Citation: Pshtiwan Othman Mohammed, Miguel Vivas-Cortez, Thabet Abdeljawad, Yenny Rangel-Oliveros. Integral inequalities of Hermite-Hadamard type for quasi-convex functions with applications[J]. AIMS Mathematics, 2020, 5(6): 7316-7331. doi: 10.3934/math.2020468
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In [1,pp. 42 and 55], we find the Maclaurin power series expansions
cosx=∞∑k=0(−1)kx2k(2k)!=1−x22+x424−x6720+x840320−⋯,x∈R |
and
lncosx=−∞∑k=122k−1(22k−1)k(2k)!|B2k|x2k=−x22−x412−x645−17x82520−⋯ | (1) |
for x2<π24, where B2k denotes the Bernoulli numbers which can be generated by
zez−1=∞∑k=0Bkzkk!=1−z2+∞∑k=1B2kz2k(2k)!,|z|<2π. |
Let
F(x)={ln2(1−cosx)x2,0<|x|<2π;0,x=0 | (2) |
and
R(x)={ln2(1−cosx)x2lncosx,0<|x|<π2;16,x=0;0,x=±π2. | (3) |
In the recent paper [2], Li and Qi obtained the following two results:
1. The function F(x) defined by (2) can be expanded into the Maclaurin power series expansion
F(x)=−∞∑n=1E2n(2n)!x2n=−x212−x41440−x690720−x84838400−⋯ | (4) |
for |x|<2π, where
E2n=−|A2n−1,1B2n−1,2n−1ωnC1,2n+1|,n≥1,ωk=(−1)k(k+1)(2k+1),k≥0,A2n−1,1=(0ω10ω2⋮0ωn−10)=(ai,j)1≤i≤2n−1j=1,n≥1,ai,1={0,1≤i=2k−1≤2n−1;ωk,2≤i=2k≤2n−2,B2n−1,2n−1=|(00)ω000⋯00(11)ω00⋯0(20)ω10(22)ω0⋯00(31)ω10⋯0⋮⋮⋮⋱⋮(2n−40)ωn−20(2n−42)ωn−3⋯00(2n−31)ωn−20⋯0(2n−20)ωn−10(2n−22)ωn−1⋯(2n−22n−2)ω0|=(bi,j)1≤i,j≤2n−1,n≥1,bi,j={0,1≤i<j≤2n−1;(ij)ωk,0≤i−j=2k≤2n−2;0,1≤i−j=2k−1≤2n−3,C1,2n−1=(0(2n−11)ωn−10(2n−13)ωn−2⋯0(2n−12n−3)ω10)=(ci,j)i=11≤j≤2n−1,n≥1,c1,j={0,1≤j=2k−1≤2n−1;(2n−12k−1)ωn−k,2≤j=2k≤2n−2. |
2. The function R(x) defined by (3) decreasingly maps [0,π2] onto [0,16].
We now introduce several new even functions as follows:
1. The first function is
F0(x)=lncosx,x∈∞⋃k=0(±2kπ−π2,±2kπ+π2). | (5) |
2. The second function is
F1(x)={ln2(1−cosx)x2,x∈R∖{±2kπ,k=1,2,…};0,x=0. | (6) |
It is clear that F1(x)=F(x) on (−2π,2π).
3. Generally, the third function we are introducing is
Fn(x)={ln((−1)n(2n)!x2n[cosx−n−1∑k=0(−1)kx2k(2k)!]),x≠00,x=0 | (7) |
for n≥2.
Since the double inequality
0<(−1)n[cosx−n−1∑k=0(−1)kx2k(2k)!]<x2n(2n)! | (8) |
is valid for n≥2 and x∈R∖{0}, see [3,p. 326], the function Fn(x) is significantly defined for n≥2 and x∈R.
As a stronger version of the double inequality (8), the following positive, nonnegative, decreasing, and concave properties of the normalized tail
CosRn(x)={(−1)n(2n)!x2n[cosx−n−1∑k=0(−1)kx2k(2k)!],x≠01,x=0 | (9) |
for x∈(0,∞) and n≥1 were discovered in the paper [4]:
(a) the normalized tail CosR1(x) is nonnegative on (0,∞) and is decreasing on [0,2π];
(b) the normalized tail CosRn(x) for n≥2 is decreasing and positive on (0,∞);
(c) the normalized tail CosR1(x) is concave on (0,x0), where x0∈(π2,π) is the first positive zero of the equation
(x2−2)sinx+2xcosx=0 |
and the normalized remainder CosRn(x) for n≥2 is concave on (0,π).
