Fourier Expansions of Polynomials and Values of Ζ

(September 13, 2011) Fourier expansions of polynomials and values of ζ

Paul Garrett [email protected] http://www.math.umn.edu/egarrett/

1. Fourier expansions of polynomials 2. Sample computations This is an example of the impact of harmonic analysis on number theory. We start from the fact that Fourier series expansions of suﬃciently good functions represent the functions pointwise. More precisely, for present o o purposes, it suﬃces to know that for piecewise-C periodic functions f, at points xo where f is C and has left and right derivatives,

X 2πinxo R 1 −2πinx f(xo) = fb(n) e (where fb(n) = 0 e f(x) dx) n∈Z and it is implicit that the two-sided series converges. Applying this to polynomials restricted to [0, 1] gives a systematic approach to summing series [1] such as

1 1 1 1 1 1 π2 ζ(2) = + + + + + + ... = 12 22 32 42 52 62 6

1 1 1 1 1 1 π4 ζ(4) = + + + + + + ... = 14 24 34 44 54 64 90 P s Small extensions of the same idea evaluate certain more complicated Dirichlet series n≥1 an/n at positive integers of suitable parity. More elementary methods can produce a few results. For example, from the geometric series

1 = 1 − x2 + x4 − x6 + x8 − x10 + ... 1 + x2 integrate to obtain x3 x5 x7 arctan x = x − + − + ... 3 5 7 Letting x = 1 gives Leibniz’ result

1 1 1 1 π 1 − + − + − ... = 3 5 7 9 4 However, elementary methods were powerless against ζ(2).

1. Fourier expansions of polynomials

On the unit interval [0, 1], there are two simple types of functions: polynomials, and exponentials e2πinx. They are very unlike each other, and expressing one in terms of the other produces interesting information.

The sawtooth function

1 f(x) = x − [[x]] − (where [[x]] is the greatest integer ≤ x) 2

[1] Euler summed these series circa 1750, although at ﬁrst he had more a heuristic than proof. Even a heuristic had eluded the Bernoullis, and Euler’s success was one of the things that made a big impression.

1 Paul Garrett: Fourier expansions of polynomials and values of ζ (September 13, 2011)

is a linear polynomial made periodic. The subtraction of the 1/2 conveniently makes the 0th Fourier coeﬃcient 0. All the other Fourier coeﬃcients are readily computed by integration by parts:

Z 1 1 1 e−2πinx 1 Z 1 e−2πinx 1 fb(n) = (x − ) e−2πinx dx = (x − ) − dx = 0 2 2 −2πinx 0 0 −2πinx −2πin

[2] This begins an inductive description of a family of polynomials B`(x), arranged to have easily computed Fourier coeﬃcients,

[1.0.1] Deﬁnition: (Beware: not quite compatible with other normalizations of Bernoulli polynomials)

1 d Z 1 B1(x) = x − Bk(x) = Bk−1(x) Bk(x) dx = 0 2 dx 0

Taking Bk(x) so that its derivative is Bk−1(x) is entirely motivated by wanting an integration by parts to work nicely: Z 1 −2πinx Z 1 −2πinx −2πinx e 1 0 e Bk(x) e dx = [Bk(x) ]0 − Bk(x) dx 0 −2πin 0 −2πinx Z 1 −2πinx Z 1 Z 1 −2πinx 1 1 0 e 1 0 0 e = [Bk(x) ]0 + Bk(x) dx = Bk(x) dx + Bk(x) dx −2πin 0 2πinx −2πin 0 0 2πinx −1 1 −1 = 0 + · = (using R 1 B (x) dx = 0) (2πin)k−1 2πin (2πin)k 0 k−1

0 Since Bk(x) = Bk−1(x), the only new thing to be determined in Bk is its constant, which is completely determined by the integral condition. Thus, this description does unambiguously specify a sequence of polynomials.

[1.0.2] Remark: In computing Fourier coeﬃcients, we think either of the restriction of Bk(x) to [0, 1], or of the periodicized version Bk(x − [[x]]).

