# A Research on a Certain Family of Numbers and Polynomials Related to Stirling Numbers, Central Factorial Numbers, and Euler Numbers

Authors: J. Y. Kang; C. S. Ryoo
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

Recently, many mathematicians have studied different kinds of the Euler, Bernoulli, and Genocchi numbers and polynomials. In this paper, we give another definition of polynomialsŨn(x). We observe an interesting phenomenon of “scattering” of the zeros of the polynomials Ũn(x) in complex plane. We find out some identities and properties related to polynomials Ũn(x). Finally, we also derive interesting relations between polynomials Ũn(x), Stirling numbers, central factorial numbers, and Euler numbers.


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## Body

## 1. Introduction

Recently, many mathematicians have studied in the areas of the Bernoulli numbers and polynomials, Euler numbers and polynomials, Genocchi numbers and polynomials, Stirling numbers, and central factorial numbers (see [1–17]). These numbers and polynomials possess many interesting properties and are arising in many areas of mathematics and physics. In this paper, we give another definition of polynomials U~n(x). We obtain some interesting identities and properties related to polynomials U~n(x). In order to study the polynomials U~n(x), we must understand the structure of the polynomials U~n(x). Therefore, using computer, a realistic study for the polynomials U~n(x) is very interesting. It is the aim of this paper to observe an interesting phenomenon of “scattering” of the zeros of the polynomials U~n(x) in complex plane.The Stirling numbers of the first kinds(n,k) define that
(1)∑k=0ns(n,k)xk=x(x-1)(x-2)⋯(x-n+1).
The generating function of (1) is as follows:(2)(log(1+x))k=k!∑n=k∞s(n,k)xnn!.
From (1) and (2), we are aware of some properties of the Stirling numbers of the first kind s(n,k) as follows (see [1, 2, 15]):
(3)s(n,k)=s(n-1,k-1)-(n-1)s(n-1,k),
with
(4)s(n,0)=0(n∈ℕ),s(n,n)=1(n∈ℤ+=ℕ∪0),s(n,1)=(-1)n-1(n-1)!(n∈ℕ),s(n,k)=0(k>nork<0).We usually define the central factorial numbersT(n,k) as the following expansion formula (see [1, 7]):
(5)∑k=0nT(n,k)x(x-12)(x-22)⋯(x-(k-1)2)=xn.
The generating function of (5) is as follows:
(6)(ex+e-x-2)k=(2k)!∑n=k∞T(n,k)x2n(2n)!.
By using (5) and (6), we are aware of some properties of the central factorial numbers T(n,k) as follows:
(7)T(0,0)=1,T(n,0)=0(n∈ℕ),T(n,1)=1(n∈ℕ),T(n,k)=T(n-1,k-1)+k2T(n-1,k),
with
(8)T(n,2)=14(4n-1-1),T(n,3)=9n360-4n60+124(n∈ℕ).
The Euler numbers En are defined by the generating function:
(9)2et+1=∑n=0∞Entnn!.
We introduce the Euler polynomials En(x) as follows:
(10)(2et+1)ext=∑n=0∞En(x)tnn!.Zhi-Hong Sun introduces the sequence{Un} similar to Euler numbers as follows:
(11)U0=1,Un=-2∑k=1[n/2](n2k)Un-2k(n≥1),
where [x] is the greatest integer not exceeding x (see [4, 5]).The outline of this paper is as follows. In Section2, this paper is to define polynomials U~n(x). In Section 3, we describe the beautiful zeros of the polynomials U~n(x) using a numerical investigation. We investigate the roots of the polynomials U~n(x). Also we carried out computer experiments for demonstrating a remarkably regular structure of the complex roots of the polynomials U~n(x). In Section 4, we derive some special relations of polynomials U~n(x) and Euler numbers.


