Google 翻译生成的粗略草稿，质量较低。术语依照术语表，但句子可能有误；之后会由仔细的译文取代。
§ 2。三角函数函数和伯努利数的欧拉展开
1. 欧拉展开 OF cot z \cot z cot z
通过VI、p. 275、数b n / n ! b_n / n! b n / n ! S / ( e S − 1 ) S/(e^S - 1) S / ( e S − 1 ) z / ( e z − 1 ) z/(e^z - 1) z / ( e z − 1 ) C \mathbf{C} C b n / n ! b_n / n! b n / n ! b n b_n b n
首先我们注
z e z − 1 = − z 2 + z 2 e z + 1 e z − 1 = − z 2 + i z 2 cot i z 2 .
\frac{z}{e^z - 1} = - \frac{z}{2} + \frac{z}{2} \frac{e^z + 1}{e^z - 1} = - \frac{z}{2} + \frac{iz}{2} \cot \frac{iz}{2}.
e z − 1 z = − 2 z + 2 z e z − 1 e z + 1 = − 2 z + 2 i z cot 2 i z . 我们将获得以下cot z \cot z cot z π \pi π z z z
对于每个复数* z z z 和每个整数 n n n
sin n z = 2 n − 1 sin z sin ( z + π n ) sin ( z + 2 π n ) … sin ( z + ( n − 1 ) π n ) .
\sin nz = 2^{n-1} \sin z \sin \left(z + \frac{\pi}{n}\right) \sin \left(z + \frac{2\pi}{n}\right) \ldots \sin \left(z + \frac{(n-1)\pi}{n}\right).
sin n z = 2 n − 1 sin z sin ( z + n π ) sin ( z + n 2 π ) … sin ( z + n ( n − 1 ) π ) . (2)
确实，可以写
sin n z = e n i z − e − n i z 2 i = e − n i z ( e 2 n i z − 1 ) 2 i = e − n i z ( e 2 i z − 1 ) ( e 2 i z − e − 2 i π / n ) … ( e 2 i z − e − 2 ( n − 1 ) i π / n ) 2 i = A sin z sin ( z + π n ) sin ( z + 2 π n ) … sin ( z + ( n − 1 ) π n )
\sin nz = \frac{e^{niz} - e^{-niz}}{2i} = \frac{e^{-niz}(e^{2niz} - 1)}{2i}
= \frac{e^{-niz}(e^{2iz} - 1)(e^{2iz} - e^{-2i\pi/n}) \ldots (e^{2iz} - e^{-2(n-1)i\pi/n})}{2i}
= A \sin z \sin \left( z + \frac{\pi}{n} \right) \sin \left( z + \frac{2\pi}{n} \right) \ldots \sin \left( z + \frac{(n-1)\pi}{n} \right)
sin n z = 2 i e ni z − e − ni z = 2 i e − ni z ( e 2 ni z − 1 ) = 2 i e − ni z ( e 2 i z − 1 ) ( e 2 i z − e − 2 iπ / n ) … ( e 2 i z − e − 2 ( n − 1 ) iπ / n ) = A sin z sin ( z + n π ) sin ( z + n 2 π ) … sin ( z + n ( n − 1 ) π ) 哪里
A = ( 2 i ) n − 1 e − i π n ( 1 + 2 + … + ( n − 1 ) ) = ( 2 i ) n − 1 e − i ( n − 1 ) π 2 = 2 n − 1 .
A = (2i)^{n-1} e^{-\frac{i\pi}{n}(1+2+\ldots+(n-1))} = (2i)^{n-1} e^{-i(n-1)\frac{\pi}{2}} = 2^{n-1}.
A = ( 2 i ) n − 1 e − n iπ ( 1 + 2 + … + ( n − 1 )) = ( 2 i ) n − 1 e − i ( n − 1 ) 2 π = 2 n − 1 .
