Fourier Series of Polygons Alain Robert The American Mathematical ...

Fourier Series of Polygons
Alain Robert
The American Mathematical Monthly, Vol. 101, No. 5. (May, 1994), pp. 420-428.
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Fri Mar 7 16:45:08 2008

Fourier Series of Polygons
Alain Robert
It is generally admitted (among mathematicians.. .) that the complex form of
Fourier series is the easiest to discuss. It has two main advantages
economical on the notational side
m
zekeik'instead of a, +
- m kr 1
(a, cos kt + b, sin kt),
straightforward for the discussion of absolute convergence
For the visualization of periodic functions however, one usually tends to
separate real and imaginary parts, and draw separately the graphs of these two
periodic functions.
But a complex Fourier series C,, ,ekeik' represents a 27-periodic map
f:R+@
and at least when it is continuous, can be viewed as a closed parametrized curve
t -f(t) E C in the complex plane. In general, this closed curve will have multiple
points (namely, f is not one-to-one). It is our purpose to illustrate this point of
view. In particular, we intend to show that for n 2 2
is the Fourier series of a regular n-gon in the complex plane.
In the first picture, the Fourier series of the pentagon is illustrated: partial sums
Ck~1(5),,k,s5n+l.. .are plotted for n = 1, 2, 3, 4 and 8.
1. THE FUNDAMENTAL FOURIER SERIES AND THE MAIN RESULT. Let
&(R/27rZ) be the space of continuous 27-periodic functions f: R + @. We shall
say that such a function f is a polygon if there is a finite subdivision
0 Ito < . .. < ti < tj+, " ' < t,-, < t, = to + 27r
of [0,271 such that f is affine linear in each subinterval [ti, ti+,[. In this case, the
image of f is a polygonal line in @ with vertices si = f(tj), and f is a parametrization
with constant speed on each side. The set of polygons is obviously a subspace
of B(R/27rZ).
Let us now determine the Fourier series of a polygon f. With the previous
notations, we compute the Fourier coefficients c,( f) given by
FOURIER SERIES OF POLYGONS

Figure 1
by integration by parts when k # 0. For a typical term
In the sum over j, the integrated terms cancel out two by two ( f is continuous and
periodic) and we obtain
With a velocity jump a, = u, - uj-, at time ti, we can write the last relation as
In spite of its appearance, this is a very simple expression indeed. Recall that a
translate rf of a function f E &([W/2rZ!)is a function of the form
(7f)(t) =f(t - a).
As the integral formula shows, the Fourier coefficients of such a translate are
ck(7f)= e-ika~k( f ). We see now that a polygon f is a linear combination of
translates of the basic Fourier series
and a constant term co= cotf 1:
fl(t) =
k#O
eikf/k2
f and co + Ca,
fl(t - t,)
have the same Fourier coefficients when a, = -a,/(2r).
Conversely, if we consider a linear combination Cjajrjflof translates of f,
where Cjaj= 0, we can construct a parametrized polygon in @ with this Fourier
series. We can indeed compute the aj = -2raj and the t,, hence we can compute
19941 FOURIER SERIES OF POLYGONS 421
j

the velocities v, starting from u, as follows
u, = u, + a,, u2 = uo + (ul + u2), etc.
We have u, = u, + (ul + . . . +an) = U,
line will be closed when
by assumption. Moreover, the polygonal
(t, -t,)u,+ ... +(t, -t,-,)u,-, =O.
But t, = 27 + to and reordering terms, we get
which fixes the value of v, = v, = u, -,+ u,, = v, -,+ u, = a, + C, ., < , t , a-./2T. ,
Hence the velocities are uniquely determined in all subintervals [t,, t,+.,[ and the
polygon itself is determined up to a translation. Without loss of generality, we can
now restrict ourselves to centered polygons, namely those in 8, = {f E B([W/~TZ):
co(f > = 01.
Theorem. The space of centered polygons V consists precisely of the linear combinations
Ca, . rj( fl) of translates of
where Ca, = 0.
As we have just seen, the study of Fourier series of polygons can be based on
the basic expansion
fl(t) = x eik'/k2 = 2 x cos kt/k2.
k#O k, 1
Its graph is easily pictured. Since the series of derivatives converges uniformly on
compact subsets of ]O,~T[ by Abel's criterium, it is legitimate to write
f;(t) = x i . eik'/k = -2 x sin kt/k for 0 < t < 27.
k # 0 kr 1
As is known, this is the Fourier series of the 2~-periodic function
From this we infer
t-T
forO

