フェルマーの最終定理の証明 (542ﾚｽ)
フェルマーの最終定理の証明 http://rio2016.5ch.net/test/read.cgi/math/1745314067/
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393: １３２人目の素数さん [] 2025/07/18(金) 07:29:57.91 ID:QNG/Z1cz lim┬(?x+i?y→0)??(u(x+?x,y+?y)+iv(x+?x,y+?y)-u(x,y)+iv(x,y))/(?x+i?y)? (1)実軸に平行に近づく場合 ?y=0かつ?x→0 f^' (z)=lim┬(?x→0)??(u(x+?x,y)+iv(x+?x,y)-(u(x,y)+iv(x,y)))/?x? =lim┬(?x→0)?((u(x+?x,y)-u(x,y))/?x+i (v(x+?x,y)-v(x,y))/?x) =∂u/∂x+i ∂v/∂x??? (2)虚軸に平行に近づく場合 ?x=0かつ?y→0 f^' (z)=lim┬(?y→0)??(u(x,y+?y)+iv(x,y+?y)-(u(x,y)+iv(x,y)))/i?y? =lim┬(?y→0)??1/i ((u(x,y+?y)-u(x,y))/?y+i (v(x,y+?y)-v(x,y))/?y)? =lim┬(?y→0)?((v(x,y+?y)-v(x,y))/?y-i (u(x,y+?y)-u(x,y))/?y) (1/i=i/i^2 =-i) =∂v/∂y-i ∂u/∂y??? ∴f^' (z)=∂u/∂x+i ∂v/∂x=∂v/∂y-i ∂u/∂y ∴∂u/∂x=∂v/∂y かつ ∂v/∂x=-∂v/∂y http://rio2016.5ch.net/test/read.cgi/math/1745314067/393
394: １３２人目の素数さん [] 2025/07/18(金) 07:30:33.59 ID:QNG/Z1cz =lim┬(?y→0)??1/i ((u(x,y+?y)-u(x,y))/?y+i (v(x,y+?y)-v(x,y))/?y)? =lim┬(?y→0)?((v(x,y+?y)-v(x,y))/?y-i (u(x,y+?y)-u(x,y))/?y) (1/i=i/i^2 =-i) =∂v/∂y-i ∂u/∂y??? ∴f^' (z)=∂u/∂x+i ∂v/∂x=∂v/∂y-i ∂u/∂y ∴∂u/∂x=∂v/∂y かつ ∂v/∂x=-∂v/∂y http://rio2016.5ch.net/test/read.cgi/math/1745314067/394
395: １３２人目の素数さん [] 2025/07/18(金) 07:31:37.93 ID:QNG/Z1cz 同様にして v(x+?x,y+?y)-v(x,y) =(v_x (x,y)+k_1 )?x+(v_y (x,y)+k_2 )?y??? (?x→0,?y→0⇒k_1→0,k_2→0) ?、?より f(z+?z)-f(z) =u(x+?x,y+?y)-u(x,y)+ i{v(x+?x,y+?y)-v(x,y)} =(u_x (x,y)+h_1 )?x+(u_y (x,y)+h_2 )?y+i{(v_x (x,y)+k_1 )?x+(v_y (x,y)+k_2 )?y} =u_x ?x+h_1 ?x+u_y ?y+h_2 ?y+i{v_x ?x+k_1 ?x+v_y ?y+k_2 ?y} =u_x ?x+u_y ?y+i?v_x ?x+iv_y ?y+h?_1 ?x+h_2 ?y+ik_1 ?x+ik_2 ?y ここで?=h_1 ?x+h_2 ?y+ik_1 ?x+ik_2 ?yとおくと =u_x ?x-v_x ?y+iv_x ?x+iu_x ?y+? (u_x=v_y , v_x=-u_y ) =u_x (?x+i?y)+iv_x (?x+i?y)+? (iv_x i?y=-v_x ?y) =u_x ?z+iv_x ?z+? f(z+?z)-f(z)=u_x ?z+iv_x ?z+? (lim)┬(?z→0)??(f(z+?z)-f(z))/?z?=(lim)┬(?z→0)?(u_x+iv_x+?/?z) ここで ?z→0⇔?x→0かつ?y→0 のとき h_1→0,h_2→0かつ k_1→0,k_2→0 であり、 |?z|^2=|?x|^2+|?y|^2 より | ?x/?z|?1, |?y/?z|?1 であるから (lim)┬(?z→0)?|?/?z|=(lim)┬(?z→0)?|(h_1 ?x+h_2 ?y+ik_1 ?x+ik_2 ?y)/?z| =(lim)┬(?z→0)?|((h_1+ik_1 )?x+(h_2+ik_2 )?y)/?z| ?(lim)┬(?z→0)?(|h_1+ik_1 ||?x/?z|+|h_1+ik_1 ||?x/?z|)=0 したがって (lim)┬(?z→0)??(f(z+?z)-f(z))/?z?=(lim)┬(?z→0)?(u_x+iv_x+?/?z)=u_x+iv_x http://rio2016.5ch.net/test/read.cgi/math/1745314067/395
