フェルマーの最終定理の証明 (965ﾚｽ)
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945: １３２人目の素数さん [] 09/29(月)22:06 ID:tsfKlyQm(1/8)
r↑(θ,φ) = ( asinθcosφ, asinθsinφ, acosθ )
∂r↑/∂θ↑= ( acosθcosφ, acosθsinφ, -asinθ ).
∂r↑/∂φ↑= ( -asinθsinφ, asinθcosφ, 0 ).
∂r↑ ∂r↑
──×──
∂θ ∂φ
= ( |acosθsinφ -asinθ| |-asinθ acosθcosφ| | acosθcosφ, acosθsinφ|
|asinθcosφ 0 |, | 0 -asinθsinφ|, |-asinθsinφ, asinθcosφ | )
= ( a^2sin^2θcosφ, a^2sin^2θsinφ, a^2cos^2φsinθcosθ+ a^2sin^2φsinθcosθ )
= ( a^2sin^2θcosφ, a^2sin^2θsinφ, a^2sinθcosθ ).
|∂r↑ ∂r↑|
|──×── | = √( a^4sin^4θcos^2φ + a^4sin^4θsin^2φ+ a^4sin^2θcos^2θ)
|∂θ ∂φ |
= √( a^4sin^4θ + a^4sin^2θcos^2θ)
= √( a^4sin^2θ(sin^θ + cos^2θ) )
= √( a^4sin^2θ) = a^2sinθ.
∬_S 1 dS
|∂r↑ ∂r↑|
= ∬_D |──×── |dθdφ
|∂θ ∂φ |
= a^2∬[D] sinθdθdφ
= a^2∫[0,2π]dφ∫[0,π]sinθdθ
= 4πa^2
946: １３２人目の素数さん [] 09/29(月)22:07 ID:tsfKlyQm(2/8)
S = |∂r↑/∂t×∂r↑/∂z|dtdz
=∫[0→3]∫[0→2π]2dtdz
=∫[0→3][2t][0→2π]dz
=∫[0→3]4πdz = 4π[z][0→3] = 12π.
S = |∂r↑/∂t×∂r↑/∂z|dtda
=∫[0→2]∫[0→2π]adtda
=∫[0→2][at][0→2π]dz
=∫[0→2]2πa]da = 2π[a^2/2][0→2] = 4π.
947: １３２人目の素数さん [] 09/29(月)22:09 ID:tsfKlyQm(3/8)
A↑= ( f(x,y,z), g(x,y,z), h(x,y,z) )
∫∫∫divA↑dV = ∬A↑･n↑dS ･･････ (#1)
V S
┌ ┐┌ ┐
│∂/∂x││f│
divA↑= ∇･A↑ = │∂/∂y││g│= ∂f/∂x + ∂g/∂y + ∂h/∂z
│∂/∂z││h│
└ ┘└ ┘
α、β、γ（方向余弦）
┌ ┐┌ ┐
│f││cosα│
A↑･n↑ = │g││cosβ│= fcosα + gcosβ + hcosγ
│h││cosγ│
└ ┘└ ┘
∫∫∫(∂f/∂x + ∂g/∂y + ∂h/∂z)dV = ∬(fcosα + gcosβ + hcosγ)dS ･･････ (#2)
V S
∫∫∫∂h/∂z dV = ∬hcosγdS
V S
∫∫∫∂h/∂z dV =∫∫∫∂h(x,y,z)/∂z dzdydx
V V
948: １３２人目の素数さん [] 09/29(月)22:10 ID:tsfKlyQm(4/8)
(sinz)^2 = (1-cos(2z))/2
1/(sinz)^2 = 2/(1-cos(2z))
= 2/{(2z)^2/2! - (2z)^4/4! + (2z)^6/6! - (2z)^8/8! + ･･･}
= (1/z^2)[1/{1-(2z)^2/4! + (2z)^4/6! - (2z)^6/8! + ･･･ }]
A = (2z)^2/4! - (2z)^4/6! + (2z)^6/8! - ･･･
1/(sinz)^2 = (1/z^2)(1+A+A^2+A^3+･･･)
1/z^2 + 1/12 - 11z^2/720 -･･･
z/(e^z-1)
= 1/(1+z/2! + z^2/3! + z^3/4! + ･･･
= 1 -(z/2!+z^2/3!+z^3/4!+･･･) + (z/2!+z^2/3!+z^3/4!+･･･)^2 - (z/2!+z^2/3!+z^3/4!+･･･)^3 + ･･･
= 1- z/2 + z^2/12 - z^4/720 + ･･･
949: １３２人目の素数さん [] 09/29(月)22:12 ID:tsfKlyQm(5/8)
tan z = sin z / cos z
cos z の零点は、z = π/2 + mπ (m は整数)。
tan z は z = π/2 + mπ で一位の極をもつ(tan z の特異点)。
tan z = a/(z - α) + b + c(z - α) + ……
(z - α)tan z = a + b(z - α) + ??……?