Comparing the definition in (9) with those in (6) and (7) leads to the following conclusions:
(a) the function F1(x) is decreasing and negative on (0,2π), and is concave on (0,x0);
(b) the function Fn(x) for n≥2 is decreasing and negative on (0,∞), and is concave on (0,π).
4. The fourth function Rm,n(x) for n>m≥0 is defined
(a) when n>m=0, by
R0,n(x)={Fn(x)F0(x),0<|x|<π2;1(n+1)(2n+1),x=0;0,x=±π2, | (10) |
(b) when n>m=1, by
R1,n(x)={Fn(x)F1(x),x∈R∖{±2kπ,k=1,2,…};6(n+1)(2n+1),x=0;0,x∈{±2kπ,k=1,2,…}, |
(c) when n>m≥2, by
Rm,n(x)={Fn(x)Fm(x),x≠0;(m+1)(2m+1)(n+1)(2n+1),x=0. |
It is easy to see that R0,1(x)=R(x).
We now propose the following three problems:
1. Is the function Fn(x) for n≥0 decreasing and concave?
2. What is the Maclaurin power series expansion of Fn(x) for n≥0 around the origin x=0?
3. Is the function Rm,n(x) for n>m≥0 decreasing?
The first problem for the case n=0 is immediate: the even function F0(x)=lncosx is decreasing and, by virtue of the series expansion (1), is concave in x∈(0,π2). The first problem for the case n≥1 was solved in the paper [4], as mentioned above. In a word, the first problem has been thoroughly solved.
The second problem for n=0 is just the Maclaurin power series expansion (1). The second problem for n=1 was solved by the Maclaurin series expansion (4), which was established in [2,Section 3].
The third problem for (m,n)=(0,1) was solved in [2,Section 4], as mentioned above.
In this paper, we will give a full answer to the second problem for all cases n≥2, solve the first problem on the interval (0,π2) once again, and discuss the third problem for the case n≥2.
In this section, we solve the second problem: what is the Maclaurin power series expansion of Fn(x) for n≥0 around the origin x=0?
Theorem 1. For n≥0, let
ei,j(n)={(−1)i/21+(−1)i21(2n+i2n),1≤i≤2m,j=1(−1)(i−j+1)/21+(−1)i−j+12(i−1j−2)(2n+i−j+12n),1≤i≤2m,2≤j≤2m |
and
D2m(n)=|ei,j(n)|(2m)×(2m). |
Then the even function Fn(x) for n≥0 can be expanded into
Fn(x)=−∞∑m=1D2m(n)(2m)!x2m,|x|<{π2,n=0;2π,n=1;∞,n≥2. | (11) |
Proof. Let u(x) and v(x)≠0 be two n-time differentiable functions on an interval I for a given integer n≥0. Then the nth derivative of the ratio u(x)v(x) is
dndxn[u(x)v(x)]=(−1)n|W(n+1)×(n+1)(x)|vn+1(x),n≥0, | (12) |
where the matrix
W(n+1)×(n+1)(x)=(U(n+1)×1(x)V(n+1)×n(x))(n+1)×(n+1), |
the matrix U(n+1)×1(x) is an (n+1)×1 matrix whose elements satisfy uk,1(x)=u(k−1)(x) for 1≤k≤n+1, the matrix V(n+1)×n(x) is an (n+1)×n matrix whose elements are vℓ,j(x)=(ℓ−1j−1)v(ℓ−j)(x) for 1≤ℓ≤n+1 and 1≤j≤n, and the notation |W(n+1)×(n+1)(x)| denotes the determinant of the (n+1)×(n+1) matrix W(n+1)×(n+1)(x). This is a slight reformulation of [5,p. 40,Exercise 5].
