A DEVELOPMENT OF EULER NUMBERS

Joseph B. Dence

The Euler polynomials, {E_{n}(x)}_{n=0}^{∞}, are the unique polynomials with the property

x^{n} = \frac{E_{n}(x + 1) + E_{n}(x)}{2}, x ∈ ℝ.

This definition leads to the differential equation (E′_{n} − nE_{n−1})(x) = 0, and hence, to a complete determination of the E_{n}(x)’s. It is known that when n is odd, 1/2 is a zero of E_{n}(x) [7]. The sequence of Euler numbers, {E_{2n}}_{n=0}^{∞}, is then defined by

E_{2n} = 2^{2n}E_{2n}(1/2).

The first 8 Euler numbers are given in Table 1 [5]. In general, the E_{n}’s increase continuously and rapidly in magnitude, and their algebraic signs alternate.

It is of some practical interest to have a closed form for the Euler numbers. The closely related Bernoulli numbers, for example, can each be represented by an infinite series of reciprocals of various integral powers of the positive integers

B_{2n} = (−1)^{n−1} \frac{2(2n)!}{(2π)^{2n}} ∑_{k=1}^{∞} 1/k^{2n}.

\begin{matrix}n & E_{2n} \\ 0 & 1 \\ 1 & −1 \\ 2 & 5 \\ 3 & −61 \\ 4 & 1385 \\ 5 & −50, 521 \\ 6 & 2, 702, 765 \\ 7 & −199, 360, 981 \end{matrix}

Table 1. The First 8 Euler Numbers.

Methods for proving this commonly employ non–elementary facts from analysis [3]; for example, one such proof begins with the infinite product formula for sin x

\frac{sin x}{x} = ∏_{k=1}^{∞}[1 − (x^{2}/k^{2}π^{2})], |x| < ∞.

There are similar non–elementary methods for deriving analogous closed forms for the Euler numbers. In this paper we show how only calculus can be used to obtain infinite series representations of the Euler numbers. Our treatment is motivated by Berndt’s work on the Bernoulli numbers [4].

1. The Even–Indexed Euler Polynomials. We shall need one basic fact about the polynomials E_{2n}(x). From the definition of the Euler polynomials, one can obtain a generating function

\frac{2e^{xt}}{e^{t} + 1} = ∑_{n=0}^{∞} E_{n}(x)\frac{t^{n}}{n!} .

In this, replacing t by −t and setting x = 0, we obtain

∑_{n=0}^{∞} E_{n}(0)\frac{(−t)^{n}}{n!} = \frac{2}{e^{−t} + 1} = \frac{2e^{t}}{e^{t} + 1} = ∑_{n=0}^{∞} E_{n}(1)\frac{t^{n}}{n!} .

Within the interval of convergence, we can subtract the two series in the last line term–by–term to give

∑_{n=0}^{∞} \frac{t^{n}}{n!} (E_{n}(1) − E_{n}(0)(−1)^{n}) = 0.

As t is arbitrary, we obtain for even indices E_{2n}(1) − E_{2n}(0) = 0. But E_{2n}(1 + x) + E_{2n}(0) = 2x^{2n} and for x = 0 this gives E_{2n}(1) + E_{2n}(0) = 0. Combination with the previous equation yields, finally, E_{2n}(0) = E_{2n}(1) = 0.

2. Two Lemmas. The general idea of the proof is to concoct a special function of E_{2n}(x) and E_{2n}, integrate this function together with sin(πkx) (k = integer), and then sum over some suitable subset of the k’s. The next two lemmas are required.

\underline{Lemma 1}. If j is a positive integer and cos θ ≠ 0, then

S = sin θ − sin 3θ + sin 5θ − ⋯ + sin(4j + 1)θ = \frac{sin(4j + 2)θ}{2 cos θ} .

\underline{Proof}.

S = Im(e^{iθ} − e^{3iθ} + e^{5iθ} − ⋯ + e^{(4j+1)iθ})

= Im(\frac{e^{iθ}(1 − (−e^{2iθ})^{2j+1})}{1 − (−e^{2iθ})})

= Im(\frac{e^{(4j+3)iθ} + e^{iθ}}{e^{2iθ} + 1} · \frac{e^{−2iθ} + 1}{e^{−2iθ} + 1})

= Im(\frac{e^{(4j+1)iθ} + e^{−iθ} + e^{(4j+3)iθ} + e^{iθ}}{2 + 2 cos 2θ} )

= \frac{sin(4j + 1)θ − sin θ + sin(4j + 3)θ + sin θ}{2(1 + cos 2θ)}

= \frac{2 sin(4j + 2)θ cos θ}{4 cos^{2} θ}

= \frac{sin(4j + 2)θ}{2 cos θ} .

The next lemma is a special case of a more general theorem known in the literature as the Riemann–Lebesgue Lemma [1, 2, 11].

\underline{Lemma 2}. If F (θ) is twice continuously differentiable on [a, b], F (a) = 0, and cos θ has no zero in (a, b], then

lim_{N→∞} ∫_{a}^{b} F (θ)\frac{sin Nθ}{cos θ} dθ = 0.

