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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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｡

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778: 09/01(月)08:57 ID:b44elzXy(1/6)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?∞

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779: 09/01(月)09:08 ID:b44elzXy(2/6)調 AAS
F(ω)=∫[-∞→∞]f(t)e^(-jωt)dt
f(t)= F^(-1) [F(ω)]=1/2π ∫[-∞→∞]F(ω) e^jωt ? dω
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
s=σ+jω　　ds=jdω　　ω: -∞ → ∞
s：σ-j∞→σ+j∞
g(t)=(1/2π)[-∞→∞]F(s)e^jωtdω
=(1/2πj)∫[σ-j∞→σ+j∞]F(s)e^jωtds
f(t)e^(-σt)=f(t)/e^σt
=(1/2πj)∫[σ-j∞→σ+j∞]F(s) e^jωtds
f(t)=(1/2πj)∫[σ-j∞→σ+j∞]F(s)e^σt e^jωtds
=(1/2πj)∫[σ-j∞→σ+j∞]F(s) e^(σ+jω)tds
f(t)=(1/2πj)∫[σ-j∞→σ+j∞]F(s) e^stds

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780: 09/01(月)09:20 ID:b44elzXy(3/6)調 AAS
f(θ)=a_0/2+?[k=1→∞](a_k cos(kθ)+b_k sin(kθ))
a_k=1/π ∫[-π→π]f(θ)cos(kθ)dθ)
b_k=1/π ∫[-π→π]f(θ)sin(kθ)dθ

e^jθ =cosθ+jsinθ
e^(-jθ)=cosθ-jsinθ
cosθ=(e^jθ+e^(-jθ))/2. sinθ=(e^jθ-e^(-jθ))/2j.
f(θ)=a_0/2+?[k=1→∞](a_k (e^jkθ+e^(-jkθ))/2+b_k (e^jkθ-e^(-jkθ))/2j)
=a_0/2+?[k=1→∞](a_k(e^jkθ+e^(-jkθ))/2+?jb?_k (e^(-jkθ)-e^jkθ)/2)
=a_0/2+?[k=1→∞]((a_k-jb_k)/2 e^jkθ) +?[k=1→∞]((a_k+jb_k)/2 e^(-jkθ) )
a_(-k)=(1/π)∫[-π→π]f(θ)cos(-kθ)dθ)
=(1/π)∫[-π→π]f(θ)cos(kθ)dθ)=a_k
b_(-k)=(1/π)∫[-π→π]f(θ)sin(-kθ)dθ
= -1/π ∫[-π→π]f(θ)sin(kθ)dθ= -b_k
f(θ)
=?[k=1→∞]((a_k+jb_k)/2 e^(-jkθ) ) +a_0/2+?[k=1→∞]((a_k-jb_k)/2 e^jkθ )
=(a_2+jb_2)/2 e^(-j2θ)+(a_1+jb_1)/2 e^(-j1θ)+a_0/2+(a_1-jb_1)/2 e^j1θ+(a_2-jb_2)/2e^j2θ+?
=(a_(-2)-jb_(-2))/2 e^j2θ+(a_(-1)-jb_(-1))/2 e^j1θ+a_0/2+(a_1-jb_1)/2 e^j1θ+(a_2-jb_2)/2 e^j2θ+?
=?_(k=-∞)^∞?((a_k-jb_k)/2 e^jkθ )
c_k=(a_k-jb_k)/2
f(θ)=?[k=-∞→∞]c_k e^jkθ
c_k=(a_k-jb_k)/2
=(1/2π)∫[-π→π]f(θ)cos(kθ)dθ-(j/2π)∫[-π→π]f(θ)sin(kθ)dθ
=(1/2π)∫[-π→π]f(θ)(cos(kθ)-jsin(kθ))dθ
=(1/2π)∫[-π→π]f(θ)(cos(-kθ)+jsin(-kθ))dθ
=(1/2π)∫[-π→π]f(θ)(cos(-kθ)+jsin(-kθ))dθ
=(1/2π)∫[-π→π]f(θ)e^(-jkθ)dθ

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781: 09/01(月)09:21 ID:b44elzXy(4/6)調 AAS
f(τ)=0 ∴f(τ)g(t-τ)=0
g(t-τ)=0 ∴f(τ)g(t-τ)=0
f*g(t)=∫_(-∞)^∞??f(τ)g(t-τ)dτ?
=∫_(-∞)^0??f(τ)g(t-τ)dτ?+∫_0^∞??f(τ)g(t-τ)dτ?=0
τ<0⇒f(τ)=0 ∴f(τ)g(t-τ)=0
0?τ?t⇒t-τ?0 ∴f(τ)=e^(-τ), g(t-τ)=t-τ
τ>t⇒t-τ<0 ∴g(t-τ)=0, f(τ)g(t-τ)=0
f*g(t)=∫_0^t??e^(-τ) (t-τ)dτ?=∫_0^t??(-e^(-τ) )^' (t-τ)dτ?
=-[?( @e^(-τ)@ )(t-τ)]_0^t-∫_0^t??-e^(-τ) (-1)dτ?
=t-∫_0^t??e^(-τ) dτ? =t+[?( @e^(-τ)@ )]_0^t=t+e^(-t)-1
τ<0⇒f(τ)=0 ∴f(τ)g(t-τ)=0
t-τ>1⇒g(t-τ)=0 ∴f(τ)g(t-τ)=0
t-τ?1 ⇒ f(τ)=e^(-τ), g(t-τ)=t-τ
f*g(t)=∫_(t-1)^t??e^(-τ) (t-τ)dτ?=∫_(t-1)^t??(-e^(-τ) )^' (t-τ)dτ?
=-[?( @e^(-τ)@ )(t-τ)]_(t-1)^t-∫_(t-1)^t??e^(-τ) dτ?
=-(0-e^(1-t) )+[?( @e^(-τ)@ )]_(t-1)^t=e^(1-t)+e^(-t)-e^(1-t)=e^(-t)

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782: 09/01(月)20:13 ID:b44elzXy(5/6)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 ?

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783: 09/01(月)20:14 ID:b44elzXy(6/6)調 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)

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上下前次1-新書関写板覧索設栞歴

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