Let
un(x)=∞∑k=0(−1)k+1(2k+2n+22n)x2k+1(2k+1)!andvn(x)=∞∑k=0(−1)k(2k+2n2n)x2k(2k)!. |
Then, straightforward differentiation yields
u(2ℓ+1)n(0)=limx→0∞∑k=0(−1)k+1(2k+2n+22n)⟨2k+1⟩2ℓ+1x2k−2ℓ(2k+1)!=(−1)ℓ+1(2ℓ+2n+22n),v(2ℓ)n(0)=limx→0∞∑k=0(−1)k(2k+2n2n)⟨2k⟩2ℓx2k−2ℓ(2k)!=(−1)ℓ(2ℓ+2n2n),u(2ℓ)n(0)=0,v(2ℓ+1)n(0)=0 |
for ℓ≥0. Considering Expression (14) and applying the derivative formula (12) for the ratio of two differentiable functions, we acquire
F(2m)n(0)=limx→0[∞∑k=0(−1)k+1(2k+2n+22n)x2k+1(2k+1)!∞∑k=0(−1)k(2k+2n2n)x2k(2k)!](2m−1)=limx→0[un(x)vn(x)](2m−1)=(−1)2m−1v2mn(0)|un(0)vn(0)0⋯0u′n(0)v′n(0)(11)vn(0)⋯0u″n(0)v″n(0)(21)v′n(0)⋯0u(3)n(0)v(3)n(0)(31)v″n(0)⋯0u(4)n(0)v(4)n(0)(41)v(3)n(0)⋯0⋮⋮⋮⋱⋮u(2m−1)n(0)v(2m−1)n(0)(2m−11)v(2m−2)n(0)⋯(2m−12m−2)v′n(0)|=−|0(00)000−1(2n+22n)0(11)000−(20)(2n+22n)0(22)01(2n+42n)0−(31)(2n+22n)0(33)0(40)(2n+42n)0−(42)(2n+22n)0⋮⋮⋮⋮⋮(−1)m−1(2n+2m−22n)0(−1)m−2(2m−31)(2n+2m−42n)0(−1)m−3(2m−33)(2n+2m−62n)0(−1)m−1(2m−20)(2n+2m−22n)0(−1)m−2(2m−22)(2n+2m−42n)0(−1)m(2n+2m2n)0(−1)m−1(2m−11)(2n+2m−22n)0(−1)m−2(2m−13)(2n+2m−42n)⋯000000⋯000000⋯000000⋯000000⋯000000⋱⋮⋮⋮⋮⋮⋮⋯(2m−32m−7)(2n+42n)0−(2m−32m−5)(2n+22n)0(2m−32m−3)0⋯0(2m−22m−6)(2n+42n)0−(2m−22m−4)(2n+22n)0(2m−22m−2)⋯−(2m−12m−7)(2n+62n)0(2m−12m−5)(2n+42n)0−(2m−12m−3)(2n+22n)0| |
for m≥1. In other words, for m≥1,
F(2m)n(0)=−D2m(n)=−|ei,j(n)|(2m)×(2m). |
Consequently, the even function Fn(x) can be expanded into
Fn(x)=∞∑k=0F(k)nxkk!=∞∑m=1F(2m)nx2m(2m)!=−∞∑m=1D2m(n)(2m)!x2m. |
The proof of Theorem 1 is completed.
Remark 1. When n=0, a direct computation gives
D4(0)=|0(00)00−10(11)00−(20)0(22)10−(31)0|=2andD2(0)=|0(00)−10|=1. |
Then, the first two terms of the Maclaurin power series expansion of the function lncosx are
−D2(0)2!x2−D4(0)4!x4=−12!x2−24!x4=−12x2−112x4, |
which coincide with the first two terms in the series expansion (1).
When n=1, straightforward computation shows
D4(1)=|0(00)00−1(42)0(11)00−(20)(42)0(22)1(62)0−(31)(42)0|=160andD2(1)=|0(00)−1(42)0|=16. |
Then, the first two terms of the Maclaurin power series expansion of the function F(x) defined by (2) are
−D2(1)2!x2−D4(1)4!x4=−112x2−11440x4, |
which coincide with the first two terms in the series expansion (4).
Remark 2. Comparing the Maclaurin series expansions (1) and (4) with the series expansion (11) reveals
|B2m|=m22m−1(22m−1)D2m(0)andE2m=D2m(1) | (13) |
for m≥1. The first formula in (13) is a new determinantal expression for the Bernoulli numbers B2m with m≥1.
Additionally, we point out that, in the papers [2,6,7,8,9,10], there have been many related results, but different from and more complicated than the first one in (13), and plenty of closely-related references on closed-form formulas and determinantal expressions for the Bernoulli numbers and polynomials B2m and Bm(x) with m∈N.
