フェルマーの最終定理の証明

フェルマーの最終定理の証明 (692ﾚｽ)
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485: 07/24(木)20:40 ID:WNOov+Jn(2/4) 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) )

486: 07/24(木)21:01 ID:WNOov+Jn(3/4) AAS
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)
(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
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

487: 07/24(木)21:02 ID:WNOov+Jn(4/4) AAS
∫_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

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