フェルマーの最終定理の証明 (726ﾚｽ)
上下前次1-新
抽出解除 必死ﾁｪｯｶｰ(本家) (べ) 自ID ﾚｽ栞 あぼーん
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693: 08/22(金)06:44 ID:aTp7UHTZ(1/9) AAS
y''(t) - 3y'(t) + 2y(t) = e^(-t) ･･･････?（初期条件）y(0) = 1/6, y'(0) = 5/6
L[y''(t)] = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - s/6 - 5/6
L[3y'(t)] = 3( sY(s) - y(0) ) = 3sY(s) - 1/2
L[2y(t)] = 2Y(s)
L[e^(-t)] = 1/(s + 1)
s^2Y(s) - s/6 - 5/6 - (3sY(s) -1/2) + 2Y(s) = 1/(s+1)
Y(s)(s^2 - 3s + 2) - s/6 -1/3 = 1/(s+1)
Y(s)(s-1)(s-2) = s/6+1/3+1/(s+1) = (s(s+1)+2(s+1)+6)/6(s+1) = (s^2 + 3s + 8)/6(s+1)
Y(s) = (s^2 + 3s + 8)/6(s+1)(s-1)(s-2) = A/(s+1) + B/(s-1) + C/(s-2)
s^2 + 3s + 8 = 6( A(s-1)(s-2) + B(s+1)(s-2) + C(s+1)(s-1) )
s = -1 のとき 1 - 3 + 8 = 6A(-2)(-3) 36A = 6 A = 1/6
s = 1 のとき 1 + 3 + 8 = 6B(2)(-1) -12B = 12 B = -1
s = 2 のとき 4 + 6 + 8 = 6C(3)(1) 18C = 18 C = 1
Y(s) = 1/6(s+1) - 1/(s-1) + 1/(s-2)
y(t) = -e^t + e^(2t) + (1/6)e^(-t)
694: 08/22(金)06:45 ID:aTp7UHTZ(2/9) AAS
k^2 -3k + 2 = (k-1)(k-2) = 0 k = 1, 2
なので
y''(t) - 3'y(t) + 2y(t) = 0
の一般解 y0 は
y0 = C1e^t + C2e^(2t)
?の特殊解をv(t)とすると
v(t) = 1/(D-1)(D-2)*e^(-t)
= 1/(D-2)*e^(-t) - 1/(D-1)*e^(-t)
= (-1/3)e^(-t) + (1/2)e^(-t) = (1/6)e^(-t)
よって?の一般解は
y(t) = C1e^t + C2e^(2t) + (1/6)e^(-t)
y(0) = C1 + C2 + 1/6 = 1/6
C1 + C2 = 0 …… ?
y'(t) = C1e^t + C2*2e^(2t) - (1/6)e^(-t)
y'(0) = C1 + C2*2 - 1/6 = 5/6
C1+ 2C2 = 1……?
??より
C1 = -1, C2= 1
初期値を満たす特殊解を改めて y とおくと
y(t) = -e^t +e^(2t) + (1/6)e^(-t)
695: 08/22(金)07:19 ID:aTp7UHTZ(3/9) AAS
D^2+1)y=1/(?cos?^3 (x) )
(D^2+1)y=0
λ^2+1=0 λ=0±i
y_0=e^(-0) (C_1 cos(x)+C_2 sin(x))=C_1 cos(x)+C_2 sin(x)
cos(x)=((e^ix+e^(-ix))/2)
1/(cos^3(x))=(2/(e^ix+e^(-ix) ))^3=8/(e^ix+e^(-ix) )^3
(D^2+1) y_s=8/(e^ix+e^(-ix) )^3
(D+i)(D-i) y_s=8/(e^ix+e^(-ix) )^3
y_s=(1/(D+i))(1/(D-i)) 8/(e^ix+e^(-ix) )^3
1/(D-i) 8/(e^ix+e^(-ix) )^3 =8e^ix 1/D e^(-ix) 1/(e^ix+e^(-ix) )^3
=8e^ix ∫e^(-ix)/(e^ix+e^(-ix) )^3 dx
e^(-ix)/(e^ix+e^(-ix) )^3 =(e^3ix e^(-ix))/(e^3ix (e^ix+e^(-ix) )^3 )=e^2ix/((e^ix )^3 (e^ix+e^(-ix) )^3 )
