Integers occurring in the coefficients of the Taylor series of 1/cosh t
In mathematics , the Euler numbers are a sequence En of integers (sequence A122045 in the OEIS ) defined by the Taylor series expansion
1 cosh t = 2 e t + e − t = ∑ n = 0 ∞ E n n ! ⋅ t n {\displaystyle {\frac {1}{\cosh t}}={\frac {2}{e^{t}+e^{-t}}}=\sum _{n=0}^{\infty }{\frac {E_{n}}{n!}}\cdot t^{n}} ,where cosh ( t ) {\displaystyle \cosh(t)} is the hyperbolic cosine function . The Euler numbers are related to a special value of the Euler polynomials , namely:
E n = 2 n E n ( 1 2 ) . {\displaystyle E_{n}=2^{n}E_{n}({\tfrac {1}{2}}).} The Euler numbers appear in the Taylor series expansions of the secant and hyperbolic secant functions. The latter is the function in the definition. They also occur in combinatorics , specifically when counting the number of alternating permutations of a set with an even number of elements.
In terms of Stirling numbers of the second kind The following two formulas express the Euler numbers in terms of Stirling numbers of the second kind : [ 1] [ 2]
E n = 2 2 n − 1 ∑ ℓ = 1 n ( − 1 ) ℓ S ( n , ℓ ) ℓ + 1 ( 3 ( 1 4 ) ℓ . ¯ − ( 3 4 ) ℓ . ¯ ) , {\displaystyle E_{n}=2^{2n-1}\sum _{\ell =1}^{n}{\frac {(-1)^{\ell }S(n,\ell )}{\ell +1}}\left(3\left({\frac {1}{4}}\right)^{\overline {\ell {\phantom {.}}}}-\left({\frac {3}{4}}\right)^{\overline {\ell {\phantom {.}}}}\right),} E 2 n = − 4 2 n ∑ ℓ = 1 2 n ( − 1 ) ℓ ⋅ S ( 2 n , ℓ ) ℓ + 1 ⋅ ( 3 4 ) ℓ . ¯ , {\displaystyle E_{2n}=-4^{2n}\sum _{\ell =1}^{2n}(-1)^{\ell }\cdot {\frac {S(2n,\ell )}{\ell +1}}\cdot \left({\frac {3}{4}}\right)^{\overline {\ell {\phantom {.}}}},} where S ( n , ℓ ) {\displaystyle S(n,\ell )} denotes the Stirling numbers of the second kind , and x ℓ . ¯ = ( x ) ( x + 1 ) ⋯ ( x + ℓ − 1 ) {\displaystyle x^{\overline {\ell {\phantom {.}}}}=(x)(x+1)\cdots (x+\ell -1)} denotes the rising factorial .
As a recursion The Euler numbers can be defined as an recursion:
E 2 n = − ∑ k = 1 n ( 2 n 2 k ) E 2 ( n − k ) , {\displaystyle E_{2n}=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2(n-k)},}
or alternatively:
1 = − ∑ k = 1 n ( 2 n 2 k ) E 2 k , {\displaystyle 1=-\sum _{k=1}^{n}{\binom {2n}{2k}}E_{2k},}
Both of these recursions can be found by using the fact that.
c o s ( x ) s e c ( x ) = 1. {\displaystyle cos(x)sec(x)=1.}
As a double sum The following two formulas express the Euler numbers as double sums [ 3]
E 2 n = ( 2 n + 1 ) ∑ ℓ = 1 2 n ( − 1 ) ℓ 1 2 ℓ ( ℓ + 1 ) ( 2 n ℓ ) ∑ q = 0 ℓ ( ℓ q ) ( 2 q − ℓ ) 2 n , {\displaystyle E_{2n}=(2n+1)\sum _{\ell =1}^{2n}(-1)^{\ell }{\frac {1}{2^{\ell }(\ell +1)}}{\binom {2n}{\ell }}\sum _{q=0}^{\ell }{\binom {\ell }{q}}(2q-\ell )^{2n},} E 2 n = ∑ k = 1 2 n ( − 1 ) k 1 2 k ∑ ℓ = 0 2 k ( − 1 ) ℓ ( 2 k ℓ ) ( k − ℓ ) 2 n . {\displaystyle E_{2n}=\sum _{k=1}^{2n}(-1)^{k}{\frac {1}{2^{k}}}\sum _{\ell =0}^{2k}(-1)^{\ell }{\binom {2k}{\ell }}(k-\ell )^{2n}.} As an iterated sum An explicit formula for Euler numbers is: [ 4]
E 2 n = i ∑ k = 1 2 n + 1 ∑ ℓ = 0 k ( k ℓ ) ( − 1 ) ℓ ( k − 2 ℓ ) 2 n + 1 2 k i k k , {\displaystyle E_{2n}=i\sum _{k=1}^{2n+1}\sum _{\ell =0}^{k}{\binom {k}{\ell }}{\frac {(-1)^{\ell }(k-2\ell )^{2n+1}}{2^{k}i^{k}k}},} where i denotes the imaginary unit with i 2 = −1 .
