由 YONG 於 星期四 一月 04, 2007 10:52 am
由 asmobia 於 星期二 二月 13, 2007 10:54 am
YONG 寫到:
1)∫π/2
0 (sinx)n次方dx={ （(n-1)...5.3.1 .(π/2)） ,n even,n>=2
n...6.4.2
(n-1)...4.2
n...5.3 } ,n odd,n>=3.these formulas are known as the wallis sine formulas.
首先假設 f( n ) = ∫π/2 (sinx)^n dx
0
然後要知道換基底公式:
∫π/2 u * dv = ( u * v )| π/2 - ∫π/2 v * du
0 0 0
令 u = (sinx) ^ (n-1), 再令 dv = sinx * dx
則算出 du = (n-1) * (sinx)^(n-2) * (cosx) * dx
又算出 v = -cosx
且算出 ( u * v )| π/2 = 0
0
所以:
f(n)
=∫π/2 (sinx)^n dx = ∫π/2 u * dv
0 0
=( u * v )| π/2 - ∫π/2 v * du (換基底公式)
0 0
= 0 - ∫π/2 v * du
0
= - ∫π/2 -cosx * (n-1) * (sinx)^(n-2) * (cosx) * dx
0
= (n-1) ∫π/2 cosx^2 * (sinx)^(n-2) * dx
0
= (n-1) ∫π/2 ( 1-sinx^2 ) * (sinx)^(n-2) * dx
0
= (n-1) ∫π/2 ( (sinx)^(n-2) - (sinx)^n ) * dx
0
= (n-1) ∫π/2 ( (sinx)^(n-2) - (sinx)^n ) * dx
0
= (n-1) ∫π/2 (sinx)^(n-2) * dx - (n-1) ∫π/2 (sinx)^n * dx
0 0
= (n-1) * f(n-2) - (n-1) * f(n)
最後一行和第一行移項:
n * f(n) = (n-1) * f(n-2)
f(n) = ( (n-1)/n ) * f(n-2)
因為:
f(n) = ( (n-1)/n ) * f(n-2)
所以:
f(n) = ( (n-1)/n ) * f(n-2)
= ( (n-1)/n ) * ( (n-3)/(n-2) ) * f(n-4)
= ( (n-1)/n ) * ( (n-3)/(n-2) ) * ( (n-5)/(n-4) )* f(n-6)
(n-1)(n-3)(n-5)......(n-p+1)
= ----------------------------- f(n-p) ,p<=n
(n)(n-2)(n-4)......(n-p+2)
將 n = 0 帶入原方程式: ∫π/2 dx = π/2
0
將 n = 1 帶入原方程式: ∫π/2 sinx * dx = -cosx | π/2 = 1
0 0
故得証.
