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单词 SymmetryOfDividedDifferences
释义

symmetry of divided differences


Theorem 1.

If y0,,yn is a permutationMathworldPlanetmath of x0,,xn, then

Δnf[x0,,xn]=Δnf[y0,,yn].
Proof.

We proceed by inductionMathworldPlanetmath. When n=1, we have, from the definition,

Δ1f[x0,x1]=f(x1)-f(x0)x1-x0=f(x0)-f(x1)x0-x1=Δ1f[x1,x0].

Since the only permutations of two elements are the identityPlanetmathPlanetmathPlanetmath and thetranspositionMathworldPlanetmath, we see that the first divided differrence is symmetricPlanetmathPlanetmath.

Now suppose that we already know that the n-th divided differenceMathworldPlanetmathis symmetric under permutation of its arguments for some n1.We will prove that the n+1-st divided difference is also symmmetricunder all permutations of its arguments.

The divided difference is symmetric under transposing x0 withx1:

Δn+1f[x0,x1,x2,,xn+1]=Δnf[x1,x2,xn+1]-Δnf[x0,x2,xn+1]x1-x0
=Δnf[x0,x2,xn+1]-Δnf[x1,x2,xn+1]x0-x1
=Δn+1f[x1,x0,x2,,xn+1]

The divided difference is symmetric under transposing x1 withx2:

Δn+1f[x0,x1,x2,,xn+1]
=Δnf[x1,x2,xn+1]-Δnf[x0,x2,xn+1]x1-x0
=(x2-x0)(Δn-1f[x2,x3xn+1]-Δn-1f[x1,x3xn+1])-(x2-x1)(Δn-1f[x2,x3xn+1]-Δn-1f[x0,x3xn+1])(x2-x1)(x1-x0)(x2-x0)
=(x1-x0)Δn-1f[x2,x3xn+1]-(x2-x0)Δn-1f[x1,x3xn+1]+(x2-x1)Δn-1f[x0,x3xn+1](x2-x1)(x1-x0)(x2-x0)
=(x1-x0)(Δn-1f[x1,x3xn+1]-Δn-1f[x2,x3xn+1])-(x1-x2)(Δn-1f[x1,x3xn+1]-Δn-1f[x0,x3xn+1])(x1-x2)(x1-x0)(x2-x0)
=Δnf[x2,x1,xn+1]-Δnf[x0,x1,xn+1]x1-x0
=Δn+1f[x0,x2,x1,,xn+1]

The divided difference is symmetric under transposingxk with xk+1 when k>1:

Δn+1f[x0,x1,x2,,xk,xk+1,,xn+1]
=Δnf[x1,x2,,xk,xk+1,,xn+1]-Δnf[x0,x2,,xk,xk+1,,xn+1]x1-x0
=Δnf[x1,x2,,xk+1,xk,,xn+1]-Δnf[x0,x2,,xk+1,xk,,xn+1]x1-x0
=Δn+1f[x0,x1,x2,,xk+1,xk,,xn+1]

Since any permutation of x0,x1,xn+1 can begenreated from the transpositions of xk with xk+1for k between 0 and n, it follows thatΔn+1f[x0,x1,,xk,xk+1,,xn+1]is symmetric under all permutaions of x0,x1,xn+1.∎

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更新时间:2025/5/4 23:55:36