A periodic function is defined by the property

$f(t+T)=f\left(t\right)$

If the function has an exponential-type power series expansion

$f\left(t\right)=\sum _{k=0}^{\infty}{c}_{k}\frac{{t}^{k}}{k!}$

then using the binomial theorem the periodic property translates to

$\begin{array}{l}f(t+T)=\sum _{k=0}^{\infty}{c}_{k}\frac{(t+T{)}^{k}}{k!}=\sum _{k=0}^{\infty}\frac{{c}_{k}}{k!}\sum _{l=0}^{k}\left(\genfrac{}{}{0ex}{}{k}{l}\right){T}^{k-l}{t}^{l}\\ \phantom{f(t+T)}=\sum _{l=0}^{\infty}\sum _{k=l}^{\infty}\frac{{c}_{k}}{l!(k-l)!}{T}^{k-l}{t}^{l}=\sum _{l=0}^{\infty}\sum _{m=0}^{\infty}\frac{{c}_{l+m}}{m!}{T}^{m}\frac{{t}^{l}}{l!}\end{array}$

Comparing this to the original power series, one must have in general

${c}_{k}=\sum _{m=0}^{\infty}\frac{{c}_{k+m}}{m!}{T}^{m}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}\sum _{m=1}^{\infty}\frac{{c}_{k+m}}{m!}{T}^{m}=0$

where the second form removes the first term from both sides of the equivalence. Since the factorial and the period are both positive, the expansion coefficients must include negative values to satisfy these conditions.

For the fundamental periodic functions with expansions

$sint=\sum _{k=0}^{\infty}(-1{)}^{k}\frac{{t}^{2k+1}}{(2k+1)!}\phantom{\rule{5em}{0ex}}cost=\sum _{k=0}^{\infty}(-1{)}^{k}\frac{{t}^{2k}}{\left(2k\right)!}$

one has immediately

$\sum _{k=0}^{\infty}(-1{)}^{k}\frac{(2\pi {)}^{2k+1}}{(2k+1)!}=0\phantom{\rule{5em}{0ex}}\sum _{k=1}^{\infty}(-1{)}^{k}\frac{(2\pi {)}^{2k}}{\left(2k\right)!}=0$

which can be confirmed with numerical approximations. To compare this with the general result above, write the expansion coefficients of the cosine function as

${c}_{2k}=(-1{)}^{k}\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}{c}_{2k+1}=0$

and insert in the general result:

$\sum _{m=1}^{\infty}\frac{{c}_{2k+m}}{m!}{T}^{m}=\sum _{m=1}^{\infty}\frac{{c}_{2k+2m}}{\left(2m\right)!}{T}^{2m}=\sum _{m=1}^{\infty}\frac{(-1{)}^{k+m}}{\left(2m\right)!}(2\pi {)}^{2m}=0$

The cosine coefficients force the index here to be even as well, reproducing the immediate result just given apart from a minus sign. Oddly enough, repeating this procedure with the coefficients of the sine expansion

${c}_{2k}=0\phantom{\rule{2em}{0ex}},\phantom{\rule{2em}{0ex}}{c}_{2k+1}=(-1{)}^{k}$

produces the same result, since again the index is forced to be even. The immediate result above for the sine expansion has a different source.

As an aside, one can take either of the immediate results, cut off the expansion at some number of terms and treat it as a polynomial estimate for $2\pi $ that can always be solved numerically. This indicates a way to tell whether a given power series expansion represents a periodic function: successively increase the number of terms used in the process to determine that the resultant numerical value converges to a stable estimate.

Derivatives of a periodic function are also periodic, which can be seen by differentiating both sides of the defining equation:

${f}^{\left(n\right)}(t+T)={f}^{\left(n\right)}\left(t\right)$

Applying this to the exponential-type power series expansion gives

${f}^{\left(n\right)}(t+T)=\sum _{k=n}^{\infty}{c}_{k}\frac{(t+T{)}^{k-n}}{(k-n)!}$

and comparing this to the original expansion at $t=0$ leads to a second general result:

${c}_{n}=\sum _{k=n}^{\infty}{c}_{k}\frac{{T}^{k-n}}{(k-n)!}\phantom{\rule{2em}{0ex}}\to \phantom{\rule{2em}{0ex}}\sum _{k=n+1}^{\infty}{c}_{k}\frac{{T}^{k-n}}{(k-n)!}=0$

With $n=1$ and the cosine expansion coefficients above, one has

$\sum _{k=2}^{\infty}\frac{{c}_{k}}{(k-1)!}{T}^{k-1}=\sum _{k=1}^{\infty}\frac{{c}_{2k}}{(2k-1)!}{T}^{2k-1}=\sum _{k=1}^{\infty}\frac{(-1{)}^{k}}{(2k-1)!}(2\pi {)}^{2k-1}=0$

which reproduces the immediate result from the sine expansion, apart from a shift in index and a subsequent minus sign. Repeating the procedure with the sine expansion coefficients apparently requires $n=2$ as the starting point.

