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多項係数の畳み込み

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ヴァンデルモンドの恒等式みたいな畳み込みは、多項係数でも成り立つのか試したらうまくいきました。
証明も出来たので見てください。

まず下降冪多項定理を証明します。
(x)nを下降冪とします。

1

(x0+x1+x2++xr)n+1=(x0+x1+x2++xrn)(x0+x1+x2++xr)n
=y0+y1+y2++yr=n((x0y0)+(x1y1)+(x2y2)++(xryr))(ny0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n(ny0y1y2yr)((x0y0)+(x1y1)+(x2y2)++(xryr))(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n(ny0y1y2yr)((x0y0)(x0)y0(x1)y1(x2)y2(xr)yr+(x1y1)(x0)y0(x1)y1(x2)y2(xr)yr+(x2y2)(x0)y0(x1)y1(x2)y2(xr)yr++(xryr)(x0)y0(x1)y1(x2)y2(xr)yr)
=y0+y1+y2++yr=n(ny0y1y2yr)((x0)y0+1(x1)y1(x2)y2(xr)yr+(x0)y0(x1)y1+1(x2)y2(xr)yr+(x0)y0(x1)y1(x2)y2+1(xr)yr++(x0)y0(x1)y1(x2)y2(xr)yr+1)
=(y0+1)+y1+y2++yr=n+1(n(y01)y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
+y0+(y1+1)+y2++yr=n+1(ny0(y11)y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
+y0+y1+(y2+1)++yr=n+1(ny0y1(y21)yr)(x0)y0(x1)y1(x2)y2(xr)yr
+y0+y1+y2++(yr+1)=n+1(ny0y1y2(yr1))(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n+1((n(y01)y1y2yr) +(ny0(y11)y2yr)+ (ny0y1(y21)yr)+ +(ny0y1y2(yr1)))(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n+1(n+1y0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr

これを踏まえてさらに進みます。

1.1

(x0+x1+x2++xr)n=y0+y1+y2++yr=n(ny0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
(x0+x1+x2++xr)!(x0+x1+x2++xr)n!=y0+y1+y2++yr=n(ny0y1y2yr)x0!x0y0!x1!x1y1!x2!x2y2!xr!xryr!
(x0+x1+x2++xr)!(x0+x1+x2++xr)n!n!=y0+y1+y2++yr=nx0!x0y0!y0!x1!x1y1!y1!x2!x2y2!y2!xr!xryr!yr!

(x0x1x2xrn)
y0+y1+y2++yr=n(x0y0) (x1y1)(x2y2)(xryr)

1.2

(x0+x1+x2++xr)n=y0+y1+y2++yr=n(ny0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
(x0+x1+x2++xr)!(x0+x1+x2++xr)n!=y0+y1+y2++yr=n(ny0y1y2yr)x0!x0y0!x1!x1y1!x2!x2y2!xr!xryr!
(x0+x1+x2++xr)!x0!x1!x2!xr!=y0+y1+y2++yr=n(ny0y1y2yr)(x0+x1+x2++xr)n!(x0y0)!(x1y1)!(x2y2)!(xryr)!

(x0+x1+x2++xr)rとすると

(rx0x1x2xr) y0+y1+y2++yr=n(ny0y1y2yr) (rn(x0y0)(x1y1)(x2y2)(xryr))

2

(x)nを上昇冪とします。

(x0+x1+x2++xr)n+1=(x0+x1+x2++xr+n)(x0+x1+x2++xr)n
=y0+y1+y2++yr=n((x0+y0)+(x1+y1)+(x2+y2)++(xr+yr))(ny0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n(ny0y1y2yr)((x0+y0)+(x1+y1)+(x2+y2)++(xr+yr))(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n(ny0y1y2yr)((x0+y0)(x0)y0(x1)y1(x2)y2(xr)yr+(x1+y1)(x0)y0(x1)y1(x2)y2(xr)yr+(x2+y2)(x0)y0(x1)y1(x2)y2(xr)yr++(xr+yr)(x0)y0(x1)y1(x2)y2(xr)yr)
=y0+y1+y2++yr=n(ny0y1y2yr)((x0)y0+1(x1)y1(x2)y2(xr)yr+(x0)y0(x1)y1+1(x2)y2(xr)yr+(x0)y0(x1)y1(x2)y2+1(xr)yr++(x0)y0(x1)y1(x2)y2(xr)yr+1)
=(y0+1)+y1+y2++yr=n+1(n(y01)y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
+y0+(y1+1)+y2++yr=n+1(ny0(y11)y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
+y0+y1+(y2+1)++yr=n+1(ny0y1(y21)yr)(x0)y0(x1)y1(x2)y2(xr)yr
+y0+y1+y2++(yr+1)=n+1(ny0y1y2(yr1))(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n+1((n(y01)y1y2yr) +(ny0(y11)y2yr)+ (ny0y1(y21)yr)+ +(ny0y1y2(yr1)))(x0)y0(x1)y1(x2)y2(xr)yr
=y0+y1+y2++yr=n+1(n+1y0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr

2.1

(x0+x1+x2++xr)n=y0+y1+y2++yr=n(ny0y1y2yr)(x0)y0(x1)y1(x2)y2(xr)yr
(x0+x1+x2++xr)+n1!(x0+x1+x2++xr)1!=y0+y1+y2++yr=n(ny0y1y2yr)(x0+y01)!x01!(x1+y11)!x11!(x2+y21)!x21!(xr+yr1)!xr1!
(x0+x1+x2++xr)+n1!(x0+x1+x2++xr)1!=y0+y1+y2++yr=n(ny0y1y2yr)(x0+y01)!x01!(x1+y11)!x11!(x2+y21)!x21!(xr+yr1)!xr1!
(x0+x1+x2++xr)+n1!(x0+x1+x2++xr)1!n!=y0+y1+y2++yr=n(x0+y01)!x01!y0!(x1+y11)!x11!y1!(x2+y21)!x21!y2!(xr+yr1)!xr1!yr!

(x0+x1+x2++xr+n1n)y0+y1+y2++yr=n(x0+y01y0) (x1+y11y1)(x2+y21y2)(xr+yr1yr)

投稿日:2024611
更新日:202475
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nakano
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