[1.0.3] Claim: The Fourier series of Bk(x) is

−1 X e2πinx B (x − [[x]]) ∼ k (2πi)k nk 06=n∈Z

For k > 1, Bk(0) = Bk(1).

[1.0.4] Remark: The equality Bk(0) = Bk(1) for k > 1 assures that the periodicized version Bk(x − [[x]]) of Bk(x) is continuous. Piecewise-polynomial functions have left and right derivatives everywhere, so the Fourier series for Bk(x) with k > 1 will converge to Bk(x − [[x]]) for all x. Thus, for even k, we evaluate ζ(k) in terms of Bk(x): [1.0.5] Corollary:  −2 ζ(2k)  B2k(0) = (for 2k even) (2πi)2k  B2k+1(0) = 0 (for 2k + 1 > 1 odd) (For k odd, in the previous claim the ±n terms cancel.) ///

[2] These polynomials are roughly Bernoulli polynomials. We are not worrying about conventional normalization or indexing, nor other deﬁnitions.

2 Paul Garrett: Fourier expansions of polynomials and values of ζ (September 13, 2011)

Proof: (of claim) First, the matching at endpoints is easy:

Z 1 Bk(1) − Bk(0) = Bk−1(t) dt = 0 (for k > 1) 0

th since, by construction, Bk−1(x) has vanishing 0 Fourier coeﬃcient. Integrating by parts, recapitulating earlier observations,

Z 1 −2πinx 1 Z 1 −2πinx −2πinx e e Bbk(n) = Bk(x) e dx = Bk(x) − Bk−1(x) dx 0 −2πin 0 0 −2πin

B (1) − B (0) −1 1 −1 = k k − · = −2πin (2πin)k−1 −2πin (2πin)k

k−1 by the inductive hypothesis that Bbk−1(n) = −1/(2πin) . ///

[1.0.6] Claim: The polynomials Bk(x) can be characterized by a generating function:

t · etx 1 + t B (x) + t2 B (x) + t3 B (x) + ... = 1 2 3 et − 1

[1.0.7] Remark: It is wise to say that we consider this as a formal power series in x and t. This is not to say that the discussion is a mere heuristic! The formal power series ring R[[x, t]] over a coeﬃcient ring R is a (projective) limit [3] lim [x, t]/In (with ideal I generated by x, t)) n C

[1.0.8] Remark: The latter relation shows that the polynomials Bk(x) have rational coeﬃcients. Rationality also follows from the recursive deﬁnition.

Proof: Let 2 3 f(t, x) = 1 + t B1(x) + t B2(x) + t B3(x) + ...

[3] Recall (!) that a projective limit X of a family ... → X2 → X1 → X0 is an object with maps X → Xi such that the natural diagram of curvy trangles commutes:

X ... , X2 /+ X1 /* X0

and, for all collections Y → Xi with commuting

Y ... , X2 /+ X1 /* X0

there exists a unique Y → X giving commuting

... , + * X X2 6/ X1 h4/ X0 `A {= mmm hhhh A { mm hhhh A {{ mmm hhhh {{ mmmhhhh A {{mmmhhh Y hmhh

The most basic naive-category-theory ideas show that the object X (and maps X → Xi) is unique up to unique isomorphism. This should be very reassuring. Unfortunately, this way of thinking is not as common as it should be, so one may be put oﬀ by this description of formal power series rings. If so, one should correct this weakness, since such descriptions/explanations are essential in higher mathematics.