## 2. Some Properties Involving a Certain Family of Numbers and Polynomials

In this section, we introduce the polynomialsU~n(x) and investigate some interesting properties and identities which are related to polynomials U~n(x). We also try to find relations between polynomials U~n(x), Stirling numbers s(n,k), and central factorial numbers T(n,k).Definition 1.
Forx∈ℝ or ℂ, the polynomials U~n(x) are defined by
(12)∑n=0∞U~n(x)tnn!=(1et+e-t-1)xext.
From Definition 1, we have Theorem 2.Theorem 2.
Letn,k,l∈ℤ+ and n≥k-l-1. Then one obtains
(13)1U~2n(x)=∑k=02n∑l=0[k/2](2n2l)(-1)k-2lU~2n+1(x)=×∑j=k-2ln-l(2j)!j!T(n-l,j)s(j,k-2l)xk,U~2n+1(x)=∑k=12n+1∑l=0[k-1/2](2n+12l-1)(-1)k-(2l+1)U~2n+1(x)=×∑j=k-(2l+1)n-l(2j)!j!T(n-l,j)s(j,k-(2l+1))xk,
where [x] is the greatest integer not exceeding x and ℤ+=ℕ∪{0}.Proof.
By Definition1, (2), and (6), we have
(14)∑n=0∞U~n(x)tnn!=(1et+e-t-1)xetx=∑j=0∞(x+j-1j)(-1)j(et+e-t-2)jetx=∑j=0∞(-1)j(x+j-1j)(2j)!∑n=j∞T(n,j)t2n(2n)!etx=∑n=0∞∑j=0n(-1)j(x+j-1j)(2j)!T(n,j)t2n(2n)!etx=∑n=0∞∑j=0n∑k=0j(-1)ks(j,k)xk(2j)!j!T(n,j)t2n(2n)!∑n=0∞xntnn!.
From now on, we have to consider odd terms and even terms by using Cauchy product. So, we get to generating terms by dividing the odd terms and the even terms, respectively,
(15)∑n=0∞U~n(x)tnn!=∑n=0∞U~2n(x)t2n(2n)!+∑n=0∞U~2n+1(x)t2n+1(2n+1)!.
The following equation is the generating even terms:
(16)U~2n(x)=∑l=0n(2n2l)∑j=0n-l(2j)!j!T(n-l,j)∑k=0j(-1)ks(j,k)xk+2l=∑k=02n∑l=0[k/2](2n2l)(-1)k-2l∑j=k-2ln-l(2j)!j!T(n-l,j)s(j,k-2l)xk,
where [x] is the greatest integer not exceeding x.
We also derive the generating odd terms:(17)U~2n+1(x)=∑l=0n(2n+12l+1)∑j=0n-l(2j)!j!T(n-l,j)∑k=0j(-1)ks(j,k)xk+(2l+1)=∑k=12n+1∑l=0[(k-1)/2](2n+12l-1)(-1)k-(2l+1)×∑j=k-(2l+1)n-l(2j)!j!T(n-l,j)s(j,k-(2l+1))xk,
where [x] is the greatest integer not exceeding x.
Thus, we complete the proof of Theorem2.Example 3.
Letn=1,2,3,4,5,6,7,8. Then we can know the following polynomials:
(18)U~1(x)=x,U~2(x)=x2-2x,U~3(x)=x3-6x2,U~4(x)=x4-12x3+12x2+10x,U~5(x)=x5-20x4+60x3+50x2,U~6(x)=x6-30x5+180x4+30x3-300x2-182x,U~7(x)=x7-42x6+420x5-490x4-2100x3-1274x2,U~8(x)=x8-56x7+840x6-2660x5-6720x4U~8(x)=+3304x3+13692x2+6970x,….Corollary 4.
