Uniform Convergence Behavior of the Bernoulli Polynomials

Home , Bernoulli polynomials

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∞ tk text B (x) = k k! et 1 Xk=0 − One use of the Bernoulli polynomials is to express sums of similar powers:

N (N + 1) km = B (N + 1) B (1) m+1 − m+1 Xk=0 In fact Ocneanu’s original question as posed in class was to consider the left side of (1), consider it as a polynomial in a complex variable and plot its roots. This his question was about the sums of like power and not necessarily Bernoulli polynomials. However, we simplify the problem silghly by not subtracting oﬀ terms.

2 History of the Problem

Given a sequence of polynomials, how do we know the roots accumulate on any curve? Futhermore, how do we compute this curve? In the October 2005 edition the American Mathemaitcal Monthly I saw this plot:

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k Figure 2: The roots of 100 (100z) = 0 ﬁt the curve similar to zez−1 =1 k=0 k! | | P

The paper, [Zem], was by Stephen Zemyan of Pennsylvania State University, Mont Alto. He wrote on the roots of the truncated exponential series. If you cut the taylor series of ex after N terms you get a polynomial

N−1 xk P (x)= N k! Xk=0 As N the roots go to since ex has no complex roots, but if you rescale by a factor of N, the roots → ∞ ∞ z−1 of PN (Nx) approach the curve ze = 1. This was proven by Gabor Szeg¨oin 1928 appearing in his famous problem book [Sze2] and in| a seperate| paper [Sze]. Basically Szeg¨o’s result has to do with the gamma function that N−1 xk x = zN ezdz k! Z Xk=0 0 However, this work seemed to come out of nowhere and I went in search of more systemattic approach to this problem.

2 Around 1935, Jean Dieudonnéproved the same result about the rescaled roots of the truncated exponential series. He shows that: n![enz P (nz)] 1 − n = [1 + λ (z)] (2) zn 1 z n − where λn(z) 0 uniformly on the z : z < 1, z 1 > ǫ . Bounding away from 1 is essential because of the pole in→ the asymptotics. In his{ paper| | Dieudonnésays| − | } that his solution is a technique which works for general classes of functions. Having read Dieudonné’s proof, the author will try this technique on the Bernoulli polynomials.

Unlike the exponential problem proved by Szegöand Dieudonne, the Bernoulli polynomials are not the truncations of any particular infinite series. The coeffients seem to “jump” around and diverge. There was no obvious way to get around this problem In 1987, Karl Dilcher showed the following formulas using properties of the Riemann Zeta funtion: Theorem 2.1. As in [Dil2], let

k k z2j z2j+1 T (z)= ( 1)j and T (z)= ( 1)j 2k − (2j)! 2k+1 − (2j + 1)! Xj=0 Xj=0 be the truncated sine and cosine series, for n N. Then ∈ (2π)n 1 ( 1)n B z + T (z) < 2−ne4π|z| − 2n! n 2 − n

Corollary 2.1. On compact subsets of C we get uniform convergence:

cos(2πz) if n is even (2π)n lim Bn(z)  n→∞ 2n! →  sin (2πz) if n is odd  This showed the Bernoulli polynomials were intimately related to the basic trigonometric functions and gave a quantative error between the limits of the respective truncations. However, if we substitute nz as an argument instead of z, we get uniform convergence on a small disc where e4π|z|/2 < 1 or z < ln 2/4π = | | 1 D 0.055 < 0.159 = 1/2π. This means we cannot show these functions are unformly close on the full disc 2π on which the roots lie. However, we do get a compelling picture

0.4

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Figure 3: The roots of the Bernoulli polynomials compared to the locus of roots of truncated cosine.

3 The Real Case

Now we will show the real roots of Bn(nz) are uniformly spaced on an interval [ 1/2πne, 1/2πne]. Since Karl Dilcher related Bernoulli polynomials to trigonometric series and exponential− functions, Fourier series may be the appropriate tool. Let’s recall a result of Hurwitz:

3 Lemma 3.1. The Fourier series of the Bernoulli polynomials are given by:

n! e2πikx B ( x )= (3) n { } −(2πi)n kn |kX|>0

Notice this Fourier series has period 1 so we use the fractional part sign x . { } Proof. The coeﬃcients can be found all at once using the generating function:

1 text t ∞ t k e−2πinxdx = = Z et 1 t 2πin − 2πin 0 − − Xk=1 For each individual k we get a Fourier series like the one above.

