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1/2 ∑(aibj -ajbi)^2 (i,j=1到n) 因为不方面用数学公式编辑器,扩后后面表示求和从多少到多少
=1/2 ∑[(aibj)^2 -2aibjajbi +(ajbi)^2 ] (i,j=1到n)
=1/2 ∑(aibj)^2 - ∑aibjajbi + 1/2 ∑(ajbi)^2 (i,j=1到n) 1/2 ∑(aibj)^2 = 1/2 ∑(ajbi)^2
= ∑(aibj)^2 - ∑aibjajbi (i,j=1到n)
= ∑ai^2 ∑bj ^2 (i,j=1到n) - ∑aibi∑ajbj (i,j=1到n)
另k=i,k=j (k=1到n) 有
最后一项= ∑ak^2 ∑bk^2 - ∑ak*bk ∑ak*bk (k=1到n)
=∑ak^2 ∑bk^2 - (∑ak*bk)^2 (k=1到n)
命题得证
1/2 ∑(aibj -ajbi)^2 (i,j=1到n) 因为不方面用数学公式编辑器,扩后后面表示求和从多少到多少
=1/2 ∑[(aibj)^2 -2aibjajbi +(ajbi)^2 ] (i,j=1到n)
=1/2 ∑(aibj)^2 - ∑aibjajbi + 1/2 ∑(ajbi)^2 (i,j=1到n) 1/2 ∑(aibj)^2 = 1/2 ∑(ajbi)^2
= ∑(aibj)^2 - ∑aibjajbi (i,j=1到n)
= ∑ai^2 ∑bj ^2 (i,j=1到n) - ∑aibi∑ajbj (i,j=1到n)
另k=i,k=j (k=1到n) 有
最后一项= ∑ak^2 ∑bk^2 - ∑ak*bk ∑ak*bk (k=1到n)
=∑ak^2 ∑bk^2 - (∑ak*bk)^2 (k=1到n)
命题得证
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