フェルマーの最終定理の証明 (791ﾚｽ)
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784: 09/03(水)02:02 ID:cpr6IQHh(1/5)調 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 ?
785: 09/03(水)02:02 ID:cpr6IQHh(2/5)調 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)
786: 09/03(水)02:03 ID:cpr6IQHh(3/5)調 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?∞
787: 09/03(水)08:20 ID:cpr6IQHh(4/5)調 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)
788: 09/03(水)08:20 ID:cpr6IQHh(5/5)調 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)
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