In this section, we give an alternative and united proof of a modification of the first problem: is the function Fn(x) for n≥0 decreasing and concave on (0,π2)?
Theorem 2. For n=0 and n≥2, the even function Fn(x) defined by (5) and (7) is decreasing and concave on (0,π2). The even function F1(x) defined in (6) is decreasing on (0,2π) and (xk,2(k+1)π) for k∈N, while it is increasing on (2kπ,xk) for k∈N, where xk∈(2kπ,2(k+1)π) for k∈N is the zero of the equation tanx2=x2 on (0,∞).
Proof. In the first section of this paper, it has been immediately verified that the function F0(x)=lncosx is decreasing and concave on (0,π2).
The derivative of F1(x) can be written as
F′1(x)=1tanx2−1x2,x≠±2kπ,k∈N. |
Therefore, the derivative F′1(x) is negative on (0,2π), is positive on (2kπ,xk), and is negative on (xk,2(k+1)π) for k∈N, where xk∈(2kπ,2(k+1)π) for k∈N is the zero of the equation tanx2=x2 on (0,∞). Accordingly, the function F1(x) is decreasing on (0,2π) and (xk,2(k+1)π), while it is increasing on (2kπ,xk) for k∈N.
On the interval (0,π2) and for n≥2, the function Fn(x) can be written as
Fn(x)=ln∞∑k=n(−1)k−n(2n)!(2k)!x2k−2n=ln∞∑k=0(−1)k(2k+2n2n)x2k(2k)!. |
Its first derivative is
F′n(x)=∞∑k=0(−1)k+1(2k+2n+22n)x2k+1(2k+1)!∞∑k=0(−1)k(2k+2n2n)x2k(2k)!. | (14) |
By virtue of [11,Theorem 7.6], or in view of the results at the site https://math.stackexchange.com/a/477549 (accessed on 18 January 2024), we derive the integral representation
cosx−n−1∑k=0(−1)kx2k(2k)!=(−1)nx2n∞∑k=0(−1)k(2n+2k)!x2k=(−1)n(2n−2)!∫x0(x−t)2n−2sintdt | (15) |
for n≥1 and x∈R. From the integral representation (15), it follows that
(−1)n(2n)!x2n[cosx−n−1∑k=0(−1)k(2k)!x2k]=∞∑k=0(−1)k(2k+2n2n)x2k(2k)!=2n(2n−1)x2n∫x0(x−t)2n−2sintdt=2n(2n−1)∫10(1−u)2n−2sin(ux)xdu>0,0<x<π | (16) |
and
[∞∑k=0(−1)k(2k+2n2n)x2k(2k)!]′=∞∑k=0(−1)k+1(2k+2n+22n)x2k+1(2k+1)!=2n(2n−1)∫10(1−u)2n−2ddx[sin(ux)x]du=2n(2n−1)∫10(1−u)2n−2cos(ux)(ux)−tan(ux)x2du<0,0<x<π2 |
for n≥1, where we used the inequalities cosx>0 and x−tanx<0 in x∈(0,π2). This means that
F′n(x)=∫10(1−u)2n−2cos(ux)(ux)−tan(ux)x2du∫10(1−u)2n−2sin(ux)xdu<0,0<x<π2 |
for n≥1. In conclusion, the function Fn(x) for n≥1 is decreasing on (0,π2).
It is known that
sinxx=∞∑k=0(−1)kx2k(2k+1)!,|x|<∞. |
Straightforward differentiating and simplifying give
(sinxx)″=∞∑k=0(−1)k+1(2k+2)(2k+1)(2k+3)!x2k=−∞∑k=0(4k+4)(4k+3)(4k+5)![(4k+1)(4k+2)(4k+5)(4k+3)−x2]x4k<0,x∈(0,√103)⊃(0,π2]. |
Therefore, a direct differentiation and simplification yield
F″n(x)=[∫10(1−u)2n−2ddx(sin(ux)x)du∫10(1−u)2n−2sin(ux)xdu]′=−1[∫10(1−u)2n−2sin(ux)xdu]2[(∫10(1−u)2n−2ddx(sin(ux)x)du)2−∫10(1−u)2n−2d2dx2(sin(ux)x)du∫10(1−u)2n−2sin(ux)xdu]=−1[∫10(1−u)2n−2sin(ux)xdu]2[(∫10u(1−u)2n−2ddx(sin(ux)ux)du)2−∫10u(1−u)2n−2d2dx2(sin(ux)ux)du∫10u(1−u)2n−2sin(ux)uxdu]<0 |
on (0,π2) for n≥1. Accordingly, the function Fn(x) for n≥1 is concave on (0,π2). The proof of Theorem 2 is thus complete.