每整数 n 人拥有*
sin π n sin 2 π n … sin ( n − 1 ) π n = n 2 n − 1 .
\sin \frac{\pi}{n} \sin \frac{2\pi}{n} \ldots \sin \frac{(n-1)\pi}{n} = \frac{n}{2^{n-1}}.
sin n π sin n 2 π … sin n ( n − 1 ) π = 2 n − 1 n . (3)
只要将(2)两边的除除以sin z \sin z sin z z z z
对于每个奇数整数 n = 2 m + 1 n = 2m + 1 n = 2 m + 1 z z z n z nz n z π \pi π
cot n z = ( − 1 ) m cot z cot ( z + π n ) … cot ( z + ( n − 1 ) π n ) .
\cot nz = (-1)^m \cot z \cot \left( z + \frac{\pi}{n} \right) \ldots \cot \left( z + \frac{(n-1)\pi}{n} \right).
cot n z = ( − 1 ) m cot z cot ( z + n π ) … cot ( z + n ( n − 1 ) π ) . (4)
事实上，(2) 中 sin n ( z + π 2 ) = sin ( n z + π 2 + m π ) = ( − 1 ) m cos n z \sin n \left( z + \frac{\pi}{2} \right) = \sin \left( nz + \frac{\pi}{2} + m\pi \right) = (-1)^m \cos nz sin n ( z + 2 π ) = sin ( n z + 2 π + mπ ) = ( − 1 ) m cos n z z z z z + π 2 z + \frac{\pi}{2} z + 2 π
cos n z = ( − 1 ) m 2 n − 1 cos z cos ( z + π n ) … cos ( z + ( n − 1 ) π n )
\cos nz = (-1)^m 2^{n-1} \cos z \cos \left( z + \frac{\pi}{n} \right) \ldots \cos \left( z + \frac{(n-1)\pi}{n} \right)
cos n z = ( − 1 ) m 2 n − 1 cos z cos ( z + n π ) … cos ( z + n ( n − 1 ) π ) (5)
公式 (2) 和 (5) 当 sin n z ≠ 0 \sin nz \neq 0 sin n z = 0
在接下来的所有内容中，我们将始终假设 n = 2 m + 1 n = 2m + 1 n = 2 m + 1
cot n z = ( − 1 ) m ∏ k = − m m cot ( z − k π n ) .
\cot nz = (-1)^m \prod_{k=-m}^{m} \cot \left( z - \frac{k\pi}{n} \right).
cot n z = ( − 1 ) m k = − m ∏ m cot ( z − n k π ) . 或者，一个人有
cot ( z − k π n ) = 1 + tan z tan k π n tan z − tan k π n
\cot \left( z - \frac{k\pi}{n} \right) = \frac{1 + \tan z \tan \frac{k\pi}{n}}{\tan z - \tan \frac{k\pi}{n}}
cot ( z − n k π ) = tan z − tan n k π 1 + tan z tan n k π 对于 tan z 有限；因此，cot n z nz n z u = tan z u = \tan z u = tan z n − 1 n - 1 n − 1 n n n n n n tan k π / n \tan k\pi / n tan k π / n
cot n z = ∑ k = − m m A k u − tan k π n
\cot nz = \sum_{k=-m}^{m} \frac{A_k}{u - \tan \frac{k\pi}{n}}
cot n z = k = − m ∑ m u − tan n k π A k 哪里
