フェルマーの最終定理の証明 (619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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453: １３２人目の素数さん [] 2025/07/22(火) 12:29:34.17 ID:UfTdyzFE log2>2/3 , log2<7/6 f(x)=(2x^2+15)log2-(4x+30)logx x?12⇒f(x)>0 f^' (x)= log2?4x-(4 logx-(4x+30)/x) = log2?4x-4 logx+4-30/x f^'' (x)=4 log2-4/x+30/x^2 4 log2>4 2/3>3 2/3=2 f^'' (x)>2-4/x+30/x^2 =(2x^2-4x+30)/x^2 =2 (x^2-2x+15)/x^2 =2 ((x-1)^2+14)/x^2 >0 http://rio2016.5ch.net/test/read.cgi/math/1745314067/453
454: １３２人目の素数さん [] 2025/07/22(火) 12:30:40.07 ID:UfTdyzFE ∫_α^β??(x-α)^m (β-x)^n ? dx=m!n!/(m+n+1)! (β-α)^(m+n+1) t=(β-α)x+α dt=(β-α)dx dx=dt/(β-α) x:0→1 t:α→β x=(t-α)/(β-α) 1-x=(β-α-(t-α))/(β-α)=(β-t)/(β-α) ∫_0^1?x^m (1-x)^n dx =∫_α^β??((t-α)/(β-α))^m ((β-t)/(β-α))^n ? dt/(β-α)=∫_α^β?((t-α)^m (β-t)^m)/(β-α)^(m+n+1) dt =1/(β-α)^(m+n+1) ∫_α^β??(t-α)^m (β-t)^m ? dt=m!n!/(m+n+1)! ∴∫_α^β??(x-α)^m (β-x)^n ? dx=m!n!/(m+n+1)! (β-α)^(m+n+1) m=1,n=1⇒∫_α^β?(x-α)(x-β) dx=-∫_α^β?(x-α)(β-x) dx =-1/6 (β-α)^3 m=2,n=1⇒∫_α^β?(x-α)(x-β) dx=-∫_α^β??(x-α)^2 (β-x) ? dx =-1/12 (β-α)^4 m=2,n=2⇒∫_α^β??(x-α)^2 (x-β)^2 ? dx=∫_α^β??(x-α)^2 (β-x)^2 ? dx =(2?2)/(5?4?3?2?1) (β-α)^5=1/30 (β-α)^5 http://rio2016.5ch.net/test/read.cgi/math/1745314067/454
455: １３２人目の素数さん [] 2025/07/22(火) 12:39:18.21 ID:UfTdyzFE Q? √(6&2^(2x^2+15)/x^(4x+30) ) (x=√2n, n?5) ･････(#12) x=e^logx 2=e^log2 2^(2x^2+15) = ?(e^log2)?^(2x^2+15)=e^((2x^2+15)log2) x^(4x+30)=?(e^logx)?^(4x+30)=e^((4x+30)logx) ここで (2x^2+15)log2 >(4x+30)logx (x?12) ･････(#14) 2^(2x^2+15)/x^(4x+30) =e^((2x^2+15)log2)/e^((4x+30)logx) =e^((2x^2+15)log2-(4x+30)logx)>e^0 √(6&2^(2x^2+15)/x^(4x+30) )>√(e^0 )=1 x=√2n?12 、つまりn?72 のとき(#15)は成り立つ。 37?n?71⇒n?73?2n 19?n?36⇒n?37?2n 10?n?18⇒n?19?2n 6?n?9⇒n?11?2n n=4,5⇒n?7?2n n=3⇒3?6?6 n=2⇒2?3?4 n=1⇒1?2?2 http://rio2016.5ch.net/test/read.cgi/math/1745314067/455