α = π/2 + mπ では、z → α の極限を取り、
(z - α)tan z = sin z / {cos z/(z - α)}
→ sin α / cos' α = sin α / (- sin α) = - 1
z-pi/2=u とおくと、
tan(z) = -cos(u)/sin(u)
= (-1/u)*{1-u^2/2!+...}/{1-u^2/3!+...}
= (-1/u)*{1+3u^2+...}.
tan(z) = sin(z) / cos(z)
sin(z) は C 上特異点なし。
cos(z) の 零点 → tan(z) の特異点
cos(z) の 零点 (1/2+n) π
cos(z) = cos ( {z-(1/2+n)π} + (1/2+n)π )
= cos (z-(1/2+n)π)cos (1/2+n)π - sin(z-(1/2+n)π)sin (1/2+n)π
= (-1)^(n+1) sin (z-(1/2+n)π)
よって、
lim[z→(1/2+n)π](z-(1/2+n)π) / cos(z)
= (-1)^(n+1) lim[z→(1/2+n)π](z-(1/2+n)π) / sin(z - (1/2+n)π)
tan(z) = i{exp(2iz)-1}/{exp(2iz)+1}
z = (1/2+n)π
A = lim{z-(1/2+n)π}tan(z)
= lim i{exp(2iz)-1}/[{exp(2iz)+1}/{z-(1/2+n)π}]
A = 1
950: １３２人目の素数さん [] 09/29(月)22:13 ID:tsfKlyQm(6/8)
∫[0→∞] (sin(x)/x)^2 dx
∫[0→∞] (sin(x)/x)^2 dx
= ∫[0→∞] (1/x^2) sin(x)^2 dx
= ∫[0→∞] (- 1/x)'sin(x)^2 dx
= [(- 1/x) sin(x)^2][0→∞] - ∫[0→∞] (- 1/x) (sin(x)^2)'dx
[(- 1/x) sin(x)^2][0→∞]
= [- sin(x)^2/x][0→∞]
= lim[x→∞](- sin(x)^2/x) - lim[x→0](- sin(x)^2/x)
= lim[x→∞](- sin(x)^2/x) - lim[x→0](- (sin(x)/x)^2 x)
= (- 0) - (- 1･0)
= 0
∫[0→∞] (- 1/x) (sin(x)^2)'dx
= ∫[0→∞] (- 1/x) 2 sin(x) cos(x) dx
= ∫[0→∞] (- 1/x) sin(2x) dx
= -∫[0→∞] sin(2x)/x dx
= -∫[0→∞] (sin(2x)/(2x))･2 dx
= -∫[0→∞] (sin(2x)/(2x)) (d (2x)/dx) dx
2x = t
∫[0→∞] (- 1/x) (sin(x)^2)'dx
= -∫[0→∞] (sin(t)/t) (dt/dx) dx
= -∫[0→∞] (sin(t)/t) dt
= -π/2
∫[0→∞] (sin(x)/x)^2 dx
= 0 - (- π/2)
= π/2
951: １３２人目の素数さん [] 09/29(月)22:15 ID:tsfKlyQm(7/8)
I=∫[0→∞] sin2x cosx (1/x2)dx = (1/2)∫[0→∞] (sin2x sinx)(1/x2)dx
=(1/2){ [sin2x sinx(-1/x)][∞,0]
- ∫[0→∞] (2cos2x sinx + sin2x cosx) (-1/x)dx }
=(1/2)[ 0+∫[0→∞] (1/2){(sin3x-sinx) + (sin3x+sinx)}/x dx ]
=(1/4)∫[0→∞] (2sin3x)/x dx = (1/2)∫[0→∞] (sin3x)/(3x) d(3x)
=π/4・・・・・?
∫[0→∞] sin3x (1/x3)dx
=[sin3x (-1/2x2)][∞,0] - ∫[0→∞] 3sin2x cosx (-1/2x2)dx
(sin3x)/x2=sinx(sinx/x)2 → 0・1=0 (x→0) )
=-0+0+(3/2)∫[0→∞] (sin2x cosx)/x2 dx
=(3/2)∫[0→∞] (sin2x cosx)/x2 dx
= (3/2)I = 3π/8 (?から)
952: １３２人目の素数さん [] 09/29(月)22:16 ID:tsfKlyQm(8/8)
?[0→∞]x^2/(x^2+1)^3dx＄
f(z) = z^2/(z^2+1)^3
Res[f(z),i]
= (1/2)lim_{z→i}{z^2/(z+i)^3}"
= (1/2)lim_{z→i}[2{z/(z+i)^3}'-3{z^2/(z+i)^4}']
= (1/2)lim_{z→i}[2{1/(z+i)^3-3z/(z+i)^4}-3{2z/(z+i)^4-4z^2/(z+i)^5}]
= (1/2)lim_{z→i}[2/(z+i)^3-12z/(z+i)^4+12z^2/(z+i)^5]
= lim_{z→i}[(z+i)^2-6z(z+i)+6z^2]/(z+i)^5
= lim_{z→i}(z^2-4iz-1)/(z+i)^5
= -i/16
∴∫_{C}f(z)dz
= i2πRes[f(z),i]
= π/8.
∫_{-R〜R}f(z)dz+∫_{Γ}f(z)dz = π/8
lim_{R→∞}∫_{Γ}f(z)dz
= lim_{R→∞}∫_{0〜π}[ie^(3it)/{Re^(2it)+1/R}^3]dt = 0
∴∫_{-∞〜∞}x^2/(x^2+1)^3dx = π/8.
∫_{0〜∞}x^2/(x^2+1)^3dx = π/16.
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