\underline{Proof}. Let g(θ) = F (θ)/ cos θ. Then

(∗)  ∫_{a}^{b} g(θ) sin Nθdθ = −g(θ)\frac{cos Nθ}{N} |_{a}^{b} + ∫_{a}^{b} g′(θ)\frac{cos Nθ}{N} dθ.

If cos θ ≠ 0 at θ = a, then g(θ) is finite there. By hypothesis, g(θ) is finite at θ = b also, so the integrated term in (∗) approaches 0 as N → ∞.

If cos a = 0, then by L’Hôpital’s Rule

lim_{θ→a^{+}} g(θ) = lim_{θ→a^{+}} \frac{F′(θ)}{− sin θ} ,

provided the latter limit exists. But this limit does exist because F (θ) is continuously differentiable at θ = a and sin a ≠ 0. Hence, the integrated term in (∗) again approaches 0 as N → ∞.

Again, if cos θ ≠ 0 at θ = a, then

g′(θ) = \frac{cos θF′(θ) + sin θF (θ)}{cos^{2} θ}

is defined there and the integrand on the right–hand side of (∗) is continuous on [a, b]. Hence, the integral approaches 0 as N → ∞.

On the other hand, if cos a = 0, we have

lim_{θ→a^{+}} g′(θ) = lim_{θ→a^{+}} \frac{− sin θF′(θ) + cos θF′′(θ) + cos θF (θ) + sin θF′(θ)}{−2 cos θ sin θ}

= lim_{θ→a^{+}} \frac{F (θ) + F′′(θ)}{−2 sin θ}

= −\frac{1}{2} \frac{F′′(a)}{sin a},

because F (θ) is twice continuously differentiable on [a, b] and sin a ≠ 0. So the integrand in (∗) is again continuous on [a, b] and the integral approaches 0 as N → ∞.

3. The Main Theorem. To begin the proof of the main result, we define the following function

G(x) = \frac{2^{2n}E_{2n}(x)}{E_{2n}} − 1,

and we choose the interval of integration to be 1/2 ≤ x ≤ 1. We note that G(1/2) = 0, and also that G(1) = −1, in view of the discussion in Section 1. We also recall (from the opening remarks) that E′_{n}(x) = nE_{n−1}(x).

\underline{Theorem 1}. If n is a positive integer, then

E_{2n} = 2(−1)^{n}(2/π)^{2n+1}(2n)! ∑_{j=0}^{∞} \frac{(−1)^{j}}{(2j + 1)^{2n+1}} .

\underline{Proof}. Let K be an odd integer; integration by parts gives

∫_{1/2}^{1} G(x) sin πKxdx = −\frac{cos πKx}{πK}G(x)|_{1/2}^{1} + \frac{2^{2n}}{E_{2n}} ∫_{1/2}^{1} \frac{cos πKx}{πK}E′_{2n}(x)dx

= \frac{−1}{πK} + \frac{2^{2n}(2n)}{πKE_{2n}} ∫_{1/2}^{1} E_{2n−1}(x) cos πKxdx

= \frac{−1}{πK} + \frac{2^{2n}(2n)}{πKE_{2n}}(\frac{sin πKx}{πK}E_{2n−1}(x)|_{1/2}^{1} − ∫_{1/2}^{1} \frac{sin πKx}{πK}E′_{2n−1}(x)dx)

= \frac{−1}{πK} −\frac{2^{2n}(2n)(2n − 1)}{(πK)(πK)E_{2n}} ∫_{1/2}^{1} E_{2n−2}(x) sin πKxdx.

After 2n − 2 more integrations by parts, we obtain

\begin{aligned}∫_{1/2}^{1} G(x) sin πKxdx & = \frac{−1}{πK} + \frac{(−1)^{n}2^{2n}(2n)!}{(πK)^{2n}E_{2n}} ∫_{1/2}^{1} E_{0}(x) sin πKxdx \\ & = \frac{−1}{πK} + \frac{(−1)^{n}2^{2n}(2n)!}{(πK)^{2n+1}E_{2n}}, \end{aligned}

because E_{0}(x) = 1.

Now let K = K(k) = (−1)^{(k−1)/2}k, and sum both sides of the last equation over all the odd k’s from k = 1 to k = 4j + 1

∫_{1/2}^{1} G(x)( ∑_{{k=1 \atop k=odd}}^{4j+1} sin πKx)dx = −\frac{1}{π} ∑_{{k=1 \atop k=odd}}^{4j+1} \frac{1}{K} + \frac{(−1)^{n}2^{2n}(2n)!}{E_{2n}} ∑_{{k=1 \atop k=odd}}^{4j+1} \frac{1}{(πK)^{2n+1}} .

Application of Lemma 1 to this yields

\frac{1}{2} ∫_{1/2}^{1} G(x)\frac{sin(4j + 2)πx}{cos πx}dx = −\frac{1}{π} ∑_{{k=1 \atop k=odd}}^{4j+1} \frac{1}{K} + \frac{(−1)^{n}2^{2n}(2n)!}{E_{2n}} ∑_{{k=1 \atop k=odd}}^{4j+1} \frac{1}{(πK)^{2n+1}} .

