一道比較陰間的級數題解答

今天晚上水課時候同學給發來一道十分陰間的題，

當然可能我這種菜雞覺得陰間有些大佬覺得已經夠簡單了，但是我還是這麼不要臉的發出來，也算水水活躍度

第一次寫出來小錯不斷，這水部落格時候又改了改，但是不免有錯，是以如果有錯請諸位大佬不吝賜教

這關鍵是裡面那個乘積挺難處理的，下面處理一下

引理1

x 2 n + 1 − 1 = ( x − 1 ) ( x − α 1 ) ( x − α 2 ) . . . . ( x − α n − 1 ) = ( x − 1 ) ∏ k = 1 n ( x 2 − 2 x cos ⁡ 2 k π 2 n + 1 + 1 ) x^{2n+1}-1=(x-1)(x-\alpha_1)(x-\alpha_2)....(x-\alpha_{n-1})\\ =(x-1)\prod_{k=1}^{n}(x^2-2x\cos \frac{2k\pi}{2n+1}+1) x2n+1−1=(x−1)(x−α1​)(x−α2​)....(x−αn−1​)=(x−1)k=1∏n​(x2−2xcos2n+12kπ​+1)

這個比較好證明，把複數根全求出來而後首位配對

同理

x 2 n − 1 = ( x − 1 ) ( x + 1 ) ∏ k = 1 n − 1 ( x 2 − 2 x cos ⁡ k π n ) x^{2n}-1=(x-1)(x+1)\prod_{k=1}^{n-1}(x^2-2x\cos \frac{k\pi}{n}) x2n−1=(x−1)(x+1)k=1∏n−1​(x2−2xcosnkπ​)

引理2

cos ⁡ π 2 n + 1 cos ⁡ 2 π 2 n + 1 . . . . cos ⁡ n π 2 n + 1 = 1 2 n \cos\frac{\pi}{2n+1}\cos \frac{2\pi}{2n+1}....\cos \frac{n\pi}{2n+1}=\frac{1}{2^n} cos2n+1π​cos2n+12π​....cos2n+1nπ​=2n1​

引理1中帶進去 x = − 1 x=-1 x=−1

− 2 = ( − 2 ) ∏ k = 1 n ( 2 + 2 cos ⁡ 2 k π 2 n + 1 ) ⇒ 1 = ∏ k = 1 n ( 2 + 2 cos ⁡ 2 k π 2 n + 1 ) ⇒ 1 2 n = ∏ k = 1 n ( 1 + cos ⁡ 2 k π 2 n + 1 ) ⇒ ∏ k = 1 n cos ⁡ k π 2 n + 1 = 1 2 n -2=(-2)\prod_{k=1}^{n}(2+2\cos \frac{2k\pi}{2n+1})\Rightarrow \\1=\prod_{k=1}^{n}(2+2\cos \frac{2k\pi}{2n+1})\Rightarrow\\ \frac{1}{2^n}=\prod_{k=1}^{n}(1+\cos \frac{2k\pi}{2n+1})\Rightarrow\\ \prod_{k=1}^{n}\cos \frac{k\pi}{2n+1}=\frac{1}{2^n} −2=(−2)k=1∏n​(2+2cos2n+12kπ​)⇒1=k=1∏n​(2+2cos2n+12kπ​)⇒2n1​=k=1∏n​(1+cos2n+12kπ​)⇒k=1∏n​cos2n+1kπ​=2n1​

下面開始變形

∏ k = 1 2 n cos ⁡ k π 2 n + 1 = ∏ k = 1 n cos ⁡ k π 2 n + 1 ∏ k = n + 1 2 n cos ⁡ k π 2 n + 1 \prod_{k=1}^{2n}\cos \frac{k\pi}{2n+1}=\prod_{k=1}^{n}\cos \frac{k\pi}{2n+1}\prod_{k=n+1}^{2n}\cos \frac{k\pi}{2n+1} k=1∏2n​cos2n+1kπ​=k=1∏n​cos2n+1kπ​k=n+1∏2n​cos2n+1kπ​

其中

∏ k = n + 1 2 n cos ⁡ k π 2 n + 1 = ∏ k = n + 1 2 n sin ⁡ ( π 2 − k π 2 n + 1 ） = sin ⁡ − π 4 n + 2 sin ⁡ − 3 π 4 n + 2 . . . sin ⁡ − ( 2 n + 1 ) π 4 n + 2 = ( − 1 ) n ∏ k = 1 n sin ⁡ ( 2 k − 1 ) π 4 n + 2 ⋅ sin ⁡ π 2 ⋅ ( − 1 ) \\\prod_{k=n+1}^{2n}\cos \frac{k\pi}{2n+1}=\prod_{k=n+1}^{2n}\sin (\frac{\pi}{2}-\frac{k\pi}{2n+1}） \\=\sin \frac{-\pi}{4n+2}\sin\frac{-3\pi}{4n+2}...\sin \frac{-(2n+1)\pi}{4n+2}\\=(-1)^n\prod_{k=1}^{n}\sin\frac{(2k-1)\pi}{4n+2}·\sin \frac{\pi}{2}·(-1) k=n+1∏2n​cos2n+1kπ​=k=n+1∏2n​sin(2π​−2n+1kπ​）=sin4n+2−π​sin4n+2−3π​...sin4n+2−(2n+1)π​=(−1)nk=1∏n​sin4n+2(2k−1)π​⋅sin2π​⋅(−1)

由引理1的做法

x 2 m + 1 = ∏ k = 1 n ( x 2 − 2 x cos ⁡ ( 2 k − 1 ) π 2 n + 1 + 1 ) x^{2m}+1=\prod_{k=1}^{n}(x^2-2x\cos\frac{(2k-1)\pi}{2n+1}+1) x2m+1=k=1∏n​(x2−2xcos2n+1(2k−1)π​+1)

帶入 x = 1 x=1 x=1

2 = ∏ k = 1 n ( 2 − 2 cos ⁡ ( 2 k − 1 ) π 2 n + 1 ) 1 2 n − 1 = ∏ k = 1 n ( 1 − ( 1 − 2 sin ⁡ 2 ( 2 k − 1 ) π 4 n + 2 ) ) 2=\prod_{k=1}^{n}(2-2\cos\frac{(2k-1)\pi}{2n+1})\\ \frac{1}{2^{n-1}}=\prod_{k=1}^{n}(1-(1-2\sin^2\frac{(2k-1)\pi}{4n+2})) 2=k=1∏n​(2−2cos2n+1(2k−1)π​)2n−11​=k=1∏n​(1−(1−2sin24n+2(2k−1)π​))

下面就好弄了

參考書：丘磚