- 324 名前：１３２人目の素数さん mailto:sage [2008/12/18(木) 02:31:50 ]
- >>315 (1) & >>319
2 > 1.96 = 1.4^2,
a_n = (1+√2)^n + (1-√2)^n,
とおくと
a_n = 2*a_(n-1) + a_(n-2), a_0 = a_1 = 2,
a_n - 1 < (1+√2)^n < a_n + 1,
を満たす。
a_5 = 82 ゆえ、81 < (1+√2)^5 < 83,
>>315 (2)
∫[0,π/2] sin(2x)/{1+sin(x)^2} dx = ∫[0,π/2] 2sin(x)cos(x)/{1+sin(x)^2} dx
= [ log{1+sin(x)^2} ](x=0,π/2)
= log(2)
= 0.69314718055994530941723212145818
cos(x) = z とおくと、
∫[0,π/2] sin(x)/{1+sin(x)^2} dx = ∫[0,1] 1/(2-z^2) dz
= (1/√8)∫[0,1] {1/(√2 -z) + 1/(√2 +z)} dz
= (1/√8) [ log{(√2 +z)/(√2 -z)} ](z=0,1)
= (1/√2) log(√2 +1)/(√2 -1)
= 0.62322524014023051339402008025057
>>316
各面は正ｍ角形、
１つの頂点に集まる面の数をｎ≧3,
とすると、
mF = 2E = nV より V-E+F= (2/n -1 +2/m)E,
{(m-2)/m}π*n < 2π より 2/m -1 +2/n > 0.
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