Given that the second general result is a set of linear equations for the constants, one might think to stop the expansion at some larger number and solve backward for all of the coefficients of a general periodic function. First compare the equations to a summation formula for Bernoulli numbers, which have alternating signs:

$\sum _{m=1}^{N-1}\left(\genfrac{}{}{0ex}{}{N}{m}\right){B}_{m}=N!\sum _{m=1}^{N-1}\frac{{B}_{m}}{(N-m)!m!}=0$

Equating the first factorial in the denominator to the one in the equations above gives the relation

${c}_{k}={c}_{N}{T}^{N-k}\frac{{B}_{N-k}}{(N-k)!}$

Since the factorial increases faster than the Bernoulli number as $N\to \infty $ , the most general periodic function is a constant. Earth shattering!

The form of the coefficients can be confirmed with the following Mathematica code:

max = 10; s[n_] := Sum[Subscript[c, n + m] T^m/m!, {m, 1, max - n}] Solve[Table[s[i] == 0, {i, 0, max - 2}], Table[Subscript[c, i], {i, 1, max - 1}]]

Given the expansion

$xcothx=\sum _{k=0}^{\infty}\frac{{B}_{2k}}{\left(2k\right)!}(2x{)}^{2k}$

one has immediately that the coefficients even in the period are identical to an expansion of $\frac{T}{2}coth\frac{T}{2}$ . The linear term for ${c}_{N-1}$ is oddly missing...

The nonpathological periodic functions used in physics can always be represented by Fourier series. If a function has the expansion

$f\left(t\right)=\sum _{n}{a}_{n}cos\left(\frac{2\pi nt}{T}\right)+{b}_{n}sin\left(\frac{2\pi nt}{T}\right)$

then consider it term by term:

$f\left(t\right)=\sum _{k=0}^{\infty}(-1{)}^{k}\frac{{t}^{2k}}{\left(2k\right)!}\phantom{\rule{.2em}{0ex}}\left[\phantom{\rule{.2em}{0ex}}\sum _{n}{a}_{n}\right(\frac{2\pi n}{T}{)}^{2k}\phantom{\rule{.2em}{0ex}}]+(-1{)}^{k}\frac{{t}^{2k+1}}{(2k+1)!}\phantom{\rule{.2em}{0ex}}\left[\phantom{\rule{.2em}{0ex}}\sum _{n}{b}_{n}\right(\frac{2\pi n}{T}{)}^{2k+1}\phantom{\rule{.2em}{0ex}}]$

This is potentially useful in correlating a known Fourier expansion with a power series expression of the same function. The coefficients in brackets are simple linear combinations of the Fourier coefficients.

Gravitational choreographies typically have constraints on the center of mass that leads to relationships of the form

$f\left(t\right)+f(t+T)+f(t-T)=0$

Again using an exponential-type power series expansion for the function, this relationship becomes

$\begin{array}{r}\sum _{k=0}^{\infty}\frac{{c}_{k}}{k!}[{t}^{k}+(t+T{)}^{k}+(t-T{)}^{k}]=0\\ \sum _{k=0}^{\infty}\frac{{c}_{k}}{k!}[{t}^{k}+\sum _{l=0}^{k}\left(\genfrac{}{}{0ex}{}{k}{l}\right)[1-(-1{)}^{k-l}]{T}^{k-l}{t}^{l}]=0\\ \sum _{k=0}^{\infty}\frac{{c}_{k}}{k!}{t}^{k}+\sum _{l=0}^{\infty}\sum _{k=l}^{\infty}\frac{{c}_{k}}{l!(k-l)!}[1-(-1{)}^{k-l}]{T}^{k-l}{t}^{l}=0\\ \sum _{k=0}^{\infty}\frac{{c}_{k}}{k!}{t}^{k}+\sum _{l=0}^{\infty}\sum _{m=0}^{\infty}\frac{{c}_{l+m}}{m!}[1-(-1{)}^{m}]{T}^{m}\frac{{t}^{l}}{l!}=0\end{array}$

Note that the double sum only contains even powers of the period. Repurposing the index, for gravitational choreographies one must have

${c}_{k}+2\sum _{m=0}^{\infty}\frac{{c}_{k+2m}}{\left(2m\right)!}{T}^{2m}=3{c}_{k}+2\sum _{m=1}^{\infty}\frac{{c}_{k+2m}}{\left(2m\right)!}{T}^{2m}=0$

This statement holds separately for the coefficients of even and odd powers of the temporal variable.

If one again assumes that the expansion can be stopped at some large number and solved backward for the coefficients of a general gravitational choreography. In this case it is less easy to see the form of the final result, so Mathematica is useful. The code here is

max = 11; s[k_] := 3 Subscript[c, k] + 2 Sum[Subscript[c, k + 2 m]/(2 m)! T^(2 m), {m, 1, (max - k)/2}] Solve[Table[s[i] == 0, {i, 0, max - 2, 2}], Table[Subscript[c, i], {i, 0, max - 2, 2}]] Solve[Table[s[i] == 0, {i, 1, max - 2, 2}], Table[Subscript[c, i], {i, 1, max - 2, 2}]]

Even and odd powers have the same numerical coefficients matching backwards from highest coefficients.

Picking any high numerator and searching on oeis.org reveals that the numerical coefficients have the same absolute value as A279120, which are the numerators in the expansion of $\frac{1}{1+2cos\sqrt{x}}$ . The denominators are in the next sequence A279121 and have a common factor of three that can be removed by simple multiplication. This is an oddly specific function: what could be its significance for gravitational choreographies?

*Uploaded 2021.04.30 — Updated 2022.11.14*
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