3 Paul Garrett: Fourier expansions of polynomials and values of ζ (September 13, 2011)

[4] 0 Applying a standard idea, diﬀerentiate in x, use B`(x) = B`−1(x), and see what happens:

∂ f(t, x) = t · 1 + t2 B (x) + t3 B (x) + ... = t · f(t, x) ∂x 1 2

Therefore, [5] tx f(t, x) = C(t) · e (for some C(t) ∈ C[[t]], independent of x) [6] R 1 The conditions 0 Bk(x) dx = 0 give, on one hand,

Z 1 Z 1 Z 1 Z 1 tx X ` C(t) · e dx = (1 + t B1(x) + ...) dx = 1 dt + t B`(x) dx = 1 0 0 0 `≥1 0

On the other hand, Z 1 Z 1 et − 1 C(t) · etx dx = C(t) etx dx = C(t) · 0 0 t Thus, t C(t) = et − 1 Therefore, t · etx 1 + t B (x) + t2 B (x) + ... = 1 2 et − 1 as claimed. ///

[1.0.9] Corollary:

2 ζ(2) 2 ζ(4) 2 ζ(6) 2 ζ(8) t t t2 · + t4 · + t6 · + t8 · + ... = 1 − − (2πi)2 (2πi)4 (2πi)6 (2πi)8 2 et − 1

1 Proof: Subtract 1 + tB1(x) = 1 + t(x − 2 ) from both sides of the identity

t · etx 1 + t B (x) + t2 B (x) + t3 B (x) + ... = 1 2 3 et − 1 evaluate at x = 0, and multiply through by −1, noting that, from above, Bk(0) = 0 for odd k > 1, while 2k B2k(0) = −2ζ(2k)/(2πi) . ///

[4] When we’ve dodged convergence issues by saying formal power series ring, we can’t talk about diﬀerentiation as the usual sort of limit. Instead, deﬁne a derivation D on C[x, t] or C[[x, t]] by requiring that D be a linear map such that Dx = 1 and Dt = 0, and satisfying Leibniz’ rule

D(fg) = Df · g + f · Dg

An induction proves Dxn = nxn−1. A bit of thought shows that the limit-taking version of derivative was not essential. Some of the properties are essential: linearity, Leibniz’ rule, annihilation of constants.

[5] The diﬀerential equation ∂f(t, x)/∂x = t·f(t, x) has the convergent power series solution etx. Further, the obvious induction on coeﬃcients proves that, given C(t) ∈ C[[t]], there is a unique solution f(t, x) with f(t, 0) = C(t).

[6] These deﬁnite integrals can be rewritten as linear functionals on vector spaces of polynomials, again avoiding any genuine limit-taking.

4 Paul Garrett: Fourier expansions of polynomials and values of ζ (September 13, 2011)

[1.0.10] Remark: Note that the above gives no information about the ζ(2k + 1) for positive odd integers 2k + 1. 2. Sample computations

Reasonable eﬃciency allows numerical computation of some small-index polynomials Bk(x).

[2.1] Symmetry/skew-symmetry We discover that it is convenient to express the polynomials B`(x) as 1 m m 1 a sum of monomials (x − 2 ) rather than x . The recursive description, starting with B1(x) = x − 2 , can be written in two steps: use an auxiliary polynomial C`(x) which needs its constant corrected to be B`(x):

Z x Z 1 C`(x) = B`−1(t) dt B`(x) = C`(x) − C`(t) dt 1 2 0

P 1 n Thus, with B`−1(x) = n cn(x − 2 ) ,

n+2 X cn X cn X cn 1 − (−1) C (x) = (x − 1 )n+1 B (x) = (x − 1 )n+1 − ` n + 1 2 ` n + 1 2 (n + 1)(n + 2) 2n+2 n n n

Looking at the beginning of the numerical computation below, we discover that when expressed in monomials 1 in x − 2 , polynomials B`(x) apparently have only odd-degree monomials for ` odd, and only even-degree monomials for ` even. That is, B`(x) apparently has the symmetry/skew-symmetry

` 1 B`(1 − x) = (−1) · B`(x) (skew/symmetry about the point x = 2 )

Indeed, the constant-term correction vanishes when B`−1(x) consists entirely of even-degree monomials in 1 x − 2 . This gives an induction in ` proving the skew/symmetric property.