456: １３２人目の素数さん [] 2025/07/22(火) 12:39:44.75 ID:UfTdyzFE C:x=x(t),y=y(t) (OP) ?=r(t)=(x(t),y(t)) (OQ) ?=r(t+?t)=(x(t+?t),y(t+?t)) ?s=??r?=?r(t+?t)-r(t)? R?θ??s,1/R=?θ/?s 1/R=lim┬(?t→0)???θ/?s?=dθ/(ds) dr/dt=r ?(t) r ?(t)=(x ?(t),y ?(t)) r ?(t+?t)=(x ?(t+?t),y ?(t+?t)) r ?(t)=r ?=(x ?,y ?) r ?(t+?t)= r ?_Q=(x ?_Q,y ?_Q) ?r ? ??r ?_Q ?sin?θ=det(r ?,r ?_Q) ?θ?sin?θ=(det(r ?,r ?_Q))/?r ? ??r ?_Q ? det(r ?,r ?_Q)=|■(x ?&(x_q ) ?@y ?&(y_q ) ? )|=x ?(y_q ) ?-(x_q ) ?(y=) ?x ?(y_q ) ?-x ?y ?+x ?y ?-(x_q ) ?y ? =x ?(y ?(t+?t)-y ?(t))-y ?(x ?(t+?t)-x ?(t)) ?r ? ??r ?_Q ?=√(x ?^2+y ?^2 ) √((x_q ) ?^2+(y_q ) ?^2 ) =√(x ?^2+y ?^2 ) √(?(x ?(t+?t))?^2+?(y ?(t+?t))?^2 ). http://rio2016.5ch.net/test/read.cgi/math/1745314067/456
462: １３２人目の素数さん [] 2025/07/22(火) 20:09:15.07 ID:UfTdyzFE L[y^'' (t)]=s^2 Y(s)-sy(0)-y^' (0) =s^2 Y(s)-2s-4 L[?4y?^' (t)]=4(sY(s)-y(0))=4sY(s)-8 L[4y(t)]=4Y(s) L[y^'' (t)]-L[?4y?^' (t)]+ L[4y(t)] =s^2 Y(s)-2s-4-4sY(s)+8+4Y(s) =Y(s)(s^2-4s+4)-2s+4 L[6te^2t ]=6L[t^1 e^2t ]=6 1!/(s-2)^2 =6/(s-2)^2 ( L[t^n e^at ]=n!/(s-a)^(n+1) ) Y(s)(s^2-4s+4)-2s+4=6/(s-2)^2 Y(s) (s-2)^2-2s+4=6/(s-2)^2 Y(s) (s-2)^2=6/(s-2)^2 +2(s-2) Y(s)=6/(s-2)^4 +2/(s-2) Y(s)= F(s-2)とおくと F(s-2)=6/(s-2)^4 +2/(s-2) ∴F(s)=6/s^4 +2/s=3!/s^(3+1) +2/s y(t)=L^(-1)[F(s-2)]=e^2t L^(-1) [F(s)] ( L^(-1) [F(s-a)]=e^at L^(-1) [F(s)]) =e^2t L^(-1) [3!/s^(3+1) +2/s] (L[t^n ]=n!/s^(n+1) ) =e^2t (t^3+2) http://rio2016.5ch.net/test/read.cgi/math/1745314067/462
463: １３２人目の素数さん [] 2025/07/22(火) 20:10:05.84 ID:UfTdyzFE E(t)=Ri(t)+L di(t)/dt L[Li^' ]=LsI(s)-Li(0) =LsI(s) L[Ri]=RI(s) L[E]=E/s E/s= LsI(s)+RI(s)=I(s)(Ls+R) I(s)=E/s(Ls+R) =E/Ls(s+R/L) 1/Ls(s+R/L) =A/Ls+B/(s+R/L) 1=A(s+R/L)+BLs s=0⇒AR/L=1 A=L/R s=-R/L⇒-BL R/L=1 B=-1/R I(s)=E(A/Ls+B/(s+R/L))=E(L/R?1/Ls-1/R?1/(s+R/L)) =E(1/Rs-1/R(s+R/L) )=E/R (1/s-1/(s+R/L)) L^(-1) [E/R (1/s-1/(s+R/L))]=E/R (L^(-1) [1/s-1/(s+R/L)])=E/R (1-e^(-R/L t) ) http://rio2016.5ch.net/test/read.cgi/math/1745314067/463
464: １３２人目の素数さん [] 2025/07/22(火) 20:11:12.14 ID:UfTdyzFE Memo 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/1745314067/464
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