Figure 2
Corollary. The subspace generated by the translates off, is dense in &(,.
Proof: Let f E Since f is uniformly continuous, for each given E > 0, one can
find a subdivision {ti)of [O, 2 ~ such [ that
If(t) -f(tj)ls&/4 foralltjst stj+,.
The polygon g going through the points f(t,) is a uniform approximation of f:
choosing the index j suitably, we can write
If(t) -g(t)I aIf(t) -f(tj)I +If(tj) -g(t)l s&/2.
Now, the mean value of g will also be small
a = ( 1/2a)/*~g(t) dt = (1/2~)/*~(f(t) - g(t)) dt satisfies la1 a &/2.
0 0
The polygon go = g - a is centered, i.e. belongs to 4,and satisfies
If(t) - go(t)I aIf(t) - g(t)l + lal 2 &,
I l f -gall = SupIf(t) -&)I 5 &.
2. FOURIER SERIES OF REGULAR N-GONS. We say that a continuous
2~-periodic function f: R -+ C has a symmetry of order n 2 2 when there is an
nth-root of unity 6 such that
f(t + 27/12) = i .f(t).
We shall only be interested in the case i = e2'"/" in this section (in Sec. 4 below,
we shall give examples of the general case).
Theorem. Let f E &(lW/27Z) be a continuous, 2~-periodic function presenting a
synmetry of order n r 2 in the sense f(t + 2 ~/n)= e2"'/"f(t) (t E R). Then the
Fourier sequence off satisfies
ck(f) = O ifk- lisnotamultipleofn,
ck(f) = (n/2~)/""/f(t)e-'"'dt
0
ifk - 1 isamultipleofn.
Proof: The nth order symmetry simply means that the function
t * e-" . f(t)
is periodic of period 2 ~/n. It can be expanded in a Fourier series according to the
19941 FOURIER SERIES OF POLYGONS 423

system (eimnt), ,,and
rn€Z rn€Z k - I(,)
f(t) = eit C C1 m eirnnt= C Cr m
ei(mn+ l)! =
C ckeikt
In particular, f E 80.Conversely, any exponential eikt where k = 1 mod n has the
n-th order symmetry property required.
In other words, the basic exponentlals (eik'), ,, , constitute a Hilbert basis
for the space of L2-functions with symmetry of order n.
We consider now more particularly the case of regular polygons in C,with
vertices at the nth roots sj of 1. Put i = 6, = e2"'/" so that
Si = e2.rrij/n = i (Osj

hence
sin n-/n
vn = fn(0) . -. .rr/n
We have obtained
sin r/n 2
from which we deduce
We have proved the following result.
Theorem. The Fourier series of the regular n-gon having the sj = e2'j"/" as vertices
-uniformly parametrized-is
Cn fn(t) = Cn
eik'/k2= C, x ei('+'")'/(l+ ln12
k= 1mod n 1€6
with normalization constant Cn = ~in~(r/n)/(r/n)~.
Corollary. We have
In particular for n = 2, we infer
C 1/k2 = (1/2) C 1/(1 + 2112 = *rr2/8.
koddr 1
IEZ
Let us observe at this point that C, < 1 tends monotonously to 1 for n + a and
fn + f = ei' (uniform parametrization of the circle) uniformly for n -+ a.
3. QUADRATIC SPACE V. The vector space V carries a quite natural quadratic
form. Here it is.
Definition. For f E V,we define Q( f ) = in-(f 'I f) = n-((l/i)f 'I f ).
Theorem. The function Q takes real values only and defines a non-degenerate real
quadratic form on V. Moreover
1. Q(f)= rxklck(f) I 2 ,
2. When f = d II is the (positively oriented) boundary of a polygonal piece
II c C,Q(f ) = S = Area(II).
19941 FOURIER SERIES OF POLYGONS
425

