How to express a power series in closed form

Question

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!

Answer 1

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:

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

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

Answer 2

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,

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

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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

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

Answer 3

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:

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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,

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

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

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

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

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

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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,

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

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

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The answers from both are of course equivalent.