フェルマーの最終定理の証明 (692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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521: １３２人目の素数さん [] 2025/07/28(月) 09:41:28.53 ID:Vsf8XHSj ∫_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/1745314067/521
522: １３２人目の素数さん [] 2025/07/28(月) 09:41:50.77 ID:Vsf8XHSj 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/1745314067/522
523: １３２人目の素数さん [] 2025/07/28(月) 09:42:26.88 ID:Vsf8XHSj I=∫_0^2023?2/(x+e^x ) dx と置く。 x?0⇒0<x<e^xであるから、 2/(2e^x )<2/(x+e^x )<2/e^x ∴1/e^x <2/(x+e^x )<2/e^x ∫_0^2023?e^(-x) dx<I<∫_0^2023??2e^(-x) ? dx [-?(■( @e^(-x) )@ )]_0^2023=-?(■( @e^(-2023) )@ )+1 1-?(■( @e^(-2023) )@ )<I<2-2e^(-2023)<2???????? f(x)=2/(x+e^x ) f^' (x)=-2(1+e^x )/(x+e^x )^2 =-(2+2e^x)/(x+e^x )^2 <0 (単調減少) f^'' (x)=-(2e^x (x+e^x )^2-2(1+e^x )2(x+e^x )(1+e^x ))/(x+e^x )^4 =(4(1+e^x )^2 (x+e^x )-2e^x (x+e^x )^2)/(x+e^x )^4 =(4(1+e^x )^2-2e^x (x+e^x ))/(x+e^x )^3 >(4(1+e^x )^2-2e^x (e^x+e^x ))/(x+e^x )^3 > (4(1+e^x )^2-4(e^x )^2)/(x+e^x )^3 >0 (下に凸) (1,f(1))におけるの接線の方程式は y- f(1)=f^' (1)(x-1) y- 2/(1+e)=-2(1+e)/(1+e)^2 (x-1) http://rio2016.5ch.net/test/read.cgi/math/1745314067/523
527: １３２人目の素数さん [] 2025/07/28(月) 12:54:47.25 ID:Vsf8XHSj ∫_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/1745314067/527
528: １３２人目の素数さん [] 2025/07/28(月) 12:55:10.57 ID:Vsf8XHSj b=t×n=1/√(a^2+c^2 ) (■(-asin?(t)@acos?(t)@c))×(■(-cos?(t)@?-sin??(t)@0)) ※外積のスカラー倍 =1/√(a^2+c^2 ) |■(i&j&k@-asin?(t)&acos?(t)&c@-cos?(t)&-sin?(t)&0)| =1/√(a^2+c^2 ) (|■(acos?(t)&c@-sin?(t)&0)|,|■(c&-asin?(t)@0&-cos?(t) )|,|■(-asin?(t)&acos?(t)@-cos?(t)&-sin?(t) )|) =1/√(a^2+c^2 ) (csin?(t), -?c?cos??(t), a) b^' (s)=db/ds=db/dt?dt/ds=1/√(a^2+c^2 ) (?c?cos??(t), csin?(t), 0) 1/√(a^2+c^2 ) =1/(a^2+c^2 ) (?c?cos??(t), csin?(t), 0) b^' (s)=-τn より 1/(a^2+c^2 ) (?c?cos??(t), csin?(t), 0)=-τ(-cos?(t), ?-sin??(t), 0) =τ(cos?(t), sin?(t), 0) 1/(a^2+c^2 ) ?c?cos??(t)=τ cos?(t) τ=c/(a^2+c^2 ) http://rio2016.5ch.net/test/read.cgi/math/1745314067/528
529: １３２人目の素数さん [] 2025/07/28(月) 12:55:39.31 ID:Vsf8XHSj 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/1745314067/529
