■学会にて(直観幾何学研究会2019,その18)

 3変数の有理式を考えてみる.

[1]1/(a−b)(a−c)+1/(b−a)(b−c)+1/(c−a)(c−b)=0

[2]a/(a−b)(a−c)+b/(b−a)(b−c)+c/(c−a)(c−b)=0

[3]a^2/(a−b)(a−c)+b^2/(b−a)(b−c)+c^2/(c−a)(c−b)=1

[4]a^3/(a−b)(a−c)+b^3/(b−a)(b−c)+c^3/(c−a)(c−b)=a+b+c

 さて,この4変数版はどうなるのだろうか?

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[1]1/(a−b)(a−c)(a−d)+1/(b−a)(b−c)(b−d)+1/(c−a)(c−b)(c−d)+1/(d−a)(d−b)(d−c)=0

[2]a/(a−b)(a−c)(a−d)+b/(b−a)(b−c)(b−d)+c/(c−a)(c−b)(c−d)+d/(d−a)(d−b)(d−c)=0

[3]a^2/(a−b)(a−c)(a−d)+b^2/(b−a)(b−c)(b−d)+c^2/(c−a)(c−b)(c−d)+d^2/(d−a)(d−b)(d−c)=0

[4]a^3/(a−b)(a−c)(a−d)+b^3/(b−a)(b−c)(b−d)+c^3/(c−a)(c−b)(c−d)+d^3/(d−a)(d−b)(d−c)=1

[5]a^4/(a−b)(a−c)(a−d)+b^4/(b−a)(b−c)(b−d)+c^4/(c−a)(c−b)(c−d)+d^4/(d−a)(d−b)(d−c)=a+b+c+d

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 一般に

  Σxj^r/Π(xj−xk)=0    (0≦r<n−1)

  Σxj^r/Π(xj−xk)=1    (r=n−1)

  Σxj^r/Π(xj−xk)=Σxj  (r=n)

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