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I am trying to express the power series $x + x^4/4! + x^7/7! + \cdots$ in closed form;

I have already worked out the power series $1 + x^3/3! + x^6/6! + \cdots$ to be $(e^x + e^{x(2\pi i/3)} + e^{x(2\pi i/3)^2})/3$ using the roots of unity and the series for $e^x$ to get there, and I think I need to somehow "cycle" through to get to my desired closed form but I am not sure how to go about doing this.

Thanks!

Michael Hardy
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Harrison W.
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    Why not integrate the series you already have? – David Quinn Dec 23 '16 at 20:47
  • I'd like to formulate it using roots of unity if possible – Harrison W. Dec 23 '16 at 20:48
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    The general technique is called series multisection. – Sophie Dec 24 '16 at 12:10
  • @Sophie Thanks for the pointer. Didn't know it by that name, but it is indeed listed as such here and here. – dxiv Dec 24 '16 at 18:53
  • @HarrisonW. What is that "roots of unity" technique you're talking about? Can you tell me some more or point to the sources where I could read more about it? Seems interesting. (Not the roots of unity themselves – I know about these – but its application to decoding power series to functions.) – BarbaraKwarc Jul 14 '17 at 14:56
  • OK never mind, found it ;) Your first example can be simplified to: $f(x) ;=; 1+\frac{x^3}{3!}+\frac{x^6}{6!}+\frac{x^9}{9!}+\frac{x^{12}}{12!} + \cdots ;=; \frac{e^{x} ,+, 2e^{-\frac{x}{2}}\cos(\frac{\sqrt{3}}{2}x)}{3}$ – BarbaraKwarc Jul 14 '17 at 19:53
  • The other example simplifies to: $f(x) ;=; x+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^{10}}{10!}+\frac{x^{13}}{13!} + \cdots ;=; \frac{e^{x} ,+, \left(\sqrt{3}\cdot\sin(\frac{\sqrt{3}}{2}x) -\cos(\frac{\sqrt{3}}{2}x) \right)\cdot e^{-\frac{x}{2}} }{3}$ – BarbaraKwarc Jul 14 '17 at 23:35

3 Answers3

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worked out the power series $1 + x^3/3! + x^6/6! + \cdots$ to be $(e^x + e^{x(2\pi i/3)} + e^{x(4\pi i/3)})/3$

Not so, with $\omega = e^{2 \pi i / 3}$ the correct expression is $\,\frac{1}{3}(e^x + e^{\omega x} + e^{\omega^2 x}) = 1 + x^3/3! \,+ x^6/6! \,+ \,\cdots\,$

Then you can take the same idea to the next step. Hint:

  • $\,\frac{1}{3}(e^x + \omega e^{\omega x} + \omega^2 e^{\omega^2 x}) = x^2/2! \,+ x^5/5! \,+ x^8/8! \,+\,\cdots\,$

  • $\,\frac{1}{3}(e^x + \omega^2 e^{\omega x} + \omega e^{\omega^2 x}) = x \,+ x^4/4! \,+ x^7/7! \,+ \,\cdots\,$

Michael Hardy
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dxiv
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    For those later readers who do not know the technique used for summing the series in the first place (like me before reading this post), see Theorem 1.1 of these online notes. +1 vote to both this answer and the OP for teaching me something new. – Paul Dec 24 '16 at 12:21
  • I have amended my question now to correct it, thanks for pointing that out, was silly of me. You set $\omega = e^{2\pi i/3}$ in your answer but I believe you meant to set $\omega = 2\pi i/3$.

    I've also attempted a full solution of my own, if you don't mind verifying I am correct in saying integration / double differentiation are both just as valid.