Proof: 1results from the Parseval formula since the Fourier coefficients of (l/i) f'
are the kck( f ). To prove 2, we use Stokes' theorem
2Tr -
Q( f) = in-( f'lf) = ia/(2a)/ ft(t)f(t) dt = i/2$ zdi
o
an
= (i/2)// dz A di = (i/2)// (dl + idy) A (dl- idy)
n
n
Corollary. Ck,, , .l/k3
= Q( fn)/n- = ((7i-/n)/sin ~/n))~ cos a/n.
Proof: The area of the regular n-gon inscribed in a circle of radius fn(0) is indeed
S = n . f:(0)
as is easily seen.
. cos 7i-/n . sin ~ / n
4. EXAMPLES. Fix the integer n > 1 and choose another integer 1 Ia < n.
Then
z eik'/k2 = (l/n) z lajf,(t -j . 27i-/n)
k-a mod n
Osj

This is a star with n vertices. The choice
A=n-6, B = n + 6
leads to the classical stars where the sides [s,,s,+~]and [s,,,, sj+,] are on the
same line when j is even. We give two more examples with MATHEMATICA~~.
fCt-I:= SumCExpC(1+20k) I t l / (1 +20k) - 2,Ck,-2,2>1
gCt-I:= SumCExpC(11+20k) I t1/ (11 +20k) - 2,Ck,-2,211
hCt-I:= f C t l - 6gCtl
uCt-l:=ReChCtll
vCt-I:= ImChCtl1
ParametricPLotCCuCtl,vCtl>,Ct,l,l+2 Pi>,AspectRatio->Automatic1
Figure 4
Figure 5
FOURIER SERIES OF POLYGONS

5. CONCLUDING COMMENTS. The interested reader may continue by programming
the Fourier series of a cross (degenerate star with four sides) and check
that Q(f = 0 for these functions.
The formula Q(f) = T C lc,( ~ f) 1
is reminiscent of the Sylvester decomposition
of the quadratic form Q. Indeed, Q is defined on the Soboleu space
' = ( ( c ) 1 ~ 1 ck12 <
which decomposes as a direct sum H'/~= H+@C $ H- with
H+= {(c,) : c, = 0 for k I 0}, H- similarly defined.
With respect to this decomposition, Q = Q+$ 0 $ - Q-, where Q +- are positive
non degenerate on H,.
In Fig. 3, Q(f) counts twice the area of the portion of the star which is the
inner pentagon. In general, when f is self intersecting, one could introduce its
algebraic area as being Q(f ). Geometrically, in the computation of this area, each
connected component of C - Image( f) is affected by a rational integer, the index
off with respect to the points in this component. Observe that the method of
determining f(0) also applies to stars since we still have a constant speed
parametrization:
f(0) = 11f 'll;
= v2 = (length/2r12.
Several generalizations of the preceding considerations can be developed:
.8' curves in C which are piecewise quadratic,
polygons in Cn or Rn (in particular in R3!).
Question. Replacing the circle group by the rotation group, is it possible to
describe in a simple way the Platonic solids by means of spherical harmonics?
Finally, let me thank R. S. Strichartz whose comments helped me to write the final version of this
paper.
REFERENCES
1. R. Edwards, Fourier Series, Springer-Verlag GTM 64, 1979.
2. A. Robert, Advanced Calculus for Users, North-Holland, 1989.
Institut de Mathkmatiques
Universitk de Neuchatel
Chantemerle 20 -
CH-2007 Neuchatel SWITZERLAND
alain.robert@maths,unine.ch
FOURIER SERIES OF POLYGONS