=e^2ix/(e^ix (e^ix+e^(-ix) ))^3 =e^2ix/(e^2ix+1)^3
∴1/(D-i) 8/(e^ix+e^(-ix) )^3 =8e^ix ∫e^(-2ix)/(e^2ix+1)^3 dx
t=e^2ix+1 dt=2ie^2ix dx dx=dt/(2ie^2ix )
∫(8e^2ix)/(e^2ix+1)^3 dx=∫(8e^2ix)/t^3 dt/(2ie^2ix )=∫4/t^3 dt/i
=-∫4i/t^3 dt=-4i∫t^(-3) dt =-4i ?-t?^(-2)/2=2it^(-2)
=2i/(e^2ix+1)^2
696: 08/22(金)07:20 ID:aTp7UHTZ(4/9) AAS
f^((k) ) (z)=(n!/2πi)?_Cf(ζ)/(ζ-z)^(k+1)dζ
?@)n=1のとき
f(z)=1/( 2πi) ?_Cf(ζ)/((ζ-z) ) dζ
f(z+h)=1/( 2πi) ?_Cf(ζ)/(ζ-(z+Δz) ) dζ
f(z+h)-f(z)=1/( 2πi) ?_Cf(ζ)/(ζ-(z+h) )-f(ζ)/((ζ-z) ) dζ
=1/( 2πi) ?_Cf(ζ)((ζ-z)-(ζ-z-h))/(ζ-z-h)(ζ-z)dζ
=1/( 2πi) ?_Cf(ζ)(ζ-z-ζ+z+h)/(ζ-z-h)(ζ-z)dζ
=1/( 2πi) ?_Cf(ζ)h/(ζ-z-h)(ζ-z)dζ
=h/( 2πi) ?_Cf(ζ)/(ζ-z-h)(ζ-z)dζ
( f(z+h)-f(z))/h=1/( 2πi) ?_Cf(ζ)/(ζ-z-h)(ζ-z)dζ
h→0
f'(z)= f^((1)) (z)=1/2πi ?_C(f(ζ))/(ζ-z)^2dζ
?A)n=k(k=1,2,3,…)のとき
f^((k)) (z)=k!/2πi ?_C(f(ζ))/(ζ-z)^(k+1)dζ ⇒f^((k+1)) (z)=(k+1)!/( 2πi) ?_Cf(ζ)/(ζ-z)^(k+2)dζ
f^((k)(z+h)- f^((k) ) (z))/h
=k!/( 2πih) ?_Cf(ζ)/(ζ-(z+h))^(k+1) -f(ζ)/(ζ-z)^(k+1)dζ
=k!/( 2πih) ?_C((ζ-z)^(k+1)-(ζ-z-h)^(k+1))/((ζ-z-h)^(k+1) (ζ-z)^(k+1) ) f(ζ)dζ??※
(a+b)^(k+1)
=(_k+1^ )C_0 a^n b^0+(_k+1^ )C_1 a^(k+1-1) b^1+(_k+1^ )C_2 a^(k+1-2) b^2+?+(_k+1^ )C_r a^(k+1-r) b^r+?+b^(k+1)
=a^(k+1)+(k+1) a^k b+(_k+1^ )C_2 a^(k-1) b^2+?+(_k+1^ )C_r a^(k+1-r) b^r+? +b^(k+1)
(ζ-z-h)^(k+1)
=(ζ-z)^(k+1)-(k+1) (ζ-z)^k h + (_k+1^ )C_2 (ζ-z)^(k-1) h^2-?+h^(k+1)
(ζ-z)^(k+1)-(ζ-z-h)^(k+1)
=(k+1) (ζ-z)^k h-(_k+1^ )C_2 (ζ-z)^(k-1) h^2+?-h^(k+1)
( f^((k) ) (z+h)- f^((k) ) (z))/h
=k!/( 2πih) ?_C((k+1) (ζ-z)^k h-(_k+1^ )C_2 (ζ-z)^(k-1) h^2+?-h^(k+1))/((ζ-z-h)^(k+1) (ζ-z)^(k+1) ) f(ζ)dζ
=(k+1)!/( 2πi) ?_Cf(ζ)/((ζ-z-h)^(k+1) (ζ-z) ) dζ-k!/( 2πi) ?_C((_k+1^ )C_2 (ζ-z)^(k-1) h-?+h^k)/((ζ-z-h)^(k+1) (ζ-z)^(k+1) ) f(ζ)dζ
h→0
f^((k+1)) (z)=(k+1)!/(2πi) ?_Cf(ζ)/(ζ-z)^(k+2)dζ
700: 08/22(金)11:01 ID:aTp7UHTZ(5/9) AAS
f(z)=1/(1-z) z=i で展開
?@) |z-i|<√2
(1-i)(1-(z-i)/(1-i))=1-i+(1-i) (z-i)/(1-i)
1/(1-z)=1/(1-i-(z-i) )=1/(1-i)?1/(1-(z-i)/(1-i))
=1/(1-i) (1+((z-i)/(1-i))+((z-i)/(1-i))^2+((z-i)/(1-i))^3+?)
=(z-i)^0/(1-i)+(z-i)^1/(1-i)^2 +(z-i)^2/(1-i)^3 +?