     – Harrison W. Dec 24 '16 at 12:40
  • @HarrisonWilde I believe you meant to set ... No, I meant just what I wrote: $\omega = e^{2\pi i / 3}$ $\ne 2 \pi i /3$ is a complex cube root of unity. – dxiv Dec 24 '16 at 16:26
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Apologies to the two earlier answers, I made a typo - as they noticed - in my original question; the question now reflects what the closed form for $1 + x^3/3! + x^6/6! + \cdots$ should be. From that we can either integrate or differentiate twice to get my desired closed form for $x + x^4/4! + x^7/7! + \cdots$ as follows:

By integration:

Notice that the infinite sequence $x + x^4/4! + x^7/7! + \cdots = \int^x_0 1 + x^3/3! + x^6/6! + \cdots dx$

For efficiency, let $\omega=2\pi i/3$, then,

$$\int^x_0 1 + x^3/3! + x^6/6! + \cdots dx = \int^x_0 \frac 1 3 (e^x+e^{\omega x} + e^{\omega^2 x}) \, dx$$

$$=[e^x/3 +e^{\omega x}/3\omega +e^{\omega^2 x}/3\omega^2]^x_0$$

$$=e^x/3 +e^{\omega x}/3\omega +e^{\omega^2 x}/3\omega^2-1/3-1/3\omega-1/3\omega^2$$

$$=\omega^2e^x/3\omega^2+\omega e^{\omega x}/3\omega^2+e^{\omega^2 x}/3\omega^2-(\omega^2+\omega+1(1/3))$$

$$=\omega^2e^x/3\omega^2+\omega e^{\omega x}/3\omega^2+e^{\omega^2 x}/3 \omega^2$$

By differentiating twice:

Notice that the infinite sequence $$x + x^4/4! + x^7/7! + \cdots = d/dx(d/dx( 1 + x^3/3! + x^6/6! + \cdots ))$$

For efficiency, let $\omega=2\pi i/3$, then,

$$\frac d {dx} \left(\frac d {dx}( 1 + x^3/3! + x^6/6! + \cdots )\right) = \frac d {dx}\left(\frac d {dx} \left(\frac 1 3(e^x+e^{\omega x} + e^{\omega^2 x}) \right) \right)$$

$$= \frac d {dx}(e^x/3+\omega e^{\omega x}/3+\omega^2 e^{\omega^2 x}/3) = e^x/3+\omega^2e^{\omega x}/3 +\omega e^{\omega^2 x}/3$$

The answers from both are of course equivalent.

Michael Hardy
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Harrison W.
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    For the starting formula to hold true you need to define $\omega$ to be a complex cube root of unity, so that $\omega^3=1$ and $1+\omega+\omega^2=0$. Then powers other than multiples of $3$ "cancel out" when you add the three series $e^x+e^{\omega x}+e^{\omega^2x}$. – dxiv Dec 24 '16 at 16:35
  • @dxiv Of course my bad, I got caught up in the closed form involving exp(something) as well, thanks for all your help – Harrison W. Dec 24 '16 at 16:44
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Following David Quinn's hint:

$$f(x)=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\dots=\frac13(e^x+e^{x(2\pi i/3)}+e^{x(4\pi i/3})$$

Notice that

$$\int_0^x1+\frac{t^3}{3!}+\frac{t^6}{6!}+\dots\ dt=x+\frac{x^4}{4!}+\frac{x^7}{7!}+\dots$$

And that

$$\int_0^x\frac13(e^t+e^{t(2\pi i/3)}+e^{t(4\pi i/3})\ dt=\frac13e^x+\frac1{2\pi i}e^{x(2\pi i/3)}+\frac1{4\pi i}e^{x(4\pi i/3)}+\frac{3i}{4\pi}-\frac13$$

Thus,

$$x+\frac{x^4}{4!}+\frac{x^7}{7!}+\dots=\frac13e^x+\frac1{2\pi i}e^{x(2\pi i/3)}+\frac1{4\pi i}e^{x(4\pi i/3)}+\frac{3i}{4\pi}-\frac13$$

Note that you probably won't be able to express it in terms of the roots of unity as nicely since it isn't of the form

$$\sum_{n=0}^\infty\frac{x^{kn}}{(kn)!}$$

Simply Beautiful Art
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