Letn,l∈ℤ+ and n≥l. Then one has
(19)1U~2n(x)=∑l=0n(2n2l)∑j=0n-l(2j)!j!T(n-l,j)∑k=0j(-1)ks(j,k)xk+2l,U~2n+1(x)=∑l=0n(2n+12l+1)∑j=0n-l(2j)!j!T(n-l,j)U~2n+1(x)=×∑k=0j(-1)ks(j,k)xk+(2l+1).Proof.
Corollary4 shows identities of U~2n(x) and U~2n+1(x). The proof of Corollary 4 is contained in Theorem 2.Settingx=1 in Corollary 4, we easily see the following corollary.Corollary 5.
Relation ofU~n(1) and central factorial numbers is shown.
Letn,l,j∈ℤ+ and n≥l+j. Then one has
(20)1U~2n(1)=∑l=0n(2n2l)∑j=0n-l(-1)j(2j)!T(n-l,j),U~2n+1(1)=∑l=0n(2n+12l+1)∑j=0n-l(-1)j(2j)!T(n-l,j).Proof.
Consider the following:(21)U~2n(x)=∑l=0n(2n2l)∑j=0n-l(2j)!j!T(n-l,j)∑k=0j(-1)ks(j,k)xk+2l=∑l=0n(2n2l)∑j=0n-l(2j)!j!T(n-l,j)(-1)2j∑k=0j(-1)ks(j,k)xk+2l=∑l=0n(2n2l)∑j=0n-l(2j)!j!(-1)-jT(n-l,j)∑k=0j(-1)k-js(j,k)xk+2l.
We find out the following equation from the above equation by substituting x=1:
(22)U~2n(1)=∑l=0n(2n2l)∑j=0n-l(2j)!j!T(n-l,j)j!U~2n(1)=∑l=0n(2n2l)∑j=0n-l(-1)j(2j)!T(n-l,j).
We also can find out U~2n+1(1) by using similar method as above:
(23)U~2n+1(x)=∑l=0n(2n+12l+1)∑j=0n-l(2j)!j!T(n-l,j)U~2n+1(x)=×∑k=0j(-1)ks(j,k)xk+2lU~2n+1(x)=∑l=0n(2n+12l+1)∑j=0n-l(2j)!j!T(n-l,j)(-1)2jU~2n+1(x)=×∑k=0j(-1)ks(j,k)xk+2lU~2n+1(x)=∑l=0n(2n+12l+1)∑j=0n-l(2j)!j!(-1)-jT(n-l,j)U~2n+1(x)=×∑k=0j(-1)k-js(j,k)xk+2l.
If we substitute x=1, then
(24)U~2n+1(1)=∑l=0n(2n+12l+1)∑j=0n-l(2j)!j!T(n-l,j)j!U~2n+1(1)=∑l=0n(2n+12l+1)∑j=0n-l(-1)j(2j)!T(n-l,j).Letn,k,l∈ℤ+ and n≥k-l-1 and
(25)111a(2n,k)=∑l=0[k/2](2n2l)(-1)k-2lb(2n+1,k)=×∑j=k-2ln-l(2j)!j!T(n-l,j)s(j,k-2l),b(2n+1,k)=∑l=0[(k-1)/2](2n+12l-1)(-1)k-(2l+1)b(2n+1,k)=×∑j=k-(2l+1)n-l(2j)!j!T(n-l,j)s(j,k-(2l+1)),
where [x] is the greatest integer not exceeding x.Then we have(26)U~2n(x)=∑k=02na(2n,k)xk,U~2n+1(x)=∑k=12n+1b(2n+1,k)xk.By usinga(2n,k) and b(2n+1,k), we obtain Theorem 6.Theorem 6.
Letn,k∈ℤ+. Then one gets
(27)U~2n(1)={1ifn=02∑k=0n-1a(2n,2k)-22n+1ifn≥k+1,U~2n+1(1)=2∑k=1nb(2n+1,2k)+22n+1-1.U~2n(1)={1ifn=02∑k=0n-1a(2n,2k+1)+2n-1ifn≥k+1,U~2n+1(1)=2∑k=12n+1b(2n+1,2k-1)-22n+1+1.Proof.
From Definition1, one easily obtains the following equation:
