Re: [中學] 求和
※ 引述《cuttlefish (無聊ing ><^> .o O)》之銘言：
: 簡單的一題, 不過我的方法很醜想問問看有無好方法?
: inf
: 求 1 + Sum (-1)^n * (2n-1)!! / (2n)!!
: n=1
: where k!! = k*(k-2)*....*2, and 1!! = 1
應該是正常解法(?) 以前玩過一兩次 不過那邊的題目自帶遞迴式還好ow o
Let c_0 = 1, c_n = (-1)^n * (2n-1)!! / (2n)!!
Then c_(n+1)/c_n = (-1) (2n+1) / (2n+2), for all n>=0
2 (n+1) c_(n+1) + 2 n c_n + c_n = 0 (*)
inf inf
Let y = sum c_n x^n, y' = sum n c_n x^(n-1)
n=0 n=1
inf
Take sum (*) x^n we have 2y' + 2y'x + y = 0
n=0
(2/y) dy + 1/(1+x) dx = 0
2 ln|y| + ln|1+x| = c
y^2 (1+x) = +-e^c = k
y = k(1+x)^(-1/2)
y(0) = k = c_0 = 1
We want = y(1) = 1/sqrt(2)
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嗯嗯ow o
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感謝 已訂正ow o
※ 編輯: Desperato (140.112.25.105), 06/16/2017 21:05:58
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2 (n+1) c_(n+1) + 2 n c_n + c_n = 0
inf inf inf
sum 2 (n+1) c_(n+1) x^n + sum 2 n c_n x^n + sum c_n x^n= 0
n=0 n=0 n=0
inf inf inf
2 sum m c_m x^(m-1) + 2x sum n c_n x^(n-1) + sum c_n x^n= 0
m=0 n=0 n=0
2 (y') + 2x (y') + y = 0
差一項的問題 應該和無窮等比差不多概念吧ow o
等比乘r相減也老是前後差一項啊XD
可是無窮等比就是只會差第一項 不會差最後一項(因為沒有這東西)
這題是因為差的第一項是0 所以看起來沒差
※ 編輯: Desperato (140.112.25.105), 06/16/2017 22:08:11
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