On the left–hand side, let θ = πx, F (θ) = G(x), N = 4j + 2; the integral becomes

\frac{1}{2π} ∫_{π/2}^{π} F (θ)\frac{sin Nθ}{cos θ} dθ.

The integrand satisfies all of the hypotheses in Lemma 2, so upon passage to the limit j → ∞, we obtain

0 = −\frac{1}{π} ∑_{{k=1 \atop k=odd}}^{∞} \frac{1}{K} + \frac{(−1)^{n}2^{2n}(2n)!}{E_{2n}} ∑_{{k=1 \atop k=odd}}^{∞} \frac{1}{(πK)^{2n+1}} .

The first series is just Gregory’s series for Tan^{−1}1 [10]. In the second series, let k = 2j + 1 and replace K by (−1)^{j}(2j + 1). We obtain

0 = −\frac{1}{4} + \frac{(−1)^{n}2^{2n}(2n)!}{π^{2n+1}E_{2n}} ∑_{j=0}^{∞} \frac{(−1)^{j}}{(2j + 1)^{2n+1}}

and the theorem follows by a simple rearrangement.

The theorem permits an asymptotic estimate of the E_{2n}’s and also of their rate of growth.

\underline{Corollary}. As n → ∞, we have |E_{2n}| ∼ 8\sqrt{(n/π)}(4n/(πe))^{2n} and

|E_{2n+2}/E_{2n}| ∼ (4(n + 1)/π)^{2}.

\underline{Proof}. Let Z_{n} = ∑_{j=0}^{∞}(−1)^{j}/(2j + 1)^{2n+1}. Clearly 1 − 1/3^{2n+1} < Z_{n} < 1, so Z_{n} ∼ 1 as n → ∞. From Stirling’s approximation [6, 8, 9] we have (2n)! ∼ \sqrt{2π(2n)}(2n)^{2n}e^{−2n} as n → ∞. Hence, using these asymptotic results in the series given in Theorem 1, we obtain

\begin{aligned}|E_{2n}| & ∼ 2\sqrt{2π(2n)}(2n)^{2n}e^{−2n}(2/π)^{2n+1} \\ & ∼ 8(n/π)^{1/2}(4n/(πe))^{2n}. \end{aligned}

Finally, making use of the result ((n + 1)/n)^{2n} ∼ e^{2}, we have

|E_{2n+2}/E_{2n}| ∼ \sqrt{(n + 1)/n}(4(n + 1)/(πe))^{2n+2}(πe/4n)^{2n}

∼ \frac{16(n + 1)^{2}}{π^{2}}

= (\frac{4(n + 1)}{π})^{2}.

\begin{matrix}n & |E_{2n+2}/E_{2n}| & (4(n + 1)/π)^{2} \\ 1 & 5 & 6.48 \\ 3 & 22.71 & 25.94 \\ 5 & 53.50 & 58.36 \\ 7 & 97.27 & 103.75 \end{matrix}

Table 2. Asymptotic Estimation of the Growth of the Euler Numbers.

Table 2 shows how well the asymptotic formula agrees with the exact values for the ratios |E_{2n+2}/E_{2n}|; at n = 7, for example, the percent error is about 6.7%. At n = 29, the percent error drops to about 1.7%.

Acknowledgment. The author thanks Dr. Thomas P. Dence (Ashland University) for assistance in the preparation of the manuscript.

\underline{References}

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2. T. M. Apostol, Mathematical Analysis, 2nd ed., Addison–Wesley, Reading, 1974, 313.

3. T. M. Apostol, Introduction to Analytic Number Theory, Springer–Verlag, New York, 1976, 266.

4. B. C. Berndt, “Elementary Evaluation of ζ(2n),” Mathematics Magazine, 48 (1975), 148–154.

5. W. H. Beyer (ed.), CRC Standard Mathematical Tables, 27th ed., CRC Press, Boca Raton, 1984, 392, 394.

6. R. C. Buck, Advanced Calculus, 3rd ed., McGraw–Hill, New York, 1978, 300– 301.

7. F. T. Howard, “Roots of the Euler Polynomials,” Pacific Journal of Mathematics, 64 (1976), 181–191.

8. R. Johnsonbaugh, “The Trapezoidal Rule, Stirling’s Formula, and Euler’s Constant,” American Mathematical Monthly, 88 (1981), 696–698.

9. V. Namias, “A Simple Derivation of Stirling’s Asymptotic Series,” American Mathematical Monthly, 93 (1986), 25–29.

10. R. Roy, “The Discovery of the Series Formula for π by Leibniz, Gregory and Nilakantha,” Mathematics Magazine, 63 (1990), 291–306.

11. E. C. Titchmarsh, The Theory of Functions, 2nd ed., Oxford University Press, Oxford, 1939, 403–404.

Joseph B. Dence Department of Chemistry University of Missouri–St. Louis St. Louis, MO 63121

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