What are the asymptotics of this periodized version of the Bernoulli polynomials? Lemma 3.2. The Bernoulli polynomials B ( x ) converge uniformly to n { } (2πi)n e2πix + e−2πix − 2n! × 2 in the unit disc D. Proof. Consider the Fourier series from the previous lemma:

n! e2πikx B ( x )= n { } −(2πi)n kn |kX|>0

Every term except k = 1 goes to zero as n goes to inﬁnity. This gives us pointwise convergence. Is this convergence uniform for ±x D? Yes ∈ e2πikx 1 ∞ dx C′ < 2 < C = kn kn Z xn 2n |kX|≥2 kX≥2 2

Now we have estimated the periodized version of the Bernoulli polynomials Bn( z ) but how do we estimate B (mx) B ( mx )? { } n − n { } Lemma 3.3. As n approaches inﬁnity B (mx) B ( mx ) approaches n − n { } x nmn−1 mtn−1dt Z0 This convergence is uniform in the unit interval [0, 1 ǫ] for any ǫ> 0. − Proof. Let’s do one at a time: B (x + 1) B (x)= nxn−1. Then n − n ⌊nx⌋−1 B (mx) B ( mx ) = B ( mx + k + 1) B ( mx + k n − n { } n { } − n { } } Xk=0 ⌊nx⌋−1 = n ( mx + k)n−1 { } Xk=0 ⌊nx⌋−1 1 mx + k n−1 = nmn−1 m { } × n m Xk=0 x nmn−1 mtn−1dt ≈ × Z0

4 For x [0, 1], does the Riemann sum above convergence uniformly in D to the following integral? ∈ x mtn−1dt = xn Z0 Since xm is monotone in x on [0, 1], the difference between the upper and lower Riemann sums is just the difference between the first term of the upper sum and the last term of the lower sum. The Riemann sum above is the lower sum. The upper sum is defined between k = 1 and k = nx . Their difference is just xn−1 [ mx /m]n−1 < xn−1. This means convergence will be uniform for x ⌊ [0,⌋1 ǫ] for ǫ> 0. − { } ∈ − How let’s take another look at (3):

∞ n! e2πinkx B ( x )= n { } −(2πi)n kn k=X−∞

We need an estimate of Stirling’s formula with errors. By considering the integral

∞ f(t)= et(ln x−x)dx Z0 it is possible using Laplace’s method to get Stirling’s esimate:

1 (1) n! = (n/e)n√2πn 1+ + O 12n n2

Thus

n√ 1 O(1) ∞ (n/e) 2πn 1+ 12n + n2 e2πinkx B ( x ) = h i n { } − (2πi)n kn k=X−∞ [n (n1/2n)]n O (e2πinx e−2πinx + (2−n)) ≈ − (2πie)n − O

2πinx −2πinx Asymptotically the roots of Bn(nx)= Bn( nx )+(Bn(nx) Bn( nx )) are the roots of e e = 0. This is basically the stategy we will use in{ passing} to the com− plex{ roots.} −

4 The Complex Case

The previous section’s result can be found in [Ves] from 1999. In this section, while the proof is new, the result was anticipated by Karl Dilcher. Let’s start writing the simple generating function relation

text te(x+yi)t eyit = et 1 × et 1 − − and rewriting it in terms of Bernoulli polynomials:

n n B (x + yi)= B (x)(yi)n−k n k k Xk=0 Let’s deﬁne a functional:

∞ ∞ n tn tn n F (x) eyit F (x) or F (x) F (x)(yi)n−k n n! → n n! n → k k Xk=0 kX=0 kX=0

This functional is linear in generating functions F (x, t) or equivalently, sequences Fn. In particular, we evaluate this functional for F (x)= B ( nx ) and F (x)= B (nx) B ( nx ). n n { } n n − n { }

5 Lemma 4.1. As n approaches inﬁnity, B (nx + niy) B ( nx + iny) approaches n − n { } 1 nn n(x + yi)n−1dx Z0 Proof.

n n ⌊nx⌋−1 n n 1 nx + k n−1 (B (nx) B ( nx ))(yi)n−k = nn n { } (yi)n−k k k − k { } k × n n kX=0 kX=0 Xk=0 ⌊nx⌋−1 1 nx + k n−1 = nn n { } + yi × n n kX=0 This cooresponds to a Riemann integral:

1 n(x + yi)n−1dx Z0 and since this function is monotone in x we get the same error bound as before:

nx n−1 n(x + yi)n−1 n { } + yi − n

As long as x + yi < 1 ǫ we get uniform convergence for ǫ> 0. | | − Now let’s give our ﬁnal major estimates of the paper: Lemma 4.2. As n approaches inﬁnity B ( nx + iny) approaches n { } [n (n1/2n(1+1/n))]n O [P (2πny)+ P ( 2πny)] − (2πie)n n n −