530: １３２人目の素数さん [] 2025/07/28(月) 16:43:49.34 ID:Vsf8XHSj y''+6y'+10y=2sin(x). D^2+6D+10=0. D=-3±i (D^2+6D+10)y=2sin(x) (D-(-3+i))(D-(-3-i))y=i(e^(-ix)-e^ix) y=1/(D-(-3+i))∙1/(D-(-3-i)) i(e^(-ix)-e^ix) a=-3+i, b = -3-i, f(x)=i(e^(-ix)-e^ix) と置くと y=1/(D-a)∙1/(D-b) f(x)=1/(D-b)∙1/(D-a) f(x) =1/(D-b) e^ax 1/D e^(-ax) f(x)=1/(D-b) e^ax ∫▒〖e^(-ax) f(x)〗 dx =e^bx 1/D e^(-bx) e^ax ∫▒〖e^(-ax) f(x)〗 dx =e^bx 1/D e^(a-b)x ∫▒〖e^(-ax) f(x)〗 dx =e^bx ∫▒(e^(a-b)x ∫▒〖e^(-ax) f(x)〗 dx) dx =e^(-(3+i)x) ∫▒(e^2ix ∫▒〖e^((3-i)x) i(e^(-ix)-e^ix)〗 dx) dx =e^(-(3+i)x) ∫▒(〖ie〗^2ix ∫▒〖e^((3-2i)x)-e^3x 〗 dx) dx =e^(-(3+i)x) i∫▒e^2ix (e^((3-2i)x)/(3-2i)-e^3x/3+A)dx =e^(-(3+i)x) i∫▒〖e^3x/(3-2i)-e^((3+2i)x)/3+A〗 e^2ix dx =e^(-(3+i)x) (〖ie〗^3x/(3(3-2i))-〖ie〗^((3+2i)x)/(3(3+2i))+A (i2e^2ix)/2i+B) =e^(-ix) e^(-3x) ((ie^3x)/(3(3-2i))-(〖ie〗^2ix e^3x)/(3(3+2i))+Ae^2ix+B) =e^(-ix) (i/(3(3-2i))-〖ie〗^2ix/(3(3+2i))+Ae^((2i-3)x)+Be^(-3x) ) =(ie^(-ix))/(3(3-2i))-(ie^ix)/(3(3+2i))+Ae^((i-3)x)+Be^(-(3+i)x) =i (3+2i)/3∙(cosx-isinx)/13-i (3-2i)/3∙(cosx+isinx)/13+e^(-3x) (Ae^ix+Be^(-ix)) =i (4icosx-6isinx)/39+e^(-3x) (Acosx+iAsinx+Bcosx-iBsinx) =(-4cosx+6sinx)/39+e^(-3x) ((A+B)cosx+i(A-B)sinx) =2sinx/13-4cosx/39+e^(-3x) (C_1 cosx+C_2 sinx) http://rio2016.5ch.net/test/read.cgi/math/1745314067/530
531: １３２人目の素数さん [] 2025/07/28(月) 16:44:14.23 ID:Vsf8XHSj L[cos(at)]=∫_0^∞??e^(-st) cos(at) ? dt=lim┬(b→∞)??∫_0^b??e^(-st) cos(at) ? dt? ∫??e^ax cos(bx) ? dx=∫??e^ax/(a^2+b^2 ) acos(bx)+bsin(bx) ? dx lim┬(b→∞)??∫_0^b??e^(-st) cos(at) ? dt? =lim┬(b→∞)??[e^(-st)/(s^2+a^2 ) (?( @?-scos??(at)+asin(at)@ ))]_0^b ? =lim┬(b→∞)?(e^(-sb)/(s^2+a^2 ) (asin?(ab)-scos(ab))-1/(s^2+a^2 ) (-s)) =e^(-sb)/(s^2+a^2 ) lim┬(b→∞)?(asin?(ab)-scos(ab))+s/(s^2+a^2 ) =1/(s^2+a^2 ) lim┬(b→∞)?((asin?(ab)-scos(ab))/e^sb )+s/(s^2+a^2 ) Asin?(ab)-Bcos(ab)=√(A^2+B^2 ) sin(ab-θ) |(asin?(ab)-scos(ab))/e^sb |=(√(s^2+a^2 ) |sin(ab-θ)|)/e^sb ?√(s^2+a^2 )/e^sb lim┬(b→∞)?((asin?(ab)-scos(ab))/e^sb )=0 ∴L[cos(at)]=s/(s^2+a^2 ) http://rio2016.5ch.net/test/read.cgi/math/1745314067/531