(28)∑n=0∞U~n(x)tnn!=∑n=0∞U~2n(x)t2n(2n)!+∑n=0∞U~2n+1(x)t2n+1(2n+1)!∑n=0∞‍U~n(x)tnn!=(etet+e-t-1)x.
We also easily get the following equation by substituting x=-1:
(29)∑n=0∞U~n(-1)tnn!=∑n=0∞U~2n(-1)t2n(2n)!∑n=0∞‍U~n(-1)tnn!=+∑n=0∞U~2n+1(-1)t2n+1(2n+1)!∑n=0∞‍U~n(-1)tnn!=(etet+e-t-1)-1∑n=0∞‍U~n(-1)tnn!=1+∑n=0∞(-1)n(2n-1)tnn!∑n=0∞‍U~n(-1)tnn!=1+∑n=0∞(-1)n(2n-1)tnn!∑n=0∞‍U~n(-1)tnn!=1+∑n=1∞(22n-1)t2n2n!∑n=0∞‍U~n(-1)tnn!=-∑n=0∞(22n+1-1)t2n+1(2n+1)!.
Thus,
(30)∑n=0∞U~2n(-1)t2n(2n)!=1+∑n=1∞(22n-1)t2n2n!,111111U~2n+1(-1)=1-22n+1.
We have to consider odd terms and even terms in (26) and the previous equations, respectively.
Odd terms are in the following form:(31)U~2n+1(-1)=∑k=12n+1b(2n+1,k)(-1)kU~2n+1(-1)=∑k=1nb(2n+1,2k)-∑k=1n+1b(2n+1,2k-1)U~2n+1(-1)=1-22n+1,U~2n+1(1)=2∑k=1nb(2n+1,2k)+22n+1-1U~2n+1(1)=2∑k=12n+1b(2n+1,2k-1)-22n+1+1,
and even terms are similar to the way of the proof process of the odd terms. Therefore, we omit the proof process of the even terms.Theorem 7.
Letn,k∈ℤ+ and k+1≤n. Then one has
(32)U~n+1(1)-U~n(1)=∑k=0n-1(nk)((1-2n-k)U~k+1(1)-U~k(1)).Proof.
From Definition1, one easily obtains the following equation:
(33)∑n=0∞U~n(1)tnn!=(etet+e-t-1).
We also easily get the following by using differential, namely,
(34)∑n=1∞U~n(1)tn-1(n-1)!=-et+2(et+e-t-1)2.
Thus,
(35)(e2t-et+1)∑n=1∞U~n(1)tn-1(n-1)!=(e2t-et+1)-et+2(et+e-t-1)2=(2-et)∑n=0∞U~n(1)tnn!.
Then, right-hand side is in the following form:
(36)(2-et)∑n=0∞U~n(1)tnn!=2∑n=0∞U~n(1)tnn!-∑n=0∞∑k=0∞(nk)U~k(1)tnn!=∑n=0∞(2U~n(1)-∑k=0n‍(nk)U~k(1))tnn!,
and left-hand side is in the following form:
(37)(e2t-et+1)∑n=1∞U~n(1)tn-1(n-1)!=∑n=0∞2ntnn!∑n=0∞U~n+1(1)tnn!-∑n!∞∑n=0∞U~n+1(1)tnn!+∑n=0∞U~n+1(1)tnn!=∑n=0∞(∑k=0n(nk)(2n-k-1)U~k+1(1)+U~n+1(1))tnn!.
By using comparing coefficients of tn/n! in the previous equations, we can represent the equation; that is,
(38)∑k=0n(nk)(2n-k-1)U~k+1(1)+U~n+1(1)=2U~n(1)-∑k=0n(nk)U~k(1).
By simple calculation, we get
(39)U~n+1(1)-2U~n(1)=∑k=0n(nk)(1-2n-k)U~k+1(1)-∑k=0n(nk)U~k(1)=∑k=0n-1(nk)((1-2n-k)U~k+1(1)-U~k(1))-U~n(1).
Therefore, we proved that
(40)U~n+1(1)-U~n(1)=∑k=0n-1(nk)((1-2n-k)U~k+1(1)-U~k(1)).