=((1+i) (z-i)^0)/2+((1+i)^2 (z-i)^1)/2^2 +((1+i)^3 (z-i)^2)/2^3 +?
=納n=0→∞]((1+i)/2)^(n+1) (z-i)^n
※(1 )/(1-i)^2 =(1/(1-i))(1/(1-i))=(1+i)/((1-i)(1+i))((1+i)/(1-i)(1+i)) =(1+i)^2/2^2
?A) |z-i|>√2の場合
|z-i|/√2=|(z-i)/(1-i)|>1
すなわち、0<|(1-i)/(z-i)|<1となるから((1-i)/(z-i))^n の級数展開を考える。
1/(1-z)=1/(1-i-(z-i) )=-1/(z-i)?1/(1-(1-i)/(z-i))
=-1/(z-i) (1+((1-i)/(z-i))+((1-i)/(z-i))^2+((1-i)/(z-i))^3+?)
=-(1/(z-i)+(1-i)/(z-i)^2 +(1-i)^2/(z-i)^3 +?)
=-(1/(z-i)+2/(1+i)(z-i)^2 +2^2/?(1+i)^2 (z-i)?^3 +?)
=-((2^0 (z-i)^(-1))/(1+i)^0 +(2^1 (z-i)^(-2))/(1+i)^1 +(2^2 (z-i)^(-3))/(1+i)^2 +?)
=-(?(1+i)^0 (z-i)?^(-1)/2^0 +?(1+i)^(-1) (z-i)?^(-2)/2^(-1) +((1+i)^(-2) (z-i)^(-3))/2^(-2) +?)
=-納n=1→∞]((1+i)/2)^(1-n) (z-i)^(-n)
※(1-i)^2=(1-i)(1-i)=(1-i)(1+i)/(1+i)?(1-i)(1+i)/(1+i)=2^2/(1+i)^2
(1-i)^n=2^n/(1+i)^n
704: 08/22(金)21:47 ID:aTp7UHTZ(6/9) AAS
y_s=1/(D+i) (2i/(e^2ix+1)^2 )=e^(-ix) 1/D e^ix 2i/(e^2ix+1)^2 =e^(-ix) ∫(2ie^2ix)/(e^2ix+1)^2 dx
t=e^2ix+1 dt=2ie^2ix dx dx=dt/(2ie^2ix )
∫?(2ie^2ix)/(e^2ix+1)^2 dx?=∫?(2ie^2ix)/t^2 dt/(2ie^2ix )?=∫t^(-2) dt=-1/t=-1/(e^2ix+1)
y_s=e^(-ix) ∫(2ie^2ix)/(e^2ix+1)^2 dx=-e^(-ix)/(e^2ix+1)
=(- e^(-ix) (e^(-ix)+e^ix-e^ix ))/(e^(-ix) (e^2ix+1) ) =(- e^(-ix) (e^(-ix)+e^ix )+1)/(e^ix+e^(-ix) )
=- e^(-ix)+1/(e^ix+e^(-ix) )=- e^(-ix)+1/2cos(x)
y=C_1 cos(x)+C_2 sin(x)- e^(-ix)+1/2cos(x)
=C_1 cos(x)+C_2 sin(x)- cos(x)+isin(x)+1/2cos(x)
=(C_1-1)cos(x)+(C_2+i)sin(x)+1/2cos(x)
=Acos(x)+Bsin(x)+1/2cos(x)
y_s=1/2cos(x)
y=C_2 cos(x)+C_1 sin(x)- 1/2 cos(2x) 1/cos(x)
=C_2 cos(x)+C_1 sin(x)- 1/2 (2?cos?^2 (x)-1) 1/cos(x)
=C_2 cos(x)+C_1 sin(x)- (?cos?^2 (x)-1/2)/cos(x)
=C_2 cos(x)+C_1 sin(x)- cos(x)+1/2 1/cos(x)
=(C_2-1)cos(x)+C_1 sin(x)+1/2cos(x)
=Acos(x)+Bsin(x)+1/2cos(x)
705: 08/22(金)21:48 ID:aTp7UHTZ(7/9) AAS
M(θ)=E[e^θX ]=∫_(-∞)^∞??e^θx f(x)dx?