A k = lim z → k π / n cot n z ( tan z − tan k π n ) = lim z → k π / n cos n z sin n z sin ( z − k π n ) cos z cos k π n
A_k = \lim_{z \to k\pi/n} \cot nz \left( \tan z - \tan \frac{k\pi}{n} \right) = \lim_{z \to k\pi/n} \frac{\cos nz}{\sin nz} \frac{\sin \left( z - \frac{k\pi}{n} \right)}{\cos z \cos \frac{k\pi}{n}}
A k = z → k π / n lim cot n z ( tan z − tan n k π ) = z → k π / n lim sin n z cos n z cos z cos n k π sin ( z − n k π ) = lim h → 0 cos n h cos k π n cos ( h + k π n ) sin h sin n h = 1 n cos 2 k π n
= \lim_{h \to 0} \frac{\cos nh}{\cos \frac{k\pi}{n} \cos \left( h + \frac{k\pi}{n} \right)} \frac{\sin h}{\sin nh} = \frac{1}{n \cos^2 \frac{k\pi}{n}}
= h → 0 lim cos n k π cos ( h + n k π ) cos nh sin nh sin h = n cos 2 n k π 1 从而，将 (6) 相应的中的项分离到 k = 0 k = 0 k = 0 k k k z z z z / n z/n z / n
cot z = 1 n tan z n + ∑ k = 1 m 2 n tan z n cos 2 k π n ( n tan z n ) 2 − ( n sin k π n ) 2
\cot z = \frac{1}{n \tan \frac{z}{n}} + \sum_{k=1}^{m} \frac{2n \tan \frac{z}{n}}{\cos^2 \frac{k\pi}{n} \left( n \tan \frac{z}{n} \right)^2 - \left( n \sin \frac{k\pi}{n} \right)^2}
cot z = n tan n z 1 + k = 1 ∑ m cos 2 n k π ( n tan n z ) 2 − ( n sin n k π ) 2 2 n tan n z 对于每个复数 z z z π / 2 \pi/2 π /2
cot z = 1 n tan z n + ∑ k = 1 ∞ v k ( n , z )
\cot z = \frac{1}{n \tan \frac{z}{n}} + \sum_{k=1}^{\infty} v_k(n, z)
cot z = n tan n z 1 + k = 1 ∑ ∞ v k ( n , z ) v k ( n , z ) = 0 v_k(n, z) = 0 v k ( n , z ) = 0 k > m k > m k > m
v k ( n , z ) = 2 n tan z n cos 2 k π n ( n tan z n ) 2 − ( n sin k π n ) 2
v_k(n, z) = \frac{2n \tan \frac{z}{n}}{\cos^2 \frac{k\pi}{n} \left( n \tan \frac{z}{n} \right)^2 - \left( n \sin \frac{k\pi}{n} \right)^2}
v k ( n , z ) = cos 2 n k π ( n tan n z ) 2 − ( n sin n k π ) 2 2 n tan n z 对于1 ≤ k ≤ m 1 \leq k \leq m 1 ≤ k ≤ m C \mathbf{C} C z z z π \pi π n n n v k ( n , z ) v_k(n, z) v k ( n , z ) n n n + ∞ +\infty + ∞ tan z n \tan \frac{z}{n} tan n z z n \frac{z}{n} n z M > 0 M > 0 M > 0 m m m z ∈ K z \in K z ∈ K ∣ n tan z n ∣ ≤ M \left| n \tan \frac{z}{n} \right| \leq M n tan n z ≤ M 0 ≤ x ≤ π / 2 0 \leq x \leq \pi/2 0 ≤ x ≤ π /2 sin x / x ≥ 1 − x 2 6 ≥ 1 2 \sin x / x \geq 1 - \frac{x^2}{6} \geq \frac{1}{2} sin x / x ≥ 1 − 6 x 2 ≥ 2 1 1 ≤ k ≤ m 1 \leq k \leq m 1 ≤ k ≤ m n sin k π n ≥ k π / 2 n \sin \frac{k\pi}{n} \geq k\pi/2 n sin n k π ≥ k π /2 m m m k k k k π / 2 > M k\pi /2 > M k π /2 > M ∣ v k ( n , z ) ∣ ≤ 8 M k 2 π 2 − 4 M 2 |v_k(n, z)| \leq \frac{8M}{k^2 \pi^2 - 4M^2} ∣ v k ( n , z ) ∣ ≤ k 2 π 2 − 4 M 2 8 M k k k v k ( n , z ) v_k(n, z) v k ( n , z ) 2 z z 2 − k 2 π 2 \frac{2z}{z^2 - k^2 \pi^2} z 2 − k 2 π 2 2 z n n n + ∞ +\infty + ∞
cot z = 1 z + ∑ n = 1 ∞ 2 z z 2 − n 2 π 2
\cot z = \frac{1}{z} + \sum_{n=1}^{\infty} \frac{2z}{z^2 - n^2 \pi^2}
cot z = z 1 + n = 1 ∑ ∞ z 2 − n 2 π 2 2 z （右侧的级数通常收敛于每个紧凑子集K ⊂ C K \subset \mathbf{C} K ⊂ C π \pi π cot z \cot z cot z
2. 欧拉展开 OF sin z \sin z sin z
对于 VI、p. 283 的每个 奇数 整数 n = 2 m + 1 n = 2m + 1 n = 2 m + 1 z z z
sin n z = ( − 1 ) m 2 n − 1 ∏ k = − m m sin ( z − k π n )
\sin nz = (-1)^m 2^{n-1} \prod_{k=-m}^{m} \sin \left( z - \frac{k\pi}{n} \right)
sin n z = ( − 1 ) m 2 n − 1 k = − m ∏ m sin ( z − n k π ) = ( − 1 ) m 2 n − 1 sin z ∏ k = 1 m sin ( z − k π n ) sin ( z + k π n ) .
= (-1)^m 2^{n-1} \sin z \prod_{k=1}^{m} \sin \left( z - \frac{k\pi}{n} \right) \sin \left( z + \frac{k\pi}{n} \right).
= ( − 1 ) m 2 n − 1 sin z k = 1 ∏ m sin ( z − n k π ) sin ( z + n k π ) . 现在，一个人有sin ( z − k π n ) sin ( z + k π n ) = sin 2 z − sin 2 k π n \sin \left( z - \frac{k\pi}{n} \right) \sin \left( z + \frac{k\pi}{n} \right) = \sin^2 z - \sin^2 \frac{k\pi}{n} sin ( z − n k π ) sin ( z + n k π ) = sin 2 z − sin 2 n k π ∏ k = 1 m sin 2 k π n = n 2 n − 1 \prod_{k=1}^{m} \sin^2 \frac{k\pi}{n} = \frac{n}{2^{n-1}} ∏ k = 1 m sin 2 n k π = 2 n − 1 n z z z z / n z/n z / n
sin z = n sin z n ∏ k = 1 m ( 1 − sin 2 z n sin 2 k π n ) .
\sin z = n \sin \frac{z}{n} \prod_{k=1}^{m} \left( 1 - \frac{\sin^2 \frac{z}{n}}{\sin^2 \frac{k\pi}{n}} \right).