1 [2.2] Rewritten recursive description Expressing B`−1(x) as a sum of monomials in x − 2 , say P 1 n B`−1(x) = cn (x − 2 ) , the odd-even property reduces the complexity:

 cn X cn 1 P (x − 1 )n+1 − · (` even)  n n + 1 2 (n + 1)(n + 2) 2n+1 B (x) = n (for B (x) = P c (x − 1 )n) ` c `−1 n 2  P n (x − 1 )n+1 (` odd) n n + 1 2

1 [2.3] Values B2`( 2 ) We saw above that

(−1)`+1ζ(2`) B (0) = 2` 22`−1 π2`

1 1 Expressing B2`(x) in powers of x − 2 makes evaluation at 2 simpler. Thus, rearrange

X −eπin (−1)` X (−1)n+1 B ( 1 ) = = 2` 2 (2πin)2` 22`−1 π2` n2` n6=0 n≥1

In the region of convergence

X (−1)n+1 1 1 = ζ(s) − 2 · · ζ(s) = 1 − · ζ(s) ns 2s 2s−1 n≥1

5 Paul Garrett: Fourier expansions of polynomials and values of ζ (September 13, 2011)

Thus, (−1)` 1 B ( 1 ) = 1 − · ζ(2`) 2` 2 22`−1 π2` 22`−1

[2.4] Numerical determination of B`, ` ≤ 6 Numerical computation of the ﬁrst few B`(x) is easy: 1 listing coeﬃcients of monomials of in x − 2 in ascending order, starting with B1(x),

0 E 1 E EE EE EE EE EE EE EE EE 1 E" E" 1 − 24 0 2 C CC EE CC CC EE CC CC EE CC CC EE C! CC EE 0 − 1 ! 0 " 1 C 24 E 6 E CC CC EE EE CC CC EE EE CC CC EE EE C CC EE EE CC ! EE E" 7 ! − 1 " 1 6!·8 0 C 48 0 E 24 CC CC EE EE CC CC CC EE EE CC C C E EE CC CC CC EE EE C C! CC E" EE C! 7 ! − 1 " 1 0 C 6!·8 0 E 144 0 C 120 CC C EE E CC A C CC EE EE C AA CC CC EE EE CC A CC CC E EE CC AA C C! EE E" C A −31 C! 7 E" 1 C! 1 8!·12 0 6!·16 0 − 576 0 720 where, as above, the constant terms for even-degree cases are determined by

X cn 1 constant coeﬃcient for B (x) = − · (for B (x) = P c (x − 1 )n) 2k (n + 1)(n + 2) 2n+1 2k−1 n 2 n  1 1 1 − · = − (going from ` = 1 to ` = 2)  (1 + 1)(1 + 2) 21+1 24 

 1 1 1 1 7  · − · = (going from ` = 3 to ` = 4) 24 · (1 + 1)(1 + 2) 21+1 6 · (3 + 1)(3 + 2) 23+1 6! · 8 and 7 1 1 1 1 1 − · + · − · 6! · 8 · (1 + 1)(1 + 2) 21+1 144 · (3 + 1)(3 + 2) 23+1 120 · (5 + 1)(5 + 2) 25+1 7 1 1 −49 + 21 − 3 = − + − = 6! · 3 · 64 6! · 64 7! · 64 7! · 3 · 64 31 = − (going from ` = 5 to ` = 6) 8! · 24

[2.5] Values of ζ Putting this together,

1 1 1 ζ(2) − = B ( 1 ) = − 1 − · ζ(2) = − 24 2 2 2 π2 2 4 · π2 which gives π2 ζ(2) = 6 6 Paul Garrett: Fourier expansions of polynomials and values of ζ (September 13, 2011)

Next, 7 (−1)2 1 7 · ζ(4) = B ( 1 ) = 1 − · ζ(4) = 6! · 8 4 2 22·2−1 π2·2 22·2−1 64 · π4 which gives π4 ζ(4) = 90 Finally, 31 (−1)3 1 − = B ( 1 ) = 1 − · ζ(6) 8! · 24 6 2 22·3−1 π2·3 22·3−1 which gives π6 π6 ζ(6) = = 33 · 5 · 7 945

All this was known long ago.