## 3. Zeros of the PolynomialsU~n(x)

In this section, we investigate the reflection symmetry of the zeros of the polynomialsU~n(x).We investigate the beautiful zeros of the polynomialsU~n(x) by using a computer. We plot the zeros of the polynomials U~n(x) for n=20,30,40,50 and x∈ℂ (Figure 1). In Figure 1(a), we choose n=20. In Figure 1(b), we choose n=30. In Figure 1(c), we choose n=40. In Figure 1(d), we choose n=50.Zeros ofU~n(x) for n=20,30,40,50.
(a)
(b)
(c)
(d)Throughout the numerical experiments, we can finally conclude that polynomialsU~n(x) have no Re(x)=0 reflection symmetry analytic complex functions. However, we observe that U~n(x) has Im(x)=0 reflection symmetry (see Figures 1, 2, and 3). The obvious corollary is that the zeros of U~n(x) will also inherit these symmetries:
(41)ifU~n(x)=0,thenU~n(x*)=0,
where * denotes complex conjugation (see Figure 1).Figure 2
Real zeros ofU~n(x) for 1≤n≤50.Figure 3
Stacks of zeros ofU~n(x) for 1≤n≤50.Plots of real zeros ofU~n(x) for 1≤n≤50 structure are presented (Figure 2).Our numerical results for approximate solutions of real zeros of theU~n(x) are displayed (Tables 1 and 2).Table 1
Numbers of real and complex zeros ofU~n(x).
Degreen
Real zeros
Complex zeros
1
1
0
2
2
0
3
3
0
4
4
0
5
5
0
6
4
2
7
5
2
8
6
2
9
5
4
10
6
4
⋮
⋮
⋮
30
20
10Table 2
Approximate solutions ofU~n(x)=x∈ℝ.
n
x
1
0
2
0, 2
3
0, 0, 6
4
−0.5346, 0, 1.73111, 10.8035
5
−0.6759, 0, 0, 4.60224, 16.0736
6
0, 1.64591338, 8.2083255, 21.6537433
7
0, 0, 4.0761207, 12.3245041, 27.4570253
8
−0.877412, 0, 1.6058053, 7.1033690, 16.8165987, 33.4298423
9
−0.968751, 0, 0, 3.8044567, 10.5908618, 21.598401, 39.536481
10
0, 1.5827926, 6.4946365, 14.442719, 26.611139, 45.751858We observe a remarkably regular structure of the complex roots of the polynomialsU~n(x). We hope to verify a remarkably regular structure of the complex roots of the polynomials U~n(x) (Table 1).Stacks of zeros ofU~n(x) for 1≤n≤50 from a 3-D structure are presented (Figure 3).Next, we calculated an approximate solution satisfyingU~n(x),x∈ℝ. The results are given in Table 2.Finally, we will consider the more general problems. Find the numbers of complex zerosCU~n(x) of U~n(x),Im(x)≠0. Since n is the degree of the polynomial U~n(x), the number of real zeros RU~n(x) lying on the real plane Im(x)=0 is then RU~n(x)=n-CU~n(x), where CU~n(x) denotes complex zeros. See Table 1 for tabulated values of RU~n(x) and CU~n(x). We plot the zeros of U~n(x), respectively (Figures 1–3). These figures give mathematicians an unbounded capacity to create visual mathematical investigations of the behavior of the roots of the U~n(x). Moreover, it is possible to create new mathematical ideas and analyze them in ways that generally are not possible by hand. The authors have no doubt that investigation along this line will lead to a new approach employing numerical methods in the field of research of U~n(x) to appear in mathematics and physics.