396: １３２人目の素数さん [] 2025/07/18(金) 07:32:53.33 ID:QNG/Z1cz e^z=e^(x+iy)=e^x e^iy=e^x (cos(y)+isin(y)) =e^x cos(y)+ie^x sin(y) u(x,y)=e^x cos(y) u_x=e^x cos(y), u_y=-e^x sin(y) v(x,y)=e^x sin(y) v_x=e^x sin(y), v_y=e^x cos(y) したがってC-R の方程式 u_x=v_y, v_x=-u_y が成り立つ。 (e^z )^'=e^z より (e^iz )^'=e^iz (iz)^'=ie^iz (e^(-iz) )^'=e^(-iz) (-iz)^'=-ie^(-iz) (sin(z))^'=((e^iz-e^(-iz))/2i)^'=(ie^iz+?ie?^(-iz))/2i=(e^iz+e^(-iz))/2=cos(z) (cos(z))^'=((e^iz+e^(-iz))/2)^'=(ie^iz-?ie?^(-iz))/2=(e^iz-e^(-iz))/2i=sin(z) (tan(z))^'=(sin(z)/cos(z) )^'=(cos(z)cos(z)+sin(z)sin(z))/(?cos?^2 (z) )=1/(?cos?^2 (z) ) http://rio2016.5ch.net/test/read.cgi/math/1745314067/396
397: １３２人目の素数さん [] 2025/07/18(金) 07:34:52.19 ID:QNG/Z1cz (1)半径方向から近づく場合 ?θ=0かつ?r→0 z+?z=re^iθ+?re^iθ=e^iθ (r+?r) f^' (z) =lim┬(?z→0)??(f(z+?z)-f(z))/?z?=lim┬(?θ→0)??(f((r+?r) e^iθ )-f(re^iθ ))/(?re^iθ )? =lim┬(?r→0)??(u(r+?r,θ)+iv(r+?r,θ)-(u(r,θ)+iv(r,θ)))/(?re^iθ )? =1/e^iθ lim┬(?r→0)??(u(r+?r,θ)+iv(r+?r,θ)-u(r,θ)-iv(r,θ))/?r? =1/e^iθ lim┬(?r→0)??(u(r+?r,θ)-u(r,θ))/?r? +i (v(r+?r,θ)-v(r,θ))/?r =1/e^iθ (∂u/∂r+i ∂v/∂r)=1/e^iθ (u_r+iv_r )??? http://rio2016.5ch.net/test/read.cgi/math/1745314067/397
398: １３２人目の素数さん [] 2025/07/18(金) 07:35:28.66 ID:QNG/Z1cz (2) θ方向から近づく場合 ?r=0かつ?θ→0 z+?z=re^i(θ+?θ) =re^iθ e^i?θ ?z=re^iθ e^i?θ-re^iθ=re^iθ (e^i?θ-1) e^i?θ=cos(?θ)+isin(?θ)?1+i?θ ?z=re^iθ (e^i?θ-1)?re^iθ (1+i?θ-1)=re^iθ i?θ f^' (z) =lim┬(?θ→0)??(f((r+?r) e^iθ )-f(re^iθ ))/(re^iθ i?θ)? =1/(re^iθ i) lim┬(?r→0)??(u(r,θ+?θ)+iv(r,θ+?θ)-(u(r,θ)+iv(r,θ)))/?θ? =1/(re^iθ i) lim┬(?θ→0)??(u(r,θ+?θ)-u(r,θ)+iv(r,θ+?θ)-iv(r,θ))/?θ? =-i/(re^iθ ) lim┬(?θ→0)??(u(r,θ+?θ)-u(r,θ))/?θ? +1/(re^iθ ) lim┬(?θ→0) (v(r,θ+?θ)-v(r,θ))/?θ =1/(re^iθ ) (∂v/∂θ-i ∂u/∂θ)=1/(re^iθ ) (v_θ-iu_θ )??? ??より 1/e^iθ u_r=1/(re^iθ ) v_θ u_r=1/r v_θ 1/e^iθ v_r=-1/(re^iθ ) u_θ v_r=-1/r u_θ したがって u_r=1/r v_θ かつ v_r=-1/r u_θ http://rio2016.5ch.net/test/read.cgi/math/1745314067/398
