フェルマーの最終定理の証明 (790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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754: １３２人目の素数さん [] 08/30(土)12:31 ID:GT1KZtG+(1/7)
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)

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755: １３２人目の素数さん [] 08/30(土)12:31 ID:GT1KZtG+(2/7)
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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759: １３２人目の素数さん [] 08/30(土)20:37 ID:GT1KZtG+(3/7)
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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760: １３２人目の素数さん [] 08/30(土)20:37 ID:GT1KZtG+(4/7)
y''+y=sin(2x)
λ^2+1=0 λ=0±i
y_0=C_1 cos(x)+C_2 sin(x)
y_1=cos(x), y_2=sin(x)
?y_1?^'=-sin(x), ?y_2?^'=cos(x)
W=|?( cos(x)@-sin(x) )?( sin(x) @ cos(x) )|
=?cos?^2 (x)+?sin?^2 (x)=1
y_s (x)=-y_1 ∫?(y_2 R(x))/W dx+y_2 ∫?(y_1 R(x))/W dx
=-cos(x) ∫?sin(x)sin(2x) dx+sin(x) ∫?cos(x)sin(2x) dx
∫?sin(2x)sin(x) dx=-1/2 ∫??cos(2x+x)-cos(2x-x) ? dx
=-1/2 ∫??cos(3x)-cos(x) ? dx=-1/2?1/3 sin(3x)+1/2 sin(x)
=-1/6 sin(3x)+1/2 sin(x)
∫?sin(2x)cos(x) dx=1/2 ∫??sin(2x+x)+sin(2x-x) ? dx
=1/2 ∫??sin(3x)+sin(x) ? dx=1/2?(-1)/3 cos(3x)+(-1)/2 cos(x)
=-1/6 cos(3x)-1/2 cos(x)
y_s (x)
=-cos(x)(-1/6 sin(3x)+1/2 sin(x))+sin(x)(-1/6 cos(3x)-1/2 cos(x))
=1/6 sin(3x)cos(x)-1/2 sin(x)cos(x)-1/6 cos(3x)sin(x)-1/2 sin(x)cos(x)
=1/6 sin(3x-x)-sin(x)cos(x)=1/6 sin(2x)-1/2 sin(2x)
=-1/3 sin(2x)
∴y=C_1 cos(x)+C_2 sin(x)-1/3 sin(2x)

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761: １３２人目の素数さん [] 08/30(土)20:37 ID:GT1KZtG+(5/7)
y''+y=sin(2x)
λ^2+1=0 λ=0±i
y_0=C_1 cos(x)+C_2 sin(x)
y_1=cos(x), y_2=sin(x)
?y_1?^'=-sin(x), ?y_2?^'=cos(x)
W=|?( cos(x)@-sin(x) )?( sin(x) @ cos(x) )|
=?cos?^2 (x)+?sin?^2 (x)=1
y_s (x)=-y_1 ∫?(y_2 R(x))/W dx+y_2 ∫?(y_1 R(x))/W dx
=-cos(x) ∫?sin(x)sin(2x) dx+sin(x) ∫?cos(x)sin(2x) dx
∫?sin(2x)sin(x) dx=-1/2 ∫??cos(2x+x)-cos(2x-x) ? dx
=-1/2 ∫??cos(3x)-cos(x) ? dx=-1/2?1/3 sin(3x)+1/2 sin(x)
=-1/6 sin(3x)+1/2 sin(x)
∫?sin(2x)cos(x) dx=1/2 ∫??sin(2x+x)+sin(2x-x) ? dx
=1/2 ∫??sin(3x)+sin(x) ? dx=1/2?(-1)/3 cos(3x)+(-1)/2 cos(x)
=-1/6 cos(3x)-1/2 cos(x)
y_s (x)
=-cos(x)(-1/6 sin(3x)+1/2 sin(x))+sin(x)(-1/6 cos(3x)-1/2 cos(x))
=1/6 sin(3x)cos(x)-1/2 sin(x)cos(x)-1/6 cos(3x)sin(x)-1/2 sin(x)cos(x)
=1/6 sin(3x-x)-sin(x)cos(x)=1/6 sin(2x)-1/2 sin(2x)
=-1/3 sin(2x)
∴y=C_1 cos(x)+C_2 sin(x)-1/3 sin(2x)

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762: １３２人目の素数さん [] 08/30(土)20:40 ID:GT1KZtG+(6/7)
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?∞

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763: １３２人目の素数さん [] 08/30(土)20:40 ID:GT1KZtG+(7/7)
∫[0→π/2]( tan(x) )^(1/n) dx　　(n≧2)
∫_0^(π/2)?(tan(x))^(1/n) dx を求める。
t=?sin?^2 x=(sin(x))^2
?sin?^2 x=1-?cos?^2 x ?cos?^2 x=1-t
dt=2sin(x)cos(x)dx=2√t √(1-t) dx
dx=dt/(2√t √(1-t))=(t^(-1/2) (1-t)^(1/2))/2 dt
(sin(x))^(1/n)=(√t)^(1/n)=t^(1/2n) (cos(x))^(1/n)=(√(1-t))^(1/n)=(1-t)^(1/2n)
∫_0^(π/2)?(tan(x))^(1/n) dx=∫_0^(π/2)?( (sin(x))^(1/n))/( (cos(x))^(1/n) ) dx=∫_0^(π/2)?( t^(1/2n))/(1-t)^(1/2n) (t^(-1/2) (1-t)^(1/2))/2 dt

=1/2 ∫_0^(π/2)???t^(1/2n) (1-t)^(-1/2n) t?^(-1/2) (1-t)^(-1/2) ? dt
=1/2 ∫_0^(π/2)??t^(1/2n-1/2) (1-t)^(-1/2n-1/2) ? dt
=1/2 ∫_0^(π/2)??t^(1/2+1/2n-1) (1-t)^(1/2-1/2n-1) ? dt
=1/2 ∫_0^(π/2)??t^(1/2+1/2n-1) (1-t)^(1/2-1/2n-1) ? dt
(1/2) B(1/2+1/(2n), 1/2-1/(2n))
= (1/2) Γ( 1/2+1/(2n) ) Γ( 1/2-1/(2n) ) / Γ( 1/2+1/(2n) + 1/2-1/(2n) )
= (1/2) Γ(z) Γ(1-z) / Γ(1)
= (1/2) ( π/sin(πz) ) / 0！
= π/( 2 sin(πz) )
= π/( 2 sin(π/2+π/(2n)) )
= π/( 2 cos(π/(2n)) ).

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上下前次1-新書関写板覧索設栞歴

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