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高斯相乘引理 (Gaussian product lemma) N(x;a,A)N(x;b,B)=N(0;a−b,A+B)N(x;aA+bB1A+1B,11A+1B)" role="presentation">N(x;a,A)N(x;b,B)=N(0;a?b,A+B)N(x;aA+bB1A+1B,11A+1B)N(x;a,A)N(x;b,B)=N(0;a?b,A+B)N(x;aA+bB1A+1B,11A+1B) 其中N(x;a,A)" role="presentation">N(x;a,A)N(x;a,A)表示以均值为a"... 阅691 转0 评0 公众公开 19-01-13 18:06 |
Λ=diag(λ1,λ2,...,λn)并且λ1≤λ2≤...≤λn B+λI=Q(Λ+λI)QTp(λ)=Q(Λ+λI)−1QTgp(λ)2=−∑1n(qjTg)2(λj+λ)2" role="presentation">B+λI=Q(Λ+λI)QTp(λ)=Q(Λ+λI)?1QTgp(λ)2=?∑1n(qTjg)2(λj+λ)2B+λI=Q(Λ+λI)QTp(λ)=Q(Λ+λI)?1QTgp(λ)2=?∑1n(qjTg)2(λj+λ)2由于是正交分解因此qiqj=1" role=&... 阅1 转自mscdj 公众公开 18-12-16 09:02 |
令导数为 0 ，求得 x" role="presentation" style="position: relative;">xx 、α" role="presentation" style="position: relative;">αα 的值后，将 x" role="presentation" style="position: relative;">xx 带入 f(x)" role="presentatio... 阅1 转自醉恋清欢 公众公开 18-12-16 09:01 |