M(θ)=E[e^θX ]=1/(√2π σ) ∫_(-∞)^∞??e^θx e^(-(x-μ)^2/(2σ^2 )) ? dx=1/(√2π σ) ∫_(-∞)^∞?e^(θx-(x-μ)^2/(2σ^2 )) dx
θx-(x-μ)^2/(2σ^2 )=1/(2σ^2 ) (2σ^2 θx-(x-μ)^2 )=-1/(2σ^2 ) (? (x-μ)?^2-2σ^2 θx )
=-1/(2σ^2 ) (? x?^2+μ^2-2μx-2σ^2 θx )
=-1/(2σ^2 ) (? x?^2-2(μ+σ^2 θ)x+μ^2 )
=-1/(2σ^2 ) ((x-(μ+σ^2 θ))^2-(μ+σ^2 θ)^2+μ^2 )
=-1/(2σ^2 ) ((x-(μ+σ^2 θ))^2-(μ^2+2μσ^2 θ+σ^4 θ^2 )+μ^2 )
=-1/(2σ^2 ) ((x-(μ+σ^2 θ))^2-(2μσ^2 θ+σ^4 θ^2 ) )
=-(x-(μ+σ^2 θ))^2/(2σ^2 )+μθ+(σ^2 θ^2)/2
M(θ)=1/(√2π σ) ∫_(-∞)^∞?e^(θx-(x-μ)^2/(2σ^2 )) dx
=1/(√2π σ) ∫_(-∞)^∞?e^((-(x-(μ+σ^2 θ))^2/(2σ^2 )+μθ+(σ^2 θ^2)/2) ) dx
=1/(√2π σ) e^(μθ+(σ^2 θ^2)/2) ∫_(-∞)^∞?e^((-(x-(μ+σ^2 θ))^2/(2σ^2 )) ) dx
t=(x-(μ+σ^2 θ))/(√2 σ) x=√2 σt+μ+σ^2 θ dx=√2 σdt
(x-(μ+σ^2 θ))^2/(2σ^2 )=((x-(μ+σ^2 θ))/(√2 σ))^2=t^2
-∞<x?∞ ⇒-∞<t?∞
706: 08/22(金)22:01 ID:aTp7UHTZ(8/9) AAS
det(r ?,r ?_Q)=|■(x ?&(x_q ) ?@y ?&(y_q ) ? )|=x ?(y_q ) ?-(x_q ) ?(y=) ?x ?(y_q ) ?-x ?y ?+x ?y ?-(x_q ) ?y ?
=x ?(y ?(t+Δt)-y Dt)-y ?(x ?(t+Δt)-x Dt)
Δr ? ?Δr ?_Q ?=√(x ?^2+y ?^2 ) √((x_q ) ?^2+(y_q ) ?^2 )
=√(x ?^2+y ?^2 ) √(?(x ?(t+Δt))?^2+?(y ?(t+Δt))?^2 ).
したがって
Δθ/Δs=(x ?(y ?(t+Δt)-y Dt)-y ?(x ?(t+Δt)-x Dt))/(√(x ?^2+y ?^2 ) √(?(x ?(t+Δt))?^2+?(y ?(t+Δt))?^2 )) 1/Δr(t+Δt)-r(t)?
=((x ?(y ?(t+Δt)-y Dt)-y ?(x ?(t+Δt)-x Dt))/Δt)/(√(x ?^2+y ?^2 ) √(?(x ?(t+Δt))?^2+?(y ?(t+Δt))?^2 )) ΔtΔr(t+Δt)-r(t)?^(-1)
=(x ? ((y ?(t+Δt)-y Dt))/Δt-y ? ((x ?(t+Δt)-x Dt))/Δt)/(√(x ?^2+y ?^2 ) √(?(x ?(t+Δt))?^2+?(y ?(t+Δt))?^2 )) ?(r(t+Δt)-r(t))/Δt?^(-1)
1/R=(lim)┬(Δt→0)??Δθ/Δs?=(x ?y ?-yx ?)/(√(x ?^2+y ?^2 ) √(x ?^2+y ?^2 )) ? Δr ? ??^(-1)
=(x ?y ?-yx ?)/(√(x ?^2+y ?^2 ) √(x ?^2+y ?^2 ) √(x ?^2+y ?^2 ))
=(x ?y ?-yx ?)/(x ?^2+y ?^2 )^(3/2)
R=(x ?^2+y ?^2 )^(3/2)/(x ?y ?-yx ? )
707: 08/22(金)22:01 ID:aTp7UHTZ(9/9) AAS
C:x=x(t),y=y(t)
OP↑=r(t)=(x(t),y(t))
OQ↑ ?=r(t+Δt)=(x(t+Δt),y(t+Δt))
Δs=|Δr|=|Δr(t+Δt)-r(t)|
RΔθ≒Δs,1/R=Δθ/Δs
1/R=lim[Δt→0](Δθ/Δs)=dθ/ds
dr/dt=rDt
r Dt=(x Dt,y Dt)
r ?(t+Δt)=(x ?(t+Δt),y ?(t+Δt))
r Dt=r ?=(x ?,y ?)
r ?(t+Δt)= r ?_Q=(x ?_Q,y ?_Q)
Δr ? ?Δr ?_Q ΔsinΔθ=det(r ?,r ?_Q)
ΔθΔsinΔθ=(det(r ?,r ?_Q))/Δr ? ?Δr ?_Q ?
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