As a sum over partitions The Euler number E 2n can be expressed as a sum over the even partitions of 2n , [ 5]
E 2 n = ( 2 n ) ! ∑ 0 ≤ k 1 , … , k n ≤ n ( K k 1 , … , k n ) δ n , ∑ m k m ( − 1 2 ! ) k 1 ( − 1 4 ! ) k 2 ⋯ ( − 1 ( 2 n ) ! ) k n , {\displaystyle E_{2n}=(2n)!\sum _{0\leq k_{1},\ldots ,k_{n}\leq n}{\binom {K}{k_{1},\ldots ,k_{n}}}\delta _{n,\sum mk_{m}}\left(-{\frac {1}{2!}}\right)^{k_{1}}\left(-{\frac {1}{4!}}\right)^{k_{2}}\cdots \left(-{\frac {1}{(2n)!}}\right)^{k_{n}},} as well as a sum over the odd partitions of 2n − 1 , [ 6]
E 2 n = ( − 1 ) n − 1 ( 2 n − 1 ) ! ∑ 0 ≤ k 1 , … , k n ≤ 2 n − 1 ( K k 1 , … , k n ) δ 2 n − 1 , ∑ ( 2 m − 1 ) k m ( − 1 1 ! ) k 1 ( 1 3 ! ) k 2 ⋯ ( ( − 1 ) n ( 2 n − 1 ) ! ) k n , {\displaystyle E_{2n}=(-1)^{n-1}(2n-1)!\sum _{0\leq k_{1},\ldots ,k_{n}\leq 2n-1}{\binom {K}{k_{1},\ldots ,k_{n}}}\delta _{2n-1,\sum (2m-1)k_{m}}\left(-{\frac {1}{1!}}\right)^{k_{1}}\left({\frac {1}{3!}}\right)^{k_{2}}\cdots \left({\frac {(-1)^{n}}{(2n-1)!}}\right)^{k_{n}},} where in both cases K = k 1 + ··· + kn and
( K k 1 , … , k n ) ≡ K ! k 1 ! ⋯ k n ! {\displaystyle {\binom {K}{k_{1},\ldots ,k_{n}}}\equiv {\frac {K!}{k_{1}!\cdots k_{n}!}}} is a multinomial coefficient . The Kronecker deltas in the above formulas restrict the sums over the k s to 2k 1 + 4k 2 + ··· + 2nkn = 2n and to k 1 + 3k 2 + ··· + (2n − 1)kn = 2n − 1 , respectively.
As an example,
E 10 = 10 ! ( − 1 10 ! + 2 2 ! 8 ! + 2 4 ! 6 ! − 3 2 ! 2 6 ! − 3 2 ! 4 ! 2 + 4 2 ! 3 4 ! − 1 2 ! 5 ) = 9 ! ( − 1 9 ! + 3 1 ! 2 7 ! + 6 1 ! 3 ! 5 ! + 1 3 ! 3 − 5 1 ! 4 5 ! − 10 1 ! 3 3 ! 2 + 7 1 ! 6 3 ! − 1 1 ! 9 ) = − 50 521. {\displaystyle {\begin{aligned}E_{10}&=10!\left(-{\frac {1}{10!}}+{\frac {2}{2!\,8!}}+{\frac {2}{4!\,6!}}-{\frac {3}{2!^{2}\,6!}}-{\frac {3}{2!\,4!^{2}}}+{\frac {4}{2!^{3}\,4!}}-{\frac {1}{2!^{5}}}\right)\\[6pt]&=9!\left(-{\frac {1}{9!}}+{\frac {3}{1!^{2}\,7!}}+{\frac {6}{1!\,3!\,5!}}+{\frac {1}{3!^{3}}}-{\frac {5}{1!^{4}\,5!}}-{\frac {10}{1!^{3}\,3!^{2}}}+{\frac {7}{1!^{6}\,3!}}-{\frac {1}{1!^{9}}}\right)\\[6pt]&=-50\,521.\end{aligned}}} As a determinant E 2n is given by the determinant
E 2 n = ( − 1 ) n ( 2 n ) ! | 1 2 ! 1 1 4 ! 1 2 ! 1 ⋮ ⋱ ⋱ 1 ( 2 n − 2 ) ! 1 ( 2 n − 4 ) ! 1 2 ! 1 1 ( 2 n ) ! 1 ( 2 n − 2 ) ! ⋯ 1 4 ! 1 2 ! | . {\displaystyle {\begin{aligned}E_{2n}&=(-1)^{n}(2n)!~{\begin{vmatrix}{\frac {1}{2!}}&1&~&~&~\\{\frac {1}{4!}}&{\frac {1}{2!}}&1&~&~\\\vdots &~&\ddots ~~&\ddots ~~&~\\{\frac {1}{(2n-2)!}}&{\frac {1}{(2n-4)!}}&~&{\frac {1}{2!}}&1\\{\frac {1}{(2n)!}}&{\frac {1}{(2n-2)!}}&\cdots &{\frac {1}{4!}}&{\frac {1}{2!}}\end{vmatrix}}.\end{aligned}}} As an integral E 2n is also given by the following integrals:
( − 1 ) n E 2 n = ∫ 0 ∞ t 2 n cosh π t 2 d t = ( 2 π ) 2 n + 1 ∫ 0 ∞ x 2 n cosh x d x = ( 2 π ) 2 n ∫ 0 1 log 2 n ( tan π t 4 ) d t = ( 2 π ) 2 n + 1 ∫ 0 π / 2 log 2 n ( tan x 2 ) d x = 2 2 n + 3 π 2 n + 2 ∫ 0 π / 2 x log 2 n ( tan x ) d x = ( 2 π ) 2 n + 2 ∫ 0 π x 2 log 2 n ( tan x 2 ) d x . {\displaystyle {\begin{aligned}(-1)^{n}E_{2n}&=\int _{0}^{\infty }{\frac {t^{2n}}{\cosh {\frac {\pi t}{2}}}}\;dt=\left({\frac {2}{\pi }}\right)^{2n+1}\int _{0}^{\infty }{\frac {x^{2n}}{\cosh x}}\;dx\\[8pt]&=\left({\frac {2}{\pi }}\right)^{2n}\int _{0}^{1}\log ^{2n}\left(\tan {\frac {\pi t}{4}}\right)\,dt=\left({\frac {2}{\pi }}\right)^{2n+1}\int _{0}^{\pi /2}\log ^{2n}\left(\tan {\frac {x}{2}}\right)\,dx\\[8pt]&={\frac {2^{2n+3}}{\pi ^{2n+2}}}\int _{0}^{\pi /2}x\log ^{2n}(\tan x)\,dx=\left({\frac {2}{\pi }}\right)^{2n+2}\int _{0}^{\pi }{\frac {x}{2}}\log ^{2n}\left(\tan {\frac {x}{2}}\right)\,dx.\end{aligned}}} Euler zigzag numbers The Taylor series of sec x + tan x = tan ( π 4 + x 2 ) {\displaystyle \sec x+\tan x=\tan \left({\frac {\pi }{4}}+{\frac {x}{2}}\right)} is
∑ n = 0 ∞ A n n ! x n , {\displaystyle \sum _{n=0}^{\infty }{\frac {A_{n}}{n!}}x^{n},} where An is the Euler zigzag numbers , beginning with
1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521, 353792, 2702765, 22368256, 199360981, 1903757312, 19391512145, 209865342976, 2404879675441, 29088885112832, ... (sequence A000111 in the OEIS ) For all even n ,
A n = ( − 1 ) n 2 E n , {\displaystyle A_{n}=(-1)^{\frac {n}{2}}E_{n},} where En is the Euler number, and for all odd n ,
A n = ( − 1 ) n − 1 2 2 n + 1 ( 2 n + 1 − 1 ) B n + 1 n + 1 , {\displaystyle A_{n}=(-1)^{\frac {n-1}{2}}{\frac {2^{n+1}\left(2^{n+1}-1\right)B_{n+1}}{n+1}},} where Bn is the Bernoulli number .
For every n ,
A n − 1 ( n − 1 ) ! sin ( n π 2 ) + ∑ m = 0 n − 1 A m m ! ( n − m − 1 ) ! sin ( m π 2 ) = 1 ( n − 1 ) ! . {\displaystyle {\frac {A_{n-1}}{(n-1)!}}\sin {\left({\frac {n\pi }{2}}\right)}+\sum _{m=0}^{n-1}{\frac {A_{m}}{m!(n-m-1)!}}\sin {\left({\frac {m\pi }{2}}\right)}={\frac {1}{(n-1)!}}.} [ citation needed ] This page is based on this
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