## 4. Some Relations of PolynomialsU~n(x) and Euler Numbers

In this section, we find out interesting relations between polynomialsU~n(x) and Euler numbers. We usually use the values of polynomials U~n(x) and Euler polynomials.Theorem 8.
Letn∈ℤ+. Then one has
(42)23nU~n(1)=En(23)+(-1)nEn,23nU~n(2)=(E(23)-E)n+2n-1(En(23)+(-1)nEn),
with the usual convention of replacing En by En.Proof.
By settingx=1 in Definition 1 and using Euler numbers, then we have
(43)∑n=0∞U~n(1)tnn!=(etet+e-t-1)∑n=0∞‍U~n(1)tnn!=11+e-3t+11+e3te2t∑n=0∞‍U~n(1)tnn!=12∑n=0∞3n((-1)nEn+En(23))tnn!.
Hence,(44)23nU~n(1)=En(23)+(-1)nEn.
We also obtain the above equation by takingx=2 and using Euler numbers:
(45)∑n=0∞U~n(2)tnn!=(etet+e-t-1)2∑n=0∞‍U~n(2)tnn!=(11+e-3t+11+e3te2t)∑n=0∞‍U~n(2)tnn!=×(11+e-3t+11+e3te2t)∑n=0∞‍U~n(2)tnn!=(12∑n=0∞3n((-1)nEn+En(23))tnn!)∑n=0∞‍U~n(2)tnn!×(12∑n=0∞3n((-1)nEn+En(23))tnn!).
By using Cauchy product in the above equation, we have
(46)∑n=0∞U~n(2)tnn!=14∑n=0∞∑k=0n(nk)3n((-1)kEk+Ek(23))∑n=0∞‍U~n(2)tnn!=×((-1)n-kEn-k+En-k(23))tnn!.
By comparing the coefficient of both sides of tn/n! and some calculations, we have the following equation:
(47)23nU~n(2)=(En(23)-E)n+2n-1(En(23)+(-1)nEn),
with the usual convention of replacing En by En.
Therefore, we consummated the proof of Theorem8.From (26), we can suppose that
(48)U~n(x)=∑k=02na(2n,k)xk+∑k=12n+1b(2n+1,k)xkU~n(x)=∑k=0nc(n,k)xk,
where
(49)c(n,k)={a(2n,k)ifn=odd,b(2n+1,k)ifn=even.
Therefore, we have Theorem 9.Theorem 9.
Letn,v1,…,vk,k∈ℕ. Then one has
(50)k!c(n,k)=3n-kn!2k×∑v1,…,vk∈ℕv1+⋯+vk=n(((Ev1-1(13)-Ev1-1(23)+2Ev1-1)∑v1,…,vk∈ℕv1+⋯+vk=n(152⋯((23)Evk-1(13)-Evk-1(23)+2Evk-1))∑v1,…,vk∈ℕv1+⋯+vk=n(155×(v1!⋯vk!)-1(((Ev1-1-(23)+2Ev1-1)).Proof.
From Definition1, we can differentiate the k-times as follows:
(51)∑n=k∞dkdxkU~n(x)|tnn!x=0tnn!=(logetet+e-t-1)k.
Here we required that
(52)logetet+e-t-1=∑n=0∞3n2(En(13)-En(23)+2En)tn+1(n+1)!=∑n=1∞3n-12(En-1(13)-En-1(23)+2En-1)tnn!.
Hence, we derive that
(53)∑n=k∞dkdxkU~n(x)|tnn!x=0tnn!=(∑n=1∞3n-12(En-1(13)-En-1(23)+2En-1)tnn!)k=∑n=k∞3n-kn!2k×∑v1,…,vk∈ℕv1+⋯+vk=n(((Ev1-1(13)-Ev1-1(23)+2Ev1-1)×∑v1,…,vk∈ℕv1+⋯+vk=n221⋯(Evk-1(13)-Evk-1(23)+2Evk-1))×∑v1,…,vk∈ℕv1+⋯+vk=n221×(v1!⋯vk!)-1((Ev1-1(13)-Ev1-1(23)+2Ev1-1))tnn!.
By comparing the coefficient of both sides of tn/n!, we have the following equation:
(54)dkdxkU~n(x)|tnn!x=0=3n-kn!2k∑v1,…,vk∈ℕv1+⋯+vk=n(((Ev1-1(13)-Ev1-1(23)+2Ev1-1)1=3n-kn!2k∑v1,…,vk∈ℕv1+⋯+vk=n(⋯(Evk-1(13)-Evk-1(23)+2Evk-1))1=3n-kn!2k∑v1,…,vk∈ℕv1+⋯+vk=n(×(v1!⋯vk!)-1((Ev1-1(13)-Ev1-1(23)+2Ev1-1)).
From (48), we easily see that
(55)dkdxkU~n(x)|tnn!x=0=k!c(n,k).
Therefore, we clear off the proof of Theorem9.Theorem 10.
Letn,v1,…,vk,k∈ℕ. Then one has
(56)k!c(n,k)=n!∑v1,…,vk∈ℕv1+⋯+vk=n((((U~(x)-3)v1-1-(U~(x)-1)v1-1)11111111111111⋯((U~(x)-3)vk-1-(U~(x)-1)vk-1))11111111111111×(v1!⋯vk!)-1(((U~(x)-3)v1-1-(U~(x)-1)v1-1)).Proof.
This proof can be proved by the similar method of Theorem9.We also can derive relation polynomialsU~n(x) and Euler numbers as follows.Theorem 11.
Letn,k∈ℤ+ and k+1≤n. Then one derives
(57)∑k=0n-1(n-1k)(-1)n-k-1(2n-k-1)U~k(1)=3n-12(En-1(13)-En-1(23)+2En-1).Proof.
By the proof of Theorem9, we can differentiate polynomials U~n(x) as follows:
(58)ddt(logetet+e-t-1)=2e-t-1et+e-t-1=etet+e-t-1(2e-2t-e-t)=∑n=0∞U~n(1)tnn!(2∑n=0∞(-2)ntnn!-∑n=0∞(-1)ntnn!)=∑n=0∞∑k=0n(nk)(-1)n-k(2n-k+1-1)U~k(1)tnn!.
By integrating from 0 to t, we deduce that
(59)logetet+e-t-1=∑n=0∞∑k=0n(nk)(-1)n-k(2n-k+1-1)U~k(1)tn+1(n+1)!=∑n=1∞∑k=0n-1(n-1k)(-1)n-(k+1)(2n-k-1)U~k(1)tnn!.
We already knew from the proof process of Theorem9 that
(60)logetet+e-t-1=∑n=1∞3n-12(En-1(13)-En-1(23)+2En-1)tnn!.
By using the above equation, we can derive that
(61)∑k=0n-1(n-1k)(-1)n-k-1(2n-k-1)U~k(1)=3n-12(En-1(13)-En-1(23)+2En-1).
Therefore, we completely demonstrated the proof of Theorem11.



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*Source: 158130-2013-10-03.xml*