在本文中，我們初步討論一下二次方程的最優(yōu)化，先貼出結(jié)論，感興趣的同學(xué)可以繼續(xù)往下看推論的過(guò)程，不想看推論的可以直接跳到最后看例子如何應(yīng)用

對(duì)于最小化方程

解是

最小值是

發(fā)現(xiàn)matlab中沒(méi)有LDU分解的函數(shù)，只有LU分解，先貼上LDU分解的代碼，需要的同學(xué)可以復(fù)制下來(lái)。然后help 一下，有說(shuō)明和例子。關(guān)于原理，這里就不贅述了，都是初等行變換

之后會(huì)稍微用到這個(gè)函數(shù)，不過(guò)喜歡手算的同學(xué)可以跳過(guò)這個(gè)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % -------------------------- %INPUT %A: matrix; %-------------------------- %OUTPUT %L: lower triangular matrix %D: diagonal matrix %U: upper triangular matrix %-------------------------- %Example 1 %A = [4 -1 -3/2; -1 3 1]; %ldu(A) % %Example 2 %syms a b c d e f %ldu([a b c; b d e; c e f]) %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%function ldu(A) nn = size(A); n = nn(1); A = [A, eye(n)]; r = 0; for j = 1:(n - 1)r = r + 1;for i = (j + 1) : nA(i, :) = A(i, :) - A(i, j)/A(r, j)*A(r, :); end end L = inv(A(1 :n, (n+1) : 2*n)) D = diag(diag(A)) U = D\A(1: n , 1: n) end

?

下面進(jìn)入正題

貼上代碼和運(yùn)行結(jié)果

syms a b c ldu([a b; b c]);

?

?所以方程變成

轉(zhuǎn)載于:https://www.cnblogs.com/Mr-ZeroW/p/7744964.html

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