sin z = n sin n z k = 1 ∏ m ( 1 − sin 2 n k π sin 2 n z ) . 我们可以将这个公式写为sin z = n sin z n ∏ k = 1 m ( 1 − w k ( n , z ) ) \sin z = n \sin \frac{z}{n} \prod_{k=1}^{m} (1 - w_k(n, z)) sin z = n sin n z ∏ k = 1 m ( 1 − w k ( n , z )) k > m k > m k > m w k ( n , z ) = 0 w_k(n, z) = 0 w k ( n , z ) = 0 1 ≤ k ≤ m 1 \leq k \leq m 1 ≤ k ≤ m w k ( n , z ) = sin 2 z n sin 2 k π n w_k(n, z) = \frac{\sin^2 \frac{z}{n}}{\sin^2 \frac{k\pi}{n}} w k ( n , z ) = s i n 2 n k π s i n 2 n z C \mathbf{C} C K K K z z z n n n w k ( n , z ) w_k(n, z) w k ( n , z ) n n n + ∞ +\infty + ∞ n sin z n n \sin \frac{z}{n} n sin n z K K K z z z M > 0 M > 0 M > 0 ∣ n sin z n ∣ ≤ M \left| n \sin \frac{z}{n} \right| \leq M n sin n z ≤ M n n n z ∈ K z \in K z ∈ K 1 ≤ k ≤ m 1 \leq k \leq m 1 ≤ k ≤ m n sin k π n ≥ k π 2 n \sin \frac{k \pi}{n} \geq \frac{k \pi}{2} n sin n k π ≥ 2 k π k k k k π / 2 ≥ M k \pi / 2 \geq M k π /2 ≥ M ∣ w k ( n , z ) ∣ ≤ 4 M 2 / k 2 π 2 |w_k(n, z)| \leq 4M^2 / k^2 \pi^2 ∣ w k ( n , z ) ∣ ≤ 4 M 2 / k 2 π 2 k k k w k ( n , z ) w_k(n, z) w k ( n , z ) z 2 / k 2 π 2 z^2 / k^2 \pi^2 z 2 / k 2 π 2 n n n + ∞ +\infty + ∞
每有复数 z z z
sin z = z ∏ n = 1 ∞ ( 1 − z 2 n 2 π 2 )
\sin z = z \prod_{n=1}^{\infty} \left( 1 - \frac{z^2}{n^2 \pi^2} \right)
sin z = z n = 1 ∏ ∞ ( 1 − n 2 π 2 z 2 )
右边的无穷乘积绝对一致地收敛于C \mathbf{C} C sin z \sin z sin z
3. 应用于伯努利数
定理 1 of VI, p. 286，表明对于 0 ≤ x < π 0 \leq x < \pi 0 ≤ x < π
2 x n 2 π 2 − x 2 ≥ 0
\frac{2x}{n^2 \pi^2 - x^2} \geq 0
n 2 π 2 − x 2 2 x ≥ 0 收敛。另一方面，对于每个复数z z z ∣ z ∣ < π |z| < \pi ∣ z ∣ < π
2 z n 2 π 2 − z 2 = 2 z n 2 π 2 ∑ k = 0 ∞ z 2 k n 2 k π 2 k
\frac{2z}{n^2 \pi^2 - z^2} = \frac{2z}{n^2 \pi^2} \sum_{k=0}^{\infty} \frac{z^{2k}}{n^{2k} \pi^{2k}}
n 2 π 2 − z 2 2 z = n 2 π 2 2 z k = 0 ∑ ∞ n 2 k π 2 k z 2 k 右边的级数绝对收敛。由此我们可以推断出“双”级数
∑ n = 1 ∞ ∑ k = 1 ∞ − 2 z 2 k − 1 n 2 k π 2 k
\sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{-2z^{2k-1}}{n^{2k} \pi^{2k}}
