【SEOテキスト】宇田雄一「古典物理学」V05の定義:∀x∈N01(N01);∀ξ∈N01;V05(ξ,x)∈F05(F05)and∀f∈F05;∀(i,j,k)∈N05;[[V05(ξ,x)](f)](i,j,k)=4排=1
4敗=1 4杯=1[∂rx(ξ)](i)・[[∂j(x-1)](x(ξ))](s)・[[∂k(x-1)](x(ξ))](t)・f(r,s,t)+4排=1[∂rx(ξ)](i)・[[∂j∂k(x-1)](x(ξ))](r),V5の定義:∀x∈N01(N01);∀b∈R;V5(x,b)∈F5(F5)and∀f∈F5;∀(ξ,i,j,4)∈N5;[[V5(x,b)](f)](x(ξ),i,j,4)=b4排=1
4敗=1[[∂i(x-1)](x(ξ))](r)・[[∂j(x-1)](x(ξ))](s)・g(ξ,r,s;f),V4,nの定義:∀x∈N01(N01);∀a∈R;∀n∈N;∀p∈Pn;V4,n(x,a,p)∈[F+4,n(x)∪F-4,n(x)]→F4,n
and【1】and【2】and【3】【1】∀f∈F+4,n(x)∪F-4,n(x);[[V4,n(x,a,p)](f)](N2,n)=[V2,n(x,p)](f(N2,n))【2】∀f∈F+4,n(x);∀ξ∈N01;[[V4,n(x,a,p)](f)](x(ξ),N3)=[V3(ξ,x,a)](f(ξ,N3))【3】∀f∈F-4,n(x);∀ξ∈N01;[[V4,n(x,a,p)](f)](x(ξ),N3)=-[V3(ξ,x,a)](f(ξ,N3)),V6,nの定義:∀x∈N01(N01);∀a∈R;∀b∈R;∀n∈N;∀p∈Pn;V6,n(x,a,b,p)∈[F+6,n(x)∪F-6,n(x)]→F6,n
and∀f∈F+6,n(x)∪F-6,n(x);【1】and【2】【1】[[V6,n(x,a,b,p)](f)](N4,n)=[V4,n(x,a,p)](f(N4,n))【2】[[V6,n(x,a,b,p)](f)](N5)=[V5(x,b)](f(N5)),V05とV5の定義より、次式を導くことが出来る。∀x∈N01(N01);∀ξ∈N01;∀b∈R;∀f∈F5;b≠0⇒Γ(x(ξ),N05;[V5(x,b)](f))=[V05(ξ,x)](Γ(ξ,N05;f))
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