フェルマーの最終定理の証明 (645ﾚｽ)
フェルマーの最終定理の証明 http://rio2016.5ch.net/test/read.cgi/math/1745314067/
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612: １３２人目の素数さん [] 2025/08/08(金) 09:05:38.87 ID:K5nrmcJ7 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) ) http://rio2016.5ch.net/test/read.cgi/math/1745314067/612
613: １３２人目の素数さん [] 2025/08/08(金) 11:49:03.02 ID:K5nrmcJ7 z^3+1=(z+1)(z^2-z+1) z^3≡-1 mod(z^2-z+1)……※ z^6≡1 mod(z^2-z+1) 以下 mod(z^2-z+1) ?n=6k (k≧1) z^2n+z^n+1≡z^2(6k) +z^6k+1≡3 ?〜?まではk≧0 ? n=6k+1 z^2n+z^n+1≡z^2(6k+1) +z^(6k+1)+1≡z^12k z^2+z^6k z+1 ≡z^2+z+1 z^2-z+1≡0⇔ z^2+1≡z ∴z^2+z+1≡2z ?n=6k+2 z^2n+z^n+1≡z^2(6k+2) +z^(6k+2)+1≡z^12k z^4+z^6k z^2+1 ≡z^4+z^2+1 z^3≡-1 z^4≡-z ∴z^4+z^2+1≡z^2-z+1≡0 ?n=6k+3 z^2n+z^n+1=z^2(6k+3) +z^(6k+3)+1=z^12k z^6+z^6k z^3+1 ≡1+z^3+1≡1 ?n=6k+4 z^2n+z^n+1=z^2(6k+4) +z^(6k+4)+1=z^12k z^6 z^2+z^6k z^4+1 ≡z^2+z^4+1≡z^2-z+1≡0 ?n=6k+5 z^2n+z^n+1=z^2(6k+5) +z^(6k+5)+1=z^12k z^10+z^6k z^5+1 ≡z^6 z^4+z^4 z+1 ≡-z-z^2+1 http://rio2016.5ch.net/test/read.cgi/math/1745314067/613
614: １３２人目の素数さん [] 2025/08/08(金) 11:50:15.22 ID:K5nrmcJ7 |∫_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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/614
615: １３２人目の素数さん [] 2025/08/08(金) 11:51:41.96 ID:K5nrmcJ7 |∫_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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/615
616: １３２人目の素数さん [] 2025/08/08(金) 11:52:51.81 ID:K5nrmcJ7 |∫_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 http://rio2016.5ch.net/test/read.cgi/math/1745314067/616
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