Remark 3. We note that a concave function must be a logarithmically concave function, but the converse is not true. However, a logarithmically convex function must be a convex function, but the converse is not true.
In [2,Section 4], the function R0,1(x)=R(x) defined by (3) or (10) for n=1 was proved to be decreasing on [0,π2] onto [0,16].
Theorem 3. The even function R0,2(x) defined by (10) for the case n=2 is decreasing on [0,π2].
Proof. For n≥2, direct differentiation gives
F′n(x)F′0(x)=[lnCosRn(x)]′(lncosx)′=−CosR′n(x)CosRn(x)cosxsinx |
and
[F′n(x)F′0(x)]′=−[CosR′n(x)CosRn(x)cosxsinx]′=−(CosRn(x)CosR″n(x)−[CosR′n(x)]2)cosxsinx−CosRn(x)CosR′n(x)[CosRn(x)sinx]2. |
Taking n=2 and simplifying lead to
[F′2(x)F′0(x)]′=−72x10[CosR2(x)sinx]2[4x(x4−6x2+12)+4(7x2−16)xcosx+16xcos(2x)−4x3cos(3x)+(3x4+4x2−16)sinx+2(x4−10x2+12)sin(2x)−(x2−2x−4)(x2+2x−4)sin(3x)+4sin(4x)]=−72x10[CosR2(x)sinx]2∞∑k=6(−1)kQ(k)x2k+1(2k+1)!=−72x10[CosR2(x)sinx]2∞∑k=3[Q(2k)Q(2k+1)(4k+3)!(4k+1)!−x2]Q(2k+1)(4k+3)!x4k+1, |
where
Q(k)=42k+2−4(4k4−28k3+107k2+61k+324)32k−3+(4k4−4k3+39k2+53k+64)22k+4(12k4−68k3−7k2−17k−20),k≥6. |
From the facts that
12k4−68k3−7k2−17k−20=12(k−6)4+220(k−6)3+1361(k−6)2+2923(k−6)+490≥490,k≥6,4k4−4k3+39k2+53k+64=4(k−6)4+92(k−6)3+822(k−6)2+3437(k−6)+5782≥5782,k≥6, |
and, by induction,
42k+2−4(4k4−28k3+107k2+61k+324)32k−3=16×32k[(43)2k−4k4−28k3+107k2+61k+324108]>0,k≥7, |
we conclude, together with Q(6)=3871296, that Q(k)≥3871296 for k≥6.
Let
Q(k)=Q(2k)Q(2k+1)(4k+3)!(4k+1)!,k≥3. |
The inequality
Q(k+1)>Q(k),k≥3 |
is equivalent to
(2k+3)(4k+7)Q(2k+1)Q(2k+2)>(2k+1)(4k+3)Q(2k)Q(2k+3) |
for k≥3, that is,
Q(k)=216[73728(k−2)9+1508352(k−2)8+13174784(k−2)7+63787136(k−2)6+185680928(k−2)5+329304964(k−2)4+345900612(k−2)3+210339955(k−2)2+94117995(k−2)+42704050]+27×24k{(142293×24k+9+93239336)k5+4(228963×24k+7−315163)k4+(1730269×24k+5−77141534)k3+(590587×24k+5−61162271)k2+31(39633×24k+3−705961)k+72(32581×16k−50315)+4096(57×24k+3+847)k8+2048(24k+7−540k−1635)k9+32k6[24k+6(3328k−3377)+1151464k+2801469]}+27×212k+7[256k(24k+5−45k5)+9×24k+10+12352k5+24016k4+33804k3+78984k2+84261k+33915]+16×34k[159784960(k−4)9+2908177408(k−4)8+31203008512(k−4)7+218332899456(k−4)6+1039043941024(k−4)5+3394763402820(k−4)4+7477586709444(k−4)3+10528669786283(k−4)2+8416641010791(k−4)+2806689073644]+8×38k+3(8192k9+9216k8+30720k7+148608k6+179232k5+77220k4+1027780k3+2247363k2+1992375k+751032)+64k[107481088(k−4)9+1948553216(k−4)8+20892092672(k−4)7+146632542496(k−4)6+702801488840(k−4)5+2321298590084(k−4)4+5179870312810(k−4)3+7367488384493(k−4)2+5849176490661(k−4)+1817172122508]+28k+334k[313600(k−13)6+20524992(k−13)5+540906544(k−13)4+7219464756(k−13)3+49643220604(k−13)2+151463526603(k−13)+98734469508]>0,k≥3. |
By virtue of the above expression for Q(k), we see that the sequence Q(k) is positive for k≥13. On the other hand, it is straightforward that