Proof. Now we transform Bn( nx ). We already know its Fourier series expansion for real arguments. Let’s use the convolution formula above{ } to get add an imaginary value iny :

n n B ( nx + iny) = B ( nx )(iny)n−m n { } m m { } mX=0 n 1 ∞ yn−m = n! k−me2πiknx − (2πi)m (n m)! mX=0 k=X−∞ − n ∞ n! (2πiy)n−m = k−me2πiknx −(2πi)n (n m)! mX=0 − k=X−∞

Therefore we get the following estimate for Bn(nx + iny). We pull out the k = 1 terms in the Fourier series, which do not decay with m. In the third line, we use our uniform bound| for| the k > 2 tail in Lemma 3.2 to give line four. | |

n! n ( 2πny)n−m ∞ B ( nx + iny) = − k−me2πiknx n { } −(2πi)n (n m)! mX=0 − k=X−∞ n n! ( 2πny)n−m = − e2πinx + ( 1)ne−2πinx + k−me2πiknx −(2πi)n (n m)! − mX=0 − |kX|≥2   n n! ( 2πny)n−m = − e2πinx + ( 1)ne−2πinx + (2−m) −(2πi)n (n m)! − O mX=0 − n! 1 = P (2πny)+ P ( 2πny)+ (P ( 4πny)) −(2πi)n n n − 2n O n ±

6 1 Having summed all the errors we get a worrisome error which might grow as 2n (Pn( 4πny)) which might O ± 1 ±4πny grow faster than the dominant terms! Indeed our proof would collapse if we estimated this as 2n (e ), 1 O but it so happens that n P (4πn y ) < P (2πn y ). Let’s double-check this: 2 n | | n | | n−1 n−1 1 1 (4πn y )k (2πn y )k P (4πn y )= | | < | | = P (2πn y ) 2n n | | 2n k! k! n | | Xk=0 kX=0

Now we use Dieudonne’s estimate to say that within the unit circle our Taylor series converges to P (2πinz) e2πinz uniformly inside the disc D/2π. n → (2πiz)n n![e2πinz P (2πinz)] = [1 + λ (z)] − n 1 2πiz n − 1 D This is to say the Pn(2πinz) converges to the exponential uniformly on the disc 2π . Also note that [xn(1 + (1)]1/n = x[1 + (1/n)]. O O n! [n (n1/2n(1+1/n))]n B ( nx + iny)= e2πny + e−2πny [1 + (1)] = O e2πny + e−2πny (4) n { } −(2πi)n O − (2πie)n and we use Stirling’s formula to get the above estimate.

Earlier we showed: x n B (nx + niy) B ( nx + iny) = [n(x + iy)]n 1+ n − n { } x + iy

This tells us this equation we are looking for is:

e2πny if y > 0 2πiez = e−2πny if y < 0

5 Acknowledgement

Many thanks to Penn State especially Adrian Ocneanu, Sergei Tabachnikov and Alberto Bressan. Also thanks to Karl Dilcher himself who wrote an entire memoir, [Dil2] on the roots of Bernoulli polynomials in 1987. He was responsive to my e-mails. Also thanks to Stepen Zemyan who showed me how to plot roots of polynomials.

7 References

[Ble] P. Bleher, R. Malisson. “Zeros of Sections of Exponential Sums” arxiv:math-ph/0605066 [Did] Jean Dieudonné. “Sur les zéros des polynomes-sections de ex, Bull. Sci. Math 70:333-351, 1935. [Dil] Karl Dilcher. “Asymptotic behaviour of Bernoulli, Euler, and generalized Bernoulli polynomials”, J. Approx. Theory 49 (1987), 321–330. [Dil2] Karl Dilcher. “Zeros of Bernoulli, generalized Bernoulli and Euler polynomials”, Mem. Amer. Math. Soc. 73 (1988), No. 386. [San] J. Sándor, B. Crstici. “Handbook of Number Theory II”. Kluwer Academic Publishers. Boston, 2004. [Sze] Gabor Szegö, “Uber¨ eine Eigenschaft der Exponentialriehe”, Collected Papers of Gabor Szegö, Vol 1. Richard Askey, ed. Birkhäuser, 1981. Boston. [Sze2] Gabor Szegö, George Polya. “Problems and Theorems in Analysis” Vol 1. D. Aeppli, trans. Springer, 1972. [Ves] A. P. Veselov, J. P. Ward. “On the Real Roots of the Hurwitz zeta-function and Bernoulli Polynomials” J. Math. Anal. Appl. 2005 [Zem] Stephen Zemyan “On the Zeros of the Nth Parital Sum of the Exponential Series. Amer. Math. Monthly 112(10):891-909, 2005.