フェルマーの最終定理の証明 (786ﾚｽ)
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574: １３２人目の素数さん [] 2025/08/03(日) 19:43:16.34 ID:FKrzG2hZ(1/4) 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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575: １３２人目の素数さん [] 2025/08/03(日) 19:43:38.45 ID:FKrzG2hZ(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) )

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576: １３２人目の素数さん [] 2025/08/03(日) 19:46:11.10 ID:FKrzG2hZ(3/4) AAS
(p⇒q∧q⇒r) ⇒ (p⇒r)
⇔¬(p⇒q∧q⇒r)∨(¬p∨r)
⇔¬{ (¬p∨q)∧(¬q∨r) }∨(¬p∨r)
⇔ {¬(¬p∨q)∨¬(¬q∨r) }∨(¬p∨r)
⇔ { (p∧¬q)∨(q∧¬r) }∨(¬p∨r)
⇔ (p∧¬q)∨(q∧¬r)∨¬p∨r
⇔ {¬p∨(p∧¬q)} ∨ {r∨(q∧¬r)}
⇔ {(¬p∨p)∧(¬p∨¬q)} ∨ {(r∨q)∧(r∨¬r)}
⇔ (¬q∨¬p)∨(q∨r)
⇔ ¬q∨q∨¬p∨r ⇔ [T]

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577: １３２人目の素数さん [] 2025/08/03(日) 19:49:45.45 ID:FKrzG2hZ(4/4) AAS
?_C^ ??f(x,y)dx?
=∫_a^b??f(x,φ_1 (x))dx?+∫_b^a??f(x,φ_2 (x))dx?
=∫_a^b??f(x,φ_1 (x))dx?-∫_a^b??f(x,φ_2 (x))dx?
=-∫_a^b??f(x,φ_2 (x))-f(x,φ_1 (x))? dx
=-∫_a^b??∫_(φ_1 (x))^(φ_2 (x))??(∂f(x,y))/∂y dy? dx?
=-∬_D^ ?(∂f(x,y))/∂y dxdy
※∫_(φ_1 (x))^(φ_2 (x))??(∂f(x,y))/∂y dy?=[?( @f(x,y)@ )]_(φ_1 (x))^(φ_2 (x))=f(x,φ_2 (x))-f(x,φ_1 (x))

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上下前次1-新書関写板覧索設栞歴

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