406: １３２人目の素数さん [] 2025/07/18(金) 16:31:24.67 ID:QNG/Z1cz F(ω)=∫_(-∞)^∞??f(t) e^(-jωt) ? dt フーリエ変換 ??? f(t)= F^(-1) [F(ω)]=1/2π ∫_(-∞)^∞??F(ω) e^jωt ? dω 逆フーリエ変換??? f(t)=1, f(t)=t, f(t)= sin(ωt) g(t)={■(0&t<0@f(t) e^(-σt)&t?0)┤ G(ω)=∫_(-∞)^∞??g(t) e^(-jωt) ? dt=∫_0^∞??g(t) e^(-jωt) ? dt=∫_0^∞??f(t) e^(-σt) e^(-jωt) ? dt =∫_0^∞??f(t) e^(-(σ+jω)t) ? dt s=σ+jω F(s)=∫_0^∞??f(t) e^(-st) ? dt ラプラス変換 ??? http://rio2016.5ch.net/test/read.cgi/math/1745314067/406
407: １３２人目の素数さん [] 2025/07/18(金) 16:32:05.10 ID:QNG/Z1cz s=σ+jω ds=jdω ω: -∞ → ∞ s? σ-j∞→σ+j∞ g(t)=1/2π ∫_(-∞)^∞??F(s) e^jωt ? dω=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^jωt ? ds f(t) e^(-σt)=f(t)/e^σt =1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^jωt ? ds f(t)=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^σt e^jωt ? ds=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^(σ+jω)t ? ds ∴f(t)=1/2πj ∫_(σ-j∞)^(σ+j∞)??F(s) e^st ? ds ラプラス逆変換 ??? http://rio2016.5ch.net/test/read.cgi/math/1745314067/407
408: １３２人目の素数さん [] 2025/07/18(金) 16:33:09.63 ID:QNG/Z1cz ∫_0^1?x^m (1-x)^n dx=1/(m+n+1)! I(m,n)=∫_0^1?x^m (1-x)^n dx I(m,n)=∫_0^1?x^m (1-x)^n dx=∫_0^1??(x^(m+1)/(m+1))^' (1-x)^n ? dx =[?( @x^(m+1)/(m+1)@ )(1-x)^n ]_0^1-∫_0^1?x^(m+1)/(m+1) (-n) (1-x)^(n-1) dx =n/(m+1) ∫_0^1?x^(m+1) (1-x)^(n-1) dx=n/(m+1) I(m+1,n-1) I(m,n)=n/(m+1) I(m+1,n-1) =n/(m+1)?(n-1)/(m+2) I(m+2,n-2) =n/(m+1)?(n-1)/(m+2)?(n-2)/(m+3) I(m+3,n-3) =n/(m+1)?(n-1)/(m+2)?(n-2)/(m+3)? ??? ?1/(m+n) I(m+n,0) n/(m+1)?(n-1)/(m+2)?(n-2)/(m+3)? ??? ?1/(m+n) =(1?2?(m-1)m?n(n-1)(n-2)? ??? ?1)/(1?2?(m-1)m(m+1)(m+2)? ??? ?(m+n) )=m!n!/(m+n)! I(m,n)=n/(m+1) I(m+1,n-1) =m!n!/(m+n)! I(m+n,0) =m!n!/(m+n)! ∫_0^1?x^(m+n) (1-x)^0 dx =m!n!/(m+n)! ∫_0^1?x^(m+n) dx=m!n!/(m+n)! [?( @x^(m+n+1)/(m+n+1)@ )]_0^1 =m!n!/(m+n+1)! ∴∫_0^1?x^m (1-x)^n dx=1/(m+n+1)! http://rio2016.5ch.net/test/read.cgi/math/1745314067/408