532: １３２人目の素数さん [] 2025/07/28(月) 17:43:24.85 ID:Vsf8XHSj 2=2?3?7 5≡1, 5^(2021^2021 )≡ 1^(2021^2021 )≡1 (mod 2) ･･････････? 5≡-1, 5^(2021^2021 )≡ (-1)^(2021^2021 )≡-1≡2 (mod 3)･･････････? 5^1≡5, 5^(2021^2021 ) (mod 7) 5^(7-1)≡5^6≡1 (mod 7) t=2021^2021, 2021^t≡5^t (mod 7) 5^t= 5^(6k+r)=5^6k 2^r≡5^r (mod 7) 5^(2021^2021 )≡5^t (mod 7) 2021≡-1 (mod 6) t=2021^2021≡(-1)^2021≡-1≡5 (mod 6) 5^5=3125=446?7+3≡3 (mod 7) 5^1≡5, 5^2≡4 (mod 7) 5^3≡20≡6 (mod 7) 5^5=5^2 5^3≡24≡3 (mod 7) ∴2021^(2021^2021 )≡5^(2021^2021 )≡5^5 ≡3 (mod 7)･･････････? x≡ 2021^(2021^2021 ) とおくと x≡1 (mod 2) ,21x≡21 (mod 42) ･･････････? x≡2 (mod 3) ,14x≡28 (mod 42) ･･････････? x≡3 (mod 7) , 6x≡18 (mod 42) ･･････････? 41x≡67 (mod 42) 42x≡42 (mod 42) ∴x≡-25≡17 (mod 42) http://rio2016.5ch.net/test/read.cgi/math/1745314067/532
533: １３２人目の素数さん [] 2025/07/28(月) 17:43:54.74 ID:Vsf8XHSj ?θ/?s=(x ?(y ?(t+?t)-y ?(t))-y ?(x ?(t+?t)-x ?(t)))/(√(x ?^2+y ?^2 ) √(?(x ?(t+?t))?^2+?(y ?(t+?t))?^2 )) 1/?r(t+?t)-r(t)? =((x ?(y ?(t+?t)-y ?(t))-y ?(x ?(t+?t)-x ?(t)))/?t)/(√(x ?^2+y ?^2 ) √(?(x ?(t+?t))?^2+?(y ?(t+?t))?^2 )) ?t?r(t+?t)-r(t)?^(-1) =(x ? ((y ?(t+?t)-y ?(t)))/?t-y ? ((x ?(t+?t)-x ?(t)))/?t)/(√(x ?^2+y ?^2 ) √(?(x ?(t+?t))?^2+?(y ?(t+?t))?^2 )) ?(r(t+?t)-r(t))/?t?^(-1) 1/R=(lim)┬(?t→0)???θ/?s?=(x ?y ?-yx ?)/(√(x ?^2+y ?^2 ) √(x ?^2+y ?^2 )) ? ?r ? ??^(-1) =(x ?y ?-yx ?)/(√(x ?^2+y ?^2 ) √(x ?^2+y ?^2 ) √(x ?^2+y ?^2 )) =(x ?y ?-yx ?)/(x ?^2+y ?^2 )^(3/2) R=(x ?^2+y ?^2 )^(3/2)/(x ?y ?-yx ? ) http://rio2016.5ch.net/test/read.cgi/math/1745314067/533
534: １３２人目の素数さん [] 2025/07/28(月) 17:44:26.20 ID:Vsf8XHSj a_1= [■(0@1@1)],a_2= [■(1@0@1)],a_3= [■(1@1@0)] a_1→u_1 u_1=a_1/?a_1 ? =a_1/√(1+1)=1/√2 [■(0@1@1)] a_2→u_2 b_1=(a_2?u_1 ) u_1=(1/√2 [■(1@0@1)]?[■(0@1@1)]) u_1=1/√2 1/√2 [■(0@1@1)]=1/2 [■(0@1@1)] b_2=a_2-(a_2?u_1 ) u_1 =[■(1@0@1)]-1/2 [■(0@1@1)]=[■(1-0@0-1/2@1-1/2)]=[■(1@-1/2@1/2)]=1/2 [■(2@-1@1)] ?b_2 ?=1/2 √(4+1+1)=√6/2 u_2=b_2/?b_2 ? =2/√6 1/2 [■(2@-1@1)]=1/√6 [■(2@-1@1)] a_3→u_3 c_1=(a_3?u_1 ) u_1=(1/√2 [■(1@1@0)]?[■(0@1@1)]) u_1=1/√2 1/√2 [■(0@1@1)]=1/2 [■(0@1@1)] http://rio2016.5ch.net/test/read.cgi/math/1745314067/534
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