フェルマーの最終定理の証明 (981ﾚｽ)
フェルマーの最終定理の証明 http://rio2016.5ch.net/test/read.cgi/math/1745314067/
上
下
前次
1-
新
通常表示
512ﾊﾞｲﾄ分割
ﾚｽ栞
抽出解除
必死ﾁｪｯｶｰ(本家)
(べ)
自ID
ﾚｽ栞
あぼーん
ﾘﾛｰﾄﾞ規制
です｡10分ほどで解除するので､
他のﾌﾞﾗｳｻﾞ
へ避難してください｡
945: １３２人目の素数さん [] 2025/09/29(月) 22:06:47.92 ID:tsfKlyQm 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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/945
946: １３２人目の素数さん [] 2025/09/29(月) 22:07:21.17 ID:tsfKlyQm 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. http://rio2016.5ch.net/test/read.cgi/math/1745314067/946
947: １３２人目の素数さん [] 2025/09/29(月) 22:09:12.52 ID:tsfKlyQm 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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/947
948: １３２人目の素数さん [] 2025/09/29(月) 22:10:51.59 ID:tsfKlyQm (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 + ･･･ http://rio2016.5ch.net/test/read.cgi/math/1745314067/948
949: １３２人目の素数さん [] 2025/09/29(月) 22:12:27.99 ID:tsfKlyQm 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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/949
950: １３２人目の素数さん [] 2025/09/29(月) 22:13:58.19 ID:tsfKlyQm ∫[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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/950
951: １３２人目の素数さん [] 2025/09/29(月) 22:15:14.02 ID:tsfKlyQm 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) http://rio2016.5ch.net/test/read.cgi/math/1745314067/951
952: １３２人目の素数さん [] 2025/09/29(月) 22:16:29.93 ID:tsfKlyQm ?[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. http://rio2016.5ch.net/test/read.cgi/math/1745314067/952
ﾒﾓ帳
(0/65535文字)
上
下
前次
1-
新
書
関
写
板
覧
索
設
栞
歴
ｽﾚ情報
赤ﾚｽ抽出
画像ﾚｽ抽出
歴の未読ｽﾚ
AAｻﾑﾈｲﾙ
Google検索
Wikipedia
ぬこの手
ぬこTOP
0.035s