Q(3)=352780960860822574080,Q(4)=249398534923066892200578048,Q(5)=61305423558540152150473900185600,Q(6)=8665426549135034591879376586377676800,Q(7)=896760345938903204692495757195177830886400,Q(8)=77679040214402381392287613848479095562672133120,Q(9)=6092826722565745296376115137707251410869680486780928,Q(10)=453017430243005416828735039714559150312936188925226332160,Q(11)=32749134150707011053456859834163170645192635726524315359872000,Q(12)=2332073744679555396812875864144181190554219273032749874293655526400. |
Consequently, the sequence Q(k) is positive for all k≥3. As a result, the sequence Q(k) is increasing in k≥3. It is immediate that Q(3)=423115=3.678…. Hence, we acquire
Q(k)≥423115=3.678…,k≥3. |
Accordingly, when
0<x≤π2=1.570⋯<√423115=3.678⋯, |
the derivative [F′2(x)F′0(x)]′ is negative, and then the derivative ratio F′2(x)F′0(x) is decreasing in x∈(0,π2].
In [12,pp. 10-11,Theorem 1.25], a monotonicity rule for the ratio of two functions was established as follows.
For a,b∈R with a<b, let p(x) and q(x) be continuous on [a,b], differentiable on (a,b), and q′(x)≠0 on (a,b). If the ratio p′(x)q′(x) is increasing on (a,b), then both p(x)−p(a)q(x)−q(a) and p(x)−p(b)q(x)−q(b) are increasing in x∈(a,b).
With the help of this monotonicity rule and in view of the decreasing property of the derivative ratio F′2(x)F′0(x) in x∈(0,π2], we derive that the ratio F2(x)F0(x)=R0,2(x) is decreasing in x∈(0,π2]. The required proof of Theorem 3 is completed.
Remark 4. How to verify the decreasing property of the function R0,n(x) for n≥3 on (0,π2], of the function R1,n(x) on (0,2π), and of the function Rm,n(x) for n>m≥2 on (0,∞)? The ideas, approaches, techniques, and methods used in the proof of Theorem 3 should not be valid again, so we need to discover new ideas, approaches, techniques, and methods for verifying the decreasing property mentioned above.
Remark 5. Let
fα(x)=∫10(1−u)αcos(ux)du,α∈R. |
Prove that the function fα(x) is positive in x∈(0,∞) if and only if α>1, while it is decreasing in x∈(0,∞) if and only if α≥2.
This paper and the articles [2,13,14] are siblings, because some results in [2] have been generalized in this paper, and the results in [13,14] are about the Maclaurin power series expansions of logarithmic expressions involving normalized tails of the tangent and sine functions.
In this paper, we presented the following results.
1. The function Fn(x) for n≥0 was expanded into the Maclaurin power series expansion (11) in Theorem 1.
2. The function Fn(x) defined by (7) for n≥0 was proved in Theorem 2 to be decreasing and concave on (0,π2).
3. A new determinantal expression (13) of the Bernoulli numbers B2m for m≥1 was derived.
4. The ratio R0,2(x) was proved in Theorem 3 to be decreasing in x∈[0,π2].
In order to verify the decreasing property of the function R0,n(x) for n≥3 on (0,π2), of the function R1,n(x) on (0,2π), and of the function Rm,n(x) for n>m≥2 on (0,∞), we need new ideas, novel approaches, creative techniques, and innovative methods.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This work was supported in part by the National Natural Science Foundation of China (Grant No. 12061033).
The authors declare there is no conflicts of interest.
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