409: １３２人目の素数さん [] 2025/07/18(金) 16:34:01.63 ID:QNG/Z1cz ∫_α^β??(x-α)^m (β-x)^n ? dx=m!n!/(m+n+1)! (β-α)^(m+n+1) t=(β-α)x+α dt=(β-α)dx dx=dt/(β-α) x:0→1 t:α→β x=(t-α)/(β-α) 1-x=(β-α-(t-α))/(β-α)=(β-t)/(β-α) ∫_0^1?x^m (1-x)^n dx =∫_α^β??((t-α)/(β-α))^m ((β-t)/(β-α))^n ? dt/(β-α)=∫_α^β?((t-α)^m (β-t)^m)/(β-α)^(m+n+1) dt =1/(β-α)^(m+n+1) ∫_α^β??(t-α)^m (β-t)^m ? dt=m!n!/(m+n+1)! ∴∫_α^β??(x-α)^m (β-x)^n ? dx=m!n!/(m+n+1)! (β-α)^(m+n+1) m=1,n=1⇒∫_α^β?(x-α)(x-β) dx=-∫_α^β?(x-α)(β-x) dx =-1/6 (β-α)^3 m=2,n=1⇒∫_α^β?(x-α)(x-β) dx=-∫_α^β??(x-α)^2 (β-x) ? dx =-1/12 (β-α)^4 m=2,n=2⇒∫_α^β??(x-α)^2 (x-β)^2 ? dx=∫_α^β??(x-α)^2 (β-x)^2 ? dx =(2?2)/(5?4?3?2?1) (β-α)^5=1/30 (β-α)^5 http://rio2016.5ch.net/test/read.cgi/math/1745314067/409
411: １３２人目の素数さん [] 2025/07/18(金) 21:14:28.81 ID:QNG/Z1cz x^'' (t)+3x^' (t)+2x(t)=e^2t , x(0)=0, x^' (0)=1 L[x^'' (t)]=s^2 X(s)-sx(0)-x^' (0)=s^2 X(s)-1 3L[x^' (t)]=3(sX(s)-x(0))=3sX(s) 2L[x(t)]=2X(s) L[x^'' (t)]+2L[x^' (t)]+ 5L[x(t)] =s^2 X(s)-1+3sX(s)+2X(s) =X(s)(s^2+3s+2)-1=X(s)(s+1)(s+2)-1 L[e^2t ]=1/(s-2) (L[e^at ]=1/(s-a)) X(s)(s+1)(s+2)-1=1/(s-2) X(s)(s+1)(s+2)=1/(s-2)+1=(1+s-2)/(s-2)=(s-1)/(s-2) X(s)=(s-1)/(s+1)(s+2)(s-2) =A/(s+1)+B/(s+2)+C/(s-2) s-1=A(s+2)(s-2)+B(s+1)(s-2)+C(s+1)(s+2) s=-1⇒-2=A(-3) A=2?3 s=-2⇒-3=B(-1)(-4) 4B=-3 B=-3?4 s=2⇒1=C(3)(4) 12C=1 C=1?12 X(s)=2/3 1/(s+1)-3/4 1/(s+2)+1/12 1/(s-2) L^(-1) [2/3 1/(s-(-1))-3/4 1/(s-(-2))+1/12 1/(s-2)]=2/3 e^(-t)-3/4 e^(-2t)+1/12 e^2t http://rio2016.5ch.net/test/read.cgi/math/1745314067/411
412: １３２人目の素数さん [] 2025/07/18(金) 21:15:23.73 ID:QNG/Z1cz L[y^'' (t)]=s^2 Y(s)-sy(0)-y^' (0) =s^2 Y(s)-2s-4 L[?4y?^' (t)]=4(sY(s)-y(0))=4sY(s)-8 L[4y(t)]=4Y(s) L[y^'' (t)]-L[?4y?^' (t)]+ L[4y(t)] =s^2 Y(s)-2s-4-4sY(s)+8+4Y(s) =Y(s)(s^2-4s+4)-2s+4 L[6te^2t ]=6L[t^1 e^2t ]=6 1!/(s-2)^2 =6/(s-2)^2 ( L[t^n e^at ]=n!/(s-a)^(n+1) ) Y(s)(s^2-4s+4)-2s+4=6/(s-2)^2 Y(s) (s-2)^2-2s+4=6/(s-2)^2 Y(s) (s-2)^2=6/(s-2)^2 +2(s-2) Y(s)=6/(s-2)^4 +2/(s-2) Y(s)= F(s-2)とおくと F(s-2)=6/(s-2)^4 +2/(s-2) ∴F(s)=6/s^4 +2/s=3!/s^(3+1) +2/s y(t)=L^(-1)[F(s-2)]=e^2t L^(-1) [F(s)] ( L^(-1) [F(s-a)]=e^at L^(-1) [F(s)]) =e^2t L^(-1) [3!/s^(3+1) +2/s] (L[t^n ]=n!/s^(n+1) ) =e^2t (t^3+2) http://rio2016.5ch.net/test/read.cgi/math/1745314067/412
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