.

$x+x^{2}+x^{4}+x^{8}+\dots=F(x)$

Let's transform $x=e^{-2^t} \tag{1}$

Where $-\infty<t<\infty$ Thus x will be in $ (0,1)$ and $F(x)$ is not divergent in this range.

$e^{-2^t}+e^{-2^{t+1}}+e^{-2^{t+2}}+\dots=F(e^{-2^t})=H(t)$

$e^{-2^t}+H(t+1)=H(t)$

$H(t+1)-H(t)=-e^{-2^t}$


The Fourier transform of both sides

$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t-\int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$

$$V(f)= \int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t$$

$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t=\int_{-\infty}^{+\infty} H(z)e^{-2πif(z-1)} \mathrm{d}z=V(f)e^{2πif}$$

$$e^{2πif}V(f)-V(f)=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$

$$V(f)=\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t$$


Now we need to take the inverse Fourier transform

$$H(z)=\int_{-\infty}^{+\infty} V(f) e^{2πifz} \mathrm{d}f=\int_{-\infty}^{+\infty} e^{2πifz}\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f $$

The closed form of $H(z)$ in integral expression: $$H(z)=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{2πifz} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f $$

$$\sum_{k=0}^\infty x^{2^k}=H(\log_2(-\ln x))=\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{2πif\log_2(-\ln x)} \frac{e^{-2^{t}-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f$$

Where $0<x<1$

$$\sum_{k=0}^\infty x^{2^k}= \int_{-\infty}^{+\infty} \frac{e^{2πif\log_2(-\ln x)}}{1-e^{2πif}} \int_{-\infty}^{+\infty} e^{-2^{t}-2πift} \mathrm{d}t\,\mathrm{d}f=\int_{-\infty}^{+\infty} e^{-2^{t}} \int_{-\infty}^{+\infty} \frac{e^{2πif(\log_2(-\ln x)-t)}}{1-e^{2πif}} \mathrm{d}f\,\mathrm{d}t$$

Note: I made the update on 07/29/2016 . Variable change was $x=e^{2^t}$ at Tag (1) and it had problems as @leonbloy 's comment below. Thanks for the comment. Please let me know if you notice something else in definitions.

Mathlover
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  • Reminds ver much the Ramanajan's style in his sums and Chandrasekharan's approach (Diophantine) with trigonometric polynomials in number theory. Absolutely beautiful! See also Winogradow's work. What if the exponents would be negative? – al-Hwarizmi May 28 '13 at 07:14
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    I must be missing something. If $F(x)$ converges for $|x|<1$, $x=e ^{2^t}$, then $H(t)$ diverges for all $t$. – leonbloy May 28 '13 at 15:08
  • Wow, updated in the year 2016, you must really love this to be caring so late :D – Simply Beautiful Art Jan 12 '17 at 02:12
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    Is the last integral convergent? $$ \int_{-\infty}^{\infty}\frac{e^{2\pi i f (\log_2(-\ln(x))-t)}}{1-e^{2\pi i f}}df $$ – Canjioh May 19 '20 at 14:49