n = 1 ∑ ∞ k = 1 ∑ ∞ n 2 k π 2 k − 2 z 2 k − 1 在开盘∣ z ∣ < π |z| < \pi ∣ z ∣ < π cot z − 1 z \cot z - \frac{1}{z} cot z − z 1 ∣ z ∣ ≤ a < π |z| \leq a < \pi ∣ z ∣ ≤ a < π 2 a 2 k − 1 / n 2 k π 2 k 2a^{2k-1} / n^{2k} \pi^{2k} 2 a 2 k − 1 / n 2 k π 2 k 2 a 2 k − 1 / n 2 k π 2 k 2a^{2k-1} / n^{2k} \pi^{2k} 2 a 2 k − 1 / n 2 k π 2 k ∑ n = 1 ∞ 2 a n 2 π 2 − a 2 \sum_{n=1}^{\infty} \frac{2a}{n^2 \pi^2 - a^2} ∑ n = 1 ∞ n 2 π 2 − a 2 2 a k k k n n n ∑ n = 1 ∞ 2 z z 2 − n 2 π 2 \sum_{n=1}^{\infty} \frac{2z}{z^2 - n^2 \pi^2} ∑ n = 1 ∞ z 2 − n 2 π 2 2 z
如果现在一个和级数 (11) 首先超过 n n n k k k ∣ z ∣ < π |z| < \pi ∣ z ∣ < π
cot z − 1 z = − 2 ∑ k = 1 ∞ S 2 k π 2 k z 2 k − 1
\cot z - \frac{1}{z} = -2 \sum_{k=1}^{\infty} \frac{S_{2k}}{\pi^{2k}} z^{2k-1}
cot z − z 1 = − 2 k = 1 ∑ ∞ π 2 k S 2 k z 2 k − 1 我们放置S k = ∑ n = 1 ∞ 1 n k S_k = \sum_{n=1}^{\infty} \frac{1}{n^k} S k = ∑ n = 1 ∞ n k 1 ∣ z ∣ < 2 π |z| < 2\pi ∣ z ∣ < 2 π
b 2 n = ( − 1 ) n − 1 ( 2 n ) ! 2 S 2 n ( 2 π ) 2 n for n ≥ 1 ,
b_{2n} = (-1)^{n-1} (2n)! \frac{2 S_{2n}}{(2\pi)^{2n}} \quad \text{for } n \geq 1,
b 2 n = ( − 1 ) n − 1 ( 2 n )! ( 2 π ) 2 n 2 S 2 n for n ≥ 1 , 公式特别表明数 S 2 n / π 2 n S_{2n}/\pi^{2n} S 2 n / π 2 n S k + 1 ≤ S k S_{k+1} \leq S_k S k + 1 ≤ S k k ≥ 2 k \geq 2 k ≥ 2 S k ≤ S 2 = π 2 / 6 ≤ 2 S_k \leq S_2 = \pi^2/6 \leq 2 S k ≤ S 2 = π 2 /6 ≤ 2
2 ( 2 n ) ! ( 2 π ) 2 n ≤ ∣ b 2 n ∣ ≤ 4 ( 2 n ) ! ( 2 π ) 2 n for n ≥ 1.
\frac{2 (2n)!}{(2\pi)^{2n}} \leq |b_{2n}| \leq 4 \frac{(2n)!}{(2\pi)^{2n}} \quad \text{for } n \geq 1.
( 2 π ) 2 n 2 ( 2 n )! ≤ ∣ b 2 n ∣ ≤ 4 ( 2 π ) 2 n ( 2 n )! for n ≥ 1. 从这些不等式中，我们可以推导出伯努利多项式的估计值
B n ( x ) = ∑ k = 0 n ( n k ) b k x n − k ; in particular, for 0 ≤ x ≤ 1 one has
B_n(x) = \sum_{k=0}^n \binom{n}{k} b_k x^{n-k}; \text{ in particular, for } 0 \leq x \leq 1 \text{ one has}
B n ( x ) = k = 0 ∑ n ( k n ) b k x n − k ; in particular, for 0 ≤ x ≤ 1 one has ∣ B n ( x ) ∣ ≤ 4 ∑ k = 0 n ( n k ) k ! ( 2 π ) k = 4 n ! ( 2 π ) n ∑ k = 0 n ( 2 π ) k k ! ≤ 4 e 2 π n ! ( 2 π ) n .
|B_n(x)| \leq 4 \sum_{k=0}^n \binom{n}{k} \frac{k!}{(2\pi)^k} = 4 \frac{n!}{(2\pi)^n} \sum_{k=0}^n \frac{(2\pi)^k}{k!} \leq 4 e^{2\pi} \frac{n!}{(2\pi)^n}.
∣ B n ( x ) ∣ ≤ 4 k = 0 ∑ n ( k n ) ( 2 π ) k k ! = 4 ( 2 π ) n n ! k = 0 ∑ n k ! ( 2 π ) k ≤ 4 e 2 π ( 2 π ) n n ! . 习题
参见习题为 § 2 。