フェルマーの最終定理の証明 (645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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612: 08/08(金)09:05 ID:K5nrmcJ7(1/5) AAS
E(t)=Ri(t)+1/C ∫?i(t) dt
i(t)=dq(t)/dt ∫?dq(t)/dt dt=q(t)
E(t)=R dq(t)/dt+q(t)/C
L[Rq^' ]=RsQ(s)-Rq(0)=RsQ(s)
L[q(t)/C]=Q(s)/C L[E]=E/s
E/s=RsQ(s)+Q(s)/C=Q(s)(Rs+1/C)
Q(s)= E/s 1/(Rs+1/C)=E/s(Rs+1/C) =(E/R)/s(s+1/CR) =E/R 1/s(s+1/CR)
1/s(s+1/CR) =A/s+B/(s+1/CR) 1=A(s+1/CR)+Bs
s=0⇒A/CR=1 A=CR
s=-1/CR⇒-B 1/CR=1 B=-CR
Q(s)=E/R (A/s+B/(s+1/CR))=E/R (CR/s-CR/(s+1/CR))=CE/s-CE/(s+1/CR)
L^(-1) [CE/s-CE/(s+1/CR)]=CE(L^(-1) [1/s-1/(s+1/CR)])=CE(1-e^(-1/CR t) )
613: 08/08(金)11:49 ID:K5nrmcJ7(2/5) AA×
614: 08/08(金)11:50 ID:K5nrmcJ7(3/5) AAS
|∫_a^b▒f(x) sin(αx)dx|
=|?_(k=1)^n▒〖∫_(x_k)^(x_(k+1))▒f(x) sin(αx)dx〗|
=|∫_(x_1)^(x_2)▒f(x) sin(αx)dx+∫_(x_2)^(x_3)▒f(x) sin(αx)dx+⋯+∫_(x_n)^(x_(n+1))▒f(x) sin(αx)dx|
≤|∫_(x_1)^(x_2)▒f(x) sin(αx)dx|+|∫_(x_2)^(x_3)▒f(x) sin(αx)dx|+⋯+|∫_(x_n)^(x_(n+1))▒f(x) sin(αx)dx|
=?_(k=1)^n▒|∫_(x_k)^(x_(k+1))▒f(x) sin(αx)dx|
=?_(k=1)^n▒|(∫_(x_k)^(x_(k+1))▒f(x) -f(x_k )+f(x_k ))sin(αx)dx|
≤?_(k=1)^n▒(|∫_(x_k)^(x_(k+1))▒( f(x)-f(x_k ) )sin(αx) dx|+|f(x_k ) ∫_(x_k)^(x_(k+1))▒sin(αx) dx|)
≤?_(k=1)^n▒(∫_(x_k)^(x_(k+1))▒|f(x)-f(x_k )||sin(αx)| dx+|f(x_k )||∫_(x_k)^(x_(k+1))▒sin(αx) dx|)
≤?_(k=1)^n▒(∫_(x_k)^(x_(k+1))▒|f(x)-f(x_k )|1 dx+M∫_(x_k)^(x_(k+1))▒|sin(αx)| dx)
∫_(x_k)^(x_(k+1))▒|sin(αx)| dx=|[(-1)/α cos(αx)]_(x_k)^(x_(k+1) ) |=1/α |cos(x_(k+1) )- cos(x_k )|
≤1/α (|cos(x_(k+1) )|+|cos(x_k )|)≤2/α
615: 08/08(金)11:51 ID:K5nrmcJ7(4/5) AAS
|∫_a^b?f(x)sin(αx)dx|
=|?_(k=1)^n??∫_(x_k)^(x_(k+1))?f(x) sin(αx)dx?|
=|∫_(x_1)^(x_2)?f(x) sin(αx)dx+∫_(x_2)^(x_3)?f(x) sin(αx)dx+?+∫_(x_n)^(x_(n+1))?f(x) sin(αx)dx|
?|∫_(x_1)^(x_2)?f(x) sin(αx)dx|+|∫_(x_2)^(x_3)?f(x) sin(αx)dx|+?+|∫_(x_n)^(x_(n+1))?f(x) sin(αx)dx|
=?_(k=1)^n?|∫_(x_k)^(x_(k+1))?f(x) sin(αx)dx|
=?_(k=1)^n?|(∫_(x_k)^(x_(k+1))?f(x) -f(x_k )+f(x_k ))sin(αx)dx|
??_(k=1)^n?(|∫_(x_k)^(x_(k+1))?( f(x)-f(x_k ) )sin(αx) dx|+|f(x_k ) ∫_(x_k)^(x_(k+1))?sin(αx) dx|)
??_(k=1)^n?(∫_(x_k)^(x_(k+1))?|f(x)-f(x_k )||sin(αx)| dx+|f(x_k )||∫_(x_k)^(x_(k+1))?sin(αx) dx|)
??_(k=1)^n?(∫_(x_k)^(x_(k+1))?|f(x)-f(x_k )|1 dx+M∫_(x_k)^(x_(k+1))?|sin(αx)| dx)
∫_(x_k)^(x_(k+1))?|sin(αx)| dx=|[(-1)/α cos(αx)]_(x_k)^(x_(k+1) ) |=1/α |cos(x_(k+1) )- cos(x_k )|
?1/α (|cos(x_(k+1) )|+|cos(x_k )|)?2/α
616: 08/08(金)11:52 ID:K5nrmcJ7(5/5) AAS
|∫_a^b?f(x) sin(αx)dx|??_(k=1)^n?(∫_(x_k)^(x_(k+1))?|f(x)-f(x_k )| dx+2M/α)
∀ε>0∃N s.t. n>N⇒|f(x)-f(x_k )|<ε (k=1,2,?,n)
|∫_a^b?f(x) sin(αx)dx|??_(k=1)^n?(∫_(x_k)^(x_(k+1))?ε dx+2M/α)
=?_(k=1)^n?ε [x]_(x_k)^(x_(k+1) )+ n 2M/α
= ε?_(k=1)^n??x +(2M n)/α=ε(b-a)+(2M n)/α
(2M n)/α?ε?(2M n)/ε?α
lim┬(α→∞) |∫_a^b?f(x) sin(αx)dx|? lim┬(α→∞) (ε(b-a)+(2M n)/α)
=ε(b-a)+ε=ε(b-a+1)
したがって任意の正数εに対し α→∞ のとき |∫_a^b?f(x) sin(αx)dx|=0
lim┬(α→∞)??∫_a